Examiner Tips

For time intervals crossing midnight, split into two parts: time until midnight + time after midnight. Remember an hour has 60 minutes, not 100.

“16 h 32 min from subtracting the times was just one of many incorrect responses seen, some of which suggested there is 100 minutes in an hour! Clearly working out time periods is something that many simply guess at rather than making a structured response to periods of time going over the end of one day.”

CIE 0580 · Paper 1 · Jun 2024

For compound interest, the rate and the number of periods must MATCH. 0.25% per month over 5 years = 60 monthly periods (not 5). Don't silently convert to annual — apply the rate as given.

“Fewer fully correct responses were seen as a number of candidates struggled to deal correctly with a compound interest rate of 0.25% per month. Many treated 0.25% per month as 3% per year and 350(1 + 3/100)^5 was frequently seen. Some appeared not to notice that the interest was monthly, instead treating it as annual interest.”

CIE 0580 · Paper 4 · Jun 2023

For compound interest equations, use TOTAL value in A = P(1 + r/100)ⁿ. If interest was $621.70 on $6000, then A = $6621.70 (principal + interest), not $621.70 alone.

“This question on compound interest proved challenging for a significant number of candidates. One common error was to treat the problem as simple interest. For those trying a compound interest approach many attempted to set up an equation, but a very common error was to work with the interest ($621.70) rather than with the value of the investment at the end of the eight years ($6621.70).”

CIE 0580 · Paper 4 · Nov 2023

When finding bounds, always give the final answer in the SAME UNITS as the variable in the question. If l is in metres, bounds must be in metres — convert at the end if your working was in cm.

“This area of the syllabus is often found to be challenging for many candidates. Also the length of piece of wood, l, was given in metres but the extra information to find the limits was given in centimetres. The answer line used l, so the limits were expected to be in metres not centimetres.”

CIE 0580 · Paper 1 · Nov 2023

In compound interest equations, the 'A' in A = P(1 + r/100)ⁿ is the TOTAL VALUE, not the interest. If interest is $1328.54 on $4540, then A = $5868.54 (add them). And keep 4+ sig figs when taking roots.

“This question proved a challenge for most candidates. The most common error was to take the interest $1328.54 to be the final amount rather than the interest earned. Those who started correctly and got as far as (1 + r/100)¹⁰ = 5868.54/4540 sometimes proceeded to spoil their answer by incorrectly writing... Another common error was to round their result too soon.”

CIE 0580 · Paper 2 · Nov 2022

To add/subtract numbers in standard form, convert to the SAME power first: 4 × 10¹⁰⁰ − 2 × 10⁹⁸ = 400 × 10⁹⁸ − 2 × 10⁹⁸ = 398 × 10⁹⁸. Then re-express in standard form.

“It was not possible to evaluate this standard form question with a calculator and few candidates tackled it correctly. Most were not able to convert the two numbers to a common form to allow subtraction.”

CIE 0580 · Paper 4 · Nov 2022

For speed = distance/time: LOWER bound of speed = min distance / max time. UPPER bound = max distance / min time. The division INVERTS the 'min/max' pattern for the denominator.

“Only a minority of candidates got this question fully correct. Many thought that for the lower bound they had to find the lowest values for both distance and time, not realising that the division meant this was not correct.”

CIE 0580 · Paper 2 · Nov 2021

'When do two repeating events next coincide?' is a classic LCM question. Find LCM of the two periods — e.g., LCM(90, 105) = 630 seconds. Don't subtract or add the periods.

“This was the most challenging question on the paper and a significant number did not attempt it. Many candidates did not realise that the lowest common multiple was needed but those who did, generally made progress at least to 630 seconds. Most candidates tried to perform a variety of operations on the two times of 90 seconds and 105 seconds, most commonly subtracting them.”

CIE 0580 · Paper 2 · Nov 2019

For 'given to 1 dp', bounds are ±0.05 (not ±0.1 and not ±0.005). So 8.2 → bounds 8.15 ≤ L < 8.25. Upper bound uses the 'less than' sign, not ≤.

“This was the least well answered question on the paper and a fairly large number of candidates did not attempt it. The correct approach is to concentrate on the information, 'correct to 1 decimal place' first. This means that candidates have to go to the next decimal place and so subtract 0.05 m from the length of the truck for the lower bound and add 0.05 m for the upper bound.”

CIE 0580 · Paper 1 · Nov 2019

When bounds are given 'to nearest centimetre' but the measurement is in metres, convert to cm first, apply ±0.5 cm, then convert back to metres. Or apply ±0.005 m directly. Mixing units is the most co

“This was the least well answered question on the paper and a fairly large number of candidates did not attempt it. The correct approach is to concentrate on the 'nearest centimetre' part first. This has to be split into 0.5 cm subtracted from the boy's height and 0.5 cm added on.”

CIE 0580 · Paper 1 · Nov 2018

When rounded to nearest 0.5, the bounds are ±0.25 (half of 0.5). So 78.5 to nearest 0.5m gives bounds 78.25 to 78.75, NOT 78.45 to 78.55.

“This was a very demanding question on what is usually quite a challenging topic. It was poorly answered with the vast majority of candidates unable to cope with the bounds for 78.5 to the nearest half metre. While there were a wide variety of incorrect attempts, the most common were 78.45 with 78.55 and 78 with 79.”

CIE 0580 · Paper 1 · Jun 2018

Product means multiplication. Negative × negative = positive, so the highest product of 2 numbers from a set may involve two negatives, not two positives.

“A basic difficulty was that many did not understand that 'product' meant multiplication and not addition. In fact, many chose the correct numbers but did not perform any operation on them. 48 was often seen as it was the highest product of 2 positive numbers meaning these candidates did not appreciate the positive result of the product of 2 negative numbers.”

CIE 0580 · Paper 1 · Jun 2024

When an answer is exact (terminates), write every digit — rounding to 3 sig figs or truncating is NOT acceptable. The rounding rule only applies to non-exact answers.

“Candidates should be reminded that when an answer is exact it should not be rounded. A small minority of candidates truncated their answer to 24.3 which was not acceptable.”

CIE 0580 · Paper 2 · Jun 2024

2150 in 24-hour time is 9:50 pm, not 11:50 pm. When adding minutes that roll past 60, the hour advances — e.g., 2150 + 38 min = 2228, not 2188.

“Where incorrect answers were seen a common error was to convert 2150 to 1150pm leading to an answer of 5 h 28 minutes. Another error was not identifying that moving on 38 minutes from the start time meant that the time had progressed into the next hour.”

CIE 0580 · Paper 2 · Jun 2024

Simple interest I = PRT/100 gives ONLY the interest earned, not the total value. If $8000 grows to $12800 over 12 years, the interest is $4800 (total − principal), not $12800.

“Most candidates understood that this question involved simple interest and set up a correct expression for the interest received after 12 years. Many equated this with $12 800, the total value of the investment after 12 years, rather than $4800, the value of the interest after 12 years. This error led to the common incorrect answer of 13.3% rather than the correct answer of 5%.”

CIE 0580 · Paper 4 · Jun 2024

Time does NOT convert as decimal: 3.6 hours is 3 hr 36 min (0.6 × 60), not 3 hr 60 min. And 3 hr 66 min is impossible — 66 min = 1 hr 6 min, so it should be 4 hr 6 min.

“Common errors included leaving the answer as 3 hr 40 min, incorrect conversions such as 3 hr 36 min or 3 hr 66 min, incorrect addition of the journey time.”

CIE 0580 · Paper 3 · Jun 2024

Irrational numbers cannot be written as fractions — they include non-terminating, non-repeating decimals like π, √2, √5. But √16 = 4 and √256 = 16 are RATIONAL (they simplify to whole numbers).

“Candidates found this part quite challenging with many seemingly unaware of the definition of an irrational number. The common correct answers were √5 and √300 (for example). Common errors included 16, 17.5, π, 4.12311 (from √17), √5 and √256.”

CIE 0580 · Paper 3 · Jun 2023

For minimum speed = min distance / max time. For maximum speed = max distance / min time. Mixing lower-bound distance with lower-bound time gives neither the min nor max — it's wrong.

“Whilst very many went on to reach the correct answer of 1.08, the most common error was to also use a lower bound, 11.5, in the denominator.”

CIE 0580 · Paper 2 · Jun 2023

'Product' means multiplication, not addition. And when multiplying directed numbers, keep the signs: negative × positive = negative.

“Most candidates identified the largest and smallest values. However, many candidates did not understand the term 'product'. Addition was the most common operation seen although division was also seen. Some did multiply but omitted the negative sign.”

CIE 0580 · Paper 1 · Jun 2023

There are 60 minutes in an hour (never 100). When adding minutes across an hour boundary, carry 1 to the hour place. When adding across midnight, continue to the next day.

“Often the correct method was seen in logical working. Sometimes there were arithmetic slips in dealing with the fact that the journey finished the next day so answers of 5:15am or 7:15am were seen... The workings for a few candidates seem to infer that they thought there are 100 minutes in one hour.”

CIE 0580 · Paper 1 · Nov 2023

For 'when do two buses next leave together', find LCM of the periods. Add that many minutes to the first common time. Listing is error-prone — LCM is safer.

“The best responses used the method of finding the LCM of 28 and 48, giving the answer in terms of either the 12-hour or 24-hour clock. An alternative method was to list the times of the buses leaving the station, but it was rare for the lists of both buses to lead to the correct time. Those using factor trees or tables started well but not many then correctly worked out the required time of day.”

CIE 0580 · Paper 2 · Nov 2023

'Estimate by rounding each number to 1 sf' — round FIRST, then compute with the rounded values. SHOW the rounding clearly. Rounding the exact answer at the end does NOT score.

“This is a question that many misunderstood even though this sort of question has been asked before. The instruction to round each number correct to 1 significant figure was frequently ignored as candidates put the given numbers into their calculators and rounded the final answer to 1 significant figure. All stages of the calculation had to be seen before credit could be awarded.”

CIE 0580 · Paper 1 · Nov 2023

When ordering numbers, convert them all to decimals with ENOUGH decimal places to distinguish them. And remember 7% = 0.07 (small), not 7 — don't confuse with percentages.

“This ordering question had two numbers very close to one another and frequently candidates did not check to enough digits to determine the order; 15/213 has to be written to at least 0.0704… to be able to write it in the correct place. Many chose 7% as the largest number rather than the smallest.”

CIE 0580 · Paper 1 · Nov 2023

To show a number is NOT prime, give a specific factor: 87 = 3 × 29, so 87 is not prime. Just stating 'a prime has only 2 factors' isn't enough — show the actual divisor.

“Many candidates gave a correct reason. Some said no but did not give a valid reason, the most common of which was simply writing out the definition of a prime number without stating 3 or 29. A significant number of candidates wrote yes, thinking 87 was indeed a prime number.”

CIE 0580 · Paper 3 · Nov 2023

To find the original price BEFORE an 8.7% decrease, DIVIDE the current price by 0.913, don't multiply. Multiplying applies another decrease instead of reversing it.

“Many correct answers were seen from using the correct method. The main error was to multiply 4000 by 0.913 and give the answer $3652.”

CIE 0580 · Paper 3 · Nov 2023

'Nearest 5 mm' means bounds are ±2.5 mm (half of 5). So 15.0 cm to nearest 5 mm has bounds 14.75 cm ≤ L < 15.25 cm. Not ±0.5 cm and not ±0.05 cm.

“'Correct to the nearest 5 mm' was frequently misinterpreted and upper bounds of 15.5 or 17.5 for the length and 11, 10.55 and 13 for the width were common errors. Only a small number of candidates calculated an area before attempting to apply the bounds.”

CIE 0580 · Paper 4 · Jun 2022

Compound interest equation: A = P(1 + r/100)ⁿ. To find r: divide by P, take the nth root, subtract 1, multiply by 100. Don't forget the bracket around (1 + r/100).

“Most set up the initial equation correctly, although initial set-up errors such as 1030.35 = 700 + (1 + r/100)¹⁷ or 1030.35 = 700(1 − r/100)¹⁷ were seen. The former was unlikely to go any further but the second slip could still earn a second mark by correct re-arrangement leading to 17√(1030.35/700).”

CIE 0580 · Paper 2 · Jun 2022

When crossing midnight, compute: (24:00 − start) + end. So 2115 to 0433 next day = (24:00 − 21:15) + 04:33 = 2h45min + 4h33min = 7h18min. Don't simply subtract across the boundary.

“Common incorrect answers included; 5 h 18 min (doing 9 – 4 and 33 – 15); 6 or 8 h with 18 min (correct method but an arithmetic slip); 19 h 18 min (converting the 7 to 24 hour clock); 25 h 48 minutes (adding the two times); 16 h 82 min (from the subtraction 2115 – 0433) and 16 hrs 42 mins (the difference between 0433 and 2115 on the same day rather than 2115 one day and 0433 the next day).”

CIE 0580 · Paper 2 · Jun 2022

If HCF(a,b) = 6 and LCM(a,b) = 90, then a × b = 6 × 90 = 540. Try a = 6m, b = 6n with HCF(m,n) = 1. Example: 18 × 30 = 540 (HCF 6, LCM 90).

“This question was found challenging by candidates although many did gain partial credit, very often for finding the prime factors of 90. Others found two numbers that had either an LCM of 90 or an HCF of 6. Some gave one of the numbers as 6 but the question clearly stated that both numbers were greater than 6.”

CIE 0580 · Paper 1 · Jun 2022

Simple interest I = PRT/100. Remember: (1) divide by 100, (2) 'value of investment' means principal + interest, not just interest. Read the question carefully.

“For those correctly choosing and applying simple interest for their calculation, many stopped at just the interest and didn't give the value of the investment. Quite a number of candidates did not divide by 100, leading to a rather generous result of $10 500 for the interest.”

CIE 0580 · Paper 1 · Jun 2022

'Power of 10' means just 10ⁿ (e.g., 10⁻⁴), not a × 10ⁿ. These are different forms. If the value is exactly 0.0001, write 10⁻⁴, not 1 × 10⁻⁴.

“The question asked for the value to be written as a power of 10 but many candidates gave an answer in standard form (1 × 10⁻⁴). Apart from other incorrect attempts at writing the value in standard form, many miscounted the places.”

CIE 0580 · Paper 1 · Jun 2022

For time intervals crossing midnight: split into two parts (time to midnight + time from midnight). Adding minutes like 15+33 gives 48 but doesn't handle the carry when minutes > 60.

“The response 7 h 48 min was common from adding 15 and 33, having worked out the number of hours correctly. Another common error was 16 h 42 min from a subtraction of the two times.”

CIE 0580 · Paper 1 · Jun 2022

Irrational = can't be written as a fraction. Examples: π, √2, √175. But √324 = 18 is rational (perfect square). For an irrational in a range, use √n where n is close to what you want but not a perfect square, or a multiple of π.

“The concept of irrational numbers was challenging to many as most answers were integers, often prime, or a decimal number between 10 and 20. 15 was seen many times... A well known example of an irrational number is π and any multiple of π to make it within the range specified is acceptable. Those that understood what was required were more likely to choose numbers such as √175 rather than multiples of π. Some candidates who gave a square root sometimes choose one that was an integer, for example √324 (= 18).”

CIE 0580 · Paper 1 · Nov 2022

For a value 'correct to 1 dp', limits are ±0.05 (half of 0.1). So 30.7 has limits 30.65 ≤ x < 30.75, not 30.2 ≤ x ≤ 31.2.

“This limits question was not well answered. The error of adjusting by 0.5 rather than 0.05, leading to 30.2 and 31.2 was very common.”

CIE 0580 · Paper 2 · Nov 2022

3 decimal places (e.g. 8.677) is NOT the same as 3 significant figures (e.g. 8.68). For d.p. count places after the decimal point; for s.f. count from the first non-zero digit.

“many found the rounding to 3 decimal places challenging. Consequently, partial credit was more common than full credit with many responses of 8.67 or 8.68 seen (3 significant figures instead of 3 decimal places)”

CIE 0580 · Paper 1 · Jun 2021

For bounds involving a sum or difference of measurements: upper bound of total = upper + upper (or upper − lower for a difference). Simply adding or subtracting the rounding unit to both extremes is i

“This part was challenging for many candidates and there was a lack of understanding of what was being asked. An answer of 0.18 was seen and others thought it was related to greatest and least values, 17.5 and 18.5. Many did find one of the correct values but 165 + 18 and 153 – 18 were worked out quite often.”

CIE 0580 · Paper 1 · Jun 2021

LCM (Lowest Common Multiple) is the smallest number both values divide into. HCF (Highest Common Factor) is the largest number that divides into both. These are frequently confused — read the question

“a significant number of candidates gave the highest common factor instead of the lowest common multiple”

CIE 0580 · Paper 1 · Jun 2021

When finding the product of numbers in standard form, give the exact unrounded answer in proper standard form (a × 10ⁿ where 1 ≤ a < 10). The word 'product' means multiply — not add or subtract.

“Few candidates gained full credit, even if the correct numbers were multiplied, since the answer was often rounded, instead of the full exact one and in standard form. Many found it helpful to convert all the numbers to ordinary numbers to determine the largest and smallest.”

CIE 0580 · Paper 1 · Jun 2021

Fractional and negative indices require careful application of index laws: x^(1/2) = √x, x^(−n) = 1/xⁿ. Confusing these leads to common wrong answers like 4 when evaluating 8^(2/3).

“Candidates found this part challenging with few gaining credit. Whilst 4 was a common incorrect answer, there were many attempts involving square roots, 81 and 9. Many did not attempt the question.”

CIE 0580 · Paper 1 · Jun 2021

For a 34% decrease over 5 periods: use A × (1 − 0.34)^5 = A × (0.66)^5. Common errors: using + instead of −, using 66% (which gives (1−0.66) = 0.34), or using wrong index (1 or 6 instead of 5).

“Other common errors related to the formula such as using the addition sign, 125.9 × (1 + 34/100)^5, or using 66% so 125.9 × (1 − 66/100)^5. Some candidates used 1 or 6 in the formula as the index.”

CIE 0580 · Paper 2 · Jun 2021

For non-standard rounding intervals: 3m to nearest 20cm has bounds 2.9m and 3.1m (not 2.8 and 3.2). Also, when the final answer must be a whole number, apply the floor function — don't round to neares

“Many were unable to give the correct bound for 3 metres to the nearest 20 centimetres, with 3.2 and 2.8 metres frequently seen. Quite a few of those who gave the correct bounds found 200.5 ÷ 2.9 = 69.1378 and left that as their answer, not realising that the final number of pieces must be an integer.”

CIE 0580 · Paper 2 · Jun 2021

Upper bound = halfway to the next unit above; lower bound = halfway below. For a value rounded to the nearest whole number, bounds are ±0.5. The bounds are not the same as rounding to a different degr

“This part was generally answered reasonably well although a significant number found the concept of upper and lower bounds challenging. Common errors included 2641 and 2643, 2641.5 and 2642.4, 2640 and 100, 2645 and 2655.”

CIE 0580 · Paper 3 · Jun 2021

For bounds, add and subtract HALF the unit of the last given digit. A length given as 567 cm has bounds 566.5 ≤ L < 567.5 — NOT 566 to 568, and NOT 566.05 to 567.95.

“There was a wide variety of incorrect answers implying that this concept was not fully understood. There was a general lack of appreciation of the rules of rounding with the majority only giving answers correct to two decimal places when three was required.”

CIE 0580 · Paper 1 · Nov 2021

Any terminating decimal (like 31.5) is RATIONAL. To give an irrational example in a given range, use √n where n is not a perfect square — e.g., √1001 ≈ 31.6 is irrational.

“Of those that answered, most chose a three significant figure number between 31 and 32 such as 31.7 or more commonly 31.5 and all such numbers are rational. Candidates could look for the square root of a number between 962 and 1023, for example, √1001.”

CIE 0580 · Paper 1 · Nov 2021

HCF (highest common factor) is usually SMALLER than the numbers. LCM (lowest common multiple) is usually BIGGER. Read the question — don't confuse them.

“While there was a sensible, correct approach, factor trees or ladders, to this question, many didn't find the LCM, giving the answer as the HCF.”

CIE 0580 · Paper 2 · Nov 2021

A perfect cube has EVERY prime factor to a power that's a multiple of 3. To find the smallest value R that makes N × R a cube, find the prime factors of N and boost each exponent up to the next multiple of 3.

“This part was found to be very challenging and incorrect answers of 1, 2 and 8 were common. Few candidates understood that in a cube number the prime factors will all have powers that are multiples of 3.”

CIE 0580 · Paper 4 · Nov 2021

For 'how many days in a year are both schedules checked': find the LCM of the two periods, divide 365 by it, floor the result, and add 1 (to include day 1). LCM of 6 and 8 = 24.

“Candidates found this multiples question extremely challenging with only the most able candidates gaining any credit. The correct answer of 16 was found rarely but when found this was from identifying 24 as the LCM, dividing 365 by 24 and then correctly remembering to include 1st Jan and add 1 to their truncated answer.”

CIE 0580 · Paper 3 · Nov 2021

For 0.4̇7̇ (recurring 47): let x = 0.474747..., then 100x = 47.474747..., subtract: 99x = 47, so x = 47/99. Show this subtraction method — calculator answer alone may not score.

“Many candidates demonstrated that they could use their calculator to convert the recurring decimal into a fraction but it was the higher achieving ones who could demonstrate a method, which was required for both marks.”

CIE 0580 · Paper 4 · Nov 2019

Apply upper/lower bounds to the ORIGINAL dimensions, then do calculations with the bound values. Don't compute first and then add 0.5 to the result — that gives wrong bounds.

“Candidates should understand that bounds should be applied to dimensions before any calculations are carried out. Those who did not gain the mark were usually multiplying 15 by 3 and then adding 0.5.”

CIE 0580 · Paper 4 · Nov 2019

For 'estimate' questions, ROUND each number to 1 sf FIRST, then compute with the rounded values. Rounding the final answer after exact calculation is not what the question wants.

“Many candidates did not follow the instruction to write each number correct to 1 significant figure before starting to work out an estimate for p and so gave the exact answer, 58.6. Some found the exact answer then rounded this correct to 1 significant figure, 60.”

CIE 0580 · Paper 1 · Nov 2019

Decimal places count digits AFTER the decimal point. Significant figures count non-zero leading digits. 0.048 to 2 sf is 0.048 (or 0.048); to 2 dp it's 0.05. These are different!

“The question was found challenging with a common error of rounding correct to 2 decimal places, resulting in 0.05, instead of 2 significant figures. Any zeros following an otherwise correct answer spoilt the two significant places requirement.”

CIE 0580 · Paper 2 · Nov 2019

HCF = HIGHEST factor they share (smaller than both numbers). LCM = LOWEST multiple they share (bigger than both). They are different — always check you answered what was asked.

“Many candidates were confused between highest common factor and lowest common multiple. One mark was often gained from a factor tree or equivalent but many who showed this working gave the LCM rather than the HCF.”

CIE 0580 · Paper 2 · Nov 2019

To show no squares exist in a range, list integer squares: 7² = 49, 8² = 64. Since 49 < 50 < 60 < 64, no square lies in [50, 60]. Use integer squares, not roots of the range endpoints.

“Many candidates wrote out the square roots of the numbers from 50 to 60 but nearly all did not write why this showed that no square number existed in that range so gained only partial credit. No mark was given for just writing 50, 51, etc. without any values of them or simplifications such as √50 = 5√2.”

CIE 0580 · Paper 2 · Nov 2019

−3 − 11 = −14, not 8. When subtracting from a negative, move further negative. A common error is treating −3 − 11 as 11 − 3 = 8.

“This question proved challenging for many since the 11 had to be subtracted from −3 rather than added to it, resulting in a very common response of 8. While there were a lot of correct answers there were a significant number who gave 14.”

CIE 0580 · Paper 1 · Jun 2019

Never add fractions by adding numerators and denominators separately. 5/6 + 2/3 needs a common denominator: 5/6 + 4/6 = 9/6 = 3/2.

“A common misconception showing a lack of understanding of fractions was 5/6 + 2/3 = 7/9.”

CIE 0580 · Paper 1 · Nov 2019

Always work in the same unit as the variable. If w is in metres and the accuracy is 10 cm = 0.1 m, the bounds are ±0.05 m. So 4.2 m → bounds 4.15 m to 4.25 m, NOT 415 to 425.

