Cambridge IGCSE Mathematics - International 0607 — 2022 Feb/March Paper 4 · Variant 2

1 (a) Find the gradient and y-intercept of the line with equation 3x + 4y = 24 . Gradient = ................................................ y-intercept = ................................................ [3] (b) y L NOT TO (8, 5) SCALE (4, 3) 0 x The diagram shows line L and the coordinates of two points on the line. (i) Show that the equation of line L is 2y - x = 2 . [3] (ii) Find the equation of the line parallel to L that passes through the point (0, 7). Give your answer in the form y = mx + c . y = ................................................ [2]

Mark scheme: Question Answer Marks Partial Marks 1(a) 3 2 M1 for isolating y oe − oe 4 6 1 1(b)(i) 5 − 3 M1 [grad = ] oe 8 − 4 Substitution of (4, 3) or (8, 5) into M1 y = (their m)x + c or y – y1 = m(x – x1) 1 A1 y = x + 1 or 2y – 6 = x – 4 or 2 2y – 10 = x – 8 leading to 2y – x = 2 without error or omission 1(b)(ii) 1 2 1 [ y = ] x + 7 B1 for [ y = ] x + k , k ≠ 1 2 2 or for [y =] jx + 7, j ≠ 0

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2 (a) Find 12 kg as a percentage of 80 kg. ..............................................% [1] (b) Find 19% of $250. $ ................................................ [2] (c) Xavier invests $500 at a rate of 1.5% per year simple interest. At the end of y years, the value of Xavier’s investment is $612.50 . Find the value of y. y = ................................................ [3] (d) Each year the value of a car decreases by 12% of its value at the beginning of that year. The original value of the car is $20 000. (i) Calculate the value of the car at the end of 3 years. Give your answer correct to the nearest dollar. $ ................................................ [3] (ii) Find the number of complete years for the value of $20 000 to decrease until it is first below $1000. ................................................. [4] (e) Each year the value of another car decreases by r % of its value at the beginning of that year. At the end of 10 years, the value has decreased from $12 000 to $4673. Find the value of r. r = ................................................ [3]

Mark scheme: 2(a) 15 1 2(b) 47.5[0] 2 19 M1 for × 250 oe 100 2(c) 15 3 500 × 1.5 × y M2 for 500 + = 612.50 oe 100 500 × 1.5 × y or M1 for oe 100 or for one year’s interest = 7.5[0] 2(d)(i) 13629 cao 3 B2 for 13630 or 13629. ...  12  3 or M1 for 20000 ×  1 −  oe  100  2(d)(ii) 24 nfww 4 B3 for 23.4 or 23.43... OR  12   1000  M3 y log  1 −  = log   oe  100   20000  or correct trials reaching 23 and 24 or good sketch indicating value between 23 and 24  12  y 1000 or M2 for  1 −  = oe  100  20000 or at least 3 correct trials or suitable graph with y > 1  12  y or M1 for 20000 ×  1 −  = 1000 oe soi by at  100  least 2 correct trials with n > 3 2(e) 9[.00] or 8.999 to 9.000... 3 4673 M2 for 10 12000 or M1 for 12 000 × (.....)10 = 4673

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3 (a) The table shows the coursework grades for 20 students. Grade 3 4 5 6 7 Frequency 1 3 6 2 8 Find (i) the mode, ................................................. [1] (ii) the range, ................................................. [1] (iii) the median, ................................................. [1] (iv) the lower quartile. ................................................. [1] (b) The table shows some information about the heights, h cm, of 100 bushes. Height (h cm) 100 1 h G 110 110 1 h G 115 115 1 h G 130 Frequency 18 37 45 Calculate an estimate of the mean height. ............................................ cm [2] (c) The table shows some information about the times, t minutes, taken by some students to read a magazine. Time (t minutes) 0 1 t G 10 10 1 t G 20 20 1 t G 30 30 1 t G 40 Frequency 3 11 n 19 When using mid-interval values, an estimate of the mean value of t is 25.4 . Find the value of n. n = ................................................ [4]

Mark scheme: 3(a)(i) 7 1 3(a)(ii) 4 1 3(a)(iii) 5.5 1 3(a)(iv) 5 1 3(b) 115.65 2 M1 for mid-values soi 3(c) 17 4 B3 for 25n + 845 = 25.4n + 838.2 oe or better or M2 for 3 × 5 + 11 × 15 + n × 25 + 19 × 35 = 25.4( n + 3 + 11 + 19) oe or M1 for 3 × 5 + 11 × 15 + n × 25 + 19 × 35 oe or25.4( n + 3 + 11 + 19) or for correct trial with integer value of n

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4 (a) a° 65° b° NOT TO 30° SCALE c° The diagram shows two straight lines crossing two parallel lines. Find the values of a, b and c. a = ................................................ b = ................................................ c = ................................................ [3] (b) L D y° 20°20° 70° E w° NOT TO SCALE K 30° u° C v° x° A z° B A, B, C, D and E are points on the circle. KL is a tangent to the circle at E. AC = AD. Find the values of u, v, w, x, y and z. u = ................................................ x = ................................................ v = ................................................ y = ................................................ w = ................................................ z = ................................................ [6]

