IGCSE Maths Past Paper Solutions: 0580/22 May/June 2024 (Paper 2 Extended)
Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 2 Extended exam paper from May/June 2024 (0580/22/M/J/24) — 24 questions, 70 marks, 1 hour 30 minutes. Attempt each question yourself first, then click “Show solution” to check your working step by step.
The temperature at midnight is −4°C. The temperature at noon is 25°C. Work out the difference between these two temperatures.
Show solution
25 − (−4) = 25+4 = 29°C
A gardener charges $6.55 for each hour he works plus a fixed charge of $15.50. Calculate the total amount he charges when he works for 4 hours.
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6.55 × 4 = 26.20. Adding the fixed charge: 26.20 + 15.50 = $41.70
A delivery driver records the number of pizzas she delivers each month for one year: 48, 44, 39, 28, 57, 22, 36, 41, 54, 57, 49, 52.
- (a) Complete the stem-and-leaf diagram. [2]
- (b) Find the median. [1]
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(a) Sorting each group of leaves in order:
| Stem | Leaves |
|---|---|
| 2 | 2 8 |
| 3 | 6 9 |
| 4 | 1 4 8 9 |
| 5 | 2 4 7 7 |
Key: 4|8 represents 48 pizzas
(b) Sorted list (12 values): 22, 28, 36, 39, 41, 44, 48, 49, 52, 54, 57, 57. Median = mean of 6th and 7th values: (44+48)÷2 = 46
Jonah has $750. He spends ¼ of this money on travel and some of this money on food. He now has $437.50. Work out the fraction of the $750 he spends on food.
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Travel spending = ¼ × 750 = $187.50. Remaining after travel = 750−187.50 = $562.50.
After also spending on food, he has $437.50 left, so food spending = 562.50−437.50 = $125.
As a fraction of the original $750: 125÷750 = 1/6
The table shows part of a tram timetable. All the trams take the same number of minutes to complete the journey from Newpoint to Westhill.
| Newpoint | Westhill |
|---|---|
| 10 30 | 11 17 |
| 12 18 | ? |
| 13 30 | 14 17 |
Complete the table.
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Comparing the first and third rows, the journey always takes 47 minutes (10:30 to 11:17, and 13:30 to 14:17 both confirm this).
For the 12:18 departure: 12:18 + 47 minutes = 13:05
Write 0.04628 correct to 2 significant figures.
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0.046
The diagram shows a Venn diagram with two overlapping circles, A and B, inside a rectangle.
On the Venn diagram, shade the region A∪B.
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Shade the entire area covered by both circles — circle A, circle B, and their overlap — leaving only the region outside both circles unshaded.
Kai invests $5000 in an account paying simple interest at a rate of r% per year. At the end of 8 years, the value of his investment is $5700. Find the value of r.
Show solution
Total interest earned = 5700−5000 = $700. Using I=Prt/100: 700 = 5000 × r × 8 ÷ 100 = 400r
r = 1.75
The grid shows triangle A (shaded, vertices at (3,1), (5,1), (3,2)) and triangle B (vertices at (5,3), (9,3), (5,5)).
- (a) Describe fully the single transformation that maps triangle A onto triangle B. [3]
- (b) On the grid, draw the image of triangle A after a translation by the vector (−4, 3). [2]
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- (a) Comparing corresponding sides: triangle A has legs of length 2 and 1, while triangle B has legs of length 4 and 2 — exactly double, so the scale factor is 2. To find the centre, note that corresponding vertex (3,1) on A maps to (5,3) on B. Using the enlargement rule (each point P maps to centre + 2×(P−centre)) and solving for the centre (h,k) using this and a second vertex pair (5,1)→(9,3) confirms a consistent centre.
Enlargement, scale factor 2, centre (1, −1) - (b) Translating each vertex of A — (3,1), (5,1), (3,2) — by (−4,3) gives the image vertices (−1,4), (1,4), (−1,5). Plot and join these three points to draw the translated triangle.
Write 174 000 in standard form.
Show solution
1.74 × 10⁵
A company surveys 40 of its employees. In the survey, 3 employees say they walk to work. The company has a total of 1240 employees. Find the expected number of employees in the company who walk to work.
Show solution
Sample proportion = 3/40. Applying to the whole company: 1240 × 3÷40 = 93
The diagram shows a right-angled triangle with hypotenuse 14 cm and one side 8.5 cm, with angle x° between them.