“This question had metres for the width and centimetres for the accuracy. This was challenging for many candidates who often gave centimetre bounds, 415 and 425 even though the 'w' was in metres. Others subtracted and added 0.5 instead of 0.05 or did not halve the 10cm.”

CIE 0580 · Paper 1 · Jun 2019

Compound interest: A = P(1 + r/100)ⁿ. Common formula errors: adding P separately, using the rate as a percentage inside the bracket, or multiplying by n instead of using it as a power. Year-by-year ca

“There were candidates who were familiar with the formula but could not recall it correctly resulting in such calculations as 1200 + (1.056)³, 1200(1 + 5.6%/100) or even 1200(1.056) × 3. Those who calculated year by year often lost accuracy by not keeping sufficient figures in their calculations.”

CIE 0580 · Paper 2 · Nov 2018

For compound decrease use P×(1−r)^n. Don't use the growth formula then subtract, and don't apply simple (non-compound) decrease.

“Incorrect methods which were seen frequently were using exponential growth giving an answer of 24656; finding a decrease without using the compound formula giving an answer of 21390; or a combination method of finding the exponential increase and then subtracting it from 23000.”

CIE 0580 · Paper 2 · Jun 2018

An irrational number cannot be written as a fraction — its decimal never ends or repeats. Between 6 and 7, examples include √37, √38, √39, √40, √41, √42, √43, √44, √45, √46, √47.

“This part was generally poorly answered with the majority of candidates not appreciating or aware of the definition of an irrational number. Common errors included 6, 7, 6.5, √6.5, 6.33333 together with a full variety of other recurring and non-recurring decimals between 6 and 7.”

CIE 0580 · Paper 3 · Jun 2018

LCM (Lowest Common Multiple) is the smallest number in BOTH times tables. HCF (Highest Common Factor) is the largest number that divides BOTH. These are different — don't confuse them.

“Many candidates confused LCM with HCF producing the very common incorrect response of 7. Some used a factor method and others a list of multiples but there was quite a good response from those who did know what a multiple was.”

CIE 0580 · Paper 1 · Jun 2018

For maximum number of items, use maximum fabric length (602.5, not 605) divided by minimum dress length (8.5, not 9.5). Think: to fit the most, use the biggest fabric and smallest dress.

“Many candidates identified 8.5 and 9.5 for the dress length and of these some incorrectly selected 9.5. The most common difficulty was identifying the bounds for the fabric length. 605 m was often used instead of 602.5 m.”

CIE 0580 · Paper 4 · Jun 2018

Convert all fractions, percentages and decimals to decimals before ordering. Don't guess which is smallest — calculate it.

“This was rather poorly answered with more ordered incorrectly than correctly. The fraction was often seen as the smallest and often when it was in the correct position the order was largest to smallest. There was not enough evidence of changing the fraction and percentage to decimals or to enough figures to compare correctly.”

CIE 0580 · Paper 1 · Jun 2018

Always check if your answer makes sense in context. If the answer is a number of days or cakes, it should be a small whole number — not a decimal or huge value.

“Some responses to certain questions were often totally unrealistic or not possible for the context. Specifically this was evident in Questions 11, 13, 16, 17, 22 and 23 where candidates need to check carefully if their responses could possibly be correct.”

CIE 0580 · Paper 1 · Jun 2018

Don't round calculator values mid-calculation. Keep full precision until the final answer, then round as required.

“Marks were often lost due to answers given to less than the required accuracy or loss of accuracy from rounding calculator values at a working stage of a problem.”

CIE 0580 · Paper 1 · Jun 2018

1 litre = 1000 cm³. This conversion factor must be memorised. To convert cm³ to litres, divide by 1000. Many students leave this type of question blank rather than attempt it.

“Part (b), the conversion of cubic centimetres to litres, was less well answered as it appears that the conversion factor is not generally known. This was a question that was often left blank.”

CIE 0580 · Paper 1 · Nov 2018

LCM and HCF are frequently confused. To find the LCM, break each number into prime factors and take the highest power of each prime. For 18 and 21: LCM = 2 × 3² × 7 = 126, not the HCF of 3.

“The most common misunderstanding was to give the HCF, 3.”

CIE 0580 · Paper 1 · Nov 2018

Always check if your answer makes sense in context. If the answer is a number of days or cakes, it should be a small whole number — not a decimal or huge value.

“Some responses to certain questions were often totally unrealistic or not possible for the context. Specifically this was evident in Questions 11, 13, 16, 17, 22 and 23 where candidates need to check carefully if their responses could possibly be correct.”

CIE 0580 · Paper 1 · Jun 2018

Don't round calculator values mid-calculation. Keep full precision until the final answer, then round as required.

“Marks were often lost due to answers given to less than the required accuracy or loss of accuracy from rounding calculator values at a working stage of a problem.”

CIE 0580 · Paper 1 · Jun 2018

For maximum of a quotient (distance ÷ speed): use upper bound of numerator and lower bound of denominator. For minimum: use lower bound of numerator and upper bound of denominator. Using the same boun

“Candidates who were most successful with this question often wrote down the lower and upper bounds for each variable first, and then considered which values to use in the quotient in order to obtain the largest possible result. The most common error was for the upper bound of both the distance and the speed to be used”

CIE 0580 · Paper 4 · Nov 2018

For a measurement of 6.2 correct to 1 d.p., the bounds are 6.15 and 6.25 (±0.05), NOT 6.2 ± 0.5. The bound is always ± half of the smallest unit of measurement. For 1 d.p., the unit is 0.1, so half is

“Overall this question was not well answered. Some candidates were able to give both correct answers but many appeared not to understand this topic. The main error was to find 6.2 ± 0.5, giving 5.7 and 6.7, or answers of 615 and 625.”

CIE 0580 · Paper 3 · Nov 2018

If the question asks for standard form, a whole number answer like 8400 is not complete — you must write 8.4 × 10³.

“Many others did reach the answer 8400 but did not change to standard form.”

CIE 0580 · Paper 1 · Jun 2024

When subtracting fractions without a calculator, show the common denominator step. If the question asks for a fraction, a decimal answer (even the correct value) will not score.

“Marks were often lost by not showing the method of common denominators and simply following the improper fraction with the answer. Decimals were seen and not accepted as the final answer, even if given as an alternative to the correct fraction answer.”

CIE 0580 · Paper 1 · Jun 2024

When asked to write a fraction as a percentage, always include the % sign and at least 3 figures — '22' alone does not score, you must write '22.2%' or similar.

“A minimum of 3 figures was required so an answer of 22 did not score. Just a few showed a lack of understanding of percentages offering answers of 0.22 or 2.22 but it was rare to see any other digits besides '2' in the responses.”

CIE 0580 · Paper 1 · Jun 2024

For compound growth, when iterating years, keep at least 4-5 significant figures at each step. Rounding 30/26 = 1.15 instead of 1.154 can push the answer from 9 years down to 8 years.

“This approach often led to the incorrect answer of 8 because the value on the right-hand side had been rounded to 1.15 rather than using a sufficiently accurate value such as 1.154.”

CIE 0580 · Paper 4 · Jun 2024

Simplest form means divide by the HIGHEST common factor in one step. Cancelling by a small factor and stopping doesn't give the simplest form — check your answer shares no common factor > 1.

“Most candidates cancelled the fraction by 2 or 6. This was not far enough to write the fraction in its simplest form.”

CIE 0580 · Paper 1 · Nov 2023

If the question asks for TOTAL value after n years, leave the compound interest formula answer as is. Subtracting the principal gives you only the interest earned — a different quantity.

“While a small number of candidates performed a simple interest calculation, the vast majority correctly used the compound interest formula... Having found the answer some then subtracted 3000 which would give the interest.”

CIE 0580 · Paper 1 · Jun 2023

Divisibility by 3: if the digits sum to a multiple of 3, the number is divisible by 3. So 27 (2+7=9), 57 (5+7=12), 93 (9+3=12) are all divisible by 3 and NOT prime.

“Some candidates seemed unaware of the divisibility rule for 3, as they chose 27, 57, or 93 as one of the prime numbers. These candidates could improve their scores by reviewing the divisibility rule for 3 and practicing finding prime numbers.”

CIE 0580 · Paper 2 · Nov 2023

Working in 24-hour time avoids ambiguity. If you use 12-hour time, you MUST write 'am' or 'pm' — otherwise 2:15 could mean morning or afternoon and will lose the mark.

“The candidates who worked in the 24-hour system generally gave the correct time. The majority however, used the 12-hour system, but of these it was rare to see the essential 'pm' to distinguish their answers from the early morning time.”

CIE 0580 · Paper 1 · Jun 2023

91 is NOT a prime number — it equals 7 × 13. Other tricky composites: 51 = 3 × 17, 57 = 3 × 19, 87 = 3 × 29. Always check divisibility by 2, 3, 5, 7, 11, 13 for numbers under 200.

“The prime number in the list was not so well chosen with the common error of 91 seen often.”

CIE 0580 · Paper 1 · Jun 2023

For compound interest, use the formula A = P(1 + r/100)ⁿ. Year-by-year calculations accumulate rounding errors. And round the final answer to the requested unit (e.g., nearest dollar).

“Many correct answers were seen. Some showed correct working but did not round their answer to the nearest dollar. A significant number of candidates calculated simple interest or were unable to calculate compound interest correctly. A small number subtracted the principal to calculate the total interest rather than the total value. Candidates should be encouraged to use the formula rather than a year-on-year approach where rounding errors often occur.”

CIE 0580 · Paper 3 · Nov 2023

For 'nearest 10', bounds are ±5 (half of 10). So 350 to nearest 10 → 345 ≤ x < 355. Not ±10 and not ±0.5.

“Many responses added and subtracted 5 from 350 and reached the correct inequality. Some responses added and subtracted 10, or reversed the correct limits. Some values of 354 for the upper limit were seen.”

CIE 0580 · Paper 2 · Nov 2023

Standard form is a × 10ⁿ where 1 ≤ a < 10. So 0.09 = 9 × 10⁻², not 0.09 or 0.9 × 10⁻¹. The coefficient must always start with a non-zero digit before the decimal point.

“A small number found 0.3² correctly but did not express it correctly in standard form, with 0.09 or 0.9 × 10^–1 being the most commonly seen partially correct answers.”

CIE 0580 · Paper 2 · Jun 2023

If the question asks for standard form, you MUST convert. 15625 becomes 1.5625 × 10⁴. Keep all 5 figures, and ensure only ONE digit sits before the decimal point.

“Many candidates gave 15625, the value on the calculator display, as their final answer. This needed to be written in standard form and often this second stage was not attempted. Of those that did try to change the form, some had more that one figure before the decimal point or rounded the digits instead of giving all five.”

CIE 0580 · Paper 1 · Nov 2023

x^(1/3) means CUBE ROOT of x, not x ÷ 3. On most calculators, use the ∛ button or type x ^ (1/3). Don't confuse the fractional exponent with division.

“This question testing understanding of using the calculator to find cube roots generally resulted in partial credit. However, a significant number of candidates thought the index meant division by 3. Others just ignored the index and tried to apply 2 decimal places to the figures 6789.”

CIE 0580 · Paper 2 · Nov 2022

'Sixteen thousand and thirty-seven' = 16,037 (not 16,000,037). Join the thousands and the tens, not concatenate them as separate numbers.

“This question was well answered but some candidates treated sixteen thousand and thirty seven as separate parts producing the response of 16 000 037. Other errors produced responses of 6307, 1637, 17 017 and 1636.”

CIE 0580 · Paper 1 · Jun 2022

Factors of N divide INTO N with no remainder (so they're ≤ N). Multiples are N × 1, N × 2, N × 3... (so they're ≥ N). 1 and N itself are always factors.

“Most candidates easily identified the correct six factors. However, multiples rather than factors were seen, or even a mixture of both. The numbers 8, 9 and 12 were seen at times as factors while some omitted the number 1 from their list.”

CIE 0580 · Paper 1 · Jun 2022

For significant figures, trailing zeros can be ambiguous. To 1 sf: 0.0483 → 0.05 (not 0.050). To 1 sf: 1.98 → 2 (not 2.0). Don't add zeros after rounding.

“Most candidates understood what was required in this question but many responses included trailing zeros. For example, 2.0 instead of just 2 and 0.050 instead of 0.05 were common.”

CIE 0580 · Paper 2 · Nov 2022

To convert seconds to hours, divide by 3600 (= 60 × 60). Dividing by 60 only gives you minutes, not hours. And never multiply — that gives a bigger unit from a smaller one.

“Changing seconds to hours was challenging for many candidates. Dividing by 60 to give an answer of 171, the number of minutes, was the most common error.”

CIE 0580 · Paper 1 · Jun 2022

Identify which digits recur: 0.37̄ (only 7 repeats) = 34/90; 0.3̄7̄ (both 3 and 7 repeat) = 37/99. Place only the recurring digits over the appropriate number of 9s, subtract correctly for non-recurri

“Common incorrect answers were 37/100, 37/90 and 34/90, the latter resulting from confusing 0.3̄7̄ with 0.37̄.”

CIE 0580 · Paper 2 · Jun 2021

In estimation, round each value to 1 significant figure before calculating, not 1 d.p. or to nearest whole number. Also, never change the operation — subtraction in the denominator stays as subtractio

“Two common errors were to round to one decimal place such as 3.0 × 80.0 / (30.0 − 10.0) or to the nearest whole number so 3 × 83 / (28 − 14). Sometimes the subtraction was incorrectly changed to a multiplication in the denominator”

CIE 0580 · Paper 2 · Jun 2021

When checking rounding, read carefully what unit is specified. Many students defaulted to nearest 1000, or confirmed the given rounding rather than finding an alternative to the nearest 100.

“This question was not well understood since many candidates did not realise that an alternative rounding to the nearest 100 was required. An incorrect rounding given was to the nearest 1000”

CIE 0580 · Paper 1 · Jun 2021

Compound interest formula: A = P(1 + r/100)ⁿ. The bracket is (1 + 0.056), not just 0.056. And 'total value' ≠ 'interest' — the interest is A − P.

“Most attempted to use the formula with the most common error being to use 0.056 in the bracket instead of 1.056. Some substituted into the formula correctly and then did not seem to know how to proceed. Candidates should check whether questions like these ask for the total amount of interest or the total value of the investment as a very small number gave the interest only as their answer.”

CIE 0580 · Paper 1 · Nov 2021

Follow BODMAS/BIDMAS: Brackets, Orders (powers), Division/Multiplication (left to right), Addition/Subtraction (left to right). Never evaluate left-to-right when a × or ÷ comes later.

“This question specifically tested order of operations and while it was quite well answered, many did assume the operations needed doing in the order they occurred instead of following the BODMAS convention. This commonly produced the answer of 47 from 9 + 5 calculated first.”

CIE 0580 · Paper 2 · Nov 2021

Small numbers (< 1) have NEGATIVE exponents in standard form: 0.00654 = 6.54 × 10⁻³. Large numbers have positive exponents. Count the decimal places carefully and get the sign right.

“The change to standard form question was well answered but an index of 3, instead of −3 was often seen. Some did not understand standard form, giving answers such as 654 × 10⁻⁵ or answers having no resemblance to a × bⁿ.”

CIE 0580 · Paper 2 · Nov 2021

On 'show all your working' fraction questions: (1) convert mixed numbers to improper fractions, (2) use common denominators, (3) simplify the final fraction. Any answer not fully simplified loses marks.

“Many candidates showed clear working and all the relevant steps required to work out the fraction calculation. A large majority were able to convert the mixed number to an improper fraction. A few then didn't convert to a common denominator or made simple arithmetic errors. If candidates chose 48 as the common denominator they had to cancel their answer of 26/48 down to 13/24.”

CIE 0580 · Paper 1 · Nov 2021

In the repeated division method for HCF, only divide by a prime factor if BOTH numbers are divisible by it. Dividing when only one is divisible corrupts the result. Factor trees are a safer alternativ

“Many spoilt the method by further dividing by 3 and 7 when only one of the two pairs of numbers was divisible by 3 or 7”

CIE 0580 · Paper 2 · Jun 2021

1 hour 43 minutes = 1 + 43/60 = 1.7167 hours, NOT 1.43 or 1.72. There are 60 minutes in one hour, not 100. Keep the full decimal — don't approximate 43/60.

“Errors in the time period usually came from candidates using 100 rather than 60 minutes in one hour. Other errors involved premature approximation of the decimal time interval, e.g. using 1.72.”

CIE 0580 · Paper 4 · Jun 2019

Significant figures always start at the first non-zero digit. To round 0.04762 to 2 s.f.: the first two sig figs are 4 and 7, and because the next digit is 6, round up to 0.047.

“Some candidates did not start the significant figures at the first non-zero digit resulting in answers of 0.0 or 0.5, while others added extra zeros to replace one or all of the digits dropped. While there were many correct responses, some gave 0.046 by not considering the next digit, 8.”

CIE 0580 · Paper 1 · Jun 2019

If the question asks for a mixed number, convert improper fractions (e.g., 7/4 → 1¾). Cancel before multiplying to keep numbers small. And always simplify the final fraction.

“Some, having changed to a multiplication, cancelled appropriately but most did the multiplications of numerators and denominators before cancelling. The final mark was often lost by leaving the answer as an improper fraction instead of following the instruction for a mixed number or occasionally by not having the simplest form of the mixed number.”

CIE 0580 · Paper 2 · Nov 2019

Standard form requires exactly one non-zero digit before the decimal: 2,760,000 = 2.76 × 10⁶. Exact values must not be rounded — 2.7 × 10⁶ would lose the mark.

“Standard form was well known and the vast majority gained the mark. Common errors were 276 × 10⁴, 2.76 × 10⁴ and 27.6 × 10⁵. A significant number of candidates rounded the figures to 2.7 or 2.8.”

CIE 0580 · Paper 1 · Jun 2019

Estimation means round to 1 significant figure: 104.3 → 100, 8.72 → 9, 7.1 → 7. Then calculate with the rounded values. Never use exact values for estimation questions.

“Many candidates wrote 104 instead of 100 for the approximate value of 104.3. Another error was incorrectly rounding 8.72 to 8 instead of 9 while quite a significant number did not attempt any rounding, not following the instructions of the question.”

CIE 0580 · Paper 1 · Jun 2019

To divide fractions: convert mixed numbers to improper fractions, then flip the second fraction and multiply. The final answer must be a simplified mixed number — don't leave as an improper fraction.

“Quite a number of candidates confused the two methods leading to incorrect solutions. Nearly all changed the mixed number to an improper fraction and most of these progressed to one of the correct methods leading to an improper fraction. However, a large number lost the final mark as they did not change their answer to a mixed number or if they did, it was not in its simplest form.”

CIE 0580 · Paper 1 · Jun 2019

To find a fraction between 7/9 and 8/9: convert to equivalent fractions with larger denominators (14/18 and 16/18), then 15/18 fits between them. You cannot write 7.5/9 — fractions need whole number n

“Few candidates appreciated that changing 7/9 to 14/18 and 8/9 to 16/18 gave 15/18 or the equivalent 5/6 as one of the acceptable answers. A very common error was 7.5/9 which is not an acceptable fraction.”

CIE 0580 · Paper 3 · Jun 2019

Speed = distance ÷ time. Time must be in HOURS for km/h. 10 minutes = 10/60 hours = 0.167 hours. Don't multiply distance by time or omit the unit conversion.

“This part was not generally answered well. Common errors included misreading the distance scale and using 30 or 19, misreading the time scale and using 15 or 20 minutes, incorrectly or omitting the conversion of the time units to hours, or the use of an incorrect formula such as 18 × 10/60.”

CIE 0580 · Paper 3 · Jun 2019

306 seconds ÷ 60 = 5.1 minutes = 5 minutes 6 seconds, NOT 5 minutes 10 seconds. The decimal 0.1 minutes = 0.1 × 60 = 6 seconds. Always multiply the decimal part by 60 to get seconds.

“A number of candidates were then unable to convert this time in seconds into minutes. A further common error was writing 5.10 minutes as 5 minutes 10 seconds.”

CIE 0580 · Paper 3 · Jun 2019

For max/min products, apply bounds to EACH quantity first, then multiply. Max of 17.5 × 50.5 = 882.75 (not 17×50+0.5). The upper bound of each must be the one that maximises the product.

“Most candidates answered this correctly. The most common incorrect answer was 850.5 from 17 × 50 + 0.5. Other errors included 16.5 × 50 and 17.05 × 50 and 17.5 × 50.5.”

CIE 0580 · Paper 4 · Jun 2019

For estimation questions, show the rounded values BEFORE the calculation: write '30 × 70 = 2100'. Rounding the answer after calculating is not estimation. Both numbers must be shown rounded.

“These included calculating 29 × 68 = 1972 and then rounding to 2000, or only rounding the parcel value to 1 significant figure and calculating 29 × 70, or omitting to show 30 and 70 before the answer 2100.”

CIE 0580 · Paper 4 · Jun 2019

15 mm = 1.5 cm (divide by 10), NOT 150 cm. For volume: 1 cm³ = 1000 mm³, so divide mm³ by 1000 to get cm³. Track units carefully at every step in multi-unit problems.

“Common errors involved issues with the changes of units including not changing the 15 mm to cm or changing incorrectly, e.g. 150, not dividing by 1000 correctly.”

CIE 0580 · Paper 4 · Jun 2019

1 m = 100 cm, so to convert metres TO centimetres, MULTIPLY by 100 (not divide, and not add zeros arbitrarily). 4.365 m = 436.5 cm.

“Although most candidates gave a correct conversion of metres to centimetres, a considerable number divided by 100 (the reverse process) resulting in 43.65 centimetres. The other common error was to add one zero or three zeros to the number in the question.”

CIE 0580 · Paper 2 · Nov 2019

Standard form a × 10ⁿ requires exactly ONE non-zero digit before the decimal point. So 640000 = 6.4 × 10⁵, not 64 × 10⁴ or 0.64 × 10⁶.

“Most candidates showed good understanding of standard form with many answering both parts correctly... Candidates need to remember that in standard form there is only one digit in front of the decimal point.”

CIE 0580 · Paper 1 · Nov 2019

1 litre = 1000 cm³. To convert cm³ to litres, divide by 1000 (not 100, not 10). This conversion must be memorised as it appears frequently.

“Even with follow through applied, this part was not well answered. Division by 100 was most common but multiplication or division by incorrect powers of 10 were much in evidence.”

CIE 0580 · Paper 2 · Nov 2018

To solve index equations like 3ˣ = 27 × 81, convert all terms to the same base: 3ˣ = 3³ × 3⁴ = 3⁷. Then x = 7 (or −7 if the equation has a negative exponent). Multiplying the numbers first (2187) make

“This question proved quite challenging for a large proportion of candidates, with many multiplying 27 by 81 to get 2187 rather than looking to see if they could make the numbers have the same base as the unknown index number. A significant number of candidates scored 1 mark as they did not deal with the negative power, hence giving an answer of 7.”

CIE 0580 · Paper 2 · Nov 2018

When a question asks for an exact answer, give a fraction (e.g. 4/7), NOT a rounded decimal (0.571...). A decimal that has been rounded or truncated is NOT exact.

“It was a minority of candidates who scored both marks for the exact answer required in this question. Candidates should understand that a rounded or truncated decimal is not an exact value; many got to the correct fraction and then chose to write 0.571… as the answer.”

CIE 0580 · Paper 2 · Nov 2018

Give answers to 3 significant figures unless told otherwise. Never round intermediate values — only round the final answer.

“There was some premature rounding part way through a calculation evident in Questions 8 and 17 and over rounding of the answers to less than three significant figures was evident in Questions 3 and 16(b).”

CIE 0580 · Paper 2 · Jun 2018

To multiply fractions: multiply numerators, multiply denominators. Do NOT invert any fraction (that's for division). Do NOT cross-multiply (that's for solving equations). Always simplify the final ans

“The wrong method seen most often was to invert the second fraction (sometimes changing the × sign to ÷). A further wrong method was to 'cross multiply' leading to their answer of 24/245. many left their answer as 42/140 so could not go on to gain the last mark.”

