Cambridge IGCSE Mathematics (9-1) 0980 — 2021 Oct/Nov Paper 4 · Variant 1

1 (a) NOT TO 5.7 cm SCALE 9.2 cm 19.4 cm The diagram shows a brick in the shape of a cuboid. (i) Calculate the total surface area of the brick. ......................................... cm2 [3] (ii) The density of the brick is 1.9 g/cm3. Work out the mass of the brick. Give your answer in kilograms. [Density = mass ÷ volume] ............................................ kg [3] (b) 9000 bricks are needed to build a house. 200 bricks cost $175. Work out the cost of the bricks needed to build 5 houses. $ ................................................ [3] (c) Saskia builds a wall using 1500 bricks. She can build at the rate of 40 bricks each hour. She works for 9 hours each day. Saskia starts work on 6 July and works every day until the wall is completed. Find the date when she completes the wall. ................................................. [3] (d) Rafa has a cylindrical tank. The cylinder has a height of 105 cm and a diameter of 45 cm. Calculate the capacity of the tank in litres. ........................................ litres [3]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 683 3 M2 for [2]((19.4 × 9.2) + (5.7 × 9.2) + (19.4 × 5.7)) oe or M1 for one of 19.4 × 9.2 or 5.7 × 9.2 or 19.4 × 5.7 1(a)(ii) 1.93[0] or 1.932 to 1.933 3 M2 for 19.4 × 9.2 × 5.7 × 1.9 or M1 for 19.4 × 9.2 × 5.7 1(b) 39 375 3 M2 for 9000 ÷ 200 × 175 × 5 175 or M1 for 9000 ÷ 200 soi or for soi 200 1(c) 10th July 3 1 B2 for 4.1 to 4.2 or 4 or 4 days 1.5 6 hours Or M2 for answer 9th July or 11th July or M1 for 1500 ÷ (9 × 40) 1(d) 167 or 166.9 to 167.0… 3 B2 for answer with figs 167 or figs 1669 to 1670.. or M1 for π× 22.5 2 × 105 oe If 0 scored SC1 for answer 668 or 667.9 to 668.1

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2 Bob, Chao and Mei take part in a run for charity. (a) Their times to complete the run are in the ratio Bob : Chao : Mei = 4 : 5 : 7. (i) Find Chao’s time as a percentage of Mei’s time. ............................................. % [1] (ii) Bob’s time for the run is 55 minutes 40 seconds. Find Mei’s time for the run. Give your answer in minutes and seconds. ....................... min ................ s [3] (b) Chao collects $47.50 for charity. (i) Bob collects 28% more than Chao. Find the amount Bob collects. $ ................................................ [2] (ii) Chao collects 60% less than Mei. Find how much more money Mei collects than Chao. $ ................................................ [3] (c) When running, Chao has a stride length of 70 cm, correct to the nearest 5 cm. Chao runs a distance of 11.2 km, correct to the nearest 0.1 km. Work out the minimum number of strides that Chao could take to complete this distance. ................................................. [4] (d) In 2015, a charity raised a total of $1.6 million. After 2015, this amount increased exponentially by 2.4% each year for the next 5 years. Work out the amount raised by the charity in 2020. $ .................................... million [2]

Mark scheme: 2(a)(i) 71.4 or 71.42 to 71.43 1 2(a)(ii) 97 [min] 25 [s] 3 B2 for 13 min 55 sec seen or 97.4 or 97.41 to 97.42 seen or 5845 seen OR M2 for 55.66… ÷ 4 × 7 oe or 3340 ÷ 4 × 7 oe or for 7/4 × 55 + 7/4 × 40 oe or M1 for 55 min 40 sec ÷ 4 oe or M1 for total time ÷ 16 soi 2(b)(i) 60.8[0] 2  28  M1 for 47.5 ×  1 +  oe  100  or B1 for 13.3[0] 2(b)(ii) 71.25 3 B2 for 118.75  60  Or M2 for 47.50 ÷  1 −  – 47.50  100   60  or M1 for x ×  1 −  = 47.50 oe or  100  better 2(c) 15 380 4 M3 for (1 120 000 – 5000) ÷ (70 + 2.5) oe or B2 for answer figs 15 379 to figs 15 380 or M2 for (1 120 000 ± 5000) ÷ (70 ± 2.5) oe or M1 for one of figs 675, 725, 1115, 1125 seen 2(d) 1.8[0] or 1.801 to 1.802 [million] nfww 2 5  2.4  M1 for figs 16 ×  1 +  oe  100 

