IGCSE Maths Past Paper Solutions: 0580/12 February/March 2024 (Paper 1 Core)
Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 1 Core exam paper from February/March 2024 (0580/12/F/M/24) — 25 questions, 56 marks, 1 hour. Attempt each question yourself first, then click “Show solution” to check your working step by step.
Write the number thirty thousand and fifty in figures.
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“Thirty thousand” = 30 000, plus “fifty” = 50: 30 050
Write 5926 correct to the nearest 10.
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The units digit is 6, which rounds the tens digit up: 5926 → 5930
Mark the midpoint of the line ST.
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Method: Measure the total length of line ST with a ruler, then mark the point exactly halfway along the line (half of the total length from either end). Alternatively, use a ruler and compasses to construct the perpendicular bisector of ST — the point where it crosses ST is the midpoint.
The diagram shows a 6×6 grid of squares (36 squares in total).
- (a) Shade &frac29; of this shape.
- (b) Write &frac29; as a percentage.
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- (a) The grid has 36 squares in total. &frac29; of 36 = 8, so shade any 8 squares of the grid.
- (b) &frac29; × 100 = 200÷9 = 22.2222…% → 22.2% (3 s.f., exact value 22&frac29;%)
A night bus runs from 21 50 to 05 18 the next day. Work out the number of hours and minutes that the night bus runs.
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From 21:50 to midnight (00:00) = 2 hours 10 minutes.
From midnight to 05:18 = 5 hours 18 minutes.
Total = 2h10min + 5h18min = 7 h 28 min
- (a) From the list 34, 55, 76, 83, 111, 121, write down all the multiples of 11.
- (b) Zaid has a non-calculator method for checking multiples of 11: subtract and add alternately the digits, and check if the result is a multiple of 11. For 919281: 9−1+9−2+8−1 = 22 = 2×11, so 919281 is a multiple of 11. Show that the number 918271937 is a multiple of 11 by using Zaid’s method.
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- (a) Checking each: 34 no, 55=11×5 yes, 76 no, 83 no, 111 no (111÷11=10.09), 121=11×11 yes.
55, 121 - (b) Digits of 918271937: 9,1,8,2,7,1,9,3,7. Subtract and add alternately, starting with +:
9−1+8−2+7−1+9−3+7 = 8+8=16, 16−2=14, 14+7=21, 21−1=20, 20+9=29, 29−3=26, 26+7=33
33 = 3 × 11, which is a multiple of 11, so 918271937 is a multiple of 11
The range of eight numbers is 31. These are seven of the numbers: 28, 36, 42, 24, 38, 16, 21. Find the two possible values of the eighth number.
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Among the seven given numbers, the minimum is 16 and the maximum is 42.
If the eighth number becomes the new minimum: 42 − new min = 31, so new min = 11.
If the eighth number becomes the new maximum: new max − 16 = 31, so new max = 47.
11 or 47
Calculate √5.76 + 2.8³.
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√5.76 = 2.4 (since 2.4²=5.76). 2.8³ = 2.8×2.8×2.8 = 21.952
2.4 + 21.952 = 24.352
Simplify 4m + 7k − m + 3k.
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Collect m-terms: 4m−m=3m. Collect k-terms: 7k+3k=10k.
3m + 10k
From the list −9, −7, −3, −1, 0, 2, 5, 6, 8, find:
- (a) the highest number possible from the product of two of the numbers
- (b) the lowest number possible from the product of three of the numbers.
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- (a) Two large positives: 6×8=48. Two large negatives (negative×negative=positive): (−9)×(−7)=63. Since 63>48: 63
- (b) To get the most negative product from three numbers, multiply one large negative by the two largest positives: (−9)×8×6 = −432
Sarah records the number of people who play golf on each of 14 days: 28, 46, 54, 71, 70, 65, 49, 50, 64, 77, 68, 72, 45, 58.
- (a) Complete the stem-and-leaf diagram.
- (b) Find the median.
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- (a) Sorting each value into its stem (tens digit) and leaf (units digit):
Stem Leaf 2 8 3 4 5 6 9 5 0 4 8 6 4 5 8 7 0 1 2 7 Key: 2|8 represents 28
- (b) With 14 values, the median is the average of the 7th and 8th values when sorted: 28, 45, 46, 49, 50, 54, 58, 64, 65, 68, 70, 71, 72, 77.
Median = (58+64)÷2 = 61
The diagram shows the net of a cuboid, with measurements 10 cm, 4 cm and 5 cm marked.
- (a) Work out the surface area of this cuboid.
- (b) Work out the volume of this cuboid.
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The net folds into a cuboid with edges 10 cm, 4 cm and 5 cm.
- (a) Surface area = 2(lw+lh+wh) = 2(10×4 + 10×5 + 4×5) = 2(40+50+20) = 2(110) = 220 cm²
- (b) Volume = 10 × 4 × 5 = 200 cm³
There are 20 cars in a car park and 3 of the cars are blue.
- (a) James wants to draw a pie chart to show this information. Find the angle of the sector for the blue cars in this pie chart.
- (b) One of the 20 cars is picked at random. Find the probability that this car is not blue.