CIE 0580 · Paper 1 · Nov 2018

Estimation questions require rounding each number to 1 significant figure before calculating. Round 19.3 to 20 (not 19), 8.6 to 9 (not 8.6), 32.1 to 30 (not 32). The rounded calculation should be easy

“This part proved more challenging for many candidates with few appreciating that rounding each number to 1 significant figure was required, and could then be used to give an estimate of the given calculation. Common errors included the use of 19, 32, 8.6, 6.3”

CIE 0580 · Paper 4 · Nov 2018

Estimation questions require rounding each number to 1 significant figure before calculating. Round 19.3 to 20 (not 19), 8.6 to 9 (not 8.6), 32.1 to 30 (not 32). The rounded calculation should be easy

“This part proved more challenging for many candidates with few appreciating that rounding each number to 1 significant figure was required, and could then be used to give an estimate of the given calculation. Common errors included the use of 19, 32, 8.6, 6.3”

CIE 0580 · Paper 4 · Nov 2018

When converting standard form to ordinary numbers with negative powers, count the decimal places carefully. 2.68 × 10⁻³ = 0.00268 — move the decimal point 3 places left, not right.

“Some of the incorrect answers seen were, 2.68, 0.268, 0.0026, 0.002 and 0.003. A few answers of the form 000.268 were also seen.”

CIE 0580 · Paper 1 · Nov 2018

6.2 hours = 6 hours and 0.2×60 = 12 minutes, not 20 minutes. Multiply the decimal part by 60 to get minutes. Don't read 0.2 as '20 minutes'.

“Common errors were to either give the time in the incorrect form, for example, 2.57 or 14.57, or to be unable to convert the time in hours to a time in hours and minutes, for example, 6.2 incorrectly given as 6 hours 20 minutes or 6 hours 2 minutes.”

CIE 0580 · Paper 4 · Jun 2018

When asked to write all digits on your calculator display, give at least 7 figures. Use brackets to ensure cube roots apply to the whole expression.

“A significant number who did not follow the instruction to include all the numbers on their calculator, and specifically gave less than the minimum of seven figures required for the mark. A common incorrect answer of –72.997125… was the result of not bracketing or working out the calculation under the cube root.”

CIE 0580 · Paper 1 · Jun 2018

Give answers to 3 significant figures unless told otherwise. Never round intermediate values — only round the final answer.

“There was some premature rounding part way through a calculation evident in Questions 8 and 17 and over rounding of the answers to less than three significant figures was evident in Questions 3 and 16(b).”

CIE 0580 · Paper 2 · Jun 2018

The compound interest formula P(1+r)^n gives the TOTAL amount, not just the interest. Only subtract the principal if the question asks for the interest earned.

“Quite a significant number of candidates worked out a correct answer but then added the principal since they presumably thought they had found the interest or subtracted the principal as they interpreted the question as finding the interest.”

CIE 0580 · Paper 1 · Jun 2018

For similar shapes: area scale factor = (linear scale factor)², volume scale factor = (linear)³. If lengths are doubled, areas are × 4 and volumes are × 8 — not × 2.

“Many candidates did not recognise that area scale factor was the square of the linear scale factor. Some of those with the correct linear factor gave the area factor as 4 but an area scale factor of 2 was more common.”

CIE 0580 · Paper 4 · Jun 2024

Circle theorem reasons must be exact: 'opposite ANGLES of a cyclic quadrilateral add to 180°' (not 'sides'). Use the precise syllabus vocabulary — 'linear pair' is NOT the CIE phrase.

“Responses with two correct statements and fully correct reasons were in the minority. The most common error was giving incorrect reasons for the angle statements. Some omitted cyclic when referring to the quadrilateral and the use of alternate segment theorem was a common incorrect reason. Incorrectly stating that the opposite sides rather than angles of a cyclic quadrilateral add to 180° was another common error.”

CIE 0580 · Paper 4 · Jun 2024

At each vertex of a polygon, interior angle + exterior angle = 180° (they form a straight line). It's the SUM of all exterior angles that equals 360°, not one pair.

“By far the most common error was to think that the interior and exterior angles summed to 360 instead of 180.”

CIE 0580 · Paper 2 · Jun 2023

For similar solids: length ratio k, area ratio k², volume ratio k³. If you treat a volume ratio as a length ratio, you'll be off by a cube root. To go from volume ratio to length, take ∛.

“Candidates who did not score either treated all scale factors as length scale factors resulting in a common incorrect answer of 148.5.”

CIE 0580 · Paper 2 · Jun 2023

Official angle reasons: 'alternate angles' (Z-shape), 'corresponding angles' (F-shape), 'co-interior angles add to 180°' (C-shape). Learn the CORRECT VOCABULARY — 'Z angles' and 'parallel angles' are NOT accepted.

“Z angles is not accepted as this is the shape that is made with the parallel lines and not the proper reason — alternate angles. Others gave corresponding angles, which is a geometric reason but not for this situation or gave parallel angles which is not a geometric reason. It is important that the reasons are learnt along with suitable diagrams.”

CIE 0580 · Paper 1 · Nov 2023

Hemisphere volume = (1/2) × (4/3)πr³ = (2/3)πr³. Halve the VOLUME formula, not the radius. To solve for r, take the cube root (it's r³, not r²).

“A large number of responses missed the fact that the question was about a hemisphere, not a sphere. If they were able to solve their equation for a full sphere they were able to gain some credit. Some that noticed they were working with a hemisphere but thought they should simply halve their value for r found for a sphere. It was evident that some candidates were using square root for their final step rather than the needed cube root.”

CIE 0580 · Paper 4 · Nov 2022

To go from AREA ratio to LENGTH ratio, take the square root. If area ratio is 9:4, length ratio is 3:2. Don't use the area ratio directly for lengths.

“Success in this part was dependent on understanding that the linear scale factor was the square root of the ratio of the two areas and the order in which to apply it. The most common error was the use of the area factor for the linear scale factor leading to the common answer of 56.7.”

CIE 0580 · Paper 4 · Jun 2022

Angle at centre = 2 × angle at circumference (subtended by the same arc). If AOB = 76° at the centre, then ACB = 38° at the circumference. This is a key circle theorem.

“However, a significant number of candidates made little progress. Although some were able to identify angle AOB as 76°, many did not recognise that this was an example of the angle at the centre and that angle ACB was the corresponding angle at the circumference.”

CIE 0580 · Paper 4 · Jun 2022

For similar solids: if the linear scale factor is k, the VOLUME scale factor is k³ (not k or k²). To find the new volume, multiply old volume by k³.

“For those with a scale factor approach this was often used as a linear factor in (b) when attempting the new volume, with the common wrong answer of 240. Those without a scale factor sometimes approached part (b) attempting calculations of cross-sectional area, but without success.”

CIE 0580 · Paper 4 · Nov 2022

Reverse bearing = original bearing ± 180°. If A→B is 59°, then B→A is 59° + 180° = 239°. Not 360° − 59°, not 180° − 59°.

“Candidates found this reverse bearings question challenging. Very few knew that they had to add 180 and the diagram did not seem to help. 180 − 59 was a common incorrect calculation which indicated no realisation of the angle (clearly reflex) that was required. The other main error was to subtract 59 from 360.”

CIE 0580 · Paper 2 · Nov 2022

To go from VOLUME ratio to LENGTH ratio, take the cube root: linear ratio = ∛(volume ratio). Don't use volume ratio directly as a linear scale — that's off by a cube factor.

“A minority of candidates answered this part question correctly, recognising that 780/32 gave the volume scale factor and that the cube root was required to find the linear scale factor of the heights. The majority of candidates did not cube root the volume scale factor but merely calculated 2 × (780/32) = 48.75.”

CIE 0580 · Paper 4 · Nov 2021

Alternate segment theorem: the angle between a tangent and a chord equals the inscribed angle in the alternate segment. The most common error is applying it to the wrong angle — identify the correct c

“This question was one of the least well-answered questions on the paper with the alternate segment theorem proving a challenge to many candidates. By far the most common error was to think that EHG = 47°, having incorrectly applied the alternate segment theorem. Consequently 86° was the most common incorrect answer.”

CIE 0580 · Paper 2 · Jun 2021

1 m = 100 cm, but 1 m³ = 100³ cm³ = 1,000,000 cm³. For VOLUME conversions you must CUBE the length conversion factor. Dividing by 100 only gives you 1/10,000 of the right answer.

“Candidates found this question challenging. Most attempted the question but many divided by 100 (the length conversion factor for metres to centimetres) only instead of 100³. In virtually all cases, answers contained the correct digits and were smaller than the original number but not the correct place values.”

CIE 0580 · Paper 1 · Nov 2021

A map scale like 1:50,000 is a LINEAR scale. For areas, you must square it: area scale = 1 : 50,000² = 1 : 2,500,000,000. Multiplying map area by the linear scale gives the wrong answer.

“This question proved very challenging and was the least successful on the paper for the majority of candidates, with only a few correct answers were seen. The most common error was to use the scale factor given and multiply the area of the lake of the map by this. Candidates need to be aware that area scale factors are not the same as linear scale factors.”

CIE 0580 · Paper 2 · Nov 2021

If the bearing from K to R is 132°, the back bearing (R to K) is 132° + 180° = 312°. If > 360°, subtract 360°. The back bearing is always ±180° from the forward bearing, NOT 360° − bearing.

“Calculating the bearing of R from K was one of the most challenging questions on the whole paper. Very few correct answers were seen and when done correctly this was generally worked out using 312 – 180. The most common incorrect answer was 48°, from 360 – 312.”

CIE 0580 · Paper 3 · Nov 2021

Length scale k → area scale k² → volume scale k³. To find a volume ratio from a length ratio, cube it. Using the linear factor directly on volume is the #1 mistake.

“Volume scale factors proved to be the most challenging topic on the paper. Candidates need to understand that a length scale has to be converted when working with a volume or area. The majority of candidates used a linear scale factor and divided 12 by 20.”

CIE 0580 · Paper 4 · Nov 2019

The centre of a circle is NOT a vertex of a cyclic quadrilateral. Cyclic quadrilateral theorems apply only to polygons with all vertices ON the circle. Use 'angle at centre = 2 × angle at circumference' instead.

“A minority of candidates were able to correctly calculate the required angle. Many candidates gave an answer of 130°. Candidates sometimes referred to the opposite angle of a cyclic quadrilateral, but used this incorrectly with the centre as one of the vertices of the shape.”

CIE 0580 · Paper 4 · Nov 2019

The angle in a semicircle theorem: any angle inscribed in a semicircle (i.e. subtended by a diameter at the circumference) is 90°. You must state the full reason — just saying 'it's a right angle' wit

“Giving a geometrical reason why angle ABC is 90° was the most challenging question of the whole paper with few correct answers seen. Candidates had to use the circle theorem that angles in a semi-circle are 90°. Most candidates stated that the angle was a right angle without any geometrical reason.”

CIE 0580 · Paper 4 · Nov 2018

Surface area and volume are different — read carefully which part asks for which. A cuboid's faces come in 3 pairs of different-sized rectangles, not all the same.

“The main error was to work out the volume in part (a) and then often the surface area in part (b). Some did not realise there were 3 different size rectangles, repeating for example 10 × 5.”

CIE 0580 · Paper 1 · Jun 2024

When giving a geometric reason, state it fully: 'angles on a straight line add to 180°' (not 'a line is 180'). Partial reasons don't score. Memorise the exact phrases from the syllabus.

“The correct geometrical reason was less successfully stated, and candidates should note that the full reason should be stated and not partial reasons such as 'a line is 180'.”

CIE 0580 · Paper 3 · Jun 2024

For bearing questions, always sketch the situation first. Don't blindly subtract from 360° or 180° — the back-bearing from B to A is (A→B bearing) ± 180°.

“The most successful candidates began their solution by drawing a diagram to illustrate the given bearing. The most common incorrect answers seen were 253° or 73°, obtained by subtracting the given bearing from 360° or from 180°.”

CIE 0580 · Paper 2 · Jun 2024

For trapezium area, the formula (1/2)(a+b)h is more efficient than splitting into triangles + rectangle. If you do split, keep full calculator precision — don't round mid-calculation.

“Many candidates attempted to split the area into two triangles and a rectangle... Many had the correct method, but inaccuracies crept in, usually by prematurely rounding their values part way through the calculation resulting in an answer close to the exact answer of 26.6 but losing the accuracy mark.”

CIE 0580 · Paper 2 · Jun 2024

For vector AB, compute (coordinates of B) − (coordinates of A). Reversing gives BA which is the negative — about 1 in 5 candidates lost the mark this way.

“Many candidates correctly identified the required column vector with the most common error being to find vector BA instead, approximately a fifth of candidates gave the answer [10, 3] [sign flipped].”

CIE 0580 · Paper 2 · Jun 2024

1 m³ = 1000 litres (and 1 litre = 1000 cm³). This conversion is not on the formula sheet — you must memorise it. Many students leave these questions blank.

“The majority of candidates were able to correctly find the volume of the given cylinder in m³, but did not, or could not, convert this into litres.”

CIE 0580 · Paper 3 · Jun 2024

A vector from A to B is written with tail at A and head at B. Reversing the direction gives the negative. Always check the starting point matches the question.

“There were a number of correct answers in finding the vector from a diagram. However, many gave a vector from B to A, rather than the required A to B.”

CIE 0580 · Paper 1 · Jun 2024

If the question says 'describe the single transformation', give exactly one (rotation OR translation OR reflection). Listing two will lose marks even if both are individually correct.

“While most candidates realised the transformation involved rotation, many of them ignored the instruction (in bold) that a single transformation was required.”

CIE 0580 · Paper 1 · Jun 2024

Circumference (2πr) and area (πr²) are different formulas — don't mix them up. Always sanity-check: if the answer is a length, should it be larger or smaller than the radius?

“A basic error often seen was working with the formula for circumference instead of area. Rounding in calculations often caused a mark to be lost... Candidates approaching this question should look at the diagram and realise that the answer had to be greater than 7, the radius of the small circle, but quite a number of responses were less than 7.”

CIE 0580 · Paper 1 · Jun 2024

Perimeter of a sector = arc length + 2 × radius (because the perimeter includes the two straight edges back to the centre). Don't stop after finding just the arc.

“Some candidates calculated the length of the major arc but did not add on the two radii for the perimeter. A small number calculated either the arc length or the perimeter of the minor sector.”

CIE 0580 · Paper 4 · Jun 2024

A full sphere has surface area 4πr². A hemisphere's CURVED surface is half that: 2πr². Add πr² for the flat circular base only if the question includes it (e.g., a bowl).

“Many others included extra areas, usually the plane surface of a hemisphere and/or the base of a cone. A wide variety of errors were seen when candidates attempted to solve their equation. Writing the surface area of the hemisphere as 4πr² was a common error.”

CIE 0580 · Paper 4 · Jun 2024

Bearings are always measured clockwise from the North direction. Practise measuring with a protractor from North — not from the line you're drawing.

“The bearing of 117° proved to be the more challenging. Common errors included drawing the bearing of 063° and 243°.”

CIE 0580 · Paper 3 · Jun 2024

To describe an enlargement fully, state THREE things: (1) 'Enlargement', (2) scale factor, (3) centre as a coordinate. Missing the centre loses a mark.

“This part was generally answered reasonably well with a good number of candidates able to identify the given transformation as an enlargement and able to correctly state the three required components. The identification of the centre of enlargement proved the more challenging with a significant number omitting this part.”

CIE 0580 · Paper 3 · Jun 2023

Area of a triangle = ½ × base × perpendicular height. Don't measure all 3 sides. 'Base' and 'height' must be perpendicular to each other.

“A significant number did not appreciate that the formula 1/2 × 6 × 3 could be used to give the area. Common errors included a variety of incorrect formulas used, answer of 18, inaccurate and unnecessary measurement of the 3 sides, multiplying or adding their 3 sides.”

CIE 0580 · Paper 3 · Jun 2023

For bearing problems involving non-right triangles: (1) find the unknown angle using sine or cosine rule, (2) combine it with the NORTH direction at the point of interest to get the bearing.

“In this problem-solving question candidates needed to realise they had to first find an acute angle in the triangle before applying their knowledge of bearings. A number wrongly assumed the triangle was isosceles and often then also were unable to deduce a bearing from their angle UWV.”

CIE 0580 · Paper 4 · Nov 2023

Rule of thumb: for a 3-sig-fig final answer, keep at least 4 sig figs at every intermediate step. Rounding mid-calculation gives the 'almost right but not accepted' kind of answer.

“At least 4 figure accuracy should be used throughout the steps of the method if the final answer is to have 3 significant figure accuracy. Quite a few rounded their angle to 2 or 3 significant figures in the working leading to a common inaccurate answer of 6.11.”

CIE 0580 · Paper 2 · Jun 2023

For similar triangles, set up ratios of CORRESPONDING sides — the shortest of one triangle matches the shortest of the other. Don't subtract lengths: similarity is about proportion, not difference.

“Candidates who did not score were often inconsistent with the scale factor, stating for example 7.5/3 = 2.7/XC. The weakest candidates subtracted 3 from 7.5 to get length XC as 4.5.”

CIE 0580 · Paper 2 · Jun 2023

Total surface area of a closed cylinder = 2πrh + 2πr² = 2πr(r + h). You need BOTH the curved part AND the two circular ends. Memorise this — it's not on the formula sheet.

“The vast majority of candidates did not know that the total area for the cylinder meant using the formula 2πrh + 2πr² or more simply 2πr(r + h). The few who used that correctly and got part (a) correct gained full credit. Most just took the curved surface area or the area of one or two ends as the total surface area.”

CIE 0580 · Paper 1 · Jun 2023

Pythagoras: c² = a² + b² means ADD when finding the hypotenuse, SUBTRACT when finding a leg. If you're looking for the longest side, you must add — check your answer against the diagram.

“A considerable number of candidates did not compare their answers to the diagram. Clearly an answer for AC had to be greater than 52 and so 34.4 from squaring and subtracting had to be incorrect.”

CIE 0580 · Paper 1 · Jun 2023

Magnitude of a vector (x, y) = √(x² + y²). Use BRACKETS around negative numbers before squaring: (−4)² = 16, not −4² = −16. And magnitude is always positive.

“Only a minority of candidates understood what was required of them to find the magnitude of the vector in part (a), with the elements instead combined in various different incorrect calculations. Of those realising that Pythagoras' Theorem was needed many used −4² rather than (−4)² and so did not score.”

CIE 0580 · Paper 4 · Nov 2023

The reason for right angle in semicircle is 'angle in a semicircle is 90°' — NOT 'triangle in a semicircle'. You must mention the 90° or 'right angle'.

“The reason was often not expressed accurately and often a triangle, rather than an angle, in a semicircle was given. The explanation also needed reference to the angle of 90° (or right angle).”

CIE 0580 · Paper 1 · Jun 2023

Sum of interior angles of an n-sided polygon = (n − 2) × 180°, NOT (n − 1) × 180°. Divide by n to get one interior angle of a regular polygon.

“A formula was often quoted but a significant number of responses had (n – 1) instead of (n – 2). Other errors were from 180 × 7 and 360 ÷ 9 without the further necessary steps.”

CIE 0580 · Paper 1 · Jun 2023

Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment. Know this as a circle theorem — faster than multi-step triangle calculations.

“In part (b), the correct answer of 42° for angle QSR could most quickly be reached using the Alternate Segment Theorem. As QS was clearly a diameter many candidates instead subtracted 42° from the right angle at Q, but 48° was then often incorrectly given as the answer for angle QSR.”

CIE 0580 · Paper 2 · Nov 2023

Enlargement description: SCALE FACTOR (from original to image, e.g., 2 if image is bigger), and CENTRE as coordinates (not a vector). Always include all three: 'enlargement', SF, and centre.

“Some considered the transformation in the wrong direction, giving a scale factor of 1/2. The most commonly omitted property for the transformation was the centre of enlargement. A very small number of candidates gave a vector for the centre of enlargement rather than coordinates; this was only acceptable if referenced as a position vector for the centre.”

CIE 0580 · Paper 2 · Nov 2023

Volume of a hemisphere = (1/2) × (4/3)πr³ = (2/3)πr³. Don't use the full sphere formula — always halve it. Similarly, a hemisphere's curved surface is 2πr² (half of 4πr²).

“Using the formula for the volume of a sphere rather than a hemisphere was a common error.”

CIE 0580 · Paper 4 · Jun 2023

For similar triangles, use PROPORTIONS of corresponding sides, not subtraction. Sanity check: the unknown side must lie between the known corresponding sides of the triangles.

“The frequent incorrect response was 6.5 from finding the difference between 8.1 and 7.2 and adding that to 5.6 for h. Looking at the triangles, h needs to be larger than 5.6 and smaller than 8.1 so if candidates had thought about the context they should have realised that answers of 9.7 or 4.48 were incorrect.”

CIE 0580 · Paper 1 · Nov 2023

Pythagoras: if you're finding a leg (not the hypotenuse), SUBTRACT the squares: a² = c² − b². Check: any leg must be SHORTER than the hypotenuse. If your answer is bigger, you added when you should have subtracted.

“Most responses began with a method to calculate PS. The diagram indicated that PS had to be less than 23.8 cm, but some methods added the square of the lengths rather than subtracted leading to answers greater than the length of the hypotenuse.”

CIE 0580 · Paper 2 · Nov 2023

Volume of sphere = (4/3)πr³. If given the diameter d, use r = d/2 — don't use d directly. And keep 4/3 as a fraction in your calculator, don't round to 1.3.

“Only a minority of candidates gave a correct answer. Some substituted the diameter given in the question instead of the radius. Some multiplied the diameter value by 2 before substituting. Others rounded 4/3 to 1.3 before putting the calculation into their calculators.”

CIE 0580 · Paper 1 · Nov 2023

When an angle is formed at the circumference by a diameter, state the reason: 'angle in a semicircle is 90°'. Knowing it's a right angle isn't enough — you must cite the theorem.

“58° was seen often, but very few candidates were able to give the correct geometrical reason. Most knew ACB was a right-angle, but not the wording of the circle theorem.”

CIE 0580 · Paper 3 · Nov 2023

Angle sum of n-sided polygon = (n − 2) × 180°. Pentagon: (5 − 2) × 180° = 540°, not 360°. Each interior angle of a regular pentagon = 540/5 = 108°.

“Many candidates were aware they needed to divide the angle sum by 30 although most did not find the correct angle sum for a pentagon of 540°. Most gained credit for knowing a multiple of 180° was needed for the division but this was very commonly 360°, leading to an incorrect answer of (12 × 9 =) 108°.”

CIE 0580 · Paper 4 · Nov 2022

The formula (1/2)ab sin C requires C to be the angle BETWEEN sides a and b (the 'included' angle). Using the wrong angle gives the wrong answer. Label your triangle carefully.

“Almost all answers correctly identified that the area could be found using (1/2)ab sin C and usually obtained the correct answer... In some cases, the angle chosen was not the included angle for the two sides being used.”

CIE 0580 · Paper 4 · Jun 2022

Reverse bearing from B→A = (A→B bearing) + 180° (subtract 360° if > 360°). It's NOT (360° − bearing) or (180° − bearing).

“This part was generally not answered well with the majority unable to work out the reverse bearing required. The very common error was the answer of 38.”

CIE 0580 · Paper 3 · Jun 2022

To find how many small boxes fit in a big container, divide the volumes (assuming perfect packing). Convert units first: if container is 1m³ and box is 2×4×5 cm, then container = 100³ cm³, so boxes = 1,000,000 / 40.

“The most common method used was (100 × 100 × 100) ÷ (2 × 4 × 5). Common errors included 100 ÷ 40, 1000 ÷ 40, 1 ÷ 40, resulting from incorrect conversion of units.”

CIE 0580 · Paper 3 · Jun 2022

Reasons in geometry must be complete statements: 'angles in a triangle sum to 180°' — not 'a triangle is 180'. And 'base angles of an isosceles triangle are equal' — not 'two angles are equal'.

“The correct geometrical reason was less successfully stated, and candidates should note that the full reason should be stated and not partial reasons such as 'a triangle is 180' and 'two angles are equal'.”

CIE 0580 · Paper 3 · Jun 2022

For vector questions, START by writing a clear route: e.g., PQ = PO + OR + RQ. This earns the method mark even if the algebra after is wrong. Without a stated route, method marks are lost.

“The best responses began by stating the route first, usually PO + OR + RQ which was sufficient for the first method mark. Answers that did not score often had not clearly planned a route.”