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3 The cumulative frequency diagram shows information about the mass, m kg, of each of 80 boys. 80 60 Cumulative frequency 40 20 0 m 30 40 50 60 70 80 90 Mass (kg) (a) m 30 40 50 60 70 80 90 Mass (kg) On the grid, draw a box-and-whisker plot to show the information in the cumulative frequency diagram. [4] (b) Use the cumulative frequency diagram to find an estimate of (i) the 30th percentile, ............................................ kg [2] (ii) the number of boys with a mass greater than 75 kg. ................................................. [2] (c) (i) Use the cumulative frequency diagram to complete this frequency table. Mass 30 1 m G 40 40 1 m G 50 50 1 m G 60 60 1 m G 70 70 1 m G 80 80 1 m G 90(m kg) Frequency 8 12 14 10 [1] (ii) Calculate an estimate of the mean mass of the boys. ............................................ kg [4] (iii) Two boys are chosen at random from those with a mass greater than 70 kg. Find the probability that one of them has a mass greater than 80 kg and the other has a mass of 80 kg or less. ................................................. [3]

Mark scheme: 3(a) Correct box-and-whisker plot 4 B1 for lowest value and highest value at 30 and 90 B1 for LQ and UQ at 50 and 72 B1 for median at 63 3(b)(i) 56 2 M1 for 24 soi 3(b)(ii) 16 2 B1 for 64 written 3(c)(i) 14, 22 1 3(c)(ii) 61.5 4 M1 for 35, 45, 55, 65, 75, 85 soi M1 for Σ fx M1 dep for their Σ fx ÷ (8 + 12 +their 14 + their 22 + 14 +10) or Σ fx ÷ 80 3(c)(iii) 35 3  10 14  oe M2 for [2]  ×  oe 69  24 23  10 14 or M1 for or oe seen 24 24 35 If 0 scored, SC1 for answer oe 72

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4 (a) Solve. (i) 6 ( 7 - 2)x = 3x - 8 x = ................................................ [3] 2x 2 (ii) = x - 5 3 x = ................................................ [3] (b) Factorise completely. (i) 2x 2 - 288y 2 ................................................. [3] (ii) 5x 2 + 17x - 40 ................................................. [2] (c) Solve x 3 + 4x 2 - 17x = x 3 - 9 . You must show all your working and give your answers correct to 2 decimal places. x = ........................ or x = ........................ [5]

Mark scheme: 4(a)(i) 10 1 3 M1 for 42 – 12x = 3x – 8 oe or or 3.33[3…] 3 x 8 3 33 or for 7 – 2x = − oe 6 6 M1 for reaching ax = b correctly FT their first step 4(a)(ii) 1 5 3 M1 for 3 × 2x = 2(x – 5) oe –2.5 or −2 or − 2 2 M1 for reaching ax = b correctly FT their first step 4(b)(i) 2(x + 12y)(x – 12y) final answer 3 B2 for (2x + 24y)(x – 12y) or (2x – 24y)(x + 12y) or for 2(x + 12y)(x – 12y) seen OR M2 for k(x + 12y)(x – 12y) or M1 for 2(x2 – 144y2) 4(b)(ii) (5x – 8) (x + 5) final answer 2 M1 for 5x(x + 5) – 8(x + 5) or x (5x – 8)+ 5(5x – 8) or for (5x + a)(x + b) where ab = – 40 or a + 5b = 17 4(c) 4x2 – 17x + 9 [= 0] oe B1 2 B2 FT their 3 term quadratic [ −− ]17 ± ( [ − ]17 ) − 4 ( 4 )( 9 ) 2 B1FT for ( [ − ]17 ) − 4 ( 4) ( 9 ) ) or better 2 × 4 2 − ]17 ) − 4 ( 4 )( 9 )  17  2 ( [ or  x −  oe or  8  4 or better [ −− ]17 + q and B1FT for or 2(4) [ −− ]17 − q or better 2(4) 17 145 17 145 or + oe or − oe or 8 64 8 64 [ −− ]17 [ −− ]17 + q − q 2 2 or 4 4 0.62 and 3.63 cao B2 B1 for each SC1 for 0.6[0] or 0.619 to 0.620 and 3.6[0] or 3.6301 to 3.6302 or 0.62 and 3.63 seen in working or –0.62 and–3.63 as final answers

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5 (a) D A NOT TO SCALE O 124° B 35° C A, B, C and D are points on a circle, centre O. Angle COD = 124° and angle BCO = 35°. (i) Work out angle CBD. Give a geometrical reason for your answer. Angle CBD = ............................. because .......................................................................... ............................................................................................................................................. [2] (ii) Work out angle BAD. Give a geometrical reason for each step of your working. Angle BAD = ............................. because .......................................................................... ............................................................................................................................................. ............................................................................................................................................. [4] (b) R 42° NOT TO S SCALE O Q 5.9 cm P P, Q, R and S are points on a circle, centre O. QS is a diameter. Angle PRS = 42° and PQ = 5.9 cm. Calculate the circumference of the circle. ............................................ cm [5]

Mark scheme: 5(a)(i) 62 2 B1 for either and Angle at centre is twice angle at circumference oe 5(a)(ii) 117 4 B2 for 117 and or B1 for [angle OCD =] 28 Isosceles [triangle] B1dep for isosceles [triangle] and and Opposite angles in a cyclic quadrilateral B1 for opposite angles in a cyclic are supplementary quadrilateral are supplementary 5(b) 24.9 or 24.94 to 24.95 5 B1 for angle PQS = 42 M2 for QS = 5.9 ÷ cos 42 oe 5.9 or M1 for cos42= oe QS M1dep for their SQ × π oe

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2 3 24(b) By drawing a suitable straight line on the grid, solve the equation x - = - 2x 2x 5 for - 3 G x G - 0.2 and 0.2 G x G 3 . x = ........................ or x = ........................ [4] 2 3 24(c) The solutions to the equation x - = - 2x are also the solutions to an equation of the 2x 5 form ax 3 + bx 2 + cx - 15 = 0 where a, b and c are integers. Find the values of a, b and c. a = ................................................ b = ................................................ c = ................................................ [4]

Mark scheme: 6(a)(i) 9.5, 4.8 and 8.5 3 B1 for each 6(a)(ii) correct curve 5 B4 for correct curve, but branches joined or touching y axis or B3FT for 9 or 10 correct plots or B2FT for 7 or 8 correct plots or B1FT for 5 or 6 correct plots AND B1 indep two separate branches not touching or cutting y-axis 6(b) 24 4 B2 for correct ruled line crossing curve y = − 2 x ruled twice 5 and or B1 for correct freehand or for short – 0.4 to – 0.2 and 1.45 to 1.7 ruled line or for line with negative gradient through (0, 4.8) or for line with gradient – 2 B1 for each value 6(c) [a =] 10 4 B3 for 10x3 – 15 = 48x – 20x2 oe or better [b =] 20 or B2 for 2 correct values [c =] – 48 or B1 for 1 correct value 2 15 or for 5 x − = 24 − 10 x or better 2 x 3 48 2 or for 2 x − 3 = x − 4 x or better 5 3 3 24 2 or for x − = x − 2 x 2 5 After 0 scored SC1 for correct elimination of a denominator of 5, x or 2x from a four term expression.

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7 (a) y 8 7 6 5 4 3 A 2 1 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 x – 1 B – 2 – 3 – 4 (i) On the grid, draw the image of (a) shape A after an enlargement, scale factor 2, centre (0, 1), [2] (b) shape A after a reflection in the line y = x - 1. [3] (ii) Describe fully the single transformation that maps shape A onto shape B. ..................................................................................................................................................... ..................................................................................................................................................... [3] (b) C B NOT TO q M SCALE O A p OABC is a trapezium and O is the origin. M is the midpoint of AB. OA = p , OC = q and OA = 2CB. Find, in terms of p and q, the position vector of M. Give your answer in its simplest form. ................................................. [3]

