IGCSE Maths Past Paper Solutions: 0580/21 May/June 2024 (Paper 2 Extended)
Full worked solutions to the Cambridge IGCSE Mathematics (0580) Paper 2 Extended exam paper from May/June 2024 (0580/21/M/J/24) — 22 questions, 70 marks, 1 hour 30 minutes. Attempt each question yourself first, then click “Show solution” to check your working step by step.
The diagram shows two sides of a parallelogram ABCD, plotted on a coordinate grid: A = (−7, 5), B = (−1, 1), and C = (3, 3).
Find the coordinates of point D.
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In parallelogram ABCD, the diagonals AC and BD bisect each other, so A + C = B + D, giving D = A + C − B.
D = (−7+3−(−1), 5+3−1) = (−3, 7)
Check: vector AB = (6,−4). Vector DC = C−D = (3−(−3), 3−7) = (6,−4) — matches AB, confirming ABCD is indeed a parallelogram.
Geetha has a box of toys. She picks a toy at random from the box. The probability that she picks a wooden toy is 0.6.
- (a) Work out the probability that she does not pick a wooden toy. [1]
- (b) The box contains three types of toys, wooden, plastic or metal.
| Type of toy | Wooden | Plastic | Metal |
|---|---|---|---|
| Number of toys | ? | 14 | 14 |
| Probability | 0.6 | ? | ? |
Complete the table. [2]
Show solution
- (a) Since picking a wooden toy and not picking one are complementary events: 1 − 0.6 = 0.4
- (b) Plastic and metal toys together = 14+14 = 28, and together they make up 1−0.6 = 0.4 of the total. So total toys = 28÷0.4 = 70. Number of wooden toys = 0.6 × 70 = 42. Probability of plastic = 14÷70 = 0.2. Probability of metal = 14÷70 = 0.2.
Wooden: 42 toys Plastic: probability 0.2 Metal: probability 0.2
The table shows some information about two sequences.
| nth term | 5th term | |
|---|---|---|
| Sequence A | 60 − 4n | ? |
| Sequence B | n² − 300 | ? |
- (a) Complete the table. [2]
- (b) Find the smallest positive number in sequence B. [2]
Show solution
- (a) Sequence A: substitute n=5 into 60−4n: 60−20 = 40. Sequence B: substitute n=5 into n²−300: 25−300 = −275
- (b) Sequence B becomes positive once n² > 300, i.e. n > √300 = 17.32. The smallest whole number satisfying this is n=18. Substituting: 18²−300 = 324−300 = 24
Find the greatest odd number that is a factor of 140 and a factor of 210.
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140 = 2² × 5 × 7, so its odd part (after removing all factors of 2) is 5×7 = 35. 210 = 2 × 3 × 5 × 7, so its odd part is 3×5×7 = 105. The greatest odd common factor is HCF(35, 105) = 35
Check: 140÷35 = 4 and 210÷35 = 6, both whole numbers.
Calculate.
- (a) ∛343 − √40.96 [1]
- (b) (192 + 4×16)1.25 [1]
Show solution
- (a) Cube root of 343 = 7 (since 7³=343). Square root of 40.96 = 6.4. 7−6.4 = 0.6
- (b) Inside the brackets: 192+4×16 = 192+64 = 256. Then 2561.25 = 2565/4 = (2561/4)&sup5; = 4⁵ = 1024
The diagram shows 5 kites that are congruent to kite ABCD. Each kite is joined to the next kite along one edge, all meeting at the shared vertex C. Angle DAB = 40° and DCE is a straight line.
Find the value of x.
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Kite ABCD is symmetric about diagonal AC, so angle ABC = angle ADC, and the quadrilateral’s four angles sum to 360°.
Step 1: Since DCE is a straight line (180°) and is filled exactly by the angle-at-C (angle BCD) of each of the 5 identical kites: 5 × angle BCD = 180°, so angle BCD = 36°
Step 2: Using the quadrilateral angle sum for one kite: angle DAB + angle BCD + 2×angle ABC = 360° (since angle ABC = angle ADC): 40 + 36 + 2×angle ABC = 360 → angle ABC = 142°
Step 3: At each point where two adjacent kites meet (other than C), the angles around that point sum to 360°: the two kite angles there (142° + 142°) plus the gap angle x° on the outside of the flower shape: 142+142+x = 360 → x = 76
The diagram shows a shape made from a triangle JKL and a semicircle with diameter JL. JKL is an isosceles right-angled triangle with JK = JL = 12.8 cm (the right angle is at J).