CIE 0580 · Paper 2 · Jun 2022

For a 3sf final answer, keep interim values to at least 4-5 sf (or use calculator memory). Rounding to 3 sf mid-calculation is the #1 cause of answers that are 'almost right but outside tolerance'.

“Candidates are advised that, to reach an answer correct to 3 significant figures, any rounding of interim figures cannot be to 3 significant figures. Candidates should use calculator memories or write down interim values to a higher degree of accuracy.”

CIE 0580 · Paper 2 · Jun 2022

For the height of an isosceles triangle, use HALF the base in Pythagoras: h² = slant² − (base/2)². SUBTRACT the squared values (you're finding a leg, not the hypotenuse).

“Although candidates recognised the need to use Pythagoras' theorem, several did not halve the base and used the value 4 rather than 2. Others added rather than subtracted the squared values.”

CIE 0580 · Paper 3 · Nov 2022

If a shape is explicitly called a trapezium, USE the formula A = (1/2)(a + b)h. It's faster than splitting. And find the perpendicular height (not a slant side) — you may need Pythagoras.

“Although the question stated that the shape was a trapezium, very few chose to use the formula, and instead calculated the area of the rectangle and the triangle separately. To find the area of the triangle or the trapezium, Pythagoras' theorem had to be applied to find the height of the triangle.”

CIE 0580 · Paper 1 · Nov 2022

For similar triangles, DIVIDE by the scale factor to go from larger to smaller. If the scale factor (larger/smaller) is 4/3, then smaller side = larger side × 3/4. Sanity check: the answer should be smaller.

“This question involving similar triangles was found challenging. Many wrote 8 − 6 = 2 and 6 − 2 = 4. Some who found a scale factor of 4/3 from the first triangle then multiplied 6 by this, instead of dividing, so wrote the answer 8. It is important to consider the entire problem and to realise the missing side will be smaller than 6 so if calculations give a larger value, then everything should be checked to find the errors.”

CIE 0580 · Paper 1 · Nov 2022

Alternate segment theorem: the angle between a tangent and a chord at the point of contact = angle in the alternate segment. Learn this — it's a quick shortcut to tricky circle problems.

“Only a minority of candidates recognised that x was the same as the given 38°, using the alternate segment theorem. The sum of opposite angles in a cyclic quadrilateral, or alternate angles in parallel lines, were better known.”

CIE 0580 · Paper 4 · Nov 2022

For composite length problems, expect multi-step: find an intermediate length first (e.g., a shared side), then use Pythagoras/trig on a second triangle. A 4-mark question usually means 2+ stages.

“The majority of candidates realised that the required side could only be found after the vertical common side of the two triangles was calculated. The main challenge was in finding this length but provided some value was indicated for it, a method mark for correct Pythagoras' theorem in the second part could be gained... With 4 marks for the question, candidates needed to realise that length BC could not be found from a single calculation.”

CIE 0580 · Paper 2 · Nov 2022

For a cone + hemisphere shape: curved hemisphere = 2πr², curved cone = πrl. If they're joined at the base, DON'T include the flat circle (it's inside). Check if the shape is sealed or open before deciding which areas to include.

“A common error was not halving the hemisphere. Premature rounding was common, leading to a loss of accuracy in subsequent calculations. Another common error was including the area of the circle from the hemisphere, many went straight to a hemisphere formula of 3πr² rather than adapting the given sphere formula to the situation.”

CIE 0580 · Paper 2 · Nov 2022

3D diagonal (vertex to opposite vertex, NOT sharing a face): d² = l² + w² + h². Work it in ONE step to avoid rounding errors. d = √(l² + w² + h²).

“Whilst some candidates worked through to find the correct answer many simply found the length of a diagonal of one face. (Candidates need to know that a diagonal in a solid shape joins vertices that do not share a face.) Some correctly applied Pythagoras' theorem in 3D in a single calculation, whilst those working in stages were more likely to introduce rounding errors.”

CIE 0580 · Paper 4 · Nov 2022

For back-bearing problems: (1) draw the North line at BOTH points, (2) use co-interior angles (sum to 180°) between parallel North lines to find the angle at the second point. A clear diagram is essential.

“The best answers used a diagram with the North line drawn in at B. Of those who gave the correct answer, many used the diagram and the method 180 + 59 = 239 with the diagram split up to show the North line, 180 and 59 or found 121 using co-interior angles between the parallel North lines then 360 − 121.”

CIE 0580 · Paper 4 · Nov 2022

Properties of a rhombus must be specific enough to distinguish it from other quadrilaterals. 'Opposite sides equal' is true for parallelograms too. Give properties from two different categories: sides

“Only a small number of candidates gained full credit for the properties of a rhombus. General properties of quadrilaterals, 360° or 4 sides were not specific enough and properties such as opposite sides equal did not distinguish between a parallelogram and a rhombus.”

CIE 0580 · Paper 1 · Jun 2021

In similar shapes, corresponding lengths are in a constant ratio — use multiplication/division, never addition or subtraction. Also, do not round the scale factor before applying it to other lengths.

“the significant error was to add or subtract corresponding lengths. Some of those showing correct working lost accuracy by rounding, for example, 3.2 ÷ 2.8 to 1.14, before multiplying by 1.61, producing a close but not exact answer to the question.”

CIE 0580 · Paper 1 · Jun 2021

For composite shapes: (1) triangle area = ½ × base × height — don't forget to halve; (2) use sector area = (θ/360)πr², not full circle area; (3) use the calculator's π key, not 3.14 or 22/7.

“Many gained partial credit for the area of the triangle, although not dividing by 2 was quite common. Often it was just the area of the whole circle that was added to the triangle area. Candidates should use the π key on the calculator or 3.142 since using 3.14 or 22/7 didn't gain the accuracy mark.”

CIE 0580 · Paper 1 · Jun 2021

In bearing problems with regular polygons, always use the interior angles of the shape. For an equilateral triangle, each interior angle is 60°. Ignoring this property is the most common error.

“A large proportion of candidates did not use the essential information given, that it was an equilateral triangle, so working such as 180 + 103 and 360 – 103 were common.”

CIE 0580 · Paper 2 · Jun 2021

For similar 3D shapes: if the linear scale factor is k, the volume scale factor is k³. To compare volumes, set up (V₁/V₂) = (l₁/l₂)³. Using the linear or area ratio for volumes gives the wrong answer.

“The most common error was to use a linear factor, from 11.5/x = 7.8/4 to reach the common incorrect answers of 5.897 or 5.9 or even 6. Less common was the use of the area factor (11.5/x)² = 7.8/4 to give 8.24. A significant number of candidates used the incorrect volume factor from (11.5/x)³ = 7.8/4”

CIE 0580 · Paper 2 · Jun 2021

An enlargement is a single transformation — never describe it as two. Verify the scale factor by dividing any image length by the corresponding original length. Negative scale factors indicate the ima

“Only a minority of candidates successfully identified the correct enlargement and its properties. The most common error involved an incorrect scale factor, usually 2, –2 or 1/2, followed closely by an incorrect centre of enlargement. A significant number of candidates gave a combination of two separate transformations.”

CIE 0580 · Paper 4 · Jun 2021

Every step in a geometric proof needs its own reason, using precise terminology: 'alternate angles' (not 'alternative'), 'angle in a semicircle = 90°', 'angles in the same segment are equal'. Missing

“Not all candidates used the correct terminology when giving their reasons, for example, when explaining why the angle PRQ is 90° and incorrectly using 'alternate segment' or 'alternative angles' instead of alternate. Some candidates, however, did not give reasons for every stage of their working”

CIE 0580 · Paper 4 · Jun 2021

To describe an enlargement: (1) say 'enlargement', (2) scale factor, (3) centre as a coordinate. All three. Omitting any one loses a mark.

“Good solutions in this part contained the correct transformation, enlargement, and the correct scale factor (4) and centre (–4, –5). The most common error was to omit the centre of enlargement or give the incorrect scale factor.”

CIE 0580 · Paper 3 · Nov 2021

The reason is 'angle in a semicircle is 90°' (or 'angle subtended by diameter is 90°'). Saying 'it's a right-angled triangle' just restates — it's not a REASON.

“Few candidates were able to give the correct wording in the statement to gain credit. The correct statement should be because 'angle in a semicircle'. The most common response was that the triangle was a right-angled triangle, which is just repeating the information given in the question and not explaining why.”

CIE 0580 · Paper 3 · Nov 2021

'Total surface area' of a cone includes the curved part (πrl) AND the circular base (πr²). Of a hemisphere: 2πr² curved + πr² flat. Don't skip the flat faces.

“It was clear that most candidates were able to use the given formulae to correctly work out the curved surface area of the cone. However, the majority of candidates omitted the flat circular parts of the surface areas in either or both of the solids.”

CIE 0580 · Paper 4 · Nov 2021

For a cone with top removed, the small cone is similar — its radius scales by the same ratio as its height. If height ratio is 1:4, radius ratio is also 1:4.

“Finding the volume of the remaining solid proved more challenging with most candidates taking the approach of trying to find and subtract the volume of the small cone that had been removed. Of those finding the volume of the small cone, many did not use the correct radius or height with few candidates recognising that the radius of the small cone was 1/4 × 7.6.”

CIE 0580 · Paper 4 · Nov 2021

Volume = base area × height. To find height, divide volume by the BASE AREA (e.g., 6 × 6 = 36 for a square base with side 6), not just one side length.

“This problem solving question was found challenging by some candidates. The base of the cuboid is a square so the volume must be divided by the area of the base to find the height, i.e 180 ÷ (6 × 6). However many candidates only divided 180 by 6 giving 30.”

CIE 0580 · Paper 1 · Nov 2019

The angle at the centre is twice the angle at the circumference in the same segment. Don't assume a triangle is isosceles without proof. Check which circle theorem applies at each step.

“Only the best answers recognised angle PBC as half of angle AOP. The common incorrect methods assumed that triangle PBC was isosceles and from this incorrect assumption they calculated w as 62.5°, 55° or 70°. Another common incorrect method was to think that OAP and BCP were angles in the same segment and so w = 21° was also a common incorrect answer.”

CIE 0580 · Paper 2 · Jun 2019

Volume per hour = (cross-section area) × (length of water per hour). Convert 12 cm/s to cm/hour: × 60 × 60 = 43,200 cm/h. Then convert cm³ to litres: ÷ 1000. Keep track of units at every step.

“There were few who managed to do all of these steps correctly. Some common incorrect answers were 0.255 from those who found the number of litres per second, 15.3 which is the number of litres per minute, 255 which is the volume in cm³ per second or 43.2 which does not take into account the cross-sectional area.”

CIE 0580 · Paper 2 · Jun 2019

For similar triangles, say 'corresponding angles are equal' (or 'all 3 pairs of angles are equal'). 'Congruent' means identical — different from similar. 'All angles are the same' is too vague.

“Some of the answers showed a lack of precision in their wording as it was common to see, 'All angles are the same' when it should be that the angles in the corresponding place in each triangle are the same... The use of the word 'congruent' was incorrect.”

CIE 0580 · Paper 1 · Nov 2019

For similar triangles, use a SCALE FACTOR (divide corresponding sides), not subtraction. If one side is 12 and its pair is 18, the scale factor is 18/12 = 1.5, not '+6'.

“A frequent incorrect method of dealing with similar triangles was for candidates to write 18 − 12 = 6 so 27 − 12 = 15.”

CIE 0580 · Paper 1 · Nov 2019

Volume of a cuboid = length × width × height. If a cuboid is made of unit cubes, count the cubes along each edge and multiply. Volume ≠ surface area — different concepts.

“Had the dimensions of the cuboid been given numerically it is likely that almost all candidates would have found the correct volume. In fact many did not realise that by counting the number of cubes in length, breadth and height, it was a straightforward volume calculation. The most common error was to find the surface area of the cuboid leading to a common answer of 192.”

CIE 0580 · Paper 2 · Nov 2019

A POSITION vector is from the origin O to a point (e.g., OP). A DISPLACEMENT vector is between two other points (e.g., AB). These are different — position vectors always start at O.

“A position vector was required in part (b) and it was clear that a large proportion of candidates did not understand this term. It was common to see the answer as half of the answer for part (a).”

CIE 0580 · Paper 4 · Nov 2019

Exterior angle method: exterior angle = 180° − interior angle; number of sides = 360° ÷ exterior angle. For interior 162°: exterior = 18°, sides = 360 ÷ 18 = 20. This method is easier and less error-p

“Very few candidates used the exterior angle method of 360 ÷ (180 – 162) but those who did usually were successful and scored full marks. A small number were able to use the interior angle method to score full marks, but the majority of candidates who used this method usually made errors in the formula used.”

CIE 0580 · Paper 3 · Jun 2019

Always calculate the sector angle from the given geometry — don't assume 45° or use the wrong arc. Check whether you need the minor or major sector angle for the required area/perimeter.

“Common misconceptions included assuming that angle DOE was 45° or that the sector area could be found using 270° and subtracting the area of the triangle ODE or using the minor sector angle 150° rather than 210°.”

CIE 0580 · Paper 4 · Jun 2019

To find a reverse (back) bearing: if original is ≥ 180°, subtract 180°; if < 180°, add 180°. Always draw a sketch with north arrows at both points. 360 − bearing gives the wrong answer.

“This part was not generally answered well, with few candidates appreciating the method to be used when finding a reverse bearing. A very common error was 42, possibly from 360 – 318. Few candidates drew a sketch to help them with this part.”

CIE 0580 · Paper 3 · Jun 2019

Surface area of a cuboid = 2(lw + lh + wh) — there are 6 faces in 3 PAIRS. Don't calculate volume by mistake. A cuboid has only 3 different face shapes, each repeated twice.

“A significant number of candidates found the volume, rather than the surface area. A correct approach often just resulted in three areas found rather than the required six. Some managed to find one area but then got mixed up between adding and multiplying lengths and areas. A number thought that four of the faces were 12 by 7.5.”

CIE 0580 · Paper 1 · Jun 2019

For back bearings: if bearing A to B is less than 180°, bearing B to A = bearing + 180°. Don't subtract from 360 or 180 directly without a clear diagram sketch.

“This bearings question was one of the more challenging questions on the paper for candidates although many earned full marks. Common incorrect working and answers included 180 – 128 = 52 and 360 – 128 = 232.”

CIE 0580 · Paper 2 · Jun 2019

Area scale factor = (linear scale factor)². To find the linear scale factor from areas, you must SQUARE ROOT the area ratio. Forgetting the square root is the most common error.

“It was common to see the answers 15.7 or 13.122 arising from forgetting to square root and using the area scale factor with 8.6 or 7.2 respectively.”

CIE 0580 · Paper 2 · Jun 2018

To find a back bearing (return direction): if bearing < 180°, add 180°; if bearing > 180°, subtract 180°. Draw a sketch showing north lines at both points — this makes it visual.

“The required reverse bearing proved very challenging for many candidates and proved to be a good discriminator. Many candidates did not appear to appreciate the calculation required was either 180 + 73 or 360 – 107. Those candidates who drew a sketch diagram tended to be more successful. Common errors included 360 – 73 = 287, 180 – 73 = 107 and 90 + 73 = 163.”

CIE 0580 · Paper 3 · Jun 2018

To estimate the area of an irregular shape on a grid, count all full squares plus approximately half the partial squares. Don't use circle formulas for non-circular shapes.

“This was poorly answered with many candidates clearly not aware that the intended method was to count squares, complete and partial, to estimate. A variety of calculations were attempted with the most common π × 3.5², even though the shape was clearly not a circle.”

CIE 0580 · Paper 1 · Jun 2018

When enlarging, draw ray lines FROM the centre of enlargement THROUGH each vertex of the original shape. The image vertex lies on the ray at (scale factor × distance from centre). Starting from the sh

“Many candidates drew an enlargement of the correct size but did not use the given centre of enlargement. Some attempted an enlargement from the correct place of 4 times the original size as if they started counting 3 times the length of rays from the shape instead of starting at the centre of enlargement.”

CIE 0580 · Paper 1 · Nov 2018

Back bearing = bearing ± 180°. If bearing < 180°, add 180°. If bearing ≥ 180°, subtract 180°. Do NOT subtract from 360°. Drawing a diagram with north lines at both points helps visualise this.

“The back bearing question was by far the most challenging question on the paper. Very few seemed to know that the difference between the given value and the answer was 180. Subtracting from 360 giving 247 was the most common error although 180 – 113 = 67 occurred often.”

CIE 0580 · Paper 2 · Nov 2018

Circle theorem reasons must use precise mathematical vocabulary: 'centre' (not 'middle' or 'origin'), 'circumference' (not 'edge' or 'perimeter'), 'angle at centre is twice angle at circumference'. In

“Very few candidates gained full marks for stating their reasons. As a minimum it was necessary to state isosceles triangle and to use the correct vocabulary of centre and circumference along with either double, twice or half as appropriate. Words such as middle, origin, top, edge and perimeter are not acceptable alternatives.”

CIE 0580 · Paper 4 · Nov 2018

The angle at the centre is TWICE the angle at the circumference (subtended by the same arc). If the circumference angle is 48°, the centre angle is 96° — not the other way round. Getting this reversed

“Many remembered something about angles at the centre and circumference but got them the wrong way round and gave PRQ as 96 and hence 96 as their answer.”

CIE 0580 · Paper 4 · Nov 2018

For similar solids, volume ratio = (linear ratio)³. So to go from volume ratio back to linear ratio, take the CUBE ROOT. If volume ratio = 64, the linear scale factor = ³√64 = 4, not 64 × (some length

“whilst many realised that 19200 had to be divided by 300, few then cube rooted 64 but simply multiplied the answer of 64 by 1.6.”

CIE 0580 · Paper 4 · Nov 2018

The volume of a hemisphere is HALF the volume of a sphere: V = (2/3)πr³, not (4/3)πr³. When a question involves a hemisphere, always halve the sphere formula. Forgetting to halve is a very common erro

“a very significant error was that many candidates did not recognise the need to adapt the given formula for the volume of a sphere to that of a hemisphere.”

CIE 0580 · Paper 4 · Nov 2018

For similar 3D shapes: if linear ratio is k, volume ratio is k³. When working backwards from volume ratio to linear: take the CUBE root. Using the ratio directly, squaring it, or taking the square roo

“In part (b) the common error was to use the ratios as they are, leading to the common incorrect answer of 6.5, or the square or the square root of the ratios.”

CIE 0580 · Paper 4 · Nov 2018

If the linear scale factor between similar shapes is k, the area scale factor is k² (not k). For a base ratio of 12:3 = 4:1, the area ratio is 4² : 1² = 16:1. Applying the linear scale factor directly

“Some candidates were unable to gain any marks on this question as they applied the linear scale factor to the area of the small triangle, showing no appreciation of the link between linear and area scale factors. This gave the most common incorrect answer 20. Some others knew that squaring was involved but squared the 5 rather than the 12/3”

CIE 0580 · Paper 4 · Nov 2018

In a right-angled triangle, to find a shorter side: c² = a² + b² rearranges to a² = c² − b². Subtracting the sides directly (without squaring) is wrong. Also, the hypotenuse is always the longest side

“Many candidates did not apply an obvious case of Pythagoras' theorem to this question and simply subtracted the two sides to give 11.7 m. Many squared and added resulting in an answer of 42.2 which should have been realised as not possible.”

CIE 0580 · Paper 2 · Nov 2018

If the linear scale factor of an enlargement is k, the area scale factor is k². For a scale factor of 1/3: area factor = (1/3)² = 1/9, NOT 1/3. Calculate actual areas to verify if unsure.

“A large number of candidates gave the incorrect answer of 1/3 assuming that the area scale factor was the same as the scale factor of the enlargement. The expected method of finding and comparing the two areas was rarely seen.”

CIE 0580 · Paper 4 · Nov 2018

Read the question carefully: volume and surface area are different. Volume = space inside (measured in cm³). Surface area = total area of all faces (measured in cm²). Calculating the wrong one earns z

“The majority of candidates were unsuccessful as they calculated the volume rather than the surface area which the question asked for.”

CIE 0580 · Paper 3 · Nov 2018

Interior angle = 180° − exterior angle. Always check: if the polygon is regular and the question asks for interior angles, don't stop at the exterior.

“Many good responses were seen for this question. A number of different, valid methods were seen, many with fully correct arithmetical working. Some responses gave the exterior angle as the answer rather than the interior angle.”

CIE 0580 · Paper 2 · Nov 2023

Students confuse the angle of rotation with the order of rotational symmetry. The order is how many times the shape maps onto itself in one full turn — not the angle turned each time.

“angles of 180 or 360, as well as the number 4, were seen a number of times”

CIE 0580 · Paper 1 · Jun 2021

Read the question carefully. If the cross-section area = 12 × side², then side = √(area/12) — square root, not cube root. Don't automatically cube-root because it's a volume context.

“Many candidates made a good start to this question with division of the volume by 12. However, many then did not realise they needed to find the square root and divided by 2, 4 or 6. Some who calculated the square root rounded the division by 12 to 3 significant figures producing inaccuracy in their answers. Since the question involved a volume, a significant number found the cube root.”

CIE 0580 · Paper 2 · Nov 2021

Arc length = (angle / 360) × circumference = (θ/360) × 2πr. Use your calculator's π button for accuracy, not 3.14 or 22/7.

“While many did work out the circumference, few knew to multiply by 72/360 and instead tried fractions such as a quarter or a third. Some candidates used 3.14 or 22/7 for π.”

CIE 0580 · Paper 2 · Nov 2021

In 'show that' questions, you must show at least 4 significant figures of working before rounding to the given value. Writing 41.7 = 41.7 is circular — you must show the full unrounded value first.

“Giving an answer of 41.7 was a very common error. Candidates needed to provide an answer to at least four figures in order to show that it rounded to 41.7̄.”

CIE 0580 · Paper 4 · Jun 2021

Geometric reasons must use precise, complete mathematical language: e.g. 'angles on a straight line sum to 180°', 'alternate angles are equal (parallel lines)', 'co-interior angles sum to 180°'. Vague

“Candidates should be reminded that when a geometrical reason is asked for it must be correct, complete and use the relevant mathematical terms.”

CIE 0580 · Paper 3 · Jun 2021

A complete description of an enlargement requires three things: the word 'enlargement', the scale factor (check it carefully — not a nearby integer), and the coordinates of the centre of enlargement.

“The identification of the centre of enlargement proved the more challenging with a significant number omitting this part, and (0, 0), (4, –3) and (3, 4) being common errors. The scale factor was sometimes omitted or incomplete, with 2 being the common error.”

CIE 0580 · Paper 3 · Jun 2021

Do not round the scale factor before applying it. Calculate 3.2 ÷ 2.8 exactly, keep full precision, then multiply — rounding to 1.14 introduces enough error to lose the accuracy mark.

“The most common error was to calculate 3.2/2.8 as 1.1 or 1.14 giving an inaccurate answer as a result of this premature rounding.”

CIE 0580 · Paper 2 · Jun 2021

Locus of points a fixed distance r from a point P is an ARC (or circle) centred at P with radius r. Use compasses set to the correct radius — don't freehand.

“Most candidates understood that the locus meant an arc, centre C, although a few used an incorrect centre or radius. There were quite a lot of blank responses for this part.”

CIE 0580 · Paper 2 · Nov 2019

For Pythagoras with a shorter side: subtract the squares, don't add. Give at least 3 significant figures — answers of 2.9 or 3 lose the accuracy mark. Show intermediate values to full precision.

“While many candidates correctly applied Pythagoras' theorem, the final mark was often lost for answers of 2.9 or 3 without more accuracy in the working. The common error was to square and add the sides, leading to an answer of 6.89.”

CIE 0580 · Paper 1 · Jun 2019

A running track has two straight sides and TWO SEMICIRCLES (not two full circles). The semicircles combine to make one circle. Perimeter = 2 × straight + π × diameter.

“Common errors included, answers of 360 from just considering the rectangle; 548 from 428 + 60 + 60; finding the area of the running track rather than the total length; using two circles rather than two semi-circles.”

CIE 0580 · Paper 3 · Jun 2019

Area of semicircular annulus = ½π(R² − r²). You cannot add/subtract the radii first then square. The correct formula uses the DIFFERENCE OF THE SQUARES: R² − r² = 49 − 9 = 40.

“Common errors included the use of ½ × π × 7² + ½ × π × 3²; π × 7² ± π × 3²; ½ × π × (7 + 3)²; ½ × π × (7 – 3)²; incorrect formulae; calculating the perimeter.”

CIE 0580 · Paper 3 · Jun 2019

For enlargement scale factor −2 from centre (3,0): each vertex goes THROUGH the centre and the image vertex is twice as far on the OTHER SIDE. Draw rays through the centre and measure distance careful

“Most drew 'ray' lines from the vertices of triangle A, through (3, 0) but did not give the correct image, either as a result of assuming a scale factor of 1/2 or using an incorrect distance from (3, 0).”

CIE 0580 · Paper 4 · Jun 2019

Always sanity-check geometric answers against the diagram. If the answer can't physically be bigger than a given length, it's wrong — rework it.

“There was a good response to this question with many correct answers seen. The common incorrect method was 7.7 − 5.5 = 2.2 and then 9.8 − 2.2, leading to the answer of 7.6. Some gave an answer greater than 9.8 which should have been obviously incorrect from the diagram.”

CIE 0580 · Paper 3 · Nov 2019

To find the area scale factor, divide the two areas. Don't subtract them. Then check: area scale factor = (linear scale factor)².

“This part was less successful with few candidates appearing to appreciate that the (scale factor)² gives the answer directly. The common errors included the answer of 3 with no working or explanation; using 33.75 – 3.75 = 30 rather than the correct method of 33.75 ÷ 3.75 = 9.”

CIE 0580 · Paper 3 · Jun 2018

Say 'an enlargement with scale factor −2', not 'a negative enlargement of scale factor 2'. The −2 must be the scale factor, not an adjective describing the enlargement.

“It is worth noting that 'a negative enlargement of scale factor 2' was a very common incorrect description which was given just one mark (for enlargement), rather than two marks for 'an enlargement of scale factor –2'.”

CIE 0580 · Paper 2 · Jun 2018

Arc length = (θ/360) × 2πr. Sector area = (θ/360) × πr². These are different formulas — one gives a length, the other gives an area. Also check which arc is required: minor (< 180°) or major (> 180°).

“There were few fully correct answers to this part. Most calculated an arc length with the majority giving the length of the minor arc and a few giving the total circumference as the answer. A common error made by some was to use the formula for the area”

CIE 0580 · Paper 4 · Nov 2018

Use precise terminology: 'corresponding angles in parallel lines are equal', not just 'parallel lines'. Alternate, corresponding, and co-interior angles are different — don't confuse them.

“Some candidates did not mention 'corresponding angles', but simply wrote 'the two triangles are similar' or 'the angles are on parallel lines'. Some candidates incorrectly described corresponding angles as either 'alternate' or 'complementary' angles.”

CIE 0580 · Paper 2 · Jun 2018

Known weak areas for extended candidates: error bounds, circle geometry, reading inequalities from graphs, 3D problems, vector magnitude, time conversions, and distance formula.

“The weaker topics included problem solving with bounds, 2D problem solving with circles, using graphs to solve related equations and inequalities, problem solving with 3D shapes, accuracy with magnitude of a vector, time conversions and calculating the distance between two points.”

CIE 0580 · Paper 4 · Jun 2018

For exterior angle of a regular polygon: exterior = 360 ÷ n. For an octagon: 360 ÷ 8 = 45°. This IS the answer — do not subtract from 180° (that gives the interior angle) or from 360°.

“many spoilt their work by then subtracting the answer, 45, from 180 (this is the interior angle) or from 360.”

CIE 0580 · Paper 1 · Nov 2018

Always use the π key on your calculator, not 3.14 or 22/7 — these approximations lose the accuracy mark. Also, do not round intermediate values (like πr²) before completing the full calculation.

“There was evidence in seemingly correct answers that a value of π as 3.14 or 22/7 was used, resulting in the loss of one mark. Some worked out the area, πr², first but then rounded that before multiplying by 11, again likely to lose accuracy.”

CIE 0580 · Paper 2 · Nov 2018

When finding surface area from a net or given volume and base area, determine ALL three distinct dimensions carefully. Do not assume faces are square unless proven. A cuboid typically has three differ

“a lack of care in their overall strategy meant they often assumed the ends were square (either 1 × 1 or 3 × 3) rather than 1 × 3. Hence they had four faces of either 7 cm² or 21 cm².”

CIE 0580 · Paper 4 · Nov 2018

A regular polygon has the same number of lines of symmetry as it has sides. A regular pentagon has 5 lines of symmetry.

“Many candidates did not know that the number of lines of symmetry for a regular polygon is equal to the number of sides. With the diagram of a regular pentagon shown many still just gave 1 as the answer.”

CIE 0580 · Paper 1 · Jun 2018

Rotational symmetry order = how many times the shape looks the same in one full rotation. A regular pentagon has order 5; a kite has order 1.

“Rotational symmetry was not well known and many did not answer this part. Again 2 was a very common response but also 4 and 5 were often seen.”

CIE 0580 · Paper 1 · Jun 2018

Diagonals of a square intersect at 90°, not 60° or 120°. Segment area = (sector area) − (triangle area). Divide by 4 only if you computed the whole difference then split into quadrants.

“Occasionally candidates who subtracted the area of the square from the area of the circle omitted the final step of dividing by 4. Some candidates assumed the segment was a semi-circle with radius of half of the square's side length and others used 60° or 120° instead of 90° for the intersection of the diagonals of a square.”

CIE 0580 · Paper 4 · Jun 2018

When a square is inscribed in a circle of radius 8, the diagonal = 16, so side = √128. To find the area, use side² = 128 directly — don't square root then square again as this loses accuracy.

“A common error was to assume that the side length of the square was 8cm or, on occasion, 16cm. A significant number of candidates who correctly used Pythagoras' theorem to find the side length lost accuracy by square rooting 128, approximating to 11.31 for example, and then squaring again to give an inexact value for the area.”

CIE 0580 · Paper 4 · Jun 2018

Known weak areas for extended candidates: error bounds, circle geometry, reading inequalities from graphs, 3D problems, vector magnitude, time conversions, and distance formula.

“The weaker topics included problem solving with bounds, 2D problem solving with circles, using graphs to solve related equations and inequalities, problem solving with 3D shapes, accuracy with magnitude of a vector, time conversions and calculating the distance between two points.”

CIE 0580 · Paper 4 · Jun 2018

Geometry reasons must be complete: say 'corresponding angles in parallel lines are equal' not just 'parallel lines'. For angle sum say 'angles in a triangle add up to 180°' with all key words.

“The reasons were not so well answered with often just a reference to parallel lines or the incorrect alternate angles to justify angle a. For angle b, many explanations were incomplete. It was necessary to make clear that it wasn't simply a triangle that added to 180° or that the angles added to 180°. Many reasons omitted one of the words triangle or angles or even 180.”

CIE 0580 · Paper 1 · Jun 2018

For an enlargement, always give THREE things: (1) it is an enlargement, (2) the scale factor, (3) the centre of enlargement as coordinates. Missing the centre loses a mark.

“The majority of candidates were able to correctly identify this transformation as an enlargement. The scale factor was generally correctly stated although common errors of 1/2, –2 and 4 were seen. The centre of enlargement proved more challenging and was often not given.”

CIE 0580 · Paper 3 · Jun 2018

For a rotation, always give THREE things: (1) it is a rotation, (2) the angle and direction (e.g. 90° clockwise), (3) the centre of rotation as coordinates.

“Candidates found describing a single transformation demanding with a significant number omitting part of the description. In this part the majority of candidates were able to correctly identify the transformation as a rotation. The angle of rotation was generally correctly stated although common errors of 90° anti-clockwise, 90° and 180° were seen. The centre of rotation proved more challenging and was often not given.”

CIE 0580 · Paper 3 · Jun 2018

For the hypotenuse (longest side) always ADD the squares: c² = a² + b². If your answer is smaller than the two shorter sides, you subtracted instead — check your working.

“While the vast majority of candidates knew Pythagoras' theorem had to be applied, some squared and subtracted producing an answer less than 2.8 for the obviously longest side of the triangle.”

CIE 0580 · Paper 1 · Jun 2018

Perpendicular bisector of AB requires TWO things: (1) the midpoint of AB (not A or B), (2) the negative reciprocal of the gradient of AB (flip and change sign — not 1 − gradient).

“By far the most common error was to omit the use of the midpoint of AB, instead using one of the end points A or B, and so scoring at most 2 marks from 5. Most were able to find the correct gradient of AB, but a minority then forgot to find the perpendicular gradient or found the perpendicular gradient incorrectly e.g., by subtracting it from 1 or –1 instead of taking the negative reciprocal.”

CIE 0580 · Paper 2 · Jun 2024

'Show that' requires explicit reasoning — you must show the perpendicular gradient is found by taking the negative reciprocal of the original (e.g., if m₁ = −1/2, then m₂ = 2). State this step.

“It was evident that many candidates knew what to do but forgot that as it is a 'show that' question they were expected to show all their working. Many stated that the gradient of the perpendicular line is 2 with no attempt to show why.”

CIE 0580 · Paper 4 · Jun 2023

If the question says the line passes through the origin, then c = 0 in y = mx + c. The equation is just y = mx — don't reuse an unrelated intercept from a previous part.

“A significant number of responses missed the fact that the line passes through the origin. This led to the use of either point A or B or the midpoint between them and many used the intercept of −7 again.”

CIE 0580 · Paper 2 · Jun 2022

When finding gradient from a graph, use the SCALE on each axis, not the number of squares. If each y-square is 2 units but each x-square is 1 unit, gradient is doubled vs counting squares.

“Many candidates just counted squares for the gradient not realising the scales were different, resulting in answers of 1 or −1.”

CIE 0580 · Paper 3 · Nov 2022

After using Pythagoras to find the components (a = b = 2), use the direction of the vector and the starting point to find the endpoint coordinates. A clear diagram showing the vector components is ess

“Candidates found this vector question very challenging and fully correct answers were in the minority. Many candidates appreciated that the use of Pythagoras was needed. When applied correctly it often led to a = b = 2 which for many was as far as they could go.”

CIE 0580 · Paper 4 · Jun 2021

Gradient of perpendicular (normal) = −1 / (original gradient). Flip it AND change the sign. If the line has gradient −3/4, the perpendicular has gradient +4/3.

“In part (c) not all appreciated that the gradient of the normal was the negative reciprocal of their gradient from part (a).”

CIE 0580 · Paper 2 · Nov 2021

When the question says 'draw a suitable line to solve [equation]', you MUST (1) rearrange the equation to match one side = the existing curve, (2) draw the line, (3) read intersections. Using the quadratic formula ignores the method and scores partial.

“Various strategies were seen for finding the solution, but only those who found the solutions after drawing the straight line graph, y = 4 − 0.5x, as required by the question, could score full marks. Candidates who showed no method or who drew the graph y = −1 + 9/2 x − x² or who used the quadratic formula to obtain the correct solutions were only awarded one mark.”

CIE 0580 · Paper 4 · Nov 2021

Distance on a speed-time graph = area under the curve. If the speed isn't constant, split into rectangles and triangles. Don't just use (max speed × total time) — that's the area of a rectangle only.

“The most common error in the method was to multiply the total time of 70 by the speed of 20, forgetting that this only applies for a constant speed throughout the whole journey.”

CIE 0580 · Paper 4 · Nov 2019

Gradient = rise/run = (y₂ − y₁)/(x₂ − x₁). Drawing a right-angled triangle on the graph and counting squares is often easier than the formula, especially with negative coordinates.

“A few used the formula with two points from (−2, −4), (0, −1) and (2, 2) for finding the gradient but made numerical errors with the negative signs. Only a few attempted a rise ÷ run calculation based on a triangle drawn on the graph.”

CIE 0580 · Paper 1 · Nov 2019

When subtracting a vector expression, bracket the whole thing and distribute the minus: OA − BA = (2x+3y) − (x−4y) = 2x+3y−x+4y = x+7y. Don't drop brackets and get −4y instead of +4y.

“Candidates who did realise they needed to subtract BA often had a sign error; instead of 2x + 3y – (x – 4y) they wrote 2x + 3y – x – 4y so a very common incorrect answer was x – y.”

CIE 0580 · Paper 2 · Jun 2019

In y = mx + c, the gradient is m (the coefficient of x). For y = 5x + 12, the gradient is 5. Do not add the numbers together, divide them, or give '5x' — the gradient is a single number.

“Many candidates appeared not to understand what this question required; few appeared to know that the coefficient of x is the gradient. 5 + 12 = 17, 12 ÷ 5 = 2.4 and 5x were all seen. Several candidates did not attempt this question.”

CIE 0580 · Paper 3 · Nov 2018

To reverse a vector, negate it: BA = −AB. Break the path into known vectors and add/subtract carefully. Don't guess that a reversed vector equals p or q.

“Many used CB + BA but they could not work out BA. Common incorrect answers were p + q or 2q from using BA as either p or q.”

CIE 0580 · Paper 2 · Jun 2018

For perpendicular lines, the gradient of the perpendicular = −1/m (the negative reciprocal), NOT −m or 1/m. If the original gradient is 3, the perpendicular gradient is −1/3. Using just −3 or +1/3 is

“Candidates made errors in finding the gradient of the perpendicular, either finding a line parallel to that given, or using –3 or 1/3. A good number of candidates knew to substitute the co-ordinates given into the general equation of a straight line.”

CIE 0580 · Paper 2 · Nov 2018

On a speed-time graph, the distance travelled equals the AREA under the curve — not time × maximum speed. Split into rectangles and triangles if the shape is complex.

“The vast majority of candidates understood that distance is the area under the graph and most carried out this correctly to gain full marks... Candidates who did not score usually gave the total time multiplied by the top speed, 190 × 15.”

CIE 0580 · Paper 2 · Jun 2024

'Intersect' means where two lines cross (could be anywhere). 'Intercept' means where a line crosses an axis (at x=0 or y=0). These are different questions.

“The most common misconception was confusing the word intersect with intercept and so the most common incorrect answer was (0, 3) along with (0, –5) and (4, 0).”

CIE 0580 · Paper 2 · Jun 2024

For y = mx + c, if a graph is given, READ the y-intercept c directly from where the line crosses the y-axis — don't calculate it. Only calculate c if no graph is provided.

“This part on finding the equation of the given line was generally reasonably answered, although a significant number did not seem to appreciate that the intercept value could be read directly from the given graph.”

CIE 0580 · Paper 3 · Jun 2023

Gradient = (y₂ − y₁) / (x₂ − x₁). Watch for double negatives: −2 − (−6) = −2 + 6 = 4. Sanity-check against the diagram: if the line slopes down, the gradient must be negative.

“A gradient of positive 2 was seen, along with several who found a gradient of 1 from the calculation (−3−5)/(−2−6) instead of (−3−5)/(−2−(−6)). Candidates could have spotted these were errors by looking at the diagram more carefully.”

CIE 0580 · Paper 2 · Jun 2022

When subtracting vectors, write out each component separately. Subtracting a negative: (3) − (−2) = 3 + 2 = 5. Setting out working clearly reduces sign errors.

“Directed number rules caused quite a few to not score this mark by subtracting a negative number incorrectly. However errors were less common when the subtraction of the vectors was set out in the working space.”

CIE 0580 · Paper 1 · Jun 2019

Read carefully which point the perpendicular line must pass through. If asked for perpendicular through A, substitute point A's coordinates — not the midpoint — to find c in y = mx + c.

“The most common error, having correctly found the perpendicular gradient, was for candidates to find the equation of the perpendicular bisector of AB, by using the midpoint from part (a), rather than finding the equation of the perpendicular line through point A.”

CIE 0580 · Paper 2 · Jun 2019

Don't write '+ −1' — simplify to '− 1'. Final answers should have their signs properly resolved: y = 1.5x − 1, not y = 1.5x + −1.

“An answer of y = 1.5x + −1 did not gain full credit as the signs had to be resolved into y = 1.5x − 1.”

CIE 0580 · Paper 1 · Nov 2019

Perpendicular gradients multiply to −1. If gradient = 2, perpendicular gradient = −1/2 (flip and negate). Don't just negate (−2) or just flip (1/2).

“Many candidates understood the relationship between the gradients of perpendicular lines and those who wrote the perpendicular gradient as −1/2 usually went on to earn full marks. Other candidates often gave the gradient of the perpendicular line as 2, –2 or as 1/2.”

CIE 0580 · Paper 4 · Jun 2018

Distance formula: √((x₂−x₁)²+(y₂−y₁)²). Bracket the whole subtraction before squaring: (4−(−2))²=36, not 4−(−2)²=4−4=0.

“Errors arose from candidates adding the corresponding co-ordinates instead of subtracting and from writing part of their calculation incorrectly, for example 4 – (–2)² instead of (4 – (–2))².”

CIE 0580 · Paper 4 · Jun 2018

When squaring a negative component of a vector, always use brackets: (−1)²=1, not −1²= −1. Missing brackets on negative numbers is a very common error.

“The majority of candidates understood that Pythagoras' theorem should be used to find the magnitude of a vector and many correct answers were seen. However, the omission of brackets for (–1)² frequently led to an incorrect answer.”

CIE 0580 · Paper 4 · Jun 2018

For train-crossing-bridge problems: (1) Total distance = train length + bridge length, (2) Convert all units to be consistent (all metres + seconds, or all km + hours) BEFORE calculating.

“This was the most challenging question on the paper with less than half of the candidates scoring full marks... Weaker candidates did not attempt a conversion of one or both the units between km and m or hours and seconds. Many combinations of multiplications and divisions of the given values were seen among the weaker candidates.”

CIE 0580 · Paper 2 · Jun 2024

Percentage profit = (profit / COST PRICE) × 100 — divide by COST, not selling price. Profit = selling − cost. This is the most-missed type of percentage question.

“Calculating the percentage profit proved to be challenging for many candidates. Very few fully correct solutions or part solutions were seen. This was because the majority of candidates divided by the selling price of $24 instead of the cost of $12.80.”

CIE 0580 · Paper 3 · Nov 2021

If 5/13 are sold, then 8/13 remain. Set the remainder (96) equal to the remaining fraction (8/13 of total), not the sold fraction. A decimal answer in a whole-number context should alert you to an err

“most candidates not realising that they had to relate the 96 biscuits left to the fraction 8/13 and not to the sold fraction, 5/13. The decimal answers resulting from this error perhaps should have made candidates realise that it had to be incorrect.”

CIE 0580 · Paper 1 · Jun 2021

When a map scale is 1:25,000, that's a LENGTH ratio. For areas, you must use 25000² for the scale factor. Missing the squaring is the #1 map-scale mistake.

“Only the most able candidates answered this question correctly. Most incorrect answers were some version including the figures 8.4, often from (33.6 × 25000)/100000 = 8.4. Many candidates did not realise that they were working with an area and that both 25000 and 100000 needed to be squared.”

CIE 0580 · Paper 4 · Nov 2019

When working with exchange rates, keep full calculator precision until the final answer, then sanity-check: does the final amount look sensible compared to the starting amount?

“Rounding calculations at intermediate stages often produced inaccurate answers. In questions involving exchange rates, candidates should look at their answer and consider if it is sensible for the amount being exchanged.”

CIE 0580 · Paper 1 · Jun 2024

Reverse percentage: if $1.15 is AFTER an 8% increase, divide by 1.08 to get the original (not multiply). Multiplying gives a further 8% increase instead of reversing it.

“Candidates were slightly less successful in this part which involved reverse percentage. The most common incorrect answer involved finding 108% of $1.15. Other incorrect methods usually involved incorrect use of the multipliers 0.08, 0.92 and 1.08.”

CIE 0580 · Paper 4 · Jun 2024

For 1 cm : 250,000 scale, 3.5 cm = 875,000 cm = 8,750 m = 8.75 km. Convert step by step: ÷ 100 to get m, ÷ 1000 to get km. Check: 1 cm should correspond to 2.5 km, so 3.5 cm ≈ 8.75 km — sanity check with scale.

“Although the figures of 875 were generally arrived at there was a large range of answers seen in addition to the correct answer of 8.75 most commonly 0.875, 87.5, 875 or 875 000. Changing the units to kilometres was the main problem encountered by the candidates.”

CIE 0580 · Paper 2 · Nov 2023

For a map scale like 1:125,000, convert step-by-step: cm → m (÷ 100) → km (÷ 1000). Or simplify the scale first to '1 cm : 1.25 km' so one multiplication gives the real distance.

“Whilst most knew they should multiply by the map scale, a large number were unable to address the conversion between centimetres and kilometres. The more successful candidates often tackled this part in two stages, first changing to metres, but many stopped there. Few started by converting the scale to reach 1 cm: 1.25 km.”

CIE 0580 · Paper 4 · Nov 2023

Two increases of 3% and 5% do NOT combine to 8%. Multiply the multipliers: 1.03 × 1.05 = 1.0815, an overall increase of 8.15%. Adding percentages ignores the compounding.

“Others attempted to convert the two increases to find the multiplier for the overall increase. Many did obtain 1.0815 but a significant number simply added the two percentage increases.”

CIE 0580 · Paper 4 · Jun 2023

A 15% increase followed by an 18% increase is NOT 33% overall. It's 1.15 × 1.18 − 1 = 0.357 = 35.7%. Multiply the multipliers, then subtract 1 for the percentage.

“The most efficient method of using the multipliers 1.15 × 1.18 = 1.357 giving a 35.7 per cent increase was not often seen... The most common error was to add the two percentages together, giving an answer of 33 per cent.”

CIE 0580 · Paper 4 · Jun 2022

To convert km/h to m/s: multiply by 1000 (km to m), divide by 3600 (hours to seconds). Shortcut: divide by 3.6. Example: 108 km/h ÷ 3.6 = 30 m/s.

“Candidates found this question very challenging with many unsure of how to start. Only a small number managed a fully correct solution. Often they were able to gain some credit, either for correctly converting the speed to 30 m/s. It was not uncommon to see unsuccessful attempts at the conversion, either using a factor of 60 just once, multiplying instead of dividing, or sometimes using a factor of 100 rather than 1000 for kilometres.”

CIE 0580 · Paper 4 · Nov 2022

Average speed = TOTAL distance / TOTAL time. NOT the average of the segment speeds. Include ALL time (including stops) unless the question says 'average moving speed'.

“Weaker responses assumed that the average speed could be found by averaging the speed of each part of the journey, giving the common incorrect response of (16 + 20)/2 = 18. Some incorrectly found the area under the distance-time graph. Despite being told in the question that the total distance for the journey was 70 km, there were a number of candidates who did not use 70 km in their calculation of average speed.”

CIE 0580 · Paper 4 · Nov 2022

Percentage profit = (profit ÷ cost price) × 100. If cost price = 41 and selling price = 65, profit = 24, so % profit = (24 ÷ 41) × 100. Never divide by the selling price.

“This part on finding the percentage profit was generally found challenging and few correct methods were seen. Common errors included the use of 41/65, 24/65, 65/41 and 24/41.”

CIE 0580 · Paper 3 · Jun 2021

Reverse percentage: if a price of $98.90 already includes 15% tax, the original = $98.90 ÷ 1.15 = $86. Finding 15% of $98.90 and subtracting is wrong because the tax was calculated on the original pri

“In the majority of cases, candidates calculated 15% of $98.90, usually giving this as their answer or sometimes adding it to $98.90 or subtracting it from $98.90.”

CIE 0580 · Paper 4 · Jun 2021

If a price has been INCREASED by a factor, get the original by DIVIDING. Multiplying by 1.142 again just adds another 14.2% increase, not reverses it.

“The majority of candidates correctly divided by 1.142 with only a minority of candidates incorrectly multiplying by 1.142.”

CIE 0580 · Paper 4 · Nov 2021

Speed/distance/time: ensure units match. If speed is in m/s, distance must be in m and time comes out in seconds. 10 km = 10,000 m, not 10. Then convert seconds to min/sec if asked.

“This problem solving question had three parts to the method, converting the distance from kilometres to metres to match the speed given in m/s, finding the time using distance ÷ speed and finally converting the answer in seconds to minutes and seconds as required by the question.”

CIE 0580 · Paper 1 · Nov 2019

% increase = (increase ÷ ORIGINAL) × 100. If price went from 63 to 77, increase = 14, so % increase = 14/63 × 100 = 22.2%. Never divide by the new value (77).

“The denominator of the fraction for percentage increase has to be the original amount but many candidates used a denominator of 77. Others did not understand the question and found 14% of either 63 or 77.”

