Cambridge IGCSE Mathematics 0580 — 2020 Oct/Nov Paper 2 · Variant 3

0580/23/O/N/20 · 25 questions · 70 marks · ≈79 min

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Cambridge IGCSE Mathematics 0580 2020 Oct/Nov Paper 2 · Variant 3 question paper, page 1 of 12
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Mark scheme7 pages

Answers below. Sit the paper first if you are practising.

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Questions as text

Q1 · Write down the cube number that is greater than 50 but less than 100

1 Write down the cube number that is greater than 50 but less than 100. ................................................. [1]

Mark scheme: Question Answer Marks Partial Marks 1 64 1

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Question 2

2 Calculate. 4 .00025 ................................................. [1]

Mark scheme: 2 80 1

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Q3 · In triangle ABC, BC = 7.6 cm and AC = 6.2 cm

3 In triangle ABC, BC = 7.6 cm and AC = 6.2 cm. Using a ruler and compasses only, construct triangle ABC. Leave in your construction arcs. The side AB has been drawn for you. A B [2]

Mark scheme: 3 Accurate triangle with correct 2 B1 for accurate triangle with construction arcs no/incorrect arcs or SC1 for accurate triangle with arcs with sides interchanged

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Question 4

4 Simplify. a 2 ' a 6 ................................................. [1]

Mark scheme: 4 1 1 a–4 or final answer a 4

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Q5 · Thor changes 40 000 Icelandic Krona into dollars when the exchange rate is 1 krona =…

5 Thor changes 40 000 Icelandic Krona into dollars when the exchange rate is 1 krona = $0.0099 . Work out how many dollars he receives. $ ................................................ [1]

Mark scheme: 5 396 1

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Q6 · A NOT TO 58° SCALE 82° B C The diagram shows triangle ABC

6 A NOT TO 58° SCALE 82° B C The diagram shows triangle ABC. The triangle is reflected in the line BC to give a quadrilateral ABDC. (a) Write down the mathematical name of the quadrilateral ABDC. ................................................. [1] (b) Find angle ACD. Angle ACD = ................................................ [2]

Mark scheme: 6(a) Kite 1 6(b) 80 2 M1 for (180 – 82 – 58) or better

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Q7 · Change 457 000 cm2 into m2

7 Change 457 000 cm2 into m2. .............................................m2 [1]

Mark scheme: 7 45.7 1

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Q8 · The length, l cm, of a line is 18.3 cm, correct to the nearest millimetre

8 The length, l cm, of a line is 18.3 cm, correct to the nearest millimetre. Complete this statement about the value of l. .................... G l 1 .................. [2]

Mark scheme: 8 18.25, 18.35 2 B1 for each or SC1 for both values correct but reversed

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Q9 · 19 Without using a calculator, work out 1 # 2

1 19 Without using a calculator, work out 1 # 2 . 7 10 You must show all your working and give your answer as a mixed number in its simplest form. ................................................. [3]

Mark scheme: 9 8 21 M1 and oe improper fractions 7 10 168 A1 oe improper fractions 70 2 A1 Dep. on first A1 2 cao final answer 5 8 21 If M0 scored SC1 for or oe 7 10 improper fractions

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Q10 · Solve the simultaneous equations

10 Solve the simultaneous equations. You must show all your working. 3x - 8y = 22 x + 4y = 4 x = ................................................ y = ................................................ [3]

Mark scheme: 10 Correctly eliminates one variable M1 [x =] 6 A2 A1 for either correct [y =] –0.5 oe If M0 scored, SC1 for 2 values satisfying one of the original equations

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Q11 · A bag contains 7 red discs, 5 green discs and 2 pink discs

11 A bag contains 7 red discs, 5 green discs and 2 pink discs. (a) Helen takes one disc at random, records the colour and replaces it in the bag. She does this 140 times. Find how many times she expects to take a green disc. ................................................. [2] (b) Helen adds 9 green discs and some pink discs to the discs already in the bag. The probability of taking a green disc is now 2. 7 Find the number of pink discs that Helen added to the bag. ................................................. [2]