- (a) Calculate the area of this shape. [3]
- (b) Calculate the perimeter of this shape. [4]
Show solution
- (a) Area of triangle JKL = ½ × JK × JL = ½ × 12.8 × 12.8 = 81.92 cm². The semicircle has diameter JL=12.8 cm, so radius = 6.4 cm: area = ½πr² = ½ × π × 6.4² = 64.34 cm². Total area = 81.92+64.34 = 146 cm² (3 s.f.)
- (b) The perimeter is made up of side JK, side KL, and the curved arc of the semicircle (the diameter JL is internal, where the triangle and semicircle join, so it is not part of the outer perimeter). Since the triangle is right-angled at J with JK=JL=12.8: KL = √(12.8²+12.8²) = 12.8√2 = 18.10 cm. Arc length = πr = π × 6.4 = 20.11 cm. Perimeter = JK + KL + arc = 12.8 + 18.10 + 20.11 = 51.0 cm (3 s.f.)
These are the first five terms of a sequence: 11, 18, 25, 32, 39. Find an expression for the nth term of the sequence.
Show solution
The common difference is 7 (18−11=7, 25−18=7, etc.), so the nth term has the form 7n+c. Using the first term (n=1, value 11): 7(1)+c=11 → c=4.
nth term = 7n + 4
The value of a car is $8000. Each year the value of the car decreases exponentially by 25%. Calculate the value of this car after 3 years.
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Each year the value is multiplied by (1−0.25)=0.75. After 3 years: 8000 × 0.75³ = 8000 × 0.421875 = $3375
Amir invests $1500 in an account. The account pays compound interest at a rate of r% per year. At the end of 8 years the value of his investment is $1656.73. Find the value of r.
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1500(1+r/100)⁸ = 1656.73 → (1+r/100)⁸ = 1656.73÷1500 = 1.104487
1+r/100 = 1.1044871/8 = 1.0125 → r/100 = 0.0125 → r = 1.25
The diagram shows a shaded (grey) region and an unshaded region R on a grid. Region R is bounded below by the horizontal line y=1 (from x=1 to x=6), on the right by the vertical line x=6 (from y=1 to y=5), above by the horizontal line y=5 (from x=5 to x=6), and on its upper-left by a dashed diagonal line y=x (from (1,1) to (5,5)).
Find the inequalities that define the unshaded region, R.
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Each boundary of R gives one inequality. The solid boundaries (included) give ≥/≤, while the dashed diagonal (excluded) gives a strict inequality:
y ≥ 1 y ≤ 5 x ≤ 6 y < x
Solve the simultaneous equations. You must show all your working.
&frac32;x + 5y = 5
4x − 3y = 46
Show solution
Multiply the first equation by 2: 3x + 10y = 10 …(1). Keep the second: 4x−3y=46 …(2)
From (1): x = (10−10y)/3. Substitute into (2): 4(10−10y)/3 − 3y = 46. Multiplying through by 3: 40−40y−9y = 138 → 40−49y = 138 → y = −2
Substituting back: x = (10−10(−2))/3 = 30/3 = 10
x = 10 y = −2
The diagram shows a cyclic quadrilateral with angles p°, 4m°, 5m°, and (4m+38)°, where 5m° and (4m+38)° are the two angles at opposite vertices of the quadrilateral, arranged so that p° and (4m+38)° are opposite each other, and 4m° and 5m° are opposite each other.
Find the value of p.
Show solution
In a cyclic quadrilateral, opposite angles sum to 180°.
Using the pair 4m° and 5m°: 4m + 5m = 180 → 9m = 180 → m = 20
So (4m+38)° = 4(20)+38 = 118°. Using the pair p° and (4m+38)°: p + 118 = 180 → p = 62
The diagram shows a circle with radius 9 cm, with a minor sector of angle 48° unshaded and the remaining major sector shaded.
Calculate the area of the shaded major sector.
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The major sector’s angle = 360−48 = 312°. Area = (312/360) × π × 9² = (312/360) × π × 81 = 221 cm² (3 s.f.)
Write 0.1̇4̇6̇ (where the digits “46” repeat, i.e. 0.146464646…) as a fraction in its simplest form. You must show all your working.
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Let x = 0.146464646…
1000x = 146.464646…
10x = 1.464646…
Subtracting: 1000x − 10x = 146.464646… − 1.464646… → 990x = 145 → x = 145/990
Dividing numerator and denominator by 5: 29/198
- (a) In the Venn diagram, shade the region M′ ∩ N′. [1]
- (b) Find n(B∩(A′∪C)). [1]
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- (a) M′∩N′ means “not in M AND not in N“, i.e. everywhere outside both circles. Shade the region of the rectangle that lies outside both circles entirely.
- (b) Using the distributive law, B∩(A′∪C) = (B∩A′) ∪ (B∩C). From the diagram: B only = 3, A∩B only (not C) = 16, B∩C only (not A) = 4, A∩B∩C = 10.
B∩A′ (in B, not in A) = B only + B∩C only = 3+4 = 7.
B∩C = B∩C only + A∩B∩C = 4+10 = 14.
Combining without double-counting the shared “4”: total = 3 (B only) + 4 (B∩C only) + 10 (A∩B∩C) = 17
In triangle ABC, AB = 6.7 cm, BC = 5.9 cm, and angle ABC = 81°.
Calculate the area of triangle ABC.
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Area = ½ × AB × BC × sin(ABC) = ½ × 6.7 × 5.9 × sin81° = 19.765 × 0.98769 = 19.5 cm² (3 s.f.)
The diagram shows the graph of y = x³+4x²−2 for −3 ≤ x ≤ 1.5.
By drawing a suitable straight line, solve the equation x³+4x²−2 = 2x for −3 ≤ x ≤ 1.5.
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Rearranging, x³+4x²−2 = 2x is exactly the curve y=x³+4x²−2 meeting the straight line y=2x. Drawing y=2x on the same grid and reading off the intersection points (within −3≤x≤1.5) gives two solutions:
x = −0.5 x = 0.9 (readings from the graph; solving the cubic exactly gives x ≈ −0.52 and x ≈ 0.88 — a third root near x=−4.35 falls outside the given domain)
Factorise completely.
- (a) 12m² − 75t² [3]
- (b) xy + 15 + 3y + 5x [2]
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- (a) First take out the common factor of 3: 12m²−75t² = 3(4m²−25t²). Then 4m²−25t² is a difference of two squares: (2m)²−(5t)² = (2m−5t)(2m+5t).
3(2m − 5t)(2m + 5t) - (b) Group terms: xy+3y + 5x+15 = y(x+3) + 5(x+3) = (x + 3)(y + 5)
Solve the equation 8sinx+6 = 1 for 0° ≤ x ≤ 360°.
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8sinx = 1−6 = −5 → sinx = −0.625
The reference angle is sin⁻¹(0.625) = 38.68°. Since sine is negative, the solutions lie in the third and fourth quadrants: x = 180+38.68 = 218.68°, and x = 360−38.68 = 321.32°
x = 218.7° x = 321.3° (1 d.p.)
The diagram shows a cuboid. HD = 4 cm, EH = 6.5 cm and EF = 9.1 cm.
Calculate the angle between CE and the base CDHG.
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E sits directly above H (since EH is a vertical edge of the cuboid, length 6.5 cm), so the projection of E onto the base is H. The angle between CE and the base is therefore angle ECH in the right-angled triangle ECH (right angle at H).
CH is the diagonal of the base rectangle CDHG, with sides DH=4 cm and DC=EF=9.1 cm (since DC is parallel to, and the same length as, the top edge EF): CH = √(9.1²+4²) = √98.81 = 9.940 cm
tan(angle) = EH/CH = 6.5÷9.940 = 0.6540 → angle = tan⁻¹(0.6540) = 33.2° (1 d.p.)
Bag A and bag B each contain red counters and blue counters only. Stephan picks a counter at random from bag A and Jen picks a counter at random from bag B. The probability that Stephan picks a red counter is 0.4. The probability that Stephan and Jen both pick a red counter is 0.25.
Find the probability that Stephan and Jen both pick a blue counter.
Show solution
Since the two picks are independent: P(Stephan red) × P(Jen red) = P(both red) → 0.4 × P(Jen red) = 0.25 → P(Jen red) = 0.25÷0.4 = 0.625
P(Stephan blue) = 1−0.4 = 0.6. P(Jen blue) = 1−0.625 = 0.375
P(both blue) = 0.6 × 0.375 = 0.225