Cambridge IGCSE Mathematics (9-1) 0980 — 2019 May/June Paper 4 · Variant 2

1 (a) The price of a newspaper increased from $0.97 to $1.13 . Calculate the percentage increase. ........................................... % [3] (b) One day, the newspaper had 60 pages of news and advertisements. The ratio number of pages of news : number of pages of advertisements = 5 : 7. (i) Calculate the number of pages of advertisements. ............................................... [2] (ii) Write the number of pages of advertisements as a percentage of the number of pages of news. ........................................... % [1] (c) On holiday Maria paid 2.25 euros for the newspaper when the exchange rate was $1 = 0.9416 euros. At home Maria paid $1.13 for the newspaper. Calculate the difference in price. Give your answer in dollars, correct to the nearest cent. $ .............................................. [3] (d) The number of newspapers sold decreases exponentially by x% each year. Over a period of 21 years the number of newspapers sold decreases from 1 763 000 to 58 000. Calculate the value of x. x = .............................................. [3] (e) Every page of the newspaper is a rectangle measuring 43 cm by 28 cm, both correct to the nearest centimetre. Calculate the upper bound of the area of a page. ........................................ cm2 [2]

Mark scheme: Question Answer Marks Partial Marks 1(a) 16.5 or 16.49... 3 M2 for 1.13 − 0.97[× 100] oe or 1.13 × 100 oe 0.97 0.97 1.13 or M1 for oe 0.97 1(b)(i) 35 2 M1 for 60 ÷ ( 5 + 7 ) 1(b)(ii) 140 1 1(c) $1.26 final answer 3 B2 for 1.259... or 1.26 but not as final answer or M1 for 2.25 ÷ 0.9416 If 0 scored, SC1 for 1.13 × 0.9416 1(d) 15[.0…] 3 58000 M2 for 21 oe 1763000 or M1 for 58000 = 1763000 ( k ) 21 1(e) 1239.75 2 B1 for 43 + 0.5 or 28 + 0.5 oe seen

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2 (a) A C 26° NOT TO SCALE F B x° D E AC is parallel to FBD, ABC is an isosceles triangle and CBE is a straight line. Find the value of x. x = .............................................. [3] (b) S P 58° 17° T NOT TO SCALE y° Q The diagram shows a circle with diameter PQ. SPT is a tangent to the circle at P. Find the value of y. y = .............................................. [5]

Mark scheme: 2(a) 103 3 M1 for angle ABC or angle ACB = 12 (180 − 26 ) oe M1 for angle ABF = 26 or angle CBD or angle FBE = 77 or exterior angle ACB = 103 correctly identified or in correct position 2(b) 75 5 B4 for 105 at a or b or 73 at c and 32 at d or B3 for 58 at m or 58 at e and 17 at k or B2 for 32 at d and 90 soi at (c+k) or 32 at d and 17 at k or 73 at c or B1 for 90 soi at (c + k) or between tangent and radius or 32 at d or 17 at k S P d 58° 17° T c m a y° b k e Q

More questions on Circle theorems II

3 The probability that Andrei cycles to school is r. (a) Write down, in terms of r, the probability that Andrei does not cycle to school. ............................................... [1] (b) The probability that Benoit does not cycle to school is 1.3 - r. The probability that both Andrei and Benoit do not cycle to school is 0.4 . (i) Complete the equation in terms of r. (.........................) # (.........................) = 0.4 [1] (ii) Show that this equation simplifies to 10r 2 - 23 r + 9 = 0 . [3] (iii) Solve by factorisation 10r 2 - 23r + 9 = 0 . r = ................... or r = ................... [3] (iv) Find the probability that Benoit does not cycle to school. ............................................... [1]

Mark scheme: 3(a) 1 – r 1 3(b)(i) (1 – r) (1.3 – r) [= 0.4] 1 FT their(a) dep on (a) being an expression in r 3(b)(ii) 1.3 – 1.3r – r + r 2 or better nfww M1 FT their (b)(i) 0.9 − 2.3r + r 2 [ = 0] M1 Strict FT their expansion to a quadratic then equating to 0.4 and then collecting to 3 terms on ‘one side’ OR OR Strict FT their expansion to a quadratic = 0.4 all 13 – 13r – 10r + 10r2 = 4 oe multiplied by 10 10 r 2 − 23r + 9 = 0 A1 no errors or omissions seen 3(b)(iii) ( 5r − 9 )( 2 r − 1) [= 0] B2 or B2 for e.g. 5r(2r – 1) – 9(2r – 1) and then 5r – 9 = 0 and 2r – 1 = 0 or B1 for 5r(2r – 1) – 9(2r – 1) [ = 0] or 2r(5r – 9) – 1(5r – 9) [ = 0] or (5r + a)(2r + b) [ = 0] where a, b are integers and ab = +9 or 2a + 5b = – 23 If 0 scored, SC1 for 5r – 9 and 2r – 1 seen but not in factorised form 9 1 B1 [r =] oe [r =] oe 5 2 3(b)(iv) 4 1 0.8 or oe 5

4 (a) The equation of a straight line is 2y = 3x + 4 . (i) Find the gradient of this line. ............................................... [1] (ii) Find the co-ordinates of the point where the line crosses the y-axis. ( ..................... , ..................... ) [1] (b) The diagram shows a straight line L. y 6 4 2 –2–2 0 22 4 6 x L –2 (i) Find the equation of line L. ............................................... [3] (ii) Find the equation of the line perpendicular to line L that passes through (9, 3). ............................................... [3] (c) A is the point (8, 5) and B is the point (- 4, 1). (i) Calculate the length of AB. ............................................... [3] (ii) Find the co-ordinates of the midpoint of AB. ( ..................... , ..................... ) [2]

Mark scheme: 4(a)(i) 1.5 oe 1 4(a)(ii) (0, 2) 1 4(b)(i) y = −2 x + 6 oe final answer 3 B2 for y = − 2 x + c oe or y = mx + 6 oe m ≠ 0 or for answer −2 x + 6 6 or B1 for [gradient =] − oe or c = + 6 soi 3 4(b)(ii) y = 0.5 x − 1.5 oe final answer 3 B1 for [gradient = ] – 1 divided by their gradient from (b)(i) evaluated soi M1 for substitution of (9, 3) into y = (their m)x + c seen in working 4(c)(i) 12.6 or 12.64 to 12.65 3 2 2 M2 for (8 −−4) + (5 − 1) oe 2 2 or M1 for ( 8 −−4 ) + ( 5 − 1) oe 4(c)(ii) (2, 3) 2 B1 for each

1 3x(c) By drawing a suitable straight line on the grid, solve the equation + - 2 = 0 . 2x 4 x = .................... or x = .................... [6] 1 x(d) Show that the graph of y = - can be used to find the value of 2 for 0.15 G x G 3.5 . 2x 4 [2]

Mark scheme: 5(a) 2.45, 0.25, − 0.25 3 B1 for each 5(b) Fully correct smooth curve 4 B3FT for 6 or 7 points or B2 FT for 4 or 5 points or B1 FT for 2 or 3 points 5(c) y = 2 – x correctly ruled B4 B3 for drawing y = 2 – x good freehand OR 1 x B2 for − = 2 − x 2 x 4 1 x 1 x or B1 for − = 2 + kx or − = p − x 2 x 4 2 x 4 M1 for line with gradient −1.05 to −0.95 or their k or y – intercept (0, 2) or (0, p) but not y = 2 or y = p OR B2FT dep for drawing their line y = ax + b, a ≠ 0, b ≠ 0 or B1dep for drawing their line good freehand or for ruled line with gradient a or for ruled line through (0, b) but not y = b Both intersections of their (b) and B2 Strict FT intersection of their curve and their their line y = ax + b line B1FT for each If 0 scored, SC1 for answers, 0.27 to 0.28 and 2.38 to 2.39 5(d) 1 x M1 Substitutes x = 2 into − 2 x 4 OR Identifies y = 0 oe OR Correctly manipulates to a single fraction 2 − x 2 eg oe seen 4 x

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6 (a) Expand and simplify. (x + 7)(x - 3) ............................................... [2] (b) Factorise completely. (i) 15p 2 q 2 - 25q 3 ............................................... [2] (ii) 4fg + 3h - 6gh - 2f ............................................... [2] (iii) 81k 2 - m 2 ............................................... [2] (c) Solve the equation. x + 2 3 (x - 4) + = 6 5 x = .............................................. [4]

Mark scheme: 6(a) x 2 + 4 x − 21 final answer 2 B1 for three of x 2, + 7 x, − 3 x, − 21 6(b)(i) 2 2 2 2 2 3 2 2 5 q 3 p − 5 q final answer B1 for 5 3 p q − 5 q or q 15 p − 25 q or ( ) ( ) ( ) q 15 p 2 q − 25 q 2 or 5 q 3 p 2 q − 5 q 2 ( ) ( ) or for correct answer seen 6(b)(ii) ( 2 f − 3h )( 2 g − 1) final answer 2 B1 for 2 g ( 2 f − 3h ) − ( 2 f − 3h ) or or (3h – 2f)(1 – 2g) final answer 2 f ( 2 g − 1) − 3h ( 2 g − 1) or 3h(1 – 2g) – 2f(1 – 2g) or 3h – 2f – 2g(3h – 2f) 6(b)(iii) ( 9 k + m )( 9 k − m ) final answer 2 M1 for (9 + m)(9 – m) or for correct answer seen 6(c) 5.5 4 M1 for 5 × 3 ( x − 4 ) + x + 2 = 5 × 6 M1 for 15 x − 60 + x + 2 = 30 FT their first step x + 2 or 3 x − 12 + = 6 5 If M0M0, SC1 for 3x – 12 + x + 2 = 30 oe M1dep for 16 x = 88 FT their previous steps

More questions on Algebraic manipulation

7 (a) Show that each interior angle of a regular pentagon is 108°. [2] (b) D E C NOT TO O SCALE M X A B The diagram shows a regular pentagon ABCDE. The vertices of the pentagon lie on a circle, centre O, radius 12 cm. M is the midpoint of BC. (i) Find BM. BM = ........................................ cm [3] (ii) OMX and ABX are straight lines. (a) Find BX. BX = ........................................ cm [3] (b) Calculate the area of triangle AOX. ........................................ cm2 [3]

Mark scheme: 7(a) 360 M2 360 180 − or or M1 for or ( 5 − 2 ) × 180 5 5 ( 5 − 2 ) × 180 ( 2 × 5 − 4 ) × 90 or 90(2 × 5 – 4) or or or 3 × 180 ÷ 5 5 5 or 6 × 90 ÷ 5 5 × 180 − 360 or 5 × 180 – 360 5 5 −×2 180 If 0 scored, SC1 for 5 7(b)(i) 7.05 or 7.053… 3 M2 for 12 × cos54 oe or M1 for implicit form or B1 for length of edge of pentagon = 14.1 to 14.11 If 0 scored, SC1 for right angle at M 7(b)(ii)(a) 22.8 or 22.81 to 22.83… nfww 3 their(b)(i) M2 for oe cos72 or M1 for implicit form oe or B1 for AX = 36.9 or 36.93 to 36.94 7(b)(ii)(b) 179 or 179.1 to 179.3… 3 M2 for 1 × 12 × their AX × sin54 oe 2 or 1 × 12 × theirOX × sin108 oe 2 or 1 × their AX × theirOX × sin18 2 or 12 × 12 2 × sin72 + areaOBX oe or 1 2 × 12 2 × sin72 + area OMB + area MBX oe or M1 for a correct method to find area of one relevant triangle AOB, OMB, MBX, OBX or ONX seen

More questions on Pythagoras’ theorem and trigonometry

8 (a) D 64° 53° 16.5 cm NOT TO SCALE 12.4 cm A 95° B C X The diagram shows two triangles ABD and BCD. DBX is a straight line. AD = 16.5 cm and BD = 12.4 cm. Angle ADB = 64°, angle BDC = 53° and angle DBC = 95°. (i) Find AB. AB = ........................................ cm [4] (ii) Find the shortest distance from C to the line BX. ......................................... cm [5] (b) y° 3.8 cm NOT TO SCALE 7.7 cm The diagram shows a sector of a circle of radius 3.8 cm. The arc length is 7.7 cm. (i) Calculate the value of y. y = .............................................. [2] (ii) Calculate the area of the sector. ........................................ cm2 [2]

Mark scheme: 8(a)(i) 15.7 or 15.70... 4 M2 for 16.5 2 + 12.4 2 −×2 16.5 × 12.4 × cos64 or M1 for implicit form A1 for 246 to 247 8(a)(ii) 18.6 or 18.61 to 18.62 5 B1 for [angle C = ]32, angle DBM = 37 or angle CBM = 58 12.4 × sin53 M2 for [BC =] oe sin32 × sin 95 or [DC = ]12.4 oe sin 32 or M1 for implicit form for either BC or DC CN M1 for sin85 = oe where N is the foot theirBC of the perpendicular from C to BX CN or for sin 53 = oe where N is the foot theirDC of the perpendicular from C to BX 8(b)(i) 116.1 or 116.08 to 116.09... 2 y M1 for × 2 × π × 3.8 = 7.7 oe 360 8(b)(ii) 14.6 or 14.61 to 14.63… 2 their (b)(i) 2 M1 for × π × 3.8 oe 360

More questions on Non-right-angled triangles

9 100 students were each asked how much money, $m, they spent in one week. The frequency table shows the results. Amount ($m) 0 1 m G 5 5 1 m G 10 10 1 m G 20 20 1 m G 30 30 1 m G 50 Frequency 16 38 30 9 7 (a) Calculate an estimate of the mean. $ .............................................. [4] (b) Complete the cumulative frequency table below. Amount ($m) m G 5 m G 10 m G 20 m G 30 m G 50 Cumulative 16 100 frequency [2] (c) On the grid, draw the cumulative frequency diagram. 100 80 60 Cumulative frequency 40 20 0 0 10 20 30 40 50 m Amount ($) [3] (d) Use your cumulative frequency diagram to find an estimate for (i) the median, $ .............................................. [1] (ii) the interquartile range, $ .............................................. [2] (iii) the number of students who spent more than $25. ............................................... [2]

Mark scheme: 9(a) 12.8[0] 4 M1 for midpoints soi M1 for use of ∑fm with m in correct interval including both boundaries M1 (dep on 2nd M1) for ∑fm ÷ 100 9(b) 54 84 93 2 B1 for 2 correct or 1 error and 2 correct or FT 9(c) Correct diagram with all points 3 B1FT their (b) for plots at 5 correct heights correctly plotted B1 for 5 points at upper ends of intervals on correct vertical line B1FT (dep on at least B1) for increasing curve or polygon through 5 points After 0 scored, SC1FT for 4 correct points plotted 9(d)(i) 9 to 9.8 final answer 1 9(d)(ii) 8.5 to 11.5 2 B1 for [UQ =] 15.5 to 17.5 or [LQ =] 6 to 7 seen 9(d)(iii) 10, 11 or 12 2 B1 for 88 to 90 seen or for answer between 10 and 12

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10 (a) A solid metal sphere of radius 9 cm is placed into an empty tank. The tank is a cylinder of radius 30 cm and height 18 cm. Water is poured into the tank until it is full. Calculate the number of litres of water poured into the tank. 4 3 [The volume, V, of a sphere with radius r is V = r r . ] 3 ...................................... litres [4] (b) A different tank is a cuboid measuring 1.8 m by 1.5 m by 1.2 m. Water flows from a pipe into this empty tank at a rate of 200 cm3 per second. Find the time it takes to fill the tank. Give your answer in hours and minutes. ........................ hours ..................... minutes [4] (c) NOT TO SCALE Area = 159.5 cm2 Area = 295 cm2 17 cm The diagram shows two mathematically similar shapes with areas 295 cm2 and 159.5 cm2. The width of the larger shape is 17 cm. Calculate the width of the smaller shape. ......................................... cm [3]

Mark scheme: 10(a) 47.8 or 47.84 to 47.85 4 B3 for answer figs 478 or figs 4784 to figs 4785 OR M1 for π × 30 2 × 18 4 3 M1 for × π × 9 3 M1dep for their volume ÷ 1000 soi 10(b) 4 [hours] 30 [ mins] nfww 4 B3 for 16200 or 4.5 or 270 figs 18 × figs 15 × figs 12 or M2 for oe figs 2 or M1 for figs 18 × figs 15 × figs 12 oe 10(c) 12.5 or 12.50… 3 159.5 M2 for 17 × oe 295 159.5 295 or M1 for or seen 295 159.5 159.5 x 2 or for = oe 295 17 2

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11 Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 The sequence of diagrams above is made up of small lines and dots. (a) Complete the table. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Number of 4 10 18 28 small lines Number of 4 8 13 19 dots [4] (b) For Diagram n find an expression, in terms of n, for the number of small lines. ............................................... [2] (c) Diagram r has 10 300 small lines. Find the value of r. r = .............................................. [2] (d) The number of dots in Diagram n is an 2 + bn + 1. Find the value of a and the value of b. a = .............................................. b = .............................................. [2]

Mark scheme: 11(a) 40 54 4 B1 for each 26 34 11(b) n 2 + 3n or n ( n + 3 ) oe 2 B1 for a quadratic expression or for 2nd common difference 2 (at least 2 shown) or for 2 correct equations seen or for subtracting n2 11(c) 100 2 M1 for their (b) = 10300 seen 11(d) 1 2 B1 for each [a = ] oe or M1 for one correct equation 2 or for 2nd difference = 1 soi (at least 2 shown) and 5 [b =] oe 2

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