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- (a) A full pie chart is 360°, shared in proportion to the 20 cars: (3÷20) × 360° = 54°
- (b) Cars that are not blue = 20−3 = 17. P(not blue) = 17÷20 = 0.85 (17/20)
Factorise: 3x³ − 7xy
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The highest common factor of 3x³ and 7xy is x:
x(3x² − 7y)
The grid shows point A at (7, 1) and point B at (−3, 4).
Write AB as a column vector.
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AB = B − A = (−3−7, 4−1) = (−10, 3)
AB = (−10, 3) (as a column vector)
Exchange rates: 1 euro = 1.05 dollars, 1 rupee = 0.013 dollars. Vani changes x euros into dollars. She then changes the dollars into 17 850 rupees. Calculate the value of x.
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First find how many dollars 17 850 rupees is worth: 17 850 × 0.013 = $232.05
This $232.05 came from converting x euros: x × 1.05 = 232.05
x = 232.05 ÷ 1.05 = 221
The line y = 2x − 5 intersects the line y = 3 at the point P. Find the coordinates of the point P.
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Substitute y=3 into the first equation: 3 = 2x−5
2x = 8, so x = 4
P = (4, 3)
The diagram shows two shapes, A and B, on a grid.
- (a) Describe fully the single transformation that maps shape A onto shape B.
- (b) On the grid, draw the image of shape A after a reflection in the line x = −1.
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- (a) Compare the size and orientation of shape A (in the upper-left region, around x: −7 to −3, y: 1 to 3) with shape B (in the lower region, around x: −2 to −1, y: −7 to −3). Since shape B is the same size as shape A but with its “wide” and “tall” dimensions swapped, this points to a 90° rotation. To find the exact centre, join at least two pairs of corresponding vertices between A and B, construct the perpendicular bisector of each pair, and the point where the bisectors cross is the centre of rotation.
- (b) To reflect any point (x, y) of shape A in the line x=−1, use the rule: new x-coordinate = −2 − x (i.e. the point moves to the same distance on the opposite side of the line x=−1), while the y-coordinate stays the same. Apply this to every vertex of shape A and join up the new points to draw the reflected image.
The diagram shows a small circle with radius 7 cm and a large circle with radius R cm. The area of 16 small circles is the same as the area of one large circle.
Calculate the value of R.
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16 × (area of small circle) = area of large circle
16 × π(7)² = πR²
Divide both sides by π: 16 × 49 = R²
R² = 784, so R = √784 = 28 cm
- (a) The nth term of a sequence is n² − 3. Find the first three terms of this sequence.
- (b) These are the first five terms of a different sequence: 2, 9, 16, 23, 30. Find the nth term of this sequence.
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- (a) n=1: 1−3=−2. n=2: 4−3=1. n=3: 9−3=6.
−2, 1, 6 - (b) Common difference = 7 (9−2=7, 16−9=7, etc). The nth term has the form 7n+k. Using the first term: 7(1)+k=2 → k=−5.
nth term = 7n − 5
The length, l m, of a rope is 18.7 m, correct to the nearest 10 centimetres. Complete this statement about the value of l.
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“Correct to the nearest 10 cm” means the nearest 0.1 m, so the true value is within half of 0.1 m (i.e. 0.05 m) of 18.7:
18.7 − 0.05 = 18.65, and 18.7 + 0.05 = 18.75
18.65 ≤ l < 18.75
6.5 × 10¹⁹ × n = 5.46 × 10²³. Calculate the value of n. Give your answer in standard form.
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n = (5.46 × 10²³) ÷ (6.5 × 10¹⁹) = (5.46÷6.5) × 10(23−19) = 0.84 × 10⁴
Rewrite in standard form (0.84 < 1, so shift the decimal point): 0.84 × 10⁴ = 8.4 × 10³
n = 8.4 × 10³
Triangle ABC is mathematically similar to triangle DEF, with AC=118.9 cm corresponding to DF=159.9 cm, and DE=97.5 cm corresponding to h=AB.
Calculate the value of h.
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Scale factor from the smaller to the larger triangle = DF÷AC = 159.9÷118.9 = 1.34483
Since h (in the smaller triangle) corresponds to DE (in the larger triangle), divide by the scale factor: h = 97.5 ÷ 1.34483
h = 97.5 × (118.9÷159.9) = 72.5 cm
Without using a calculator, work out 1¼ − ⅚. You must show all your working and give your answer as a fraction in its simplest form.
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Convert to an improper fraction: 1¼ = &frac54;
Find a common denominator for &frac54; and ⅚ (LCM of 4 and 6 is 12): &frac54; = &frac{15}{12}, ⅚ = &frac{10}{12}
&frac{15}{12} − &frac{10}{12} = 5/12
The highest common factor (HCF) of two numbers is 6. The lowest common multiple (LCM) of the two numbers is 90. Both numbers are greater than 6. Work out the two numbers.
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Write the two numbers as 6a and 6b, where a and b share no common factor. Since HCF×LCM = product of the two numbers: 6 × 90 = 6a × 6b → ab = (6×90)÷36 = 15.
Coprime pairs multiplying to 15: (1,15) or (3,5). Using (1,15) gives numbers 6 and 90 — but 6 is not greater than 6, so this is rejected.
Using (3,5): numbers = 6×3=18 and 6×5=30. Check: HCF(18,30)=6 ✓, LCM(18,30)=90 ✓, both >6 ✓
18 and 30