Mark scheme: 4(a) [ a = ] 65 3 B1 for each [ b = ] 85 [ c = ] 95 4(b) [u = ] 70 6 B1 for each [v = ] 30 [w = ] 80 FT 180 – their u – their v [x = ] 20 [y = ] 50 FT 150 – their x – their w [ z = ] 60 FT 110 – their y

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5 (a) (i) Expand and simplify 2x + 3 . .................................................. [2] 2 2 (ii) The equation 4x + 12x + 5 = 0 can be written as 2x + 3 = k . ` j Find the value of k. k = ................................................ [1] (iii) Use your answer to part(ii) to solve the equation 4x 2 + 12x + 5 = 0 . x = ................... or x = .................. [2] (b) x varies inversely as the square root of (w – 1). When w = 10, x = 2. (i) Find x in terms of w. x = ................................................ [2] (ii) Find x when w = 3.25 . x = ................................................ [1] (iii) Find w in terms of x. w = ................................................ [3]

Mark scheme: 5(a)(i) 4 x 2 + 12 x + 9 final answer 2 B1 for three of 4x2, 6x, 6x, 9 or for correct answer seen 5(a)(ii) 4 1 FT their 9 – 5 5(a)(iii) 2 x + 3 = ± their 4 M1 their 4 > 0 1 1 B1 −2 , − oe 2 2 5(b)(i) 6 2 k final answer M1 for [x = ] oe w − 1 w − 1 5(b)(ii) 4 1 FT only incorrect k 5(b)(iii) 2 2 3 M1 for correct multiplication of term in w 36 36 + x  6 + + 1 or or   1 M1 for correct squaring 2 2 x x  x  M1 for correctly isolating w final answer Max M2 if incorrect answer

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6 In this question all lengths are in centimetres. NOT TO SCALE 2x – 1 7 5x + 1 13 – x The area of the larger rectangle is 84 cm 2 greater than the area of the smaller rectangle. (a) Show that 5x 2 + 2x - 88 = 0 . [4] (b) Factorise 5x 2 + 2x - 88 . ................................................. [2] (c) Find the area of the smaller rectangle. ......................................... cm2 [2]

Mark scheme: 6(a) (5x + 1)(2x – 1) – 7(13 – x) = 84 oe M1 Correct first statement without brackets expanded 10x2 – 5x + 2x – 1 B1 – 91 + 7x B1 on LHS or 91 – 7x on RHS 10 x 2 + 4 x − 176 = 0 oe A1 leading to 5 x 2 + 2 x − 88 = 0 with no errors or omissions 6(b) ( 5 x + 22 )( x − 4 ) 2 B1 for (5x + a)(x + b) with ab = –88 or a + 5b = 2 or for 5x(x – 4) + 22(x – 4) or for x(5x + 22) – 4(5x + 52) 6(c) 63 2 M1 for solving their factorised quadratic, allowing omission of negative root. FT (13 – their positive root) × 7 1 if < x < 13 2

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7 y 12 x – 4 0 4 – 12 f ( )x = 4 - x 2 for - 4 G x G 4 (a) On the diagram, sketch the graph of y = f ( x) . [2] (b) Write down the zeros of f(x). ................................................. [2] (c) Write down the coordinates of the local maximum. (...................... , ......................) [1] (d) The equation 4 - x 2 = k has 4 solutions and k is an integer. Write down a possible value of k. k = ................................................ [1] (e) (i) On the diagram, sketch the graph of y = 2x . [1] (ii) Solve the equation 4 - x 2 = 2x . ............................................................. [2] (iii) On the diagram, shade the regions where y H 0, y G 2x and y G 4 - x 2 . [2]

Mark scheme: 7(a) Correct sketch 2 10101010 5555 M1 for a modulus graph or 4444 -2-2-2-2 0000 0000 2222 4444 for graph of y = 4 – x2 -5-5-5-5 -10-10-10-10 7(b) –2, 2 2 B1 for each If 0 scored, SC1 for (2, 0) and (–2, 0) 7(c) (0, 4) 1 7(d) 1 or 2 or 3 1 7(e)(i) Correct sketch 1 10101010 5555 4444 -2-2-2-2 0000 0000 2222 4444 -5-5-5-5 -10-10-10-10 7(e)(ii) 1.24 or 1.236..., 3.24 or 3.236... 2 B1 for each or B1 for both seen e.g. 1.24 ⩽ x ⩽ 3.24 or with y coords included or for 1.23 and 3.23 7(e)(iii) Two correct regions above x-axis and 2 B1 for one correct and no wrong or for one correct below both graphs and one incomplete

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8 f ( x) = 2 x + 1 g ( x) = 3 - 2 x h ( x) = log ( x + 1) (a) Find the value of (i) f(12), ................................................. [1] (ii) g(f(12)). ................................................. [1] (b) Find the value of x when f ( x) = g ( x) . x = ................................................ [2] (c) Find f(g(x)), giving your answer in its simplest form. ................................................. [2] (d) Find g -1 ( x) . g -1 ( x) = ................................................ [2] (e) Find x when h (x) = f ( 0 .5 ) . x = ................................................ [2] (f) Find h -1 ( x) . h -1 ( x) = ................................................ [2]

