Cambridge IGCSE Mathematics (US) 0444 — 2020 May/June Paper 2 · Variant 1

1 Rectangle A measures 3 cm by 8 cm. 8 cm NOT TO 3 cm A SCALE Five rectangles congruent to A are joined to make a shape. NOT TO SCALE Work out the perimeter of this shape. ............................................ cm [2]

Mark scheme: Question Answer Marks Partial Marks 1 86 2 M1 for correct method to find the perimeter e.g. (8 + 3) × 2 × 5 – 3 × 8 If 0 scored, SC1 for answer 98

2 Find the highest odd number that is a factor of 60 and a factor of 90. ................................................. [1]

Mark scheme: 2 15 1

3 Mrs Salaman gives her class two mathematics tests. The scatter diagram shows information about the marks each student scored. 70 60 50 40 Test 2 30 20 10 0 0 10 20 30 40 50 60 70 Test 1 (a) Write down the highest mark scored on test 1. ................................................. [1] (b) Write down the type of correlation shown in the scatter diagram. ................................................. [1] (c) Draw a line of best fit on the scatter diagram. [1] (d) Hamish scored a mark of 40 on test 1. He was absent for test 2. Use your line of best fit to find an estimate for his mark on test 2. ................................................. [1]

Mark scheme: 3(a) 66 1 3(b) Positive 1 3(c) Ruled line of best fit 1 3(d) 46 to 50 1 FT their line of best fit if a positive gradient

4 A bag contains blue, red, yellow, and green balls only. A ball is taken from the bag at random. The table shows some information about the probabilities. Color Blue Red Yellow Green Probability 0.15 0.2 0.43 (a) Complete the table. [2] (b) Abdul takes a ball at random and replaces it in the bag. He does this 200 times. Find how many times he expects to take a red ball. ................................................. [1]

Mark scheme: 4(a) 0.22 oe 2 M1 for 0.15 + 0.2 + ? + 0.43 = 1 or better 4(b) 40 1

5 (a) The nth term of a sequence is 60 - 8n . Find the largest number in this sequence. ................................................. [1] (b) Here are the first five terms of a different sequence. 12 19 26 33 40 Find an expression for the nth term of this sequence. ................................................. [2]

Mark scheme: 5(a) 52 1 5(b) 7n + 5 oe final answer 2 B1 for 7n + a or bn + 5 b ≠ 0

6 The diagram shows a trapezoid. ( 92- