Cambridge IGCSE Mathematics (US) 0444 — 2019 May/June Paper 4 · Variant 1

1 y 10 9 8 7 6 5 4 3 T 2 1 0 1 2 3 4 5 6 7 8 9 10 x - 1 (a) (i) Translate shape T by the vector . c 6 m Label the image A. [2] (ii) Rotate shape T about the point (5, 3) through 180°. Label the image B. [2] (iii) Describe fully the single transformation that maps shape A onto shape B. .................................................................................................................................................... .................................................................................................................................................... [3] (b) (i) Reflect shape T in the line y = x. [2] Label the image C. (ii) Shape C can be mapped onto shape A by a rotation about the point (1, 7) followed by a reflection. Write down (a) the angle of rotation, .................................................... [1] (b) the equation of the line of reflection. .................................................... [1]

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) Image at (1, 7), (4, 7), (4, 9), (3, 9) 2  −1  k B1 for translation by   or   k  6 1(a)(ii) Image at (5, 3), (6, 3), (8, 5), (5, 5) 2 B1 for 180˚ rotation with wrong centre 1(a)(iii) Rotation 3 B1 for rotation 180˚ B1 for 180° (4.5, 6) B1FT for centre from their (a)(i) OR Enlargement, B1 for enlargement [factor] – 1 B1 for – 1 (4.5, 6) B1FT for centre from their (a)(i) 1(b)(i) Image at (1, 2), (1, 5), (3, 5), (3, 4) 2 B1 for y = x drawn or for 3 correct points 1(b)(ii)(a) 90˚ [anticlockwise] 1 1(b(ii)(b) x = 3.5 1

2 The table shows some values for y = x 3 + 3x 2 + 2 . x -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 y -4.1 5.1 6 5.4 4 2.6 2.9 12.1 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 3 + 3x 2 + 2 for - 3.5 G x G 1.5 . y 16 12 8 4 0 0.5 1 1.5 x -3.5 -3 -2.5 -2 -1.5 -1 -0.5 -4 -8 [4] (c) Use your graph to solve the equation x 3 + 3x 2 + 2 = 0 for - 3.5 G x G 1.5 . x = ................................................... [1] (d) By drawing a suitable straight line, solve the equation x 3 + 3x 2 + 2x + 2 = 0 for - 3.5 G x G 1.5 . x = ................................................... [2] (e) For - 3.5 G x G 1.5 , the equation x 3 + 3 x 2 + 2 = k has three solutions and k is an integer. Write down a possible value of k. k = ................................................... [1]

Mark scheme: 2(a) 2, 2, 6 3 B1 for each 2(b) Correct graph 4 B3FT for 10 or 11 correct plots or B2FT for 8 or 9 correct plots or B1FT for 6 or 7 correct plots 2(c) –3.3 to –3.1 1 FT their graph 2(d) y = –2x ruled M1 or B1 for y = −2 x stated –2.6 to –2.45 A1 2(e) 3 or 4 or 5 1 FT their graph Allow more than one correct value

3 North C D 170 m 120 m NOT TO 150 m SCALE E 50 m A 100 m B The diagram shows a field ABCDE. (a) Calculate the perimeter of the field ABCDE. ................................................. m [4] (b) Calculate angle ABD. Angle ABD = .................................................... [4] (c) (i) Calculate angle CBD. Angle CBD = .................................................... [2] (ii) The point C is due north of the point B. Find the bearing of D from B. .................................................... [2] (d) Calculate the area of the field ABCDE. Give your answer in hectares. [1 hectare = 10 000 m2] ...................................... hectares [4]

