Cambridge IGCSE Mathematics 0580 — 2017 Oct/Nov Paper 4 · Variant 2

0580/42/O/N/17 · 10 questions · 130 marks · ≈146 min

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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 · Alex has $20 and Bobbie has $25

1 (a) Alex has $20 and Bobbie has $25. (i) Write down the ratio Alex’s money : Bobbie’s money in its simplest form. ....................... : ...................... [1] 1 (ii) Alex and Bobbie each spend of their money. 5 Find the ratio Alex’s remaining money : Bobbie’s remaining money in its simplest form. ....................... : ...................... [1] (iii) Alex and Bobbie then each spend $4. Find the new ratio Alex’s remaining money : Bobbie’s remaining money in its simplest form. ....................... : ...................... [2] (b) (i) The population of a town in the year 1990 was 15 600. The population is now 11 420. Calculate the percentage decrease in the population. .............................................% [3] (ii) The population of 15 600 was 2.5% less than the population in the year 1980. Calculate the population in the year 1980. ................................................. [3] (c) Chris invests $200 at a rate of x% per year simple interest. At the end of 15 years the total interest received is $48. Find the value of x. x = ................................................ [2] (d) Dani invests $200 at a rate of y% per year compound interest. At the end of 10 years the value of her investment is $256. Calculate the value of y, correct to 1 decimal place. y = ................................................ [3]

Mark scheme: Question Answer Marks Partial marks 1(a)(i) 4 : 5 1 1(a)(ii) 4 : 5 1 1(a)(iii) 3 : 4 2 B1 for 12 : 16 or answer 4 : 3 1(b)(i) 26.8 or 26.79… 3 15600 − 11420 11420 M2 for [× 100 ] or × 100 15600 15600 11420 or M1 for 15600 1(b)(ii) 16 000 nfww 3 100 M2 for 15600 × oe 100 − 2.5 or M1 for 15600 associated with 97.5[%] seen 1(c) 8 2 200 × x × 15 1.6 or M1 for = 48 oe 5 100 or M1 for figs 16 1(d) 5 3 B2 for 2.49[9…] or 102.4[99…] or 1.024[99…] 2.5 or cao nfww or 2.50 or 102.5 or 1.025 2 256 or M2 for 10 oe 200 or M1 for 256 = 200 ( x )10 seen

More questions on Ratio and proportion

Q2 · NOT TO r 2r SCALE A sphere of radius r is inside a closed cylinder of radius r and height…

2 (a) NOT TO r 2r SCALE A sphere of radius r is inside a closed cylinder of radius r and height 2r. r r .] [The volume, V, of a sphere with radius r is V = 43 3 (i) When r = 8 cm, calculate the volume inside the cylinder which is not occupied by the sphere. ..........................................cm3 [3] (ii) Find r when the volume inside the cylinder not occupied by the sphere is 36 cm3. r = .......................................... cm [3] (b) 12 cm NOT TO SCALE 5 cm The diagram shows a solid cone with radius 5 cm and perpendicular height 12 cm. (i) The total surface area is painted at a cost of $0.015 per cm2. Calculate the cost of painting the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] $ ................................................ [4] (ii) The cone is made of metal and is melted down and made into smaller solid cones with radius 1.25 cm and perpendicular height 3 cm. Calculate the number of smaller cones that can be made. ................................................. [3]

Mark scheme: 2(a)(i) 1070 or 1072. .. 3 M1 for π × 82 × 2 × 8 4 3 M1 for × π × 8 3 or M2 for 23 πr 3 or M1 for π r 2 2 r − 43 π r 3 2(a)(ii) 2.58 or 2.580 to 2.581 3 3 36 × 3 B2 for r = or better 2π 2 4 3 or M1 for π × r × 2 × r – × π × r = 36 oe 3 2(b)(i) 4.24 or 4.241 to 4.242 4 2 2 2 M3 for (π × 5 + π × 5 × 5 + 12 ) or M2 for π × 5 × 5 2 + 12 2 or M1 for 5 2 + 12 2 or π × 52 2(b)(ii) 64 cao final answer 3 [ kπ ] × 5 2 × 12 M2 for [ kπ ] × 1.25 2 × 3 or M1 for 13 × π × 5 2 × 12 or 13 × π × 1.25 2 × 3 OR 3  1  3 M2 for 4 or   seen  4  1 or M1 for factor 4 or soi 4

More questions on Surface area and volume

Q3 · B North NOT TO 70 m SCALE 100 m C A 40° 110 m D The diagram shows a field ABCD

3 B North NOT TO 70 m SCALE 100 m C A 40° 110 m D The diagram shows a field ABCD. (a) Calculate the area of the field ABCD. ............................................m2 [3] (b) Calculate the perimeter of the field ABCD. ............................................. m [5] (c) Calculate the shortest distance from A to CD. ............................................. m [2] (d) B is due north of A. Find the bearing of C from B. ................................................. [3]

Mark scheme: 3(a) 7040 or 7035. … 3 1 M1 for × 100 × 70 oe 2 1 M1 for × 100 × 110 × sin 40 oe 2 3(b) 374 or 375 or 374.4 to 374.5…. 5 2 2 M2 for 110 + 100 −×2 110 × 100 × cos40 oe or M1 for implicit form A1 for 5250 or 5247. … (or 72.4 or 72.43 to 72.44) M1 for 70 2 + 100 2 3(c) 64.3 or 64.27 to 64.28 nfww 2 distance M1 for sin40 = oe 100 3(d) 235 3 B2 for [angle ACB = ] 34.99 to 35 or [angle ABC = ] 55[.0…] 70 or M1 for tan[ ACB ] = 100 100 or tan[ ABC ] = or equivalent trig ratio 70

More questions on Non-right-angled triangles

Q4 · Y 6 5 4 3 2 1 x –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –1 F –2 –3 –4 –5 –6 –7 Draw the image…

4 (a) y 6 5 4 3 2 1 x –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –1 F –2 –3 –4 –5 –6 –7 Draw the image of 6 (i) flag F after translation by the vector [2] f- 2p, (ii) flag F after rotation through 180° about (– 2, 0), [2] (iii) flag F after reflection in the line y = x. [2] (b) y 5 4 P 3 2 Q 1 x 0 1 2 3 4 5 (i) Describe fully the single transformation that maps triangle P onto triangle Q. ...................................................................................................................................................... ...................................................................................................................................................... [3] (ii) Find the matrix that represents this transformation. [2] f p 4u (c) The point A is translated to the point B by the vector . f3up AB = 12.5 Find u. u = ................................................ [3]

Mark scheme: 4(a)(i) Correct translation 2 6  k  B1 for translation  or   k  − 2  4(a)(ii) Correct rotation 2 B1 for rotation 180˚ but other centre 4(a)(iii) Correct reflection 2 B1 for reflection in y = − x 4(b)(i) Enlargement 3 B1 for each 1 [factor] or 0.5 2 [centre] (0, 0) oe 4(b)(ii)  1  2  k 0   0  B1 for matrix of form   oe, k ≠ 0 or 1 2   oe  0 k   1   0   2  4(c) ± 2.5 3 B2 for 25u 2 = 156.25 or 5u = [±]12.5 or M1 for (4u ) 2 + (3u ) 2

More questions on Transformations

Q5 · X 3 25 y = - 2 , x =Y 0 8 x (a) Complete the table of values

x 3 25 y = - 2 , x =Y 0 8 x (a) Complete the table of values. x 0.5 1 1.5 2 2.5 3 3.5 y – 8.0 – 1.9 – 0.5 0.5 1.6 [2] (b) y 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 –9 x 3 2 The graph of y = - 2 for - 3.5 G x G - 0. 5 has already been drawn. 8 x x 3 2 On the grid, draw the graph of y = - 2 for 0.5 G x G 3.5 . [4] 8 x x 3 2(c) Use your graph to solve the equation - 2 = 0 . 8 x x = ................................................ [1] x 3 2(d) - 2 = k and k is an integer. 8 x x 3 2 Write down a value of k when the equation - 2 = k has 8 x (i) one answer, k = ................................................ [1] (ii) three answers. k = ................................................ [1] (e) By drawing a suitable tangent, estimate the gradient of the curve where x =- 3 . ................................................. [3] x 3 2(f) (i) By drawing a suitable line on the grid, find x when - 2 = 6 - x . 8 x x = ................................................ [3] x 3 2 5 3 2 (ii) The equation - 2 = 6 - x can be written as x + ax + bx + c = 0 . 8 x Find the values of a, b and c. a = ................................................ b = ................................................ c = ................................................ [4]

