Cambridge IGCSE Mathematics 0580 — 2017 May/June Paper 2 · Variant 3
0580/23/M/J/17 · 24 questions · 70 marks · ≈79 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper12 pages
Mark scheme5 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · Calculate (1 - cos 48 °)
11 Calculate (1 - cos 48 °) . 2 ................................................... [1]
Mark scheme: Question Answer Marks Part Marks 1 0.407 or 0.4067... 1
Question 2
2 Factorise completely. 4x2 - 8xy ................................................... [2]
Mark scheme: 2 4 x ( x − 2 y ) final answer 2 2 M1 for 4 x − 2 xy or x ( 4 x − 8 y ) ( ) or 2 2 x 2 − 4 xy or 2x (2x – 4y) ( )
Q3 · Find the lowest common multiple (LCM) of 20 and 24
3 Find the lowest common multiple (LCM) of 20 and 24. ................................................... [2]
Mark scheme: 3 120 2 M1 for finding a correct product of prime factors or correctly listing a minimum of 3 multiples of 20 and 24 or for answer 23 × 3 × 5 oe or 120k where k is an integer > 1
Q4 · Make a the subject of the formula
4 Make a the subject of the formula. x = y + a a = ............................................ [2]
Mark scheme: 4 ( x − y ) 2 oe final answer 2 M1 for x − y = a or their (x − y) squared
Q5 · Calculate the volume of a hemisphere with radius 3.2 cm
5 Calculate the volume of a hemisphere with radius 3.2 cm. 4 [The volume, V, of a sphere with radius r is V = rr3 .] 3 ............................................cm3 [2]
Mark scheme: 5 68.6 or 68.62 to 68.64 2 1 4 3 M1 for × π × 3.2 2 3 If zero scored, SC1 for final answer 137 or 137.2 to 137.3
Q6 · The probability that Pedro scores a goal in any match is
26 The probability that Pedro scores a goal in any match is . 5 Calculate the probability that Pedro scores a goal in each of the next two matches. ................................................... [2]
Mark scheme: 6 4 2 2 2 oe M1 for × oe or denominator 52 oe 25 5 5
Q7 · Y is inversely proportional to x2
7 y is inversely proportional to x2. When x = 2, y = 8. Find y in terms of x. y = ............................................ [2]
Mark scheme: 7 32 2 k or 32x−2 final answer M1 for y = oe x 2 x 2 or [k =] 32
Question 8
8 Simplify. 12 c a8 31 m ................................................... [2]
Mark scheme: 8 2 2 2 k or 2a−4 final answer B1 for oe or oe (k ≠ 0) final answer a 4 a k a 4
Q9 · 9 (a) GH = c- 4 m Find (i) 5 GH , [1] f p (ii) HG
6 9 (a) GH = c- 4 m Find (i) 5 GH , [1] f p (ii) HG. [1] f p 6 2 8 (b) + = c 7 m c y m c 3 m Find the value of y. y = ............................................ [1]
Mark scheme: 9(a)(i) 30 1 −20 9(a)(ii) −6 1 4 9(b) −4 1
Q10 · The three angles in a triangle are 5x°, 6x° and 7x°
10 The three angles in a triangle are 5x°, 6x° and 7x°. 7x° NOT TO SCALE 6x° 5x° (a) Find the value of x. x = ............................................ [2] (b) Work out the size of the largest angle in the triangle. ................................................... [1]
Mark scheme: 10(a) 10 2 180 M1 for 5 x + 6 x + 7 x = 180 oe or 5 + 6 + 7 or B1 for angles 50, 60 and 70 10(b) 70 1FT FT 7 × their (a) provided 0 < their answer < 180
Q11 · Y 14 12 10 8 6 4 2 x 0 –2 2 4 6 8 10 12 14 16 18 20 22 –2 By shading the unwanted regions…
11 y 14 12 10 8 6 4 2 x 0 –2 2 4 6 8 10 12 14 16 18 20 22 –2 By shading the unwanted regions of the grid above, find and label the region R that satisfies the following four inequalities. x H 0 x + y H 7 y H x x + 2y G 20 [3]
Mark scheme: 11 Correct region 3 1 2 2 3 1 1 2 2 1 1 1 0 1 2 SC1 for R not marked and reverse shading
Q12 · F(x) = 3 + 4x g(x) = 6x + 7 Find, in its simplest form, (a) f(3x)…
12 f(x) = 3 + 4x g(x) = 6x + 7 Find, in its simplest form, (a) f(3x), ................................................... [1] (b) fg(x). ................................................... [2]
Mark scheme: 12(a) 3 + 12x final answer 1 12(b) 24 x + 31 final answer 2 M1 for 3 + 4(6 x + 7)
Q13 · Two bottles and their labels are mathematically similar
13 Two bottles and their labels are mathematically similar. The smaller bottle contains 0.512 litres of water and has a label with area 96 cm2. The larger bottle contains 1 litre of water. Calculate the area of the larger label. ............................................cm2 [3]
Mark scheme: 13 150 3 2 2 1 3 0.512 3 M2 for oe or oe 0.512 1 1 1 1 3 0.512 3 or M1 for scale factor oe or oe 0.512 [1]
Q14 · Write the recurring decimal .063o o as a fraction in its lowest terms
14 Write the recurring decimal .063o o as a fraction in its lowest terms. You must show all your working. ................................................... [3]
Mark scheme: 14 10k+2 × [0].63& & − 10k × [0].63& & oe M1 where k > 1 63 A1 6300 7 or equivalent fraction e.g. but not 99 9900 11 7 B1 11
Q15 · 39 cm NOT TO SCALE x° 24 cm 71.8° Find the value of x