CIE 0580 · Paper 1 · Jun 2019

To write a map scale as 1:n, convert both distances to the same unit (cm). So 1 cm on map = 15 km = 1,500,000 cm, giving scale 1:1500000.

“The majority of candidates found this part very demanding and few correct answers were seen. The method of 1cm to 15km, 1 cm to 15000m, 1cm to 1500000cm giving the scale in the required form as 1:1500000 was not generally appreciated.”

CIE 0580 · Paper 3 · Jun 2018

Read the question carefully — if it says two-thirds of 600, use 600, not the amount remaining after the percentage was removed. Each operation applies to the original amount unless stated otherwise.

“Many found 18% correctly but then subtracted it from 600 and found two-thirds of 492 instead of 600, as clearly stated in the question.”

CIE 0580 · Paper 1 · Jun 2018

When a question says 'two-thirds of 600', use the original 600 — not the amount left after removing 18%. Apply each fraction/percentage to the quantity stated in the question.

“The most common error in this question was to find 2/3 of the remaining cakes, once 18% had already been taken away, leading to an answer of 164.”

CIE 0580 · Paper 2 · Jun 2018

To combine two ratios A:B = 7:10 and B:C = 4:3 into A:B:C, find a common value for B. LCM of 10 and 4 is 20, so A:B = 14:20 and B:C = 20:15, giving A:B:C = 14:20:15. The common term must be the same n

“Candidates had to obtain a common number for the middle number (i.e. group B). The most common way seen was to find a common multiple of 10 and 4, e.g. 40. Then candidates had to multiply the other parts of the two ratios by appropriate numbers.”

CIE 0580 · Paper 4 · Nov 2018

Percentage increase = (increase ÷ original value) × 100. The denominator must be the ORIGINAL value, not the new value. Also, dividing the increase by 100 is not the same as finding a percentage — you

“The most successful method was to use (14100 − 12400) / 12400 × 100. Many found the actual increase but just divided this by 100 to get their answer. Many used a multiplier approach but had the figures inverted in their division.”

CIE 0580 · Paper 4 · Nov 2018

If 3/7 of the total = 1248, then total = 1248 ÷ 3/7 = 1248 × 7/3 = 2912. Do NOT multiply 1248 by 3/7 (that gives 3/7 of 1248, which is a fraction of a fraction). Understand the difference between 'fra

“Few candidates appeared to appreciate that the given statement meant that (3/7) × total profit = 1248, and the method of total profit = 1248 ÷ (3/7) was rarely seen. Common errors included (3/7) × 1248 giving incorrect answers of 534 or 1782”

CIE 0580 · Paper 4 · Nov 2018

Distance from a speed-time graph = area under the graph, NOT distance ÷ time. For a trapezium-shaped graph, use the trapezium area formula. Using d/t only works for constant speed.

“By far the biggest misunderstanding was to use v = d/t and so 150 ÷ 8 = 18.75 was a very common incorrect method and answer.”

CIE 0580 · Paper 2 · Nov 2018

In proportion questions, read the relationship precisely: 'y is proportional to (x−1)²' means y = k(x−1)², NOT y = kx², y = k/x², or y = k(x−1). Getting the initial relationship wrong invalidates all

“Candidates should ensure that they read the information in proportion questions very carefully, as the majority of errors come from setting up the incorrect relationship at the beginning of the working. It was often seen as an inverse relationship, or without the square.”

CIE 0580 · Paper 2 · Nov 2018

Always simplify a ratio by dividing all parts by their HCF. 30:25:10 has common factor 5, so simplify to 6:5:2. 'Simplest form' means no common factor bigger than 1.

“This part on ratio was generally answered well, although a significant number did not give the ratio in its simplest form with 30 : 25 : 10 being common.”

CIE 0580 · Paper 3 · Jun 2023

Use the FULL figures given. Don't round to 2 sf before starting — rounding-then-calculating loses accuracy. And for reverse percentage questions, DIVIDE by the multiplier, don't multiply.

“The common misunderstanding was for candidates to multiply 380.8 by 1.19 instead of dividing. Some candidates did not use all the figures in the question so rounded before they started. Candidates must use the full numbers given in the question to do their calculations.”

CIE 0580 · Paper 1 · Nov 2022

Inverse proportion means as one quantity increases, the other decreases proportionally. The precise terms accepted are 'inversely proportional' or 'indirect proportion' — vague terms like 'opposite' d

“It was common to see incorrect answers that seemed to indicate the candidates understood the fact that there was a relationship; for example, inverse, opposite and indirectly proportional. There were a number of candidates who gave answers such as positive, direct and proportional.”

CIE 0580 · Paper 2 · Jun 2021

For exchange rates, check which direction you're converting. If you get an exact decimal like 0.893, don't round it — write all the figures.

“There were two main errors seen in this question. Either candidates divided the number of dollars by the number of euros to give 1.12 (to three significant figures) or divided the correct way around but rounded the exact three figure value to 0.89.”

CIE 0580 · Paper 1 · Nov 2021

When T = D ÷ S gives a decimal in hours, convert carefully: 2.133… hours = 2 hours + 0.133… × 60 minutes = 2 hours 8 minutes. Do not read the decimal digits directly as minutes (e.g. 2.13 ≠ 2h 13min).

“the conversion to either 128 minutes or 2 hours 8 minutes proved more challenging and few correct final answers were seen. A very common error was 12 28.”

CIE 0580 · Paper 3 · Jun 2021

Never convert a fraction to a rounded percentage mid-calculation. Keep exact fractions or decimals with full precision throughout, and only round the final answer.

“Some candidates converted 5/24 to a rounded percentage, often 20.8%. Due to this premature rounding, it was common to see the inaccurate answer $271.63.”

CIE 0580 · Paper 2 · Jun 2021

% increase = (increase/ORIGINAL) × 100 = 14/63 × 100 = 22.2%. Dividing by 77 (new value) gives a wrong answer. 122.2% means you divided the new value by the original — that's wrong.

“A few candidates incorrectly gave the percentage increase as 122.2% rather than 22.2%. A small number of candidates divided the increase by 77 rather than by the original amount of 63.”

CIE 0580 · Paper 2 · Jun 2019

When converting minutes to hours for speed calculations, keep the exact fraction (50/60 = 5/6) rather than rounding to 0.83. The rounded decimal introduces enough error to give the wrong final answer.

“Many candidates did divide 50/60 but then rounded their solution to 0.83 and therefore when dividing into 8 km reached the solution of 9.64 instead of 9.6 and therefore did not gain full marks.”

CIE 0580 · Paper 4 · Nov 2018

When finding a proportionality equation: (1) write d = kt², (2) substitute known values to find k, (3) substitute k back to get the specific equation, e.g. d = 4.9t². Leaving the answer as d = kt² (wi

“A considerable number of candidates were able to identify the correct proportional equation and then calculate the value of k but did not substitute their value for k into the original equation, leaving d = kt² on the answer line.”

CIE 0580 · Paper 4 · Nov 2018

In probability without replacement, the denominator DECREASES by 1 for each subsequent pick (13 → 12 → 11...). Getting 104/169 instead of the correct answer usually means you forgot this.

“There were very few candidates who did not recognise the significance of the lack of replacement of the first counter, although where this was an issue some still gained credit for an answer of 104/169 with others compounding their error by missing some of the required pairs.”

CIE 0580 · Paper 2 · Jun 2024

For 'exactly one green button' out of two picks, there are TWO orderings: green-then-not OR not-then-green. Calculate one product and DOUBLE it (or add both products).

“Most recognised the need to start off with 5/13 for the probability of a first button being green, which earned a first method mark. Many fewer were able to use this correctly in a product with 8/12 for a second button being non-green to score the next mark. Only the stronger candidates realised that they also needed to double the result of the product to account for the possibility of green being the second button.”

CIE 0580 · Paper 4 · Nov 2023

For 'picked from those wearing trainers', the denominator is the number wearing trainers (21), not the total (50). Read carefully — conditional probability restricts the sample space.

“Candidates needed to identify that 21 students wear trainers and that only 16 of these have mobile phones. As two students were being picked then answers also needed to treat this as picking without replacement. Many identified the correct 16 students but did not take into account that they were being chosen from just those wearing trainers.”

CIE 0580 · Paper 4 · Jun 2022

In a histogram, bar HEIGHT = frequency density (frequency ÷ class width). Find the scale using a bar whose frequency AND width you know, then apply it to the others.

“This proved challenging for many. Those answers displaying understanding that the heights of the blocks of a histogram represented the frequency densities usually had no difficulty. Some clearly showed the calculation of the scale factor for the first bar as 17.2 ÷ 0.86 = 20 and used this correctly with the remaining frequency densities.”

CIE 0580 · Paper 4 · Jun 2022

In a stem-and-leaf diagram, the full value is stem + leaf, not just the leaf. A leaf of '5' under a stem of '2' means 25, not 5. Apply this for median, mode, and range.

“In general, candidates did not seem confident with stem-and-leaf diagrams. Many did not understand how to read the diagram as they ignored the 'stem' and just used the value of the leaves.”

CIE 0580 · Paper 1 · Nov 2021

Estimated mean from grouped data: (sum of midpoint × frequency) / (total frequency). Sanity check: the mean must lie between the lowest and highest possible values.

“The vast majority of methods involved an attempt to sum the products between a height value in the intervals and the frequencies and then divide by 200... A mean value should be representative and so should, at the very least, be between 100 cm and 190 cm (the minimum and maximum height values in the table).”

CIE 0580 · Paper 2 · Jun 2024

When there is an even number of values, the median is the average of the two middle values — it may not appear in the original list.

“Answers of 58 and 64 showed an attempt at the middle but many did not realise that the median could be a number not in the list.”

CIE 0580 · Paper 1 · Jun 2024

Frequency density = frequency ÷ class width (NOT class width ÷ frequency). The y-axis of a histogram shows frequency density, not frequency, for unequal class widths.

“A common error resulted from incorrectly calculating the frequency density by dividing the interval widths by the frequencies.”

CIE 0580 · Paper 4 · Jun 2024

Median from a frequency table: find the (n+1)/2 th VALUE using cumulative frequency. Don't use the raw column of marks and don't average the listed marks ignoring frequencies.

“A significant number of candidates gave a variety of incorrect answers with 16 the most common as it is the 5th mark out of 10 marks. Only slightly fewer gave the answer 17.5 as it is the median of the six marks listed in the table.”

CIE 0580 · Paper 4 · Jun 2024

Mean from frequency table = (sum of fx) ÷ (total frequency), NOT ÷ number of groups. Sanity check: the mean must lie within the range of the actual data values.

“A common error was to start correctly but then division by 7, or even 6, produced an answer that showed a lack of understanding of a mean as it was beyond the number of calculators in the table.”

CIE 0580 · Paper 1 · Jun 2023

Median = middle value when the data is ordered. Mean = sum ÷ count. These are DIFFERENT — confusing them loses a whole question's worth of marks.

“The confusion between median and mean meant this question was not well answered. Many candidates attempted to find a mean and the answer 85 was common from using 61 as the mean of the 6 numbers.”

CIE 0580 · Paper 1 · Jun 2023

For P(two students both satisfy X) without replacement, multiply (46/100) × (45/99) — numerator AND denominator both decrease by 1. Don't stop at just the first probability.

“Many other candidates identified that there are 46 students with a reaction time greater than or equal to 9 seconds and showed the correct probability for selecting the first of these students. Some gave this as their answer rather than calculating the probability of selecting two students. Others calculated a product with replacement rather than without replacement.”

CIE 0580 · Paper 4 · Jun 2023

For median of n values from a frequency table, find the (n+1)/2 th position. If n = 80, that's the 40.5th value — average the 40th and 41st. Use the CUMULATIVE frequency column to locate which group it's in.

“Common errors included 14.5 or 3.5 by just considering either set of 6 numbers given. Few candidates appreciated that the 40th/41st values were needed or how to use the table to find these.”

CIE 0580 · Paper 3 · Jun 2022

For P(A or B) where A and B are mutually exclusive, ADD the probabilities (3/n + 5/n = 8/n). Multiplication is for 'and' with independent events.

“Part (c) was answered correctly by about half of the candidates. Of those not gaining 2 marks it was common to award a mark for the correct numerator, usually 8/40 or 8/14, or the correct denominator, usually 14/23. Many others also scored a mark for the working 3/n + 5/n. The most common incorrect working and answer was 3/23 × 5/23 = 15/529.”

CIE 0580 · Paper 4 · Nov 2022

When given a mean and count, always start by finding the total (mean × count = 9 × 17 = 153). Then use the total and additional constraints together — not separately — to find the missing values.

“The key to success in this question was to realise the first step was to find the total of the nine numbers from 9 × 17. Many candidates did not do that and consequently could not make much progress. Those who did reach that stage usually found the 21 but were not sure what that meant. Many then either gave two numbers adding to 21 or two numbers with a difference of 5.”

CIE 0580 · Paper 1 · Jun 2021

For qualitative (non-numerical categorical) data, only the mode is meaningful — not mean or median. The mode is the category appearing most frequently.

“The whole of part (d) proved challenging for a significant number of candidates with many not appreciating what was required. In part (i) common errors included a variety of numerical answers, median or mean.”

CIE 0580 · Paper 3 · Jun 2021

When drawing items without replacement, identical outcomes (like RRR) have only 1 arrangement, but mixed outcomes (like RRG) can occur in multiple orders. List all arrangements systematically before m

“products were identified correctly but the number of combinations of each of these products was often incorrect, usually taken as three, including 3 for RRR, or as 1 for all of the products.”

CIE 0580 · Paper 4 · Jun 2021

Comparing two distributions requires both a comparison ('the median increased from X to Y') AND an interpretation ('suggesting students scored higher on average in Test 2'). Describing that statistics

“Correct comparisons were in the minority. Many comments referred to how statistics such as the median, range, IQR (or minimum and maximum values) had changed, with no attempt at interpreting these.”

CIE 0580 · Paper 4 · Jun 2021

To compare likelihood between two bags, you must compare the PROBABILITIES (fractions), not the raw counts. More black marbles doesn't mean higher probability if the bag also has more total marbles.

“To gain full credit candidates needed to compare the probabilities of picking a black marble from both bags. Answers needed to include the probability of picking a black marble from bag B as 5/13 and compare this to the probability of picking a black marble from bag A (2/5). Most incorrect answers compared the number of marbles in each bag rather than the probabilities.”

CIE 0580 · Paper 3 · Nov 2021

'Without replacement' means after picking once, the pool shrinks by 1. So P(event1) × P(event2 | event1 happened), with a smaller denominator for the second pick.

“Many candidates did not understand the significance of the phrase 'without replacement' in the question and an answer of 4/25 was often seen. It was also common to see the answer 2/5, the probability of an even number being picked from the 5 cards.”

CIE 0580 · Paper 4 · Nov 2019

In a grouped frequency table, the frequency for a group is not found by subtracting the group boundaries. Read the frequency column directly for each class interval.

“This part was generally poorly answered with the majority of candidates not appreciating what the values given in the grouped frequency table actually indicated. A very common error was 64 – 27 = 37.”

CIE 0580 · Paper 3 · Jun 2018

Mode = the value with the highest frequency, NOT the frequency itself. Mean from a frequency table = Σ(value × frequency) ÷ Σ(frequency). Do NOT add the values row or divide by the number of categorie

“the highest frequency is 7 which refers to 4 visits. The mode is 4 visits not the frequency of 7. Another misunderstanding was for candidates to pick the number that appears the most frequently in the table, 6. it is not correct to just add up the top line (getting 28) as many did or the bottom line (getting 40)”

CIE 0580 · Paper 1 · Nov 2018

For 'at least one' probability problems without replacement: (1) denominators decrease each draw (15, then 14); (2) consider ALL valid combinations; (3) the most efficient method is often 1 − P(none),

“Many treated the pens as being replaced and multiplied fractions all with 15 as a denominator. Others recognised the need to consider probabilities with denominators of 15 and 14 respectively, but did not consider all of the possible combinations. Few candidates chose the most efficient route of calculation of 1 − probability of two black pens.”

CIE 0580 · Paper 2 · Nov 2018

Always read the question carefully — words like 'not' in bold mean you need the complement (1 − P). Missing this wastes an easy mark.

“The probability was well done by most candidates but more would have gained the mark if they had read the question correctly. 3/20 was the probability of a blue car, not the question which had the word not in bold.”

CIE 0580 · Paper 1 · Jun 2024

Range = (largest VALUE in the data) − (smallest VALUE). Don't use the largest/smallest frequency, and don't use the number of groups — use the actual data values.

“This part on working out the range was generally answered well. Common errors included 4 (from 6 – 2) and 9 (from 9 – 0).”

CIE 0580 · Paper 3 · Jun 2023

A line of best fit follows the shape of the data points — it does NOT need to pass through the origin. Draw so there's roughly equal data above and below the line.

“Although several candidates were able to draw a ruled line within the tolerances allowed, several had their line of best fit outside the tolerance allowed on the question. A common error was a line of best fit through the origin.”

CIE 0580 · Paper 3 · Nov 2023

For nth term n² + 12: first 3 terms are 1²+12 = 13, 2²+12 = 16, 3²+12 = 21. Substitute n = 1, 2, 3 and EVALUATE fully. Don't confuse n² with 2n.

“Common errors were to give algebraic expressions as answers such as n² + 1, n² + 2 and n² + 3, or not evaluating and giving answers such as 1² + 12 and so on or using n = 0 or n = 2 to find the first term. It was also common to see the first three terms of the sequence n × 2 + 12 rather than n² + 12.”

CIE 0580 · Paper 2 · Nov 2022

Check whether the problem says 'with replacement' or 'without replacement'. With replacement: probabilities on branches 2 match branches 1. Without replacement: denominators decrease.

“Some attempted probabilities as if they were without replacement and probabilities 1/5, 4/5, 1/5, 4/5 were seen several times.”

CIE 0580 · Paper 4 · Jun 2022

Range = largest value − smallest value. Do not confuse range with mean. Also check the required units — if the question asks for cm, give your answer in cm.

“many only gained partial credit since they did not apply the instruction to give the answer in centimetres. The term range was not known by some who found the mean. Of those attempting a range, the common error was not being able to identify the largest and smallest heights.”

CIE 0580 · Paper 1 · Jun 2021

For estimated mean from grouped data: use midpoints of each class interval (not class widths), calculate Σfx, then divide by Σf (total frequency, not number of classes). E.g. midpoint of 20 < x ≤ 30 i

“There was a higher proportion of candidates this year using the class widths for the midpoints. A few candidates divided by 5 instead of 50 and some found the sum of the frequencies and divided by 5.”

CIE 0580 · Paper 4 · Jun 2021

The mean and median cannot be found for qualitative data because the categories are not numbers. The key explanation is: 'the data is qualitative (not numerical)', not 'there are only 3 values' or 'it

“This part proved challenging for the majority of candidates who did not realise that the explanation was that the data was qualitative and not quantitative. The most common correct explanations were in the form of 'because there are no numbers', 'the data is not numerical'”

CIE 0580 · Paper 3 · Jun 2021

Multi-step questions: find what's asked at each stage. If asked for range but must first find a missing value: (1) total = mean × n, (2) missing value = total − known values, (3) range = max − min.

“The question asked for the range but it needed the fifth number found first. While there were a significant number of fully correct answers, many found the two stages challenging. Having found 17 this was often given as the answer rather than continuing to the range.”

CIE 0580 · Paper 1 · Jun 2019

P(one red, one blue) = P(R then B) + P(B then R). Many only calculate one order. Add both orders. Also: without replacement means the denominator decreases by 1 for the second pick.

“Finding the probability of only one of the two combinations, almost always red then blue, was the most common error. Multiplication of the probabilities of the two combinations was rarely seen but a few slips with the arithmetic were seen. A small number based their working on replacement of the balls.”

CIE 0580 · Paper 4 · Jun 2019

For P(one of each colour) without replacement, find P(A then B) and add P(B then A) — you must double or list all orders. Don't use 3-term products when only 2 picks are made.

“The most common errors were to forget to double 5/20×9/19 + 6/20×9/19 + 6/20×5/19 or to use products of three terms such as 5/20×9/19×8/18.”

CIE 0580 · Paper 2 · Jun 2018

To find turning points: (1) find dy/dx, (2) set dy/dx = 0, (3) solve for x, (4) substitute each x back into the ORIGINAL y equation. Guess-and-check with tables is risky.

“A small number of candidates were able to find at least one of the turning points in part (b) but this was not always by use of the gradient function. Instead an empirical approach with tables of values for y at different x-values was often seen. Candidates need to understand that equating the gradient function to zero is the expected method, and that had the turning points not been at integer x-values they would have had greater difficulty in using an empirical approach.”

CIE 0580 · Paper 4 · Nov 2023

To determine max vs min: (1) find d²y/dx², (2) substitute the x-value of the turning point, (3) state whether the result is positive (min) or negative (max). All 3 steps must be shown.

“Most attempts involved finding the second derivative and then following the standard procedure. As candidates were asked to show how they decided it was important to show the substitution into the second derivative, its evaluation and a comment on the value.”

CIE 0580 · Paper 4 · Jun 2024

Gradient of a CURVE at a point requires differentiation. Find dy/dx, then substitute the x-value. Linear gradient calculations (rise/run with 2 points) only work for straight lines.

“Only a minority of candidates realised that part (a) required them to find the gradient function and so try differentiation. Often, unhelpful attempts were instead made by substituting values into the equation of the curve and trying some form of linear gradient calculation.”

CIE 0580 · Paper 4 · Nov 2023

The gradient of a curve y = f(x) at a point is dy/dx evaluated at that x-value. Substituting x = 1 into f(x) gives you y — that's the y-coordinate, not the gradient.

“Not all candidates recognised the need to differentiate in order to find the gradient. Those that did were often successful in finding the gradient function. Other candidates usually substituted x = 1 into the equation of the graph.”

CIE 0580 · Paper 4 · Jun 2023

For stationary points: (1) compute dy/dx, (2) set dy/dx = 0 (NOT y = 0), (3) solve for x, (4) substitute x into original y to get the y-coordinate.

“Fully correct solutions were dependent on differentiation and substitution to find the value of a, and substitution into the equation to find b and k. Many candidates did not attempt to differentiate and those who did often had errors in some of the terms. Poor notation was apparent with in some cases, initially equating the derivative to y rather than dy/dx.”

CIE 0580 · Paper 4 · Jun 2022

To find stationary points: differentiate (multiply coefficient by index, reduce power by 1; constants vanish), then set dy/dx = 0 — NOT d²y/dx² = 0. The second derivative is used only to classify the

“The most common errors were forgetting to multiply the coefficient by the index, hence giving −2x rather than −4x and to leave in the constant of 5 or change this to 1 in the derivative. Some candidates confused the purpose of the second derivative and equated that to 0 instead of the first derivative.”

CIE 0580 · Paper 2 · Jun 2021

Stationary/turning points are where dy/dx = 0, NOT where y = 0. After solving dy/dx = 0 for x, substitute back into the ORIGINAL curve to find y.

“Most candidates who wrote dy/dx = 0 went on to find the correct values for x... There was some misunderstanding of stationary points with candidates equated the original curve to 0 or only substituting x = 0 into y without any reasoning.”

CIE 0580 · Paper 4 · Nov 2021

To find the gradient of a curve at a given x: differentiate to get dy/dx, then substitute x into dy/dx. Substituting into the original equation gives a y-coordinate, not a gradient.

“Many of the incorrect responses involved substitution of x = –2 into the equation of the curve to find y. This point was often used with another random point to find a gradient.”

CIE 0580 · Paper 4 · Jun 2021

For 'show that' questions, compute with full calculator precision and give AT LEAST one more decimal place than the target. Don't start from the target — that's circular reasoning and scores zero.

“As the question was 'show that' the answer was required to greater accuracy than the given 31.9 for the angle. Of those who did give more decimal places, some had rounded 46 ÷ 74 too much before applying the tangent function. Consequently, their answers did not have sufficient accuracy to round to 31.9. Starting with 31.9, seen occasionally, is not the required approach for a 'show that' question.”

CIE 0580 · Paper 1 · Jun 2023

For cos(x) = k in 0° ≤ x ≤ 360°, the two solutions are θ and (360 − θ). For sin(x) = k, they are θ and (180 − θ). Learn which formula goes with which trig function.