Mark scheme: 5(a) 3.2 or 3.15 or 3.152 to 3.153 2 B1 for each 5.2 or 5.19 or 5.20 or 5.196… 5(b) Correct graph for 0.5 ⩽ x ⩽ 3.5 4 B3FT for 6 or 7 correct points or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points 5(c) 1.7 to 1.8 1FT FT their graph if one answer 5(d)(i) Any integer k ⩾ −1 1 5(d)(ii) Any integer k < –1 1 5(e) Tangent ruled at x = –3 B1 2.5 to 4 B2 dep on tangent drawn at x = –3 or close attempt at tangent at x = –3 M1 for rise/run also dep on tangent at x = –3 or close attempt at tangent at x = –3 5(f)(i) y = 6 – x ruled accurately M2 M1 for correct line but freehand or ruled line gradient –1.1 to –0.9, or through (0, 6) but not y = 6 2.85 ⩽ x ⩽ 3 A1 5(f)(ii) [a = ] 8 4 B3 for 2 correct [b = ] –48 or x 5 + 8 x 3 − 48 x 2 − 16 = 0 seen [c = ] –16 5 3 2 or − x − 8 x + 48 x + 16 = 0 seen or M2 for correct multiplication by 8x2 or B1 for answers ± 8, ± 48, ± 16 x 2 × x 3 −×8 2 or M1 for = 6 − x x 2 × 8 or M1 for correct multiplication by 8 or M1 for correct multiplication by x2

More questions on Differentiation

Q6 · There are 100 students in group A

6 (a) There are 100 students in group A. The teacher records the distance, d metres, each student runs in one minute. The results are shown in the cumulative frequency diagram. 100 90 80 70 60 Cumulative 50 frequency 40 30 20 10 0 d 100 200 300 400 Distance (metres) Find (i) the median, ............................................. m [1] (ii) the upper quartile, ............................................. m [1] (iii) the inter-quartile range, ............................................. m [1] (iv) the number of students who run more than 350 m. ................................................. [2] (b) There are 100 students in group B. The teacher records the distance, d metres, each of these students runs in one minute. The results are shown in the frequency table. Distance 100 1 d G 200 200 1 d G 250 250 1 d G 280 280 1 d G 32 0 320 1 d G 400(d metres) Number of 20 22 30 16 12students (i) Calculate an estimate of the mean distance for group B. ............................................. m [4] (ii) Complete the histogram to show the information in the frequency table. 1 0.8 0.6 Frequency density 0.4 0.2 0 d 100 200 300 400 Distance (metres) [4] (c) For the 100 students in group B, the median is 258 m. Complete the statement. On average, the students in group A run ............................... than the students in group B. [1]

Mark scheme: 6(a)(i) 280 1 6(a)(ii) 320 1 6(a)(iii) 90 1 6(a)(iv) 10 2 M1 for 90 written 6(b)(i) 250.2 nfww cao 4 M1 for at least 4 correct mid-values M1 for Σfx M1 dep on second M1 for Σfx ÷ 100 6(b)(ii) Correct completion of 4 B1 for each correct block histogram If zero scored, then SC1 for correct frequency densities seen 6(c) [22 m] further oe 1

More questions on Interpreting statistical data

Q7 · 0 1 0 1 1 2 A B The diagram shows two fair dice

7 0 1 0 1 1 2 A B The diagram shows two fair dice. The numbers on dice A are 0, 0, 1, 1, 1, 3. The numbers on dice B are 1, 1, 2, 2, 2, 3. When a dice is rolled, the score is the number on the top face. (a) Dice A is rolled once. Find the probability that the score is not 3. ................................................. [1] (b) Dice A is rolled twice. Find the probability that the score is 0 both times. ................................................. [2] (c) Dice A is rolled 60 times. Calculate an estimate of the number of times the score is 0. ................................................. [1] (d) Dice A and dice B are each rolled once. The product of the scores is recorded. (i) Complete the possibility diagram. 3 0 0 2 0 0 2 0 0 Dice B 2 0 0 1 0 0 1 0 0 1 1 1 3 0 0 1 1 1 3 Dice A [2] (ii) Find the probability that the product of the scores is (a) 2, ................................................. [1] (b) greater than 3. ................................................. [1] (e) Eva keeps rolling dice B until 1 is scored. Find the probability that this happens on the 5th roll. ................................................. [2]

Mark scheme: 7(a) 5 1 6 7(b) 4 2 2 2 oe M1 for × 36 6 6 7(c) 20 1 7(d)(i) Diagram completed correctly 2 B1 for 3 correct columns or for 4 correct rows x x 3 3 3 9 x x 2 2 2 6 x x 2 2 2 6 x x 2 2 2 6 x x 1 1 1 3 7(d)(ii)(a) 9 1FT FT their (d)(i) oe 36 7(d)(ii)(b) 4 1FT FT their (d)(i) oe 36 7(e) 512 2  4 k 2 oe M1 for oe k = 3, 4 or 5 only  × 7776  6  6