Mark scheme: 3(a) 530 4 B3 for [DE] = 130 m and [DC] = 80 m or B2 for [DE] = 130 m or [DC] = 80 m or M1 for 502 + 1202 or 1702 – 1502 3(b) 52.9 or 52.89… 4 100 2 + 150 2 − 120 2 M2 for 2 × 100 × 150 or M1 for 1202 = 1002 + 1502 – 2 × 100 × 150cos(…) 181 A1 for 0.603 or 0.6033…or 300 3(c)(i) 28.1 or 28.07… 2 15 M1 for cos = oe 17 3(c)(ii) 331.9 or 331.9… 2 FT 360 – their (c)(i) M1 for 360 – their (c)(i) oe 3(d) 1.5[0] or 1.498… nfww 4 1 M1 for × 50 × 120 oe 2 1 M1 for × 100 × 150sin(their(b)) oe 2 1 M1 for × 150 ×theirCD oe 2 1 or × 150 × 170 × sin their(c)(i) 2 If 0 scored, SC1 for dividing their area by 10 000

4 (a) The test scores of 14 students are shown below. 21 21 23 26 25 21 22 20 21 23 23 27 24 21 (i) Find the range, mode, median, and mean of the test scores. Range = ................................................... Mode = ................................................... Median = ................................................... Mean = ................................................... [6] (ii) A student is chosen at random. Find the probability that this student has a test score of more than 24. .................................................... [1] (b) Petra records the score in each test she takes. The mean of the first n scores is x. The mean of the first (n – 1) scores is (x + 1). Find the nth score in terms of n and x. Give your answer in its simplest form. .................................................... [3] (c) During one year the midday temperatures, t°C, in Zedford were recorded. The table shows the results. Temperature (t°C) 0 1 t G 10 10 1 t G 15 15 1 t G 20 20 1 t G 25 25 1 t G 35 Number of days 50 85 100 120 10 (i) Calculate an estimate of the mean. ............................................... °C [4] (ii) Complete the histogram to show the information in the table. 25 20 15 Frequency density 10 5 0 0 5 10 15 20 25 30 35 t Temperature (°C) [4]

Mark scheme: 4(a)(i) range = 7 1 mode = 21 1 median = 22.5 2 M1 for evidence of middle value mean = 22.7 or 22.71… 2 M1 for use of Σ x ÷ 14 4(a)(ii) 3 1 oe 14 4(b) x − n + 1 final answer 3 M2 for nx − ( n − 1)( x + 1) or M1 for ( n − 1)( x + 1) 4(c)(i) 16.6 or 16.60 to 16.61 nfww 4 M1 for 5, 12.5, 17.5, 22.5, 30 soi M1 for Σfx where x is in correct interval, including boundaries M1 dep on second M1 for Σ fx 50 + 85 + 100 + 120 + 10 4(c)(ii) Correct histogram 4 B1 for each correct block If 0 scored, SC1 for 5, 20, 24, 1 seen

5 NOT TO SCALE 1.2 m 3 m The diagram shows the surface of a garden pond, made from a rectangle and two semicircles. The rectangle measures 3 m by 1.2 m. (a) Calculate the area of this surface. ............................................... m2 [3] (b) The pond is a prism and the water in the pond has a depth of 20 cm. Calculate the number of liters of water in the pond. ............................................ liters [3] (c) After a rainfall, the number of liters of water in the pond is 1007. Calculate the increase in the depth of water in the pond. Give your answer in centimeters. ............................................... cm [3]

Mark scheme: 5(a) 4.73 or 4.730 to 4.731... 3 M2 for 3 × 1.2 + π × 0.6 2 oe 2 1 2 or M1 for π × 0.6 or × π × 0.6 2 or 3 × 1.2 5(b) 946 or 946.0 to 946.2... 3 M2 for their (a) × 0.2 × 1000 oe or M1 for their (a) × 0.2 or 20 implied by figs 946[0] to 9462 5(c) 1.28 or 1.29 or 1.284 to 1.290 3 (1007 − their (b)) ÷ 1000 M2 for × 100 their (a) oe 1007 − their (b) or for × 20 oe their (b) 1007 − their (b) or M1 for figs or their (a) 1007 figs their (a) 1007 −their (b) or for or their (b) 1007 × 20 oe their (b)

6 (a) (i) s = ut + at 2 2 Find s when t = 26.5, u = 104.3 and a = -2.2 . Give your answer in scientific notation, correct to 4 significant figures. s = ................................................... [4] 1 2 (ii) s = ut + at 2 Solve for a. a = ................................................... [3] (b) NOT TO SCALE (x – 1) cm (x – 2) cm (2x + 3) cm (x + 1) cm The difference between the areas of the two rectangles is 62 cm2. (i) Show that x 2 + 2x - 63 = 0 . [3] (ii) Factor x 2 + 2x - 63 . .................................................... [2] (iii) Solve the equation x 2 + 2x - 63 = 0 to find the difference between the perimeters of the two rectangles. ............................................... cm [2]

Mark scheme: 6(a)(i) 1.991 × 103 4 B3 for 1991 or 1.99 × 103 or 1.991… × 103 or B2 for 1990 or 1991. … OR M1 for 1 2 104.3 × 26.5 + × ( −2.2) × 26.5 oe 2 B1 for their seen value correctly rounded to 4 sf B1 for their seen value correctly converted into standard form 6(a)(ii) 2( s − ut ) 3 M1 for correct multiplication by 2 oe oe final answer 2 M1 for correct rearrangement to t isolate term with a M1 for correct division by t2 for 3 marks e.g. cannot have a fraction in denominator nor ÷t 2 in numerator 6(b)(i) (2 x + 3)( x − 1) − ( x + 1)( x − 2) = 62 M1 2 x 2 + 3 x − 2 x − 3 oe B1 or x 2 + x − 2 x − 2 oe x 2 + 2 x − 63 = 0 A1 Established with no errors or omissions 6(b)(ii) ( x + 9)( x − 7) 2 B1 for ( x + a )( x + b ) where ab = – 63 or a + b = 2 or for x ( x − 7) + 9( x − 7) or for x ( x + 9) − 7( x + 9) 6(b)(iii) 20 2 FT 2 × their positive root + 6 M1 for substituting their positive root into four lengths or for stating 2 x + 6

7 (a) The price of a book increases from $2.50 to $2.65 . Calculate the percentage increase. ................................................ % [3] (b) Scott invests $500 for 14 years at a rate of 1.5% per year simple interest. Calculate the value of his investment at the end of the 14 years. $ .................................................... [3] (c) Marie invests $500 for 14 years at a rate of 1.5% per year compound interest. Calculate the value of her investment at the end of the 14 years. $ ................................................... [2] (d) Pedro invests $500 at a rate of r % per year compound interest. At the end of 14 years the value of his investment is $586.80 . Find the value of r. r = ................................................... [3]

Mark scheme: 7(a) 6 nfww 3 2.65 − M2 for 2.50[ ×100] or for 2.50 2.65 × 100 2.50 2.65 or M1 for 2.50 7(b) 605 3 B2 for 105 500 × 1.5 × 14 M2 for + 500 oe 100 1.5 or M1 for 500 × [× 14] oe 100 7(c) 616 or 615.88 2 14  1.5  M1 for 500 ×  1 +  oe  100  7(d) 1.15 or 1.149 to 1.150 3 586.80 M2 for 14 or better 500 or M1 for 500 × x14 = 586.80 oe or better

8 (a) Solve the equation 2x 2 + 3x - 4 = 0 . Show all your working and give your answers correct to 2 decimal places. x = ........................ or x = ........................ [4] (b) Solve the following equations. (i) x - 1 = 1 - 2 x x = .................................................... [2] (ii) 5 x - 3 = 1 x = ................................................... [1]

Mark scheme: 8(a) 2 B2 2 −±3 (3) − 4(2)( −4) B1 for ( 3 ) − 4 ( 2) ( − 4 ) ) 2 × 2 −+3 q −−3 q B1 for or 2 × 2 2 × 2 –2.35 and 0.85 final answer cao B2 B1 for each If 0 scored, SC1 for − 2.4 or – 2.351 to – 2.350 and 0.9 or 0.850 to 0.851 or for – 0.85 and 2.35 or for – 2.35 and 0.85 seen but not final answers 8(b)(i) 4 2 B1 for x + 2 x = 1 + 1 or better 9 8(b)(ii) 3 1