Mark scheme: 7(a)(i)(a) Shape at (–2, 1) ( –4, 1) ( –4, 7) (0, 7) 2 B1 for 3 correct points or for enlargement SF2 from any centre 7(a)(i)(b) Shape at (2, –2) (2, –3) (5, –1) (5, –3) 3 B2 for correct orientation but wrong position or for 3 correct points or B1 for y = x – 1 drawn 7(a)(ii) Rotation 3 B1 for each 90 [anticlockwise] oe (0, 0) oe 7(b) 3 1 1 3 p + 2q 3  1  1  p + q or ( 3p + 2q ) or M2 for AM = AM =  - p + q + p  oe 4 2 4 4 2  2   final answer or M1 for correct route for AB oe soi by –½ p + q  or for OM soi

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8 (a) f ( )x = 3 - 5 x (i) Find x when f ( )x =- 5 . x = ................................................ [2] (ii) Find f - 1 ( )x . f - 1 ( )x = ................................................ [2] (b) g ( )x = 18 - 3x - x 2 (i) Write g ( )x in the form b - ( a + x) 2 . ................................................. [3] (ii) Sketch the graph of y = g ( x) . On your sketch, show the coordinates of the turning point. y O x [3] (iii) Find the equation of the tangent to the graph of y = 18 - 3 x - x 2 at x = 4 . Give your answer in the form y = mx + c . y = ................................................ [6]

Mark scheme: 8(a)(i) 1.6 oe 2 M1 for 3 – 5x = – 5 8(a)(ii) 3 − x 2 y 3 oe final answer M1 for x = 3 – 5y or = − x or better, 5 5 5 or y – 3 = – 5x oe 8(b)(i) 2 3 Method 1 20.25 − (1.5 + x ) 2 B1 for ( ±1.5 ± x ) seen B1 for [b =] 18 + their 1.52 OR Method 2 B1 for b − a 2 − 2 ax − x 2 or for b = 20.25 B1 for a = 1.5 8(b)(ii) Correct sketch with max in correct 3 2 FT their 20.25 − ( their1.5 + x ) provided in quadrant at ( − 1.5, 20.25 ) that form B1 for ∩shape or for ∪shape if in form 2 c + ( d + x ) in part (b)(i) B1 for TP at ( − 1.5, k ) or ( k , 20.25 ) FT 2 their 20.25 ± ( their 1.5 + x ) or for (–1.5, 20.25) seen 8(b)(iii) [y =] 34 – 11x 6 B2 for –3 – 2x or B1 for either kx –3, k ≠ 0 or –2x + n or for 18 – 3 – 2x M1dep for gradient = their (–3 – 2(4) ) B1 for y-value at x = 4, is –10 M1dep for their –10 = (their –11)4 + c oe

More questions on Functions

9 (a) NOT TO x cm SCALE ( x + 3)cm This rectangle has perimeter 20 cm. Find the value of x. x = ................................................ [3] (b) M y° NOT TO SCALE 20° This rhombus has perimeter 20 cm and angle y is obtuse. M is the midpoint of one of the sides. Find the value of y. y = ................................................ [5] (c) r cm NOT TO SCALE z cm 40° This sector of a circle has radius r and perimeter 20 cm. Find the value of z. z = ................................................ [6]

Mark scheme: 9(a) 3.5 oe 3 M1 for 2(x + x + 3) = 20 oe M1 for correct ax = b for their linear equation 9(b) 116.8 or 116.83 to 116.85 nfww 5 5sin 20 M2 for sin p = 2.5 2.5 5 or M1 for = sin 20 sin p A1 for 43.2 or 43.15 to 43.17 M1dep for 180 – (20 + their 43.2) After 0 scored, SC1 for length of side = 5 9(c) 5.07 or 5.068 to 5.071 6 B3 for 7.41 or 7.412 to 7.413 40 or M2 for r + r + × 2 × π× r = 20 oe 360 40 or M1 for × 2 × π× r oe seen 360 M2 for 2 × 7.41 × sin 20 oe or 7.412 + 7.412 – 2(7.412) cos 40 oe 7.41sin 40 or oe sin70 or M1 for implicit version

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