Calculate the value of x.
Show solution
Since 8.5 cm is adjacent to angle x and 14 cm is the hypotenuse: cos(x) = 8.5÷14 = 0.6071
x = cos⁻¹(0.6071) = 52.7° (1 d.p.)
Without using a calculator, work out 2¼ ÷ 1⅞. You must show all your working and give your answer as a mixed number in its simplest form.
Show solution
Convert to improper fractions: 2¼ = &frac94;, 1⅞ = &frac{15}{8}
Dividing is the same as multiplying by the reciprocal: &frac94; × &frac{8}{15} = &frac{72}{60} = &frac65;
As a mixed number: 1 1/5
A is the point (0, 2) and B is the point (8, 6).
Find the equation of line AB. Give your answer in the form y=mx+c.
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Gradient = (6−2)/(8−0) = 4/8 = 0.5. Since A=(0,2) lies on the y-axis, the y-intercept is c=2.
y = 0.5x + 2
Three towns, A, B and C, are equidistant from each other (forming an equilateral triangle). The bearing of C from A is 104°.
Calculate the bearing of B from C.
Show solution
Since ABC is equilateral, each internal angle is 60°. Working around the triangle consistently (matching the layout shown, with B below and to the south of both A and C), the bearing of B from A is 104°+60°=164°.
Setting up coordinates from these two bearings and computing the direction from C to B gives a bearing of 224°
Check: the back bearing of C from A (i.e. A from C) is 104+180=284°. Since angle ACB=60° (equilateral triangle), the bearing of B from C is 284−60=224° — consistent.
The speed–time graph shows information about a car journey: rising from 0 to 16 m/s over the first 30 seconds, constant at 16 m/s from 30s to 240s, then decreasing to 0 by 320s.
- (a) Find the deceleration of the car between 240 and 320 seconds. [1]
- (b) Calculate the total distance the car travels during the 320 seconds. [3]
Show solution
- (a) Change in speed = 0−16 = −16 m/s over 320−240=80 s. Deceleration = 16÷80 = 0.2 m/s²
- (b) Distance = area under the graph, in three sections:
Rising (0 to 30s): ½ × 30 × 16 = 240 m
Constant (30 to 240s): 210 × 16 = 3360 m
Falling (240 to 320s): ½ × 80 × 16 = 640 m
Total = 240+3360+640 = 4240 m
W = {students who walk to school}, G = {students who wear glasses}. There are 20 students in a class: 8 walk to school, 3 wear glasses and walk to school, 2 do not wear glasses and do not walk to school.
Complete the Venn diagram.
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W∩G (wear glasses and walk) = 3. W only (walk, no glasses) = 8−3 = 5. Outside both (neither) = 2.
Total accounted for so far: 5+3+2 = 10, so G only (glasses, don’t walk) = 20−10 = 10
W only: 5 W∩G: 3 G only: 10 Neither: 2
The graph of y=f(x) is drawn on the grid, a curve rising from near the origin to (4.5, 18).
- (a) Draw the tangent to the graph at the point x=3. [1]
- (b) Use your tangent to find an estimate for the gradient of the curve at the point x=3. [2]
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- (a) Draw a straight line that just touches the curve at x=3 without crossing through it there, matching the curve’s direction at that exact point.
- (b) Reading the tangent’s slope (rise÷run) using two convenient points on your drawn tangent line gives an estimate of gradient ≈ 5.3 (accept any value in the range 5 to 6, since this is a graphical estimate)
- (a) y is directly proportional to (x−1)². When x=4, y=3. Find y when x=7. [3]
- (b) m is inversely proportional to the square root of p. Explain what happens to the value of m when the value of p is multiplied by 9. [1]
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- (a) y = k(x−1)². Substituting x=4, y=3: 3 = k(3)² = 9k → k=⅓. When x=7: y = ⅓(7−1)² = ⅓(36) = y = 12
- (b) m = k/√p. If p is multiplied by 9, √p is multiplied by √9=3, so m is divided by 3
Two parcels are mathematically similar. The larger parcel has volume 80 cm³ and height 5.2 cm. The smaller parcel has volume 33.75 cm³. Calculate the height of the smaller parcel.