9 f (x) = 7x - 2 g (x) = x 2 + 1 h (x) = 3x (a) Find g(h(2)). .................................................... [2] (b) Find f – 1(x). f – 1(x) = ................................................... [2] (c) g (g (x)) = ax 4 + bx 2 + c Find the values of a, b, and c. a = ................................................... [3] b = ................................................... [3] c = ................................................... [3] (d) Find x when h(f(x)) = 81. x = ................................................... [3]

Mark scheme: 9(a) 82 2 M1 for (3x)2+1 soi by (32)2+1 or g(9) isw 9(b) x + 2 2 y 2 final answer M1 for y + 2 = 7x or = x − or 7 7 7 x = 7y – 2 9(c) [a =] 1, [b =] 2, [c =] 2 3 B2 for x 4 + x 2 + x 2 + 1 + 1 or M1 for ( x 2 + 1) 2 + 1 9(d) 6 3 M2 for 7x – 2 = 4 oe or M1 for 3x = 81 soi f(x) = 4 7 or for 37 x −=2 81 or better

10 The volume of each of the following solids is 1000 cm3. Calculate the value of x for each solid. (a) A cube with side length x cm. x = ................................................... [1] (b) A sphere with radius x cm. x = .................................................. [3] (c) NOT TO x 5cm SCALE x cm A cone with radius x cm and sloping edge x 5cm. x = .................................................. [4] (d) x NOT TO cm 2 SCALE x cm 27x cm 2 A prism with a right-angled triangle as its cross-section. x = .................................................. [4] Question 11 is printed on the next page.

Mark scheme: 10(a) 10 1 10(b) 6.2[0] or 6.203 to 6.204 3 4 M2 for [x3 = ] 1000 ÷ π oe or better 3 4 3 or M1 for π x = 1000 3 10(c) 7.82 or 7.815 to 7.816 4 3 1 B3 for [ x = ]1000 ÷ π ÷ 2 oe or better 3 2 or M1 for x 5 − x 2 soi by 4x2 or 2x ( ) 1 2 M1dep for π × x × their h [ = 1000] 3 27 3 x oe or10(d) 6 23 or 6.67 or 6.666 to 6.667 4 B3 for [ x 3 = ]1000 ÷ = 10 8 2 or better 1 x 27 x or M2 for × x × × =1000 oe 2 2 2 1 x or M1 for × x × 2 2 If 0 scored, SC2 for answer 5.29 or 5.291..

11 Brad traveled from his home in New York to Chamonix. • He left his home at 16 30 and traveled by taxi to the airport in New York. This journey took 55 minutes and had an average speed of 18 km/h. • He then traveled by plane to Geneva, departing from New York at 22 15. The flight path can be taken as an arc of a circle of radius 6400 km with a sector angle of 55.5°. The local time in Geneva is 6 hours ahead of the local time in New York. Brad arrived in Geneva at 11 25 the next day. • To complete his journey, Brad traveled by bus from Geneva to Chamonix. This journey started at 13 00 and took 1 hour 36 minutes. The average speed was 65 km/h. The local time in Chamonix is the same as the local time in Geneva. Find the overall average speed of Brad’s journey from his home in New York to Chamonix. Show all your working and give your answer in km/h. .......................................... km/h [11]

Mark scheme: 11 [Total time =]16 h 6 min or 16.1 h 2 B1 for 22 h 6 min or 22.1h or 966 mins If 0 scored, SC1 for 9 h 41 min [Distance to airport in New York 2 M1 for 18 × 55 =] 16.5 [Arc length =] 3 55.5 M2 for × 2 × π × 6400 6200 or 6199 to 6200. … 360 55.5 or M1 for or 2 × π × 6400 360 [Distance Geneva to Chamonix = ] 2 M1 for 65 × 1.6 or 65 × 96 oe 104 392 to 393 2 6316 to 6322.4 M1 for their 1 6.1 Must be correct value in numerator