Cambridge IGCSE Mathematics - International 0607 — 2018 Oct/Nov Paper 4 · Variant 1

1 (a) Solve the following equations. (i) 12 - x = 4 x = .................................................... [1] (ii) 9x - 4 = 6x + 8 x = .................................................... [2] 12 (iii) + 5 = 9 x x = .................................................... [2] (b) (i) Solve 6x 2 - 5x + 1 = 0 . x = .................... or x = .................... [3] (ii) Use your answer to part (b)(i) to solve 6 sin 2 x - 5 sin x + 1 = 0 for 0° G x G 90 ° . x = ................... or x = .................... [3]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 8 1 1(a)(ii) 4 2 M1 for correct 1st step 1(a)(iii) 3 2 M1 for correct 1st step 1(b)(i) 1 1 3 M2 for (3 x − 1)(2 x − 1) [ = 0] [ x = ] , 2 3 or M1 for ( ax ± 1)(bx ± 1) where ab = 6 or a + b = –5 or 3 x (2 x − 1) − 1(2 x − 1) or 2 x (3 x − 1) − 1(3 x − 1) OR M2 for correct sketch or M1 for any U-shaped parabola crossing x-axis twice OR 5 ± ( −5) 2 − 4 × 6[× 1] M2 for 2 × 6 b 2 or M1 for or b − 4 ac correct 2 a 1(b)(ii) 30, 19.5 or 19.47… 3 B2 FT for one correct answer 1  or M1 for sin x = their   2  1  sin x = their    3 

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2 The table shows the marks for 75 students in a test. Mark 0 1 2 3 4 5, 6 or 7 8 Number of students 6 18 16 8 15 5 7 (a) Write down the mode. .................................................... [1] (b) Find the range. .................................................... [1] (c) Find the median. .................................................... [1] (d) Find the inter-quartile range. .................................................... [2] (e) Calculate an estimate of the mean. .................................................... [2] (f) Give a reason why your answer to part (e) is an estimate. ............................................................................................................................................................. ............................................................................................................................................................. [1] (g) Two of these students are chosen at random. Find the probability that the highest mark of these students is 2. .................................................... [3]

Mark scheme: 2(a) 1 1 2(b) 8 1 2(c) 2 1 2(d) 3 2 B1 for either [LQ =] 1, or [UQ =] 4 2(e) 2.93 or 2.933… 2 M1 for ‘ ∑ fx ’ values 2(f) Assumed all scored 6 oe 1 2(g) 1008 3 M2 for 0.182 or 0.1816... or oe 16 15 16 24 24 16 5550 × + × + × oe 75 74 75 74 75 74 or M1 for one correct product 1560 52 SC1 for = = 0.28108... 5550 185

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3 y 4 x 0 –90 90 –4 f (x) = 1 - 2 sin (2x - 10) ° (a) On the diagram sketch the graph of y = f (x) , for - 90 G x G 90 . [3] (b) Write down the co-ordinates of the x-intercepts. (................ , ................) (................ , ................) [2] (c) Write down the co-ordinates of the local maximum. (................ , ................) [1] x ° (d) The graph of y =- intersects the graph of y = 1 - 2 sin (2x - 10 ) three times. 60 Find the value of the x co-ordinate at each point of intersection. x = .......................... or x = .......................... or x = .......................... [3]

Mark scheme: 3(a) Correct sketch 3 Intersections with x-axis both positive and not 90 and maximum below 4 B1 correct sine graph shape B1 max and min in correct quadrant 3(b) (20, 0) (80, 0) 2 B1 for each 3(c) (–40, 3) 1 3(d) –80.2 or –80.16… 3 B1 for each 28.9 or 28.90… 56.7 or 56.71 to 56.72

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4 y A NOT TO SCALE D x O B C ABCD is a rectangle. The equation of the line AB is 4x + 3y = 24 . (a) Find the co-ordinates of (i) point A, (................ , ................) [1] (ii) point B, (................ , ................) [1] (iii) the midpoint of AB. (................ , ................) [2] (b) Rearrange the equation 4x + 3y = 24 to make y the subject. y = .................................................... [2] (c) Find the equation of the line BC. Give your answer in the form y = mx + c . y = .................................................... [3] (d) Find the co-ordinates of (i) point C, (................ , ................) [1] (ii) point D. (................ , ................) [3]

Mark scheme: 4(a)(i) (0, 8) 1 4(a)(ii) (6, 0) 1 4(a)(iii) (3, 4) 2 FT their (i) and (ii) B1FT for each co-ordinate 4(b) 4 [ y = ] − x + 8 oe 2 M1 for correct isolating y term or for 3 correct division 4(c) 3 FT their (a)(ii) y = x − 4.5 oe 3 3 4 B2 for y = x + k , k ≠ 0 4 or M1 for gradient = 0.75 oe and M1 for correct subst of their (a)(ii) into y = mx + c 4(d)(i) (0, –4.5) 1 Strict FT their (c) and only if in form y = mx + c 4(d)(ii) (–6, 3.5) 3 FT their (a), (d)(i) B2 for one correct co-ordinate  − 6   6  or M1 for   or   soi  −4.5   4.5 