More questions on Probability of combined events

Q8 · The cost of 1 apple is a cents

8 (a) The cost of 1 apple is a cents. The cost of 1 pear is p cents. The total cost of 7 apples and 9 pears is 354 cents. (i) Write down an equation in terms of a and p. ................................................. [1] (ii) The cost of 1 pear is 2 cents more than the cost of 1 apple. Find the value of a and the value of p. a = ................................................ p = ................................................ [3] (b) Rowena walks 2 km at an average speed of x km/h. (i) Write down an expression, in terms of x, for the time taken. .............................................. h [1] (ii) Rowena then walks 3 km at an average speed of (x – 1) km/h. The total time taken to walk the 5 km is 2 hours. (a) Show that 2x 2 - 7x + 2 = 0 . [3] (b) Find the value of x. Show all your working and give your answer correct to 2 decimal places. x = ................................................ [4]

Mark scheme: 8(a)(i) 7a + 9p = 354 oe final answer 1 8(a)(ii) [a = ] 21 3 M1 for correctly eliminating one variable [p = ] 23 A1 for a = 21 A1 for p = 23 8(b)(i) 2 1 x 8(b)(ii)(a) 2 3 M1 + = 2 x x − 1 2( x − 1) + 3 x = 2 x ( x − 1) oe M1dep Both sides of the equation could be over x(x – 1) at this stage Dep on M1 or 3 term equation with fractions but one sign error 2 x − 2 + 3 x = 2 x 2 − 2 x oe A1 Answer reached with one correctly expanded line 2 seen and no errors seen 2 x − 7 x + 2 = 0 8(b)(ii)(b) 2 B1 2 ( −7) − 4(2)(2)  7  or for  x −   4  −− 7 + q −− 7 − q B1 2 7  7  or or for + or −−+1   2 × 2 2 × 2 4  4  3.19 only B2 B1 for 3.19 with other root or for 3.2 or 3.186… isw other root or for 0.31 or 0.314 or 0.3138 to 0.3139

More questions on Equations

Q9 · F ()x = 1 - 2x g ()x = x + 4 h ()x = x 2 + 1 (a) Find (f-1)

9 f ()x = 1 - 2x g ()x = x + 4 h ()x = x 2 + 1 (a) Find (f-1) . ................................................. [1] (b) Solve the equation. 2f ( x) = g ( x) x = ................................................ [2] (c) Find fg ()x . Give your answer in its simplest form. ................................................. [2] (d) Find hh(2). ................................................. [2] (e) Find f - 1 ()x . f - 1 ()x = ................................................ [2] (f) hgf ()x = 4x 2 + px + q Find the value of p and the value of q. p = ................................................ q = ................................................ [4] Question 10 is printed on the next page.

Mark scheme: 9(a) 3 1 9(b) 2 2 M1 for 2(1 − 2 x ) = x + 4 − oe 5 9(c) −2 x − 7 final answer 2 M1 for 1 – 2(x + 4) 9(d) 26 2 B1 for h(5) soi 2 or M1 for x 2 + 1 + 1 ( ) 9(e) 1 −x 2 M1 for x = 1 – 2y or 2x = 1 – y or oe final answer 2 y 1 = − x or y – 1 = – 2x 2 2 9(f) [p = ] – 20 4 B3 for [hgf(x)] = 4 x 2 − 20 x + 26 seen and not [q = ] 26 spoilt by further working or M1 for (1 – 2x) + 4 M1 dep for ( their (5 − 2 x ) ) 2 + 1 B1FT dep for 25 – 10x – 10x + 4x2

More questions on Functions

Q10 · 270° O x cm NOT TO SCALE 2x cm The diagram shows a sector of a circle, a triangle and a…

10 270° O x cm NOT TO SCALE 2x cm The diagram shows a sector of a circle, a triangle and a rectangle. The sector has centre O, radius x cm and angle 270°. The rectangle has length 2x cm. The total area of the shape is kx2 cm 2. (a) Find the value of k. k = ................................................ [5] (b) Find the value of x when the total area is 110 cm2. x = ................................................ [2]

Mark scheme: 10(a) 5.68 or 5.684 to 5.685 5 2 2 2 M2 for 2 x x + x oe or 2 × 2 × x or M1 for x 2 or x 2 + x 2 oe soi 270 2 M1 for × π × x oe 360 M1 for 0.5 x² oe 10(b) 4.4[0] or 4.398 to 4.401 2 dep on a correct value for k in (a) 2 110 M1 for  x =   their k

More questions on Equations

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Cambridge’s own grade thresholds for 2017 Oct/Nov, Paper 4 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.

A108/130
B91/130
C74/130
D64/130
E54/130