Mark scheme: 8(a)(i) 25 1 8(a)(ii) –47 1 FT 3 – 2 × their 25 8(b) 1 2 M1 for 2x + 1 = 3 – 2x or better oe 2 8(c) 7 – 4x final answer 2 M1 for 2(3 – 2x) + 1 8(d) 3 −x 2 M1 for y + 2x = 3 or better oe final answer y 3 2 or = − x 2 2 or x = 3 – 2y 8(e) 99 2 M1 for log( x + 1) = 2(0.5) + 1 or better 8(f) 10 x − 1 2 M1 for 10 y = x + 1 or x = log( y + 1)

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9 (a) x cm 4 cm NOT TO SCALE 40° Calculate the value of x. x = ................................................ [3] (b) C 8 cm 9 cm NOT TO SCALE A B 10 cm (i) Calculate angle ABC. Angle ABC = ................................................ [3] (ii) T is the point on AB that is the shortest distance from C. Calculate BT. BT = ........................................... cm [3] (c) Another triangle PQR has QR = 12 cm, PR = 7 cm and angle PQR = 35°. Calculate the difference between the two possible values of angle QPR. ................................................. [5]

Mark scheme: 9(a) 6.22 or 6.222 to 6.223 3 4 M2 for oe sin40 4 or M1 for sin 40 = oe x 9(b)(i) 49.5 or 49.45 to 49.46 3 9 2 + 10 2 − 8 2 M2 for [cos=] oe 2.9.10 or M1 for 82 = 92 + 102 – 2 × 9 × 10 cos(...) 9(b)(ii) 5.85 or 5.845 to 5.851... 3 BT M2 for = cos(their(b)(i)) oe or better 9 or M1 for CT drawn and right angle at T 9(c) 21[.0] or 20.98 to 21.00 5 12sin35 M2 for 7 7 12 or M1 for = oe sin35 sin P A1 for 79.5 or 79.50 to 79.51 M1 for 180 – their 79.5 If 0 scored, SC1 for diagram showing the two angles

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10 When Zena wears a sweatshirt, the probability that she goes for a walk is 10. 9 When Zena does not wear a sweatshirt, the probability that she goes for a walk is . 10 On any day, the probability that she wears a sweatshirt is 1. 5 (a) Complete the tree diagram. Wears a sweatshirt Goes for a walk Yes .......... Yes 1 5 .......... No Yes .......... .......... No .......... No [3] (b) (i) Find the probability that on one day Zena does not wear a sweatshirt and she goes for a walk. ................................................. [2] (ii) Find the probability that on one day Zena goes for a walk. ................................................. [2] (c) In the tree diagram below, the value of J is the answer to part (b)(i) and the value of K is the answer to part (b)(ii). Goes for a walk Wears a sweatshirt No J .......... Yes K .......... Yes No .......... .......... No .......... Yes (i) Find the probability that Zena does not wear a sweatshirt when she goes for a walk. ................................................. [2] (ii) Complete the tree diagram above. [3]

Mark scheme: 10(a) 4 3 oe B1 for each pair of branches 5 7 3 , oe 10 10 9 1 , oe 10 10 10(b)(i) 18 2 4 9 oe M1 for their ×their 25 5 10 10(b)(ii) 43 2 1 7 oe M1 for their(b)(i) + ×their oe 50 5 10 10(c)(i) 36 2 M1 for their(b)(ii) × p = their(b)(i) or better oe 43 10(c)(ii) their(b)(ii), 1 – their (b)(ii) oe 3 B1 for each pair of branches their (c)(i), 1 – their(c)(i) oe 4 3 , oe 7 7

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11 (a) 2r NOT TO r SCALE y° The diagram shows a sector of a circle with radius r and angle y°. The length of the arc of the sector is 2r. Calculate the value of y. y = ................................................ [3] (b) NOT TO 8 cm SCALE x° The diagram shows a sector of a circle with radius 8 cm and angle x°. The area of the shaded segment is A cm 2. 8x (i) Show that A = r - 32 sin x . 45 [2] (ii) Find the value of A when x = 90. ................................................. [1] 8x (iii) By sketching the graph of A = r - 32 sin x , find the value of x when A = 5.5 . 45 A 20 0 90 x x = ................................................ [3]

Mark scheme: 11(a) 114.6 or 114.5 to 114.6 3 y M2 for × 2πr = 2 r oe 360 y or M1 for × 2 πr oe 360 11(b)(i) x 2 1 2 M2 x 2 1 2 × π× 8 − × 8 × sin x = A M1 for × π× 8 or × 8 × sin x 360 2 360 2 11(b)(ii) 18.3 or 18.26 to 18.27... 1 11(b)(iii) Correct sketch of curve and line B2 B1 for correct shape of curve 6666 4444 2222 0000 0000 20202020 40404040 60606060 80808080 58.9 or 58.90 to 58.92 1

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