Show solution
Volume ratio (small:large) = 33.75÷80 = 0.421875. The ratio of lengths is the cube root of the volume ratio: ∛0.421875 = 0.75 (since 0.75³=0.421875)
Height of smaller parcel = 5.2 × 0.75 = 3.9 cm
Solve the simultaneous equations. You must show all your working.
4y + 3x = 13
y = x² − 18
Show solution
Substituting y=x²−18 into the first equation: 4(x²−18)+3x = 13 → 4x²−72+3x = 13 → 4x²+3x−85 = 0
Using the quadratic formula: discriminant = 3²−4(4)(−85) = 9+1360 = 1369 = 37². x = (−3±37)/8
x = 34/8 = 4.25, or x = −40/8 = −5
For x=4.25: y = 4.25²−18 = 18.0625−18 = 0.0625. For x=−5: y = 25−18 = 7
x = 4.25, y = 0.0625 x = −5, y = 7
- (a) For each sketch, put a ring around the correct type of function shown.
- A curve that dips down then rises steeply, passing through the origin with an S-shape. [1]
- A curve with two separate branches, one in the upper-left and one in the lower-right, approaching but never touching the axes. [1]
- (b)
- On the grid, sketch the curve y=sinx for 0° ≤ x ≤ 360°. [2]
- Solve the equation sinx+0.4=0 for 0° ≤ x ≤ 360°. [3]
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- (a)(i) An S-shaped curve through the origin, dipping then rising, is characteristic of a cubic function
- (a)(ii) Two separate branches approaching the axes without touching is characteristic of a reciprocal function
- (b)(i) The sine curve starts at (0°,0), rises to a maximum of 1 at 90°, returns to 0 at 180°, falls to a minimum of −1 at 270°, and returns to 0 at 360° — a smooth wave shape.
- (b)(ii) sinx = −0.4. The reference angle is sin⁻¹(0.4) = 23.58°. Since sine is negative, the solutions lie in the third and fourth quadrants: x = 180+23.58 = 203.58°, and x = 360−23.58 = 336.42°
x = 203.6° x = 336.4° (1 d.p.)
The Venn diagram shows information about the number of students in a class studying English (E), French (F), and Spanish (S): E only = 8, E∩F only = 1, F only = 6, E∩S only = 2, E∩F∩S = 1, F∩S only = 3, S only = 4, outside all three = 5.
- (a) Find n((E∪F)′∪S). [1]
- (b) One student is picked at random from those who study Spanish. Find the probability that this student studies exactly two languages. [2]
Show solution
- (a) (E∪F)′ = outside both E and F = S only (4) + outside all (5) = regions {4,5}. S (all Spanish students) = S only (4) + E∩S only (2) + F∩S only (3) + E∩F∩S (1) = regions {4,2,3,1}. Taking the union (avoiding double-counting the shared region “4”): 5+4+2+3+1 = 15
- (b) Total studying Spanish = 4+2+3+1 = 10. Of these, studying exactly two languages: E∩S only (2) + F∩S only (3) = 5 (excluding S only, which is just one language, and E∩F∩S, which is three languages).
Probability = 5÷10 = 1/2
O is the origin and OPQR is a parallelogram. M is the midpoint of PQ and N divides QR in the ratio 2:1. OP→ = a and OR→ = b.
- (a) Find MN→. Give your answer in terms of a and/or b and in its simplest form. [2]
- (b) The lines MN and OR are extended to meet at S. Find the position vector of S. Give your answer in terms of a and/or b and in its simplest form. [3]
Show solution
Since OPQR is a parallelogram: P=a, Q=a+b (since PQ is parallel and equal to OR=b), R=b.
- (a) M = midpoint of PQ = P + ½(Q−P) = a+½b. N divides QR in ratio 2:1 (i.e. QN:NR=2:1): N = Q + ⅔(R−Q) = (a+b) + ⅔(−a) = ⅓a+b.
MN→ = N−M = (⅓a+b) − (a+½b) = −⅔a + ½b - (b) Points on line OR extended have the form tb (no a component). Parametrising line MN as M+s(MN→) and setting the a-coefficient to zero: 1 + s(−⅔) = 0 → s=&frac32;. Substituting into the b-coefficient: ½ + &frac32;×½ = ½+¾ = &frac54;.
Position vector of S = 5/4 b