Mark scheme: 11(a) 50 2 5 M1 for [× 140] 7 + 5 + 2 140 or [× 5] 7 + 5 + 2 11(b) 26 2 5 + 9 2 M1 for = oe n 7 5 + 9 2 or = oe p + 7 + 5 + 2 + 9 7

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Q12 · A straight line, l, has equation y = 5 x + 12

12 A straight line, l, has equation y = 5 x + 12 . (a) Write down the gradient of line l. ................................................. [1] (b) Find the coordinates of the point where line l crosses the x-axis. ( ....................... , ......................) [2] (c) A line perpendicular to line l has gradient k. Find the value of k. k = ................................................ [1]

Mark scheme: 12(a) 5 1 12(b) 12 2 M1 for 5x + 12 = 0 ( − oe, 0) 5 12(c) 1 1 1 − oe FT 5 −their ( a )

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Q13 · A B Use set notation to describe the shaded region

13 A B Use set notation to describe the shaded region. ................................................. [1]

Mark scheme: 13 A ' B 1

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Q14 · N = 2 4 # 3 # 7 5 PN = K , where P is an integer and K is a square number

14 N = 2 4 # 3 # 7 5 PN = K , where P is an integer and K is a square number. Find the smallest value of P. P = ................................................ [2]

Mark scheme: 14 21 2 B1 for 3 × 7 soi or 24 × 32 × 76 oe or answer of 21 × k2

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Q15 · X15 m = 2p + y Make x the subject of this formula

x15 m = 2p + y Make x the subject of this formula. x = ................................................ [3]

Mark scheme: 15 [x =] y(m – 2p)2 nfww 3 M1 for subtract 2p or their term in p to isolate a term in x or [x =] y(m2 – 4mp + 4p2) final answer M1 for squaring M1 for multiplying by their term in y Maximum of 2 marks for an incorrect answer

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Q16 · A paperweight has height 4 cm and volume 38.4 cm3

16 A paperweight has height 4 cm and volume 38.4 cm3. A mathematically similar paperweight has height 7 cm. Calculate the volume of this paperweight. .......................................... cm3 [3]

Mark scheme: 16 205.8 3 3  7  M2 for 38.4 ×   oe  4   7  3  4  3 or M1 for   or   oe or  4   7  7 v = 3 oe 4 38.4

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Q17 · Adil and Brian are paid the same wage

17 Adil and Brian are paid the same wage. Adil is given a 7% pay decrease and his new wage is $427.80 . Brian is given a 7% pay increase. Work out Brian’s new wage. $ ................................................ [3]

Mark scheme: 17 492.2[0] 3 B2 for 32.2[0] OR  7  M1 for x ×  1 −  = 427.8[0] oe or  100  better  7  M1 for their 460 ×  1 +  oe  100  7 or their 460 × correctly evaluated 100

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Question 18

18 (a) Simplify. 3 4xy 2 ` j ................................................. [2] (b) 25 = 125k Find the value of k. k = ................................................ [1]

Mark scheme: 18(a) 64x3y6 final answer 2 B1 for kx3y6 or 64xky6 or 64x3yk final answer or correct answer then spoilt 18(b) 2 1 3

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Q19 · V + 10 NOT TO SCALE Speed v (m/s) 0 0 24 40 Time (seconds) The diagram shows the…

19 v + 10 NOT TO SCALE Speed v (m/s) 0 0 24 40 Time (seconds) The diagram shows the speed–time graph for the final 40 seconds of a car journey. At the start of the 40 seconds the speed is v m/s. (a) Find the acceleration of the car during the first 24 seconds. ......................................... m/s2 [1] (b) The total distance travelled during the 40 seconds is 1.24 kilometres. Find the value of v. v = ................................................ [4]