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5 The number of fish in a lake decreases by 4% each year. In January 2018 there are 30 000 fish in the lake. (a) Calculate the number of fish in the lake in (i) January 2019, .................................................... [2] (ii) January 2029, .................................................... [3] (iii) January 2017. .................................................... [3] (b) Find the last year in which there were at least 50 000 fish in the lake. .................................................... [4] (c) Philip runs a fishing business and he works 50 weeks every year. In 2018, he catches 800 kg of fish in each of these weeks. He sells all the fish he catches at a price of $3.50 for each kilogram. (i) Calculate the total amount he receives in 2018. $ .................................................... [3] (ii) For each of the 50 weeks, Philip’s business costs $2240 to run. Calculate his profit as a percentage of $2240. ................................................% [3] (d) In 2019, Philip’s business costs 8% more to run than in 2018. The selling price of fish decreases by 10%. Find the amount of fish, in kilograms, Philip will need to catch each week to keep the percentage profit found in part (c)(ii) the same. ............................................... kg [4]

Mark scheme: 5(a)(i) 28 800 2 100 − 4 M1 for 30000 × oe 100 5(a)(ii) 19 147 or 19 100 nfww 3 FT their 0.96, must be <1 and not 0.04 M2 for 30000 × (their 0.96)11 or 28800 × (their 0.96)10 or M1 for 30000 × ( their 0.96) k , k > 1 oe 5(a)(iii) 31 250 3 M2 for 30000 ÷their (0.96) or M1 for 30000 = their 0.96[ x ] 5(b) 2005 nfww 4 30000 M3 for n log(their 0.96) = log oe 50000 or M2 for (their 0.96) n = 0.6 oe or M1 for 50000 × (0.96) n = 30000 oe OR M3 for T and I with ‘12 and13’ seen or M2 for at least 3 correct trials or M1 for 50000 × (0.96) n = 30000 oe 5(c)(i) 140 000 3 M2 for 800 × 50 × 3.5 or M1 for multiplying any two 5(c)(ii) 25 3 their ( i ) − 2240 × M2 for 50[× 100] oe 2240 × 50 their ( i ) or × 100 oe 2240 × 50 800 × 3.5 − 2240 or [× 100 ] oe 2240 800 × 3.5 or × 100 2240 or M1 for their ( i ) − 2240 × 50 their ( i ) or 2240 × 50 or 800 × 3.5 − 2240 800 × 3.5 or 2240 5(d) 960 4 2240 × 1.08 × 1.25 M3 for oe 3.5 × 0.9 x× 3.5 × 0.9 − 2240 × 1.08 or for 2240 × 1.08 their ( c )( ii ) = oe 100 or B1 for 3.15 or 157.50 and B1 for 2419.2 or 120 960 or 3024

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6 y 9 8 7 6 5 4 A 3 2 1 x 0 –9 –8 –7 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8 9 –1 –2 –3 –4 –5 –6 –7 –8 –9 (a) Reflect triangle A in the line x =- 2 . Label the image B. [2] (b) Rotate triangle A through 180° about (-2, -1). Label the image C. [2] (c) Describe fully the single transformation that maps triangle C onto triangle B. ............................................................................................................................................................. ............................................................................................................................................................. [2] (d) Enlarge triangle A with centre of enlargement (1, 2) and scale factor 2. Label the image D. [2]

Mark scheme: 6(a) Vertices (–5, 3) (–8, 3) (–8, 5) 2 B1 for reflection in x = k , k ≠−2 or B1 for reflection in y = –2 6(b) Vertices (–5, –5) (–8, –5) (–8, –7) 2 B1 for correct rotation with incorrect centre of rotation. 6(c) Reflection 2 B1 for each y = –1 cao 6(d) Vertices (1, 4) (7, 4) (7, 8) 2 M1 for enlargement SF2 different centre or for enlargement different SF, correct centre

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7 (a) Find an expression for the nth term for each of these sequences. (i) 80, 77, 74, 71, … .................................................... [2] (ii) 128, 64, 32, 16, … .................................................... [2] (b) The nth term of a sequence is n 2 - 1. Find the first four terms of this sequence. ................ , ................ , ................ , ................ [2] (c) The nth term of a sequence is n - 3 . Find the first four terms of this sequence. ................ , ................ , ................ , ................ [2] (d) The nth term of a sequence is n 2 + n + 41. (i) Find the first three terms of this sequence. ................ , ................ , ................ [2] (ii) Show that when n = 41 the number in this sequence is not prime. [1]