15 39 cm NOT TO SCALE x° 24 cm 71.8° Find the value of x. x = ............................................ [3]
Mark scheme: 15 35.8 or 35.77…. 3 24 × sin71.8 M2 for [sin =] 39 39 24 or M1 for = oe sin71.8 sin x
Question 16
16 (a) Solve the inequality. x + 13 H 3x + 7 ................................................... [2] (b) List the positive integers that satisfy the inequality in part (a). ................................................... [1]
Mark scheme: 16(a) x ⩽ 3 final answer 2 M1 for 13 − 7 ⩾ 3x − x oe 16(b) 1, 2, 3 1FT correct answer or FT their answer to (a)
Q17 · P NOT TO p SCALE Z O Q q O is the origin, OP = p and OQ = q
17 P NOT TO p SCALE Z O Q q O is the origin, OP = p and OQ = q . Z is a point on PQ such that PZ : ZQ = 5 : 2. Work out, in terms of p and q, the position vector of Z. Give your answer in its simplest form. ................................................... [3]
Mark scheme: 17 2 5 3 5 2 p + q M1 for PZ = (q − p) oe or QZ = (p − q) oe 7 7 7 7 M1 for correct route from O to Z or identifying OZ
Q18 · The diagram shows a speed-time graph for the journey of a car
18 The diagram shows a speed-time graph for the journey of a car. 12.5 NOT TO SCALE Speed (m/s) 0 0 20 220 280 Time (s) Calculate the total distance travelled. ...............................................m [3]
Mark scheme: 18 3000 3 1 M2 for 12.5 × ( 200 + 280 ) oe 2 or M1 for part area
Q19 · 3 219 Without using your calculator, work out - - 12 c 4 3 m
11 3 219 Without using your calculator, work out - - 12 c 4 3 m. You must show all your working and give your answer as a fraction in its simplest form. .............................................. [4]
Mark scheme: 19 common denominator 12 B1 accept k × 12 throughout 9 8 M1 9 k 8 k one correct from or oe accept or 12 12 12 k 12 k A2 10 10 k 5 A1 for or cao 12 12 k 6
Question 20
20 Simplify. (a) 6w0 .............................................. [1] (b) 5x3 − 3x3 .............................................. [1] (c) 3y6 # 5y-2 .............................................. [2]
Mark scheme: 20(a) 6 1 20(b) 2x 3 final answer 1 20(c) 15y 4 final answer 2 B1 for 15 ky or ky 4 as final answer (k ≠ 0)
Q21 · Solve the equation 5x 2 + 10x + 2 = 0
21 Solve the equation 5x 2 + 10x + 2 = 0 . You must show all your working and give your answers correct to 2 decimal places. x = ........................... or x = ........................... [4]
Mark scheme: 21 2 B1 If completing the square: 10 − 4 × 5 × 2 oe or better B1 for ( x + 1) 2 oe 2 2 B1 for −+1 1 − or −−1 1 − oe 5 5 −10 + q −10 − q B1 or oe 2(5) 2(5) − 0.23, −1.77 final ans cao B1B1 SC1 for − 0.2 or − 0.225... and −1.8 or −1.774... or −1.775 or 0.23 and 1.77 as answer or − 0.23 and −1.77 seen in working Maximum score without working is 2
Q22 · The diagram shows a cube ABCDEFGH of side length 26 cm
22 The diagram shows a cube ABCDEFGH of side length 26 cm. H G E NOT TO F SCALE D C A B Calculate the angle between AG and the base of the cube. ................................................... [4]
Mark scheme: 22 35.3 or 35.26… 4 26 M3 for [tan =] oe 26 2 + 26 2 or M1 for [ AC 2 = ] 26 2 + 26 2 oe and M1 for [tan =] 26 ÷ their AC oe or for angle CAG indicated
Question 23
23 (a) Simplify. 4 (x - 6) 2 ( x - 6) ................................................... [1] (b) Expand the brackets and simplify. (x + 4) 2 + 5 (3x + 2) ................................................... [3]
Mark scheme: 23(a) 4 ( x − 6 ) or 4x − 24 as final answer 1 23(b) x 2 + 23 x + 26 final answer 3 B2 for x 2 + 4 x + 4 x + 16 or better or B1 for 15 x + 10
Q24 · Marcel invests $2500 for 3 years at a rate of 1.6% per year simple interest
24 Marcel invests $2500 for 3 years at a rate of 1.6% per year simple interest. Jacques invests $2000 for 3 years at a rate of x% per year compound interest. At the end of the 3 years Marcel and Jacques receive the same amount of interest. Calculate the value of x correct to 3 significant figures. x = ............................................ [5]
Mark scheme: 24 1.96 cao 5 2500 × 1.6 × 3 + 2000 M4 for ( ( 3 100 − 1 ) − 1) [× 2000 100] oe or 1.96…or [0].0196... or 101.96… or 1.0196… or 2500 × 1.6 × 3 + 2000 3 100 M3 for 2000 or B2 for [SI =] 120 or [CI total=] 2120 2500 × 3 × 1.6 or M1 for 100 and 3 M1 for 2000 × ( k )
What was in this paper
The subtopics covered by these 24 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
3Equations2Fractions, decimals and percentages2Ratio and proportion2Vectors in two dimensions2Angles1Exact trigonometric values1Functions1Indices II1Inequalities1Non-right-angled triangles1Percentages1Powers and roots1Probability of combined events1Rates1Right-angled triangles1Surface area and volume1Types of number1What you needed in this session
Cambridge’s own grade thresholds for 2017 May/June, Paper 2 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.