Mark scheme: 19(a) 5 1 or 0.417 or 0.4166 to 0.4167 12 19(b) 32.5 4 M3 for 1 1 ( v + v + 10 ) × 24 + × 16 ( v + 10 ) = 1240 2 2 oe OR 1 M2 for ( v + v + 10 ) × 24 oe and 2 1 × 16 ( v + 10 ) oe 2 or M1 for one area expression M1 for correctly solving their (av + b = 1240) oe (a ≠ 0, b ≠ 0)

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Question 20

20 Factorise. 3x + 8y - 6ax - 16ay ................................................. [2]

Mark scheme: 20 (3x + 8y)(1 – 2a) 2 M1 for 3x(1 – 2a) + 8y(1 – 2a) or 3x + 8y – 2a(3x + 8y) or better

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Q21 · B NOT TO SCALE 8 cm 30° O P A OAB is the sector of a circle, centre O

21 B NOT TO SCALE 8 cm 30° O P A OAB is the sector of a circle, centre O. OB = 8 cm and angle AOB = 30°. BP is perpendicular to OA. (a) Calculate AP. AP = ........................................... cm [3] (b) Work out the area of the shaded region APB. .......................................... cm2 [3]

Mark scheme: 21(a) 1.07 or 1.071 to 1.072 3 M2 for [8 −] 8 cos 30 oe OP or M1 for = cos30 oe 8 21(b) 2.9[0] or 2.895 to 2.901 3 30 2 M1 for × π× 8 oe 360 1 M1 for × 8 × their 6.93 × sin30 oe 2 1 or × 8cos30 × 4 oe 2

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Q22 · The table shows information about the times, t seconds, taken by each of 100 students to…

22 The table shows information about the times, t seconds, taken by each of 100 students to solve a puzzle. Time (t seconds) 0 1 t G 10 10 1 t G 15 15 1 t G 20 20 1 t G 40 40 1 t G 75 Frequency 9 18 22 30 21 (a) Calculate an estimate of the mean time. ............................................... s [4] (b) Emmanuel draws a histogram to show this information. The table shows the heights, in cm, of some of the bars for this histogram. Complete the table. Time (t seconds) 0 1 t G 10 10 1 t G 15 15 1 t G 20 20 1 t G 40 40 1 t G 75 Height of bar (cm) 3.6 14.4 17.6 [3]

Mark scheme: 22(a) 27.625 4 M1 for 5, 12.5, 17.5, 30, 57.5 M1 for where x is in correct fx interval including boundaries Σfx M1 dep on second M1 for 100 22(b) 6 and 2.4 3 B2 for either correct or M1 for [fd =] 1.5 or 0.6 oe or B1 for [multiplier] 4

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Q23 · Y is inversely proportional to the square root of x

23 y is inversely proportional to the square root of x. When y = 7 , x = 2.25 . Write y in terms of x. y = ................................................ [2]

Mark scheme: 23 10.5 2 k y = oe final answer M1 for y = x x

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Question 24

24 Simplify. x 2 - 25 x 2 - 17x + 60 ................................................. [4] Question 25 is printed on the next page.

Mark scheme: 24 x + 5 4 B1 for (x + 5) (x – 5) nfww final answer B2 for (x – 12) (x – 5) x − 12 or B1 for x(x – 5) – 12 (x – 5) or x(x – 12) – 5(x – 12) or for (x + a)(x + b) where ab = –60 or a + b = –17

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Q25 · Solve 3 tanx =- 4 for 0° G x G 360°

25 Solve 3 tanx =- 4 for 0° G x G 360° . x = .................. or x = ................... [3]

Mark scheme: 25 126.9 or 126.86 to 126.87 and 306.9 or 3 B2 for one correct 306.86 to 306.87 4 or M1 for tan x = − 3 if 0 scored then SC1 for two answers with a difference of 180°

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Cambridge’s own grade thresholds for 2020 Oct/Nov, Paper 2 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.

A47/70
B36/70
C26/70
D20/70
E14/70