Mark scheme: 7(a)(i) − 3n + 83 oe 2 B1 for − 3n + k or − kn + 83 7(a)(ii) n −1 2 k  1  8 −n  1  k − n kn + c 128   oe or 2 oe B1 for 128   or 2 or 2 oe  2   2  7(b) 0, 3, 8, 15 2 B1 for 3 correct no extras 7(c) 2, 1, 0, 1 2 B1 for 3 correct no extras 7(d)(i) 43, 47, 53 2 B1 for 2 correct no extras 7(d)(ii) 41(41 + 1 + 1) oe 1

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8 S A NOT TO SCALE B O D T C A, B, C and D lie on a circle, centre O. ST is a tangent to the circle at A. ODT is a straight line that bisects angle AOC. (a) Complete the statement. Angle OAT = ............ because ........................................................................................................... ............................................................................................................................................................. [2] (b) DT = OC Find angle ABC. Angle ABC = .................................................... [4]

Mark scheme: 8(a) 90 2 B1 for each Angle [between] tangent and radius oe 8(b) 60 4 B3 for angle AOT = 60 1 or M2 for sin(OAT ) = 2 1 or cos( AOT ) = oe 2 or M1 for OT = 2OA oe

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9 y 50 x 0 –3 5 –50 f ()x = x 3 - 3x 2 - 4x + 1 for - 3 G x G 5 . (a) On the diagram, sketch the graph of y = f ( x) . [2] (b) Write down the co-ordinates of the local minimum. (..................... , .....................) [2] (c) Find the range of values of k so that f ()x = k has only one solution. ....................................................................... [2] (d) g (x) = 3x 2 - 6x - 4 for - 3 G x G 5 . The graph of y = f ( x) intersects the graph of y = g (x) twice. Solve f (x) 2 g (x) . .................................................... [2]

Mark scheme: 9(a) Correct sketch 2 B1 for cubic graph with max/min incorrect 9(b) (2.53, –12.1) 2 B1 for each co-ordinate 9(c) k < –12.1 2 B1 for each k > 2.13 FT their –12.1 9(d) –0.726 < x < 1.26 2 B1 for both critical values seen or for [k <] x < 1.26 or for –0.726 < x [ < k]

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10 A NOT TO SCALE B X O D C OAC is a triangle with AB : BC = 1 : 2 and OD : DC = 1 : 2. The lines OB and AD intersect at X. OA = 6a and OC = 6c . (a) Find an expression, in terms of a and/or c, for (i) AC, AC = .................................................... [1] (ii) BC, BC = .................................................... [1] (iii) BD, giving your answer in its simplest form. BD = .................................................... [2] (b) Use your answer to part (a)(iii) to explain why OA and BD are parallel. ............................................................................................................................................................. [1] (c) Explain why triangle OAX and triangle BDX are similar. ............................................................................................................................................................. ............................................................................................................................................................. [2] (d) Find an expression, in terms of a and c, for (i) AD, AD = .................................................... [2] (ii) XD, giving your answer in its simplest form. XD = ................................................... [2] (e) Find the ratio area AXO : area BXD. ................ : ................ [2] Question 11 is printed on the next page.

Mark scheme: 10(a)(i) –6a + 6c oe 1 10(a)(ii) 2 1 FT their (a)(i) if a vector (–6a + 6c ) oe 3 10(a)(iii) –4a 2 2 JJJG M1 for their (a)(ii) + CO 3 or correct unsimplified route 10(b) Both multiples of a oe 1 Depends on (a)(iii) being a multiple of a 10(c) AngleOAX = angle BDX 2 Two correct statements Angle OXA = angle BXD B1 for one correct statement JJJG JJJG 10(d)(i) –6a + 2c oe 2 B1 for a correct route eg AO + OD or for –6a + kc or for ka + 2c , k ≠ 0 10(d)(ii) 1 2 (–12a + 4c ) oe JJJG JJJG 5 M1 for 3 XD = 2( − 6 a + 2 c − XD ) 2 or (their(d)(i)) 5 10(e) 9 : 4 oe 2 B1 for 3 : 2 oe soi or for 1.5 2  2  2 or   seen  3 

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11 B NOT TO 8.6 cm SCALE A 9.3 cm C The area of triangle ABC = 23.5 cm2. (a) Show that angle BAC = 36.0°, correct to 1 decimal place. [2] (b) Use the cosine rule to find BC. BC = ............................................... cm [3] (c) All the angles in triangle ABC are acute. Use the sine rule to find the largest angle in the triangle ABC. .................................................... [3]

Mark scheme: 11(a) 0.5 × 8.6 × 9.3 × sin A = 23.5 M1 35.99...[ = 36.0] A1 11(b) 2 2 or M1 for [ x = ] 8.6 + 9.3 −×2 8.6 × 9.3 × cos36 M2 [ x 2 = ] 8.6 2 + 9.32 − 2 × 8.6 × 9.3 × cos36 5.57 or 5.569 to 5.571... A1 M1 with correct answer scores full marks 11(c) 9.3 × sin36 M2 9.3 their ( b ) sin B = or M1 for = their ( b ) sin B sin36 78.8 or 78.9 or 79.[0] A1 M1 with correct answer scores full marks or 78.77 to 78.98...

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