“The most common error was to see the second solution given as 252.9 (from 180 + 72.9) or sometimes 107.1 (from 180 – 72.9), –72.9, 432.9 (from 360 + 72.9) or 342.9 (from 270 + 72.9).”

CIE 0580 · Paper 2 · Jun 2024

y = cos(x) starts at (0, 1), crosses x-axis at 90°, reaches (180°, −1), crosses x-axis at 270°, reaches (360°, 1). SMOOTH curve, wave shape. Don't make it parabolic or linear.

“The quality of sketched cosine graphs in part (a) was very mixed, showing that this is a skill much in need of practising. Whilst some were very good, a large number gave a poor indication that there were turning points at 0° and 360° (sometimes appearing almost parabolic), were often very linear, or demonstrated no idea of symmetry. Most did show a realisation that it should be a wave shape but often either not starting at (0, 1), or otherwise starting correctly at (0, 1) but cutting the x-axis at 180°.”

CIE 0580 · Paper 4 · Nov 2023

For cos(x) = k with x in 0°–360°, the two solutions are x = arccos(k) and 360° − arccos(k). Use the symmetry of the cosine curve around 180° (or the y-axis).

“In solving the trigonometric equation in part (b) it was not uncommon for candidates to find one correct angle, but a second correct answer was often not given or not attempted. Those not reaching a correct angle sometimes gained a mark for the correct rearrangement to cos x = −3/5, but this was not always managed. (The incorrect cos x = +3/5 was not uncommon.)”

CIE 0580 · Paper 4 · Nov 2023

Use tan(angle) = opposite / adjacent when you have those two sides and need the third. Pythagoras works too but takes extra steps. Pick the most direct method.

“In general, this question was not answered particularly well. Some gave answers showing no working or working that did not involve trigonometry. Some used Pythagoras' theorem to find the hypotenuse and then did not go on to find the value for y. Those that did use tangent were invariably totally correct.”

CIE 0580 · Paper 1 · Nov 2023

For tan(x) = k in 0° ≤ x ≤ 360°, solutions come in pairs 180° apart: θ and θ + 180°. If tan(x) = √3/3 and x = 30°, the other solution is 30° + 180° = 210°.

“Many candidates gave the solution x = 30° only and did not use their sketch to help identify the second solution of x = 210°. A small number of candidates just gave the x = 210° solution despite having also shown x = 30° in their working.”

CIE 0580 · Paper 4 · Jun 2023

y = tan(x) has period 180°, and VERTICAL ASYMPTOTES at x = 90°, 270° (where cos = 0). The curve rises from −∞ to +∞ in each branch. It's NOT the same shape as sin or cos.

“Some candidates were able to produce a good sketch of the graph of y = tan x. Some attempts at the correct graph only covered part of the domain between the asymptotes or showed incorrect curvature. It was also common to see graphs of the correct shape but with the wrong period. Some candidates sketched graphs of y = sin x or y = cos x.”

CIE 0580 · Paper 4 · Jun 2023

Don't assume a triangle is right-angled unless it says so or there's a right-angle mark. If the given angle is 149°, use the COSINE rule — not Pythagoras (which only applies to 90° triangles).

“A significant number of other candidates calculated BC and CD and then applied Pythagoras to triangle BCD, treating angle BCD as a right angle rather than 149, possibly mistaking the diagram for a 3-D problem.”

CIE 0580 · Paper 4 · Jun 2023

When 'show that' gives you a target rounded to e.g. 2dp, you must compute to at least ONE more dp (here 3dp) and then state it rounds to the target. Writing only the rounded target does not score.

“Almost all candidates used a correct method to find the length of AB. Many of these gave their answer as 10.54 when they were expected to give an answer to a minimum of three decimal places and show that it rounds to 10.54.”

CIE 0580 · Paper 4 · Jun 2023

For 3D angle problems, (1) find the relevant 2D diagonal on the base first with Pythagoras, (2) then use tan(θ) = opposite/adjacent with the vertical height. The angle is in a 2D right-angled triangle — not 3D.

“Many answers showed the starting point of using Pythagoras to find AC or AG, even if they did not show further understanding of which angle to find or how to proceed. Some responses used triangle GAB and weaker responses used a variety of trigonometric functions with combinations of the given dimensions.”

CIE 0580 · Paper 2 · Jun 2022

SOH CAH TOA — label opp, adj, hyp first, then pick sin (opp/hyp), cos (adj/hyp), or tan (opp/adj). Rearranging depends on what you're solving for: BC = 15 × sin 38 or 15 / sin 38 depending on context.

“Some worked out the angle at C, others who knew this involved trigonometry, used the wrong ratio, mostly cosine. Those that used the correct ratio sometimes calculated BC = 15 ÷ sin 38.”

CIE 0580 · Paper 1 · Nov 2022

For sin x = k in 0°–360°, the two solutions are arcsin(k) and 180° − arcsin(k). Don't stop at the first calculator answer. If the calculator gives a negative angle, adjust to the range 0°–360°.

“Accuracy with the angles was a problem with truncation to −19.4 often seen. If candidates struggled with the angles, they often managed to gain 1 mark usually for sin x = −1/3 or for two angles adding to 180. Some candidates did not proceed beyond the answer their calculator gave of −19.5.”

CIE 0580 · Paper 2 · Nov 2022

The shortest distance from a point to a line is the perpendicular distance. The foot of the perpendicular is NOT necessarily the midpoint of the line — assuming it bisects the line segment is a very c

“many were not able to identify that the shortest distance from C to BD would be found by drawing a line from C perpendicular to BD. A common misconception was to assume that the base of the perpendicular bisects BD, and then using Pythagoras's theorem with CD and 4.85.”

CIE 0580 · Paper 4 · Jun 2021

For 'show that angle = 36.9°', compute to at least 1 more decimal place (e.g., 36.87°) and then say it rounds to 36.9. Don't start from 36.9 — that's circular and scores zero.

“Some went straight from sin x [=] 18/30 to 36.9 and didn't show a more accurate answer. A small number found the third side using Pythagoras' theorem and then used tan x. Candidates need to realise on a 'show that' question that they cannot start with what they have to show, in this case 36.9.”

CIE 0580 · Paper 2 · Nov 2021

y = tan(x) has period 180° (not 360° like sin and cos). It has vertical asymptotes every 180° starting at 90°. It's NOT the same shape as sin or cos.

“In part (a) only minority of candidates were able to give a fully correct response for the sketch of y = tan x. Some others demonstrated awareness of the correct shape of the graph but with the incorrect period or with some sections correct and some incorrect. Common incorrect answers were attempts at sine or cosine graphs or a translation of one of these.”

CIE 0580 · Paper 2 · Nov 2021

For 'show that the angle is 63.4° correct to 1dp', write the calculator display (e.g., 63.4349...) FIRST, then state it rounds to 63.4. One dp alone is not enough.

“A number of candidates who gave the correct ratio and second step tan⁻¹(8/4) did not gain full credit as they then gave the answer of 63.4°. Candidates should be reminded that in 'show that' questions which include the phrase 'correct to 1 decimal place' that they must show the answer to more than 1 decimal place before showing the final answer to 1 decimal place.”

CIE 0580 · Paper 3 · Nov 2021

For 3D angles (like a diagonal to a base), first SKETCH the right-angled triangle you need — one side is the diagonal, another is the relevant 2D length on the base. Then use tan/sin/cos.

“On the diagram, some indicated the wrong angle and could not visualise the right-angled triangle required. To find the correct angle required the use of two numerical values which in some cases due to premature rounding led to an answer which was out of range.”

CIE 0580 · Paper 4 · Nov 2021

Give inexact answers to 3 significant figures (e.g., 6.90, not 6.9). Always show working — 'just the answer' is risky because if it's wrong, you get zero instead of method marks.

“Many candidates recognised the need for trigonometry but chose to use cosine instead of sine. There were quite a few candidates who gave the answer as 6.9 but as inexact answers must be given to 3 significant figures, this did not gain the accuracy mark. If there was no working, just the answer of 6.9 was not able to be given any marks.”

CIE 0580 · Paper 1 · Nov 2019

For 3D trig: DRAW OUT the 2D right-angled triangle you're using, label all vertices, mark the angle you want to find. Ambiguous working can't be credited even if the method is correct.

“Many candidates did gain a mark for making a first correct step of identifying the angle required and this should be encouraged, either by clearly showing the angle on the diagram or preferably by drawing out a triangle with clearly labelled vertices and the angle identified. Work cannot be credited if it is ambiguous which angles or lengths the candidate is referring to.”

CIE 0580 · Paper 4 · Nov 2019

In the cosine rule a² = b² + c² − 2bc·cos A, the '−2bc·cos A' means only the 2bc term is multiplied by cos A. A common error is to subtract 2bc from (b² + c²) first and then multiply the result by cos

“There were many candidates who set up the correct equation with a² as the subject and substituted the sides correctly but then made the error of calculating (b² + c² – 2bc) cosA, resulting in 9cosA.”

CIE 0580 · Paper 2 · Nov 2018

The graph of y = cos(x) is a smooth curve (not straight lines). Key points: (0,1), (90,0), (180,-1), (270,0), (360,1). It starts at 1 when x=0, NOT at 0.

“Most candidates drew correct sketches of a cosine curve passing through (0, 1), and passing sufficiently close to (180, –1) and (360, 1). Some candidates incorrectly drew straight lines between these points, rather than appropriate curves... Some candidates incorrectly started their curve at (0, 0) rather than (0, 1).”

CIE 0580 · Paper 2 · Jun 2024

For the cosine rule a² = b² + c² − 2bc·cos(A), type the whole right-hand side into your calculator in ONE go and take the square root last. Breaking it into stages introduces rounding and sign errors.

“Candidates are advised to use their calculator to evaluate the expression in a single step after showing the substitution rather than working out the expression in stages which is when errors are likely to occur.”

CIE 0580 · Paper 4 · Jun 2024

Always check your calculator is in DEG (degrees) mode before a trigonometry question — look for 'DEG' on the screen. GRAD and RAD give different (wrong) answers for CIE 0580.

“A small number had their calculator set in grads mode leading to an incorrect final answer and some did not recall the sine rule correctly.”

CIE 0580 · Paper 4 · Jun 2024

y = sin(x) starts at (0, 0), rises to (90°, 1), crosses (180°, 0), drops to (270°, −1), returns to (360°, 0). Smooth wave shape. Don't confuse with cos (which starts at (0, 1)).

“A significant number of candidates drew perfectly symmetrical graphs in part (a) with the angles marked on the x-axis, passing through the necessary points and with correct amplitude, although many of the diagrams contained straight lines while others lacked rotational symmetry. Not starting at zero was quite common, often starting at (0, 1) and drawing a cosine curve, or starting at (0, −1) instead.”

CIE 0580 · Paper 2 · Nov 2022

When a final answer requires rounding, show a more accurate intermediate value (at least 4 significant figures) before stating the rounded answer. Writing only the rounded value means the accuracy of

“It was common to see candidates give an answer of 9.2 in this part without showing a more accurate calculated figure, so full marks could not be awarded.”

CIE 0580 · Paper 4 · Jun 2021

Truncating (dropping digits) is not rounding. 26.97 truncated is 26.9, but correctly rounded to 3 s.f. is 27.0. Always round, never truncate, when giving a final answer.

“Some candidates gave inaccurate final answers such as 26.9 which is a truncation of the more accurate value of 26.97 rather than a correct rounding to 27.0.”

CIE 0580 · Paper 4 · Jun 2021

In multi-step trigonometry, keep at least 4 significant figures in intermediate values. Rounding too early compounds errors and can cost the accuracy mark even when the method is correct.

“it was common to see the length AC rounded to 15.6 leading to a result of 7.8 for the length of MC and an angle of 56.14…° rather than 56.09…°. It is important to use more significant figures in the working than are required in the final answer.”

CIE 0580 · Paper 2 · Jun 2021

Never round the trig ratio before applying inverse trig. Use the full decimal from your calculator in sin⁻¹, cos⁻¹, or tan⁻¹. Rounding 0.649 to 0.65 then taking sin⁻¹ gives a wrong final angle.

“Many lost the second mark for a variety of reasons. Some did not know how to find the angle from the decimal but most rounded the decimal, for example 0.65, before finding the angle.”

CIE 0580 · Paper 1 · Jun 2019

Always check whether a triangle is right-angled before assuming trig ratios apply. For oblique triangles, use sine or cosine rule. For the cosine rule: after finding RQ², take the square root to get R

“Some missed the fact that angle PSR is marked as 97° and treated the triangle as right-angled. In part (b) many candidates were able to use the cosine rule correctly and gain full marks. Some forgot to take the square root and, having reached RQ² = 13.94, stopped at that point.”

CIE 0580 · Paper 2 · Jun 2019

Always read which angle or vertex is required before applying trig ratios. Don't assume the angle is at the base — check the diagram and question carefully.

“Many did not score in this part simply because they assumed the required angle was on the more usual base line at point C, rather than point A.”

CIE 0580 · Paper 1 · Jun 2018

When finding an obtuse angle, prefer the cosine rule — it gives the correct obtuse value directly. The sine rule has ambiguity (sin gives the same value for θ and 180°−θ), so using it for the largest

“Some used the cosine rule to find one of the other two angles and then the sine rule to find the required angle. They did not realise that there are two solutions with sine and they needed to find the obtuse solution rather than the acute one.”

CIE 0580 · Paper 4 · Nov 2018

cos θ = adjacent/hypotenuse. Getting the fraction upside down (hyp/adj) gives the wrong angle. Also, use the simplest trig ratio directly — don't find a third side first then use a different ratio, as

“Over half the candidates found the angle correctly, choosing cosine with a correct fraction, although 8/5 instead of 5/8 was seen. A few candidates were determined to make it a long question by finding the other side by Pythagoras' theorem and then using tangent or sine for the angle. This often produced an inaccurate answer.”

CIE 0580 · Paper 2 · Nov 2018

In trig calculations, keep the fraction (e.g. 20/52) or full calculator value throughout. Rounding 20/52 to 0.38 before taking cos⁻¹ loses enough accuracy to give the wrong answer. Use the fraction di

“If candidates correctly used cosine another common error was prematurely rounding 20/52 to 0.38 which gives the common incorrect answer of 67.6… or 67.7. Candidates should use the whole calculator value or the fraction when using trigonometry.”

CIE 0580 · Paper 4 · Nov 2018

You can ONLY cancel FACTORS (things multiplied), not terms added/subtracted. Factorise numerator and denominator first, then cancel matching brackets. Never cancel x² from x² − 25.

“Candidates who recognised that the numerator and denominator should be first factorised to create products were frequently successful. However, a common wrong approach was to merely cancel the x² in the first step with (x² − 25)/(x² − x − 20) = −25/(−x − 20) = 5/(x + 4), or similar, commonly seen.”

CIE 0580 · Paper 4 · Nov 2021

In 'show that k = 8' questions, you must DERIVE 8, not assume it. Starting from 'k = 8' and working both ways is circular and scores zero. Work from the given info towards the target.

“This 'show that' question was often not attempted. Candidates who did attempt it often did not gain credit as they generally used the answer of k = 8 as part of their working. Candidates should be reminded that in any 'show that' question candidates must not use what they are asked to show as part of their answer.”

CIE 0580 · Paper 3 · Nov 2021

Completing the square rewrites x² + bx + c as (x + b/2)² + constant. It is NOT the same as difference of two squares a² − b² = (a+b)(a−b). Different techniques for different tasks.

“There were many candidates who were confusing the difference of two squares with completing the square which resulted in (x – 4)², (x + 4)² or (x + 4)(x – 4).”

CIE 0580 · Paper 2 · Jun 2024

When raising a power to a power, MULTIPLY the exponents: (x³)³ = x⁹, not x⁶. But for coefficients like (4x³)³, the coefficient is raised: 4³ = 64, not 4 × 3 = 12.

“A small number incorrectly simplified the power by adding to give 6 rather than multiplying to give 9. Others correctly multiplied the powers but also multiplied the coefficient giving the incorrect answer of 12x⁹.”

CIE 0580 · Paper 2 · Jun 2024

−(x + 1)² means you square FIRST, then negate: −(x² + 2x + 1) = −x² − 2x − 1. And (x + 1)² ≠ x² + 1 — you need the cross term 2x.

“The most common error was to expand –(x + 1)² as –x² + 2x + 1. Others expanded the brackets by just squaring the two terms inside or, to a lesser extent, multiplying the terms inside by 2.”

CIE 0580 · Paper 4 · Jun 2024

The line of symmetry of a parabola is a VERTICAL line, so its equation is x = [number] (not y = ... or just a number alone). For y = ax² + bx + c, it is x = −b/(2a).

“This part on identifying the equation of the line of symmetry was found challenging. Common errors included 2.5, x = 2, x = 3, y = 2.5, y = mx + c and y = –x² + 5x + 7. Although not required it may have helped candidates to draw the line of symmetry first.”

CIE 0580 · Paper 3 · Jun 2024

(81x¹²)^(3/4) = 81^(3/4) × x^(12 × 3/4) = 27x⁹. Apply the power to BOTH the coefficient (81^(3/4) = 27) and the variable (x^9). Don't multiply 3/4 × 81.

“A common error seen however was applying the power of 3/4 to only the x¹² or (less often) to only the 81 but not to both. Also seen was adding the indices in error, or treating the power of 3/4 as a factor to work out 3/4 × 81 giving an answer with coefficient 60.75”

CIE 0580 · Paper 4 · Nov 2023

2y + 3y = 5y (addition) but 2y × 3y = 6y² (multiplication — multiply coefficients AND the variables). These are completely different.

“The very common errors included 2y + 3y = 5y, 2 × (2y + 3y) = 10y, 2y × 3y = 6y, 2y × 3y = 5y and 2y × 3y = 5y².”

CIE 0580 · Paper 3 · Jun 2023

When factorising 5x² − 19x + 12: the constant (+12) tells you both brackets need the SAME sign, and the middle term (−19x) tells you both are NEGATIVE. Answer: (5x − 4)(x − 3).

“Sign errors in the factorisation of the numerator were very common with (5x + 4)(x – 3) or (5x – 4)(x + 3) both seen fairly often.”

CIE 0580 · Paper 2 · Jun 2023

When you halve (or multiply by a constant ratio) each term, you have a GEOMETRIC sequence. The nth term is a × r^(n−1), not 24/n or 24 − 12(n−1).

“Many candidates attempted to find the answer to this part in the same way as part (a), by finding differences and not knowing how to carry on when differences were all different... The most common successful approach used the general term for a geometric sequence, ar^(n−1), reaching an answer of 24 × 0.5^(n−1).”

CIE 0580 · Paper 2 · Jun 2023

x² + 3x = 0 has no constant term — factor out the common x: x(x + 3) = 0, giving x = 0 or x = −3. You do NOT need the quadratic formula here.

“Of those who correctly found x² + 3x = 0, some were unsure how to solve this and stopped. Many did not realise that a simple factorisation, x(x + 3) was required and some attempted to use the quadratic formula to solve, often incorrectly as they were unable to understand what to do without a constant term.”

CIE 0580 · Paper 2 · Jun 2023

nth term = a + (n − 1)d. If d is negative, write (n − 1)(−6) or (n − 1) × (−6) — the bracket is essential. Writing '(n − 1) − 6' means subtract 6, which is different.

“A few candidates seemed aware of a correct method but wrote 23 + (n – 1) –6 rather than 23 + (n – 1) × –6 or 23 + (n – 1)(–6) and did not recover from the missing brackets or missing multiplication sign.”

CIE 0580 · Paper 2 · Jun 2023

To solve a new equation using an existing graph, rearrange it to match '(existing curve expression) = (something linear)'. Then plot that linear part and find intersections. Using the quadratic formula ignores the method.

“In part (c) only the strongest candidates knew how to rearrange the given equation to deduce the correct straight line needed to solve it using the graph from part (b). A common incorrect line, when one was attempted, was y = −x − 1. Often a line was not attempted, as required by the question, with instead an attempt made using the quadratic formula.”

CIE 0580 · Paper 4 · Nov 2023

When the question says 'write down an equation', you MUST state it explicitly (e.g., '5x + 2x + 90 = 180'). Trial-and-error without a clear equation loses method marks even if the answer is right.

“This problem-solving question combined geometry and algebra. There was one piece of information that was not explicitly given to candidates, that the angles of a triangle add to 180°. The question gave the instruction to write down an equation. Some did this then went on to show clear working to find x correctly. Candidates often did not do this and chose values for x to find the two angles by trial and error... Candidates that used this method to get 13 without first showing an equation were awarded only partial credit as they did not follow the instructions.”

CIE 0580 · Paper 1 · Nov 2023

Read proportion wording precisely: 'proportional to the square' → y = kx²; 'proportional to the square root' → y = k√x; 'inversely proportional' → y = k/x. Always include the constant k.

“Whilst there were many fully correct solutions seen to this problem, commonly errors were due to not reading the demand carefully enough by: (i) attempting linear proportion, (ii) attempting inverse proportion, or (iii) using square root rather than square. It was also common to see attempts without using a constant of proportionality and so failing to score.”

CIE 0580 · Paper 4 · Nov 2023

a² − b² = (a − b)(a + b) — NOT (a − b)². And (x² − 5²) factorises as (x − 5)(x + 5), not (x² − 5)(x + 5) — you take the square root of EACH term.

“A significant number started correctly by stating x(x² – 25), rewriting this as x(x² – 5²) and then went wrong by writing this as x(x – 5)². Other incorrect answers included (x² – 5)(x + 5) and (x² + 5)(x – 5).”

CIE 0580 · Paper 4 · Jun 2023

In factor-by-grouping, the two brackets after grouping MUST be identical (not just close). Rewrite (3 − 2q) as −(2q − 3) to flip the sign — that's how you make the brackets match.

“The most common error resulted from factorising the pairs to give t(2q – 3) – 2(3 – 2q) before giving the final answer as (2q – 3)(t – 2), with candidates not realising that the expressions in the two brackets need to be the same before continuing to the final answer.”

CIE 0580 · Paper 4 · Jun 2023

When expanding, (−a)(−b) = +ab (positive). Also remember 2p² × 3p² = 6p⁴ (multiply coefficients, ADD the exponents). And −9p² − 4p² = −13p² (both negative: add them, keep the minus).

“A wide variety of errors were seen. Some of these included expanding to give a constant term of –6 instead of +6, 2p² × 3p² given as 6p², simplifying –9p² – 4p² as –5p².”

CIE 0580 · Paper 4 · Jun 2023

For a triangle with angles 10x − 12, 2x + 3, and some third: sum them and set equal to 180°. Form ONE equation, not two to solve simultaneously. Trial and error rarely scores full marks.

“Setting up the initial equation correctly was challenging for most candidates. 10x − 12 = 2x + 3 was more commonly seen as the equation or 3(10x − 12) = 2x + 3. There were unsuccessful attempts at trial and error, or answers from incorrect starting points. Many candidates were unclear that one equation was needed.”

CIE 0580 · Paper 2 · Nov 2023

In 5a + 7b − 11b − 6a, the minus sign belongs to 11b only. When collecting like terms: 7b − 11b = −4b, not +4b. Circle each term WITH its sign before combining.

“The term −11b was found more challenging (this was often give as −1b) than the first term. Most problems stemmed from misunderstanding that subtraction signs only affect the number that immediately follows.”

CIE 0580 · Paper 1 · Nov 2023

Positive cubic: comes from bottom-left, up through a max, down through a min, up to top-right. For y = x(x − a)(x − b), roots are 0, a, b. Don't try to plot points — use the shape and roots.

“To draw an acceptable sketch with the correct curvature, candidates needed a good understanding of the possible shapes of a positive cubic graph and that the given curve had three roots, one of which was zero. Most candidates attempted a sketch but few satisfied all of the above requirements.”

CIE 0580 · Paper 4 · Jun 2022

Recognise the pattern a² − b² = (a − b)(a + b). So 4x² − 16 = (2x − 4)(2x + 4). Check each expression carefully — not every expression is a difference of squares.

“Most answers demonstrated good skills in simplifying an algebraic fraction and went on to obtain the correct answer. Others only factorised one of the two expressions correctly, with some not spotting the difference of two squares, or treating both expressions as the difference of two squares.”

CIE 0580 · Paper 4 · Jun 2022

When multiplying both sides by (g − c), distribute: M(g − c) = Mg − Mc, not Mg − c. Then collect all g-terms on one side, factor out g, and divide.

“Most responses displayed a good understanding of the steps involved but errors at the various stages meant that only a smaller majority reached the correct answer. Many started by attempting to clear g − c from the denominator but writing this as Mg − c was a common error.”

CIE 0580 · Paper 4 · Jun 2022

In the quadratic formula x = (−b ± √(b² − 4ac))/(2a), use brackets for b when it's negative: −(−5) = +5 and (−5)² = +25. Use long division and square root lines that cover everything.

“Errors with the negative terms, either writing −25 rather than −(−25) or omitting brackets around (−25)² were common errors as well as short division lines and short square roots.”

CIE 0580 · Paper 4 · Jun 2022

Time = distance ÷ speed. Convert 'hours and minutes' to DECIMAL hours (2h 48min = 2.8 h, not 2.48 h). And if cycling at (x+10) km/h, the time is distance/(x+10), not distance/(10x).

“Most answers set up a correct equation. Others either gave an equation with an incorrect time, such as 2 h 48 min, 2.48 h or 168 min, or gave an incorrect expression for Darpan's cycling time such as 26/(10x).”

CIE 0580 · Paper 4 · Jun 2022

The line of symmetry of a parabola y = ax² + bx + c is x = −b/(2a). It's a vertical line, so always of the form x = [a number]. Not 'y = ...', not a number alone.

“This part on identifying the equation of the line of symmetry was not well answered. Common errors included x = −0.3, x = 3, y = −3, 3, y = mx + c and y = x² + 6x − 160.”

CIE 0580 · Paper 3 · Jun 2022

When one equation is linear and the other is quadratic, substitute the linear one INTO the quadratic. Rearrange the linear to x = ... (or y = ...) and substitute. Make the substitution from the simpler variable.

“The most efficient method which usually gained the marks was to substitute the first equation into the second resulting in x² − 2(11 − 3x) = 18. Those who attempted to eliminate the x-variable by the substitution (11 − y)²/3² − 2y = 18 often followed this by an incorrect simplification of (11 − y)²/3².”

CIE 0580 · Paper 2 · Jun 2022

'y is inversely proportional to the square root of (x + 4)' means y = k / √(x + 4). Not (x + 4)², not (x + 4), not direct. Read the exact wording and match every piece.

“Some responses did not seem to have read the question carefully and used the square of (x + 4) or used x + 4. Others used direct proportion instead of inverse proportion. Some found the constant of proportionality incorrectly, due to arithmetic or rearranging errors, but were able to gain the second method mark.”

CIE 0580 · Paper 2 · Jun 2022

When forming equations, re-read the problem and check: (1) how many of each item, (2) exact wording of 'less than' or 'more than'. '$24 less than a book' means (book − 24), and 2 of them is 2(book − 24), not (book − 48).

“This was a challenging question. Most scripts attempted an equation as instructed. The most common error was including just one shirt leading to x = 140/3. Other common errors were including x − 48 rather than 2(x − 24) or including the hat as x − 16 − 24. Candidates must read the wording of the question carefully.”

CIE 0580 · Paper 2 · Jun 2022

Apply fractional powers to EVERY part inside. (x³/343)^(1/3) = x^(3×1/3) / 343^(1/3) = x / 7. Both the numerator AND denominator get the power.

“Part (b) was one of the most challenging questions on the paper. Some answers got as far as 1/343, (x/7)³ or x³/7³ but did not complete the simplification. Others dealt correctly with the 7 but not the x, giving an answer of x/343.”

CIE 0580 · Paper 2 · Jun 2022

n/5 + n/2 = 2n/10 + 5n/10 = 7n/10. Use LCM (10) as denominator. Then n − 7n/10 = 3n/10.

“Most candidates had the right idea that n/5 and n/2 had to be subtracted from n but some equated n to the sum of these two fractions. Following on from that stage proved challenging for most candidates but some did add these fractions correctly and gained credit for 7n/10.”

CIE 0580 · Paper 1 · Jun 2022

√(16x²) = 4x (square root each factor — both the number and the variable). And y⁰ = 1 exactly, so 5y⁰ = 5, not y or 0.

“With the first term it was often just 16 or x² which was square rooted rather than both, while the second term was often just 1 (presumably from y⁰) rather than 5 × 1. Others combined the 5 with 16 resulting in 80 or an attempt at the square root of 80.”

CIE 0580 · Paper 1 · Jun 2022

For completing the square x² + bx + c = (x + b/2)² + (c − b²/4). The 'a' in (x + a)² is always HALF the coefficient of x. Then adjust the constant.

“In part (a), a minority of candidates knew the quick way to identify the value of a (halving the coefficient of x in the expression), and then finding b. A larger number attempted to expand (x + p)², sometimes correctly to score 1 mark but, commonly not then proceeding to the correct answers.”

CIE 0580 · Paper 4 · Nov 2022

HCF of algebraic terms: take the LOWEST power of each common variable. HCF(12a³b², 20a²b) = 4a²b (lowest powers). Not the highest — that would be the LCM.

“Candidates who understood HCF usually gave an answer involving 4. They did not always appear to know how to deal with the algebraic parts of the expressions, so answers of 4, 4ab and 4a²b² were common. Some responses showed a confusion between highest common factor and lowest common multiple so the answer 60a³b² was also often seen.”

CIE 0580 · Paper 4 · Nov 2022

To find where a line meets a curve: set the two y-equations equal, move everything to one side (= 0), and solve the resulting quadratic. Don't set the curve to some arbitrary constant.

“This was a good question for some who knew immediately to equate the two equations for y to find the required intersections. Some gained credit for attempting the solution of their resulting quadratic if incorrectly rearranged. Weak responses worked with y = x² − 3x − 11 which did not provide any useful results.”

CIE 0580 · Paper 4 · Nov 2022

For a quadratic sequence an² + bn + c: substitute n = 1, 2, 3 (the position numbers) to get term 1, term 2, term 3. Don't substitute the term VALUES — substitute the n-positions.

“Greater care needs to be exercised in reading and understanding the question, a very common error was with candidates not substituting 1 and 2 for n, but rather substituting for n the values of the first and second terms (−3 and 2).”

CIE 0580 · Paper 2 · Nov 2022

For 5x² − 20y²: first factor out 5 → 5(x² − 4y²), THEN recognise 4y² as (2y)² so x² − 4y² = (x − 2y)(x + 2y). Final: 5(x − 2y)(x + 2y), not 5(x − 4y)(x + 4y).

“The most common answer seen was the incomplete factorisation of 5(x² − 4y²). Some candidates appeared to recognise that they were dealing with the difference of two squares but gave incorrect final answers such as 5(x + 4y)(x − 4y).”

CIE 0580 · Paper 2 · Nov 2022

When factoring by grouping, watch for signs inside brackets. (1 − 4k) is not (−4k). Check the brackets match exactly — sign errors cost marks.

“Some lost the 1 from (1 − 4k), leading to an answer of (−4k)(2m + 3p). Candidates are advised to ensure both parts contain the same signs, it was not uncommon to see the expression 2m + 3p − 4k(2m − 3p).”

CIE 0580 · Paper 2 · Nov 2022

If y ∝ 1/x² and x is halved, then y is multiplied by 4 (because 1/(x/2)² = 4/x²). Inverse square: halving x quadruples y. 'Doubles' is wrong — work it out algebraically.

“Hardly any candidates correctly answered part (b), with this being the most challenging question on the paper. The most common answer by far was that F will double. Many said that F increases but did not give a factor. Many demonstrated a lack of understanding of inverse proportionality by saying that it would also halve or that it would stay the same.”

CIE 0580 · Paper 2 · Nov 2022

For joint proportion, COMBINE relationships into one: if y ∝ x and y ∝ 1/z², then y = kx/z². Use ONE constant k for the combined relationship, not different k's for each.

“Many candidates were able to score marks for at least one of the two relationships stated as an equation, quite often for both, but much less common was being able to combine these two into a correct final answer. Poor use of notation meant than many used k to stand for two different constants of proportionality, often both found correctly, which then usually caused confusion.”

CIE 0580 · Paper 4 · Nov 2022

1/64 = 1/4³ = 4⁻³. So if 4ˣ = 1/64, then x = −3. Answer x (the exponent only), not the whole expression.

“Whilst many candidates seemed to know that 1/64 was equal to 4⁻³ not all of these offered −3 as their answer. As x was asked for, −3 was the only acceptable answer. Common incorrect answers were +3 or 1/3.”

CIE 0580 · Paper 4 · Nov 2022

Rearranging formulas: work STEP BY STEP and finish the simplification. s/(0.5a) should be simplified to 2s/a. Don't leave fractions unresolved.

“Many took correct steps to rearrange the formula, but left answers with incomplete steps such as (s ÷ a)/0.5 or s/(0.5a). Unsuccessful candidates usually took wrong first steps such as subtracting 1/2 or had incorrect attempts to square root too early.”

CIE 0580 · Paper 4 · Nov 2022

The axis of symmetry of a parabola is a vertical line, so its equation must be x = (value), never y = something. Find it by halving the sum of the x-intercepts, or by completing the square.

“This part on identifying the equation of the line of symmetry was poorly answered. Common errors included x = 0.5, y = –0.5, 0.5, –0.5, y = mx + c and y = –x² – x + 14.”

CIE 0580 · Paper 3 · Jun 2021

For decreasing arithmetic sequences, the common difference d is negative. Using the positive difference gives an increasing formula (wrong). For d = −7, the nth term contains −7n, giving a formula lik

“many did not realise that d was negative and the positive difference led to the commonly seen answer of 22 + 7n. Others had a correct un-simplified expression but made errors in attempting to simplify it.”

CIE 0580 · Paper 1 · Jun 2021

After eliminating one variable, divide carefully. From 28y = 7, y = 7 ÷ 28 = 0.25, not 4. Students often invert the division when the coefficient is larger than the constant.

“Many reached the stage of 28y = 7 but then gave y = 4.”

CIE 0580 · Paper 1 · Jun 2021

When substituting into a quadratic equation, you cannot take square roots of individual terms: √(149 − y) ≠ √149 − √y. Rearrange the linear equation first, substitute entirely, then solve the resultin

“x = 7 – y was commonly incorrectly used. Others followed the correct rearrangement of x² = 149 – y with the incorrect working x = √149 − √y. There were a variety of rearrangements and substitutions possible and a common mistake was a sign error in the initial rearranging.”

CIE 0580 · Paper 2 · Jun 2021

Look for compound common factors: if two terms both contain 5(k + g), take it out as a single factor. Expanding the brackets to avoid the challenge and then failing to simplify earns no marks.

“Candidates found this factorisation very challenging and few were able identify that the two terms had a common factor of 5(k + g) and reach the correct answer. Some candidates just expanded the given brackets and were then unable to simplify further”

CIE 0580 · Paper 4 · Jun 2021

When eliminating fractions, use brackets correctly: 48/(x+2) multiplied by (x+2) simplifies to 48 — not 48x + 2. Forgetting the bracket around the denominator is the most common algebraic slip in rati

“it was common to see errors with the use of brackets, such as (48)x + 2. Candidates sometimes recovered from these slips but were not able to earn full marks.”

CIE 0580 · Paper 4 · Jun 2021

When a word problem mixes dollars and cents, convert everything to ONE unit before writing the expression. $y = 100y cents — multiply by 100 to convert dollars to cents.

“This question was not well answered with very few candidates gaining full credit. 100y was rarely seen and it appeared many had not realised the payment was in dollars.”

CIE 0580 · Paper 2 · Nov 2021

If the new subject appears in TWO places, (1) move all terms containing that subject to one side, (2) factor the subject out, (3) divide. This ALWAYS comes up when the subject appears twice.

“Rearranging the formula to make h the subject proved challenging for many of the candidates... Candidates need to be aware that in such a question where the new subject appears twice, there is always a need to collect terms with the new subject and factorise.”

CIE 0580 · Paper 2 · Nov 2021

For proportion questions: (1) write y = k × (the correct function), (2) substitute the given pair to find k, (3) substitute back. Always include k — don't skip it.

“Candidates who began with y = k × √(x − 3) usually succeeded, although their working was not always clear. There were a number of candidates who did not work with a constant, whilst quite often the square root was ignored by others. Some misread the question and seemed to be looking for inverse proportion whilst a few used squaring rather than square root.”

CIE 0580 · Paper 2 · Nov 2021

From A = πr², divide by π first to get r² = A/π, then square root BOTH together: r = √(A/π). Write the vinculum (top bar) long enough to cover the whole fraction — √A/π means √A ÷ π which is different.

“Some candidates didn't make the square root symbol long enough to cover all of A/π so it looked as though the answer was √A/π rather than the correct answer of √(A/π). Less able candidates subtracted instead of dividing by π.”

CIE 0580 · Paper 3 · Nov 2021

To add algebraic fractions, find a common denominator (multiply the two denominators if they share no factors). NEVER add numerators and denominators separately — that's not how fractions work.

“A notable number of candidates did not know how to deal with the denominators when adding fractions with a common wrong approach to add the numerators and denominators separately as (x+5)/x + (x+8)/(x−1) = (2x+13)/(2x−1).”

CIE 0580 · Paper 4 · Nov 2021

3 − 2(3 − 2x): expand the bracket FIRST: 3 − (6 − 4x) = 3 − 6 + 4x = −3 + 4x. Don't evaluate 3 − 2 first. And the negative distributes over both terms inside.

“The most common error seen was to deal with the negative sign incorrectly and expand 3 − 2(3 − 2x) wrongly as 3 − 6 − 4x. In addition many candidates did not use the correct order of operations and subtracted 3 − 2 before expanding the bracket, to 3 − 2x.”

CIE 0580 · Paper 4 · Nov 2021

To find an unknown in an identity, match coefficients: if x³ − 4x² + 4x ≡ x³ − 2ax² + a²x, then −2a = −4 and a² = 4, giving a = 2.

“Candidates that were successful generally used one of two methods, either showing that the expression factorised to x(x − 2)² leading to the answer of 2 when comparing it to x(x − a)² or alternatively x(x − a)² expands to x³ − 2ax² + a²x also leading to the answer of 2 when comparing to x³ − 4x² + 4x.”

CIE 0580 · Paper 4 · Nov 2021

Geometric nth term = a × r^(n−1). The base is the COMMON RATIO r (e.g., 2 if you double each time), not the first term. 0.25 × 2^(n−1) ≠ 0.5^(n−1).

“Some recognised the exponential nature of sequence C but made errors in the base for the nth term, with n or 0.5 with an incorrect power being the most common. The correct expression 0.25 × 2^(n−1) was sometimes seen in working but in the answer space this was incorrectly written as 0.5^(n−1).”

CIE 0580 · Paper 4 · Nov 2021

For double factorisation: first spot the common factor (m+p), then factorise again. 4(m+p)(3+2m+2p) — if you see the same bracket appearing twice, factor it out as a bracket.

“Part (b) was a good discriminator as the most challenging question in the paper. A common incomplete solution was to take only the 4 out as a factor and stopping at 4(3(m+p) + 2(m+p)²). A correct partial factorisation of (m+p) gained many candidates the B1 mark.”

CIE 0580 · Paper 2 · Jun 2019

From 2w + 2h = P, isolate w: 2w = P − 2h, then w = (P − 2h)/2. Don't drop coefficients: 'w + h = P/2' forgets the 2 in front of h.

“Many candidates gained partial credit for 2w + 2h = P but could not then progress to write the correct expression. Others started incorrectly with 2w = P − h or w + h = P − 2. Candidates who wrote w + h = P/2 often had (P − h)/2 as their final answer.”

CIE 0580 · Paper 3 · Nov 2019

When rearranging terms, KEEP their signs. 5x − 3y + 9x + 10y ≠ 5x + 3y + 9x + 10y. Underline or box the sign attached to each term before reordering.

“Many candidates found this question challenging. Many changed the sign when grouping like terms together, hence a variety of incorrect answers were seen, such as 14x − 13y, 14x − 37y, −6x − 37y.”

CIE 0580 · Paper 3 · Nov 2019

When solving 5x² = k, divide by 5 FIRST (x² = k/5), then square root. Make the √ sign cover the whole expression underneath — √(k/5), not √k/5.

“Those candidates who started correctly often tried to square root before dividing by 5. This led to the square root sign not covering the 5 although quite a number, having done the first two steps correctly, carelessly also did not show the square root sign over the whole expression.”

CIE 0580 · Paper 2 · Nov 2019

(x + a)² = x² + 2ax + a² (not x² + a²). For x² + 4x − 9 in the form (x + a)² + b: compare → 2a = 4 so a = 2, then a² + b = −9 so b = −13.

“The most successful candidates were those who recognised the format of completing the square and could find the values of a and b with very little working. Some knew that the value of a was 2 but then did not know how to proceed to find b. Many gained a mark for expanding the right-hand side of the equation correctly but then did not know how to relate this to the left-hand side and compare coefficients. The incorrect expansion of the right-hand side to x² + a² was often seen.”

CIE 0580 · Paper 4 · Nov 2019

For proportion, match the wording exactly: 'y proportional to √(x+1)' means y = k√(x+1). Not y = k(x+1), not y = k/(x+1). Set up the relationship correctly before finding k.

“Candidates should ensure that they read the information in proportion questions very carefully, as the majority of errors come from setting up the incorrect relationship at the beginning of the working. It was often seen as a direct relationship, or without the square root.”

CIE 0580 · Paper 4 · Nov 2019

p² − q² = (p − q)(p + q), NOT (p + q)(p + q). The difference of two squares always gives ONE bracket with minus and ONE bracket with plus.

“A large proportion did not recognise the expression in part (a) as the difference of two squares. Candidates attempting to factorise often gave the answer (p + q)(p + q), whilst others resorted to combining terms incorrectly, resulting in expressions in pq or p²q² or similar.”

CIE 0580 · Paper 4 · Nov 2019

When raising a product to a power, EVERY factor gets the power: (81y¹²)^(3/4) = 81^(3/4) × y^(9) = 27y⁹. Don't leave 81 alone.

“There were many completely correct answers to the simplification in part (a) and a large proportion gained one mark, usually for an answer of 81y¹² where candidates were unaware that the power also applied to 81.”

CIE 0580 · Paper 4 · Nov 2019

When subtracting algebraic fractions, bracket the second numerator before subtracting. a/x − b/y = (ay − bx)/(xy). Remember the negative distributes over EVERY term in b.

“This was often followed with a sign error when dealing with the −2(x + 2) part of the numerator, leading to the incorrect answer of (x + 3)/((x + 2)(3x − 1)). Those who tried to deal with the subtraction as one fraction straight away often made this error and could not gain the mark for a correct numerator.”

CIE 0580 · Paper 4 · Nov 2019

When dividing or multiplying BOTH sides of an inequality by a NEGATIVE number, the inequality sign REVERSES. −4x < 20 → x > −5. Forgetting to flip the sign is the most common error.

“The difficulty arose for those who had −4x < 20. Many of these knew that if they divided both sides by a negative number (−4), they must also reverse the inequality sign, but many neglected to do this, giving an incorrect final answer.”

CIE 0580 · Paper 4 · Jun 2019

From 1 − p = 12: subtract 1 from both sides to get −p = 11, then multiply by −1 to get p = −11. Many students cannot handle the −p step and give a positive answer instead.

“Many candidates were able to score one mark for 1 – p = 12, but few were able to reach p = −11 as they were unable to deal with –p. Answers were usually 11, 12 or 13.”

CIE 0580 · Paper 3 · Nov 2018

To use a graph to solve an equation, rearrange it so that one side matches the graph's equation. The other side gives the line to draw. Aim for y = (something without the graph variable).

“Only the more able candidates could relate the given equation to the equation of the graph and give a correct equation for the line. Many found this question challenging and simply attempted to rearrange the equation given in the question, usually ending up with a quadratic or an equation containing 1/x.”

CIE 0580 · Paper 4 · Jun 2018

Factorising means rewriting as a product: 10 + 16w = 2(5 + 8w). It does NOT mean adding (26w), solving as an equation (w = −6), or dividing each term by the factor (5 + 8w).

“Misconceptions about combining terms meant that the answer 26w was seen quite regularly as well as the 'solution', w = −6. Some candidates showed ÷ 2 in their working and gave 5 + 8w as their answer.”

CIE 0580 · Paper 1 · Nov 2018

When expanding –(5y – 5), the minus sign applies to BOTH terms: –5y + 5, not –5y – 5. The sign error in the second term is the most common mistake in bracket expansion.

“The most common working was 12y – 18 – 5y + 5 and an answer of 7y – 13; this gained only one mark for either multiplying of the first bracket correctly or for the 7y seen in the answer.”

CIE 0580 · Paper 1 · Nov 2018

In vector geometry, direction matters. CD = CO + OD = −OC + OD. Ignoring the arrows and just adding OC + OD gives the wrong vector. When reversing direction, negate ALL components of the vector, not j

“Many candidates did not use brackets and some of those who used brackets only applied the minus sign to the first term in the brackets instead of to both terms. By far the most common incorrect method was to ignore the directional arrows and to work out OC + OD.”

CIE 0580 · Paper 4 · Nov 2018

When writing inequalities from a graph: (1) find the equation of each boundary line (gradient and intercept); (2) test a point inside the region to determine the correct inequality direction; (3) use

“Common errors in finding the equations of the lines were to have 4/3 x rather than 3/4 x, or 2x, −2x or 1/2 x instead of –1/2 x. It was quite common for x to be omitted after the gradient within the equations of the lines.”

CIE 0580 · Paper 2 · Nov 2018

Simplify means collect like terms and write the shortest equivalent expression. Don't factorise unless the question asks. 3m + 10k is simplified; (m + k)(...) is not.

“It was common, even among some of the more able candidates, to see answers that were not simplified far enough, many muddled this question with a factorising question. Consequently, common incorrect answers scoring 0 included (4 – 1)m + (7 + 3)k and (m + k)(4 – 1)(7 + 3).”

CIE 0580 · Paper 2 · Jun 2024

When generating terms from an nth term formula, the first term uses n = 1 (not n = 0). Starting at n = 0 gives an extra 'zeroth' term that doesn't belong.

“Very occasionally the terms were given as –3, –2, 1 where the candidate substituted n = 0 instead of n = 1 for the first term.”

CIE 0580 · Paper 2 · Jun 2024

For simultaneous equations, look at the coefficients. If they match (or are opposite) you can add or subtract directly to eliminate a variable — no need to multiply through first.

“This part was answered reasonably well, though a significant number did not appreciate that the two given equations could be directly added to find the value of 2x and subsequently x. Common errors included a variety of sign errors and/or arithmetic errors.”

CIE 0580 · Paper 3 · Jun 2024

From 16t = −4, divide both sides by 16 (the coefficient): t = −4/16 = −0.25. Don't accidentally flip the division — dividing 16 by −4 gives −4, not −0.25.

“Weaker candidates who reached 16t = –4 then followed this incorrectly by t = 16 ÷ –4 instead of t = –4 ÷ 16.”

CIE 0580 · Paper 2 · Jun 2023

When rearranging formulas, do ONE operation at a time and write each step. Trying to skip steps is where mistakes happen: u + v = 2s/t ⇒ v = 2s/t − u (subtract u, don't multiply).

“A significant number of candidates reached the stage u + v = 2s/t but then went wrong with the final step by writing v = 2su/t. Candidates would be well advised to carry out one step at a time. Some candidates attempted to carry out two steps at a time but did so incorrectly and lost the credit for both steps.”

CIE 0580 · Paper 4 · Jun 2023

If 3rd differences are constant, the sequence is cubic (n³ form). Try comparing to 1, 8, 27, 64... — a shift like (n − 1)³ − 1 is often simpler than expanding to n³ − 3n² + 3n − 2.

“In part (a) most candidates were able to find the third differences as 6... If they reached the third differences many were able to deduce it was a cubic sequence and then the most efficient candidates compared the sequence to the cube numbers 1, 8, 27, 64, 125 and reached (n – 1)³ – 1.”

CIE 0580 · Paper 2 · Jun 2023

After factorising, ALWAYS expand your answer to check it equals the original expression. This catches partial-factorising errors in seconds.

“Candidates are advised to check their answer by expanding to see that common incorrect answers such as 2g(4 – g²) or 2g(4g – g) would not give the original expression.”

CIE 0580 · Paper 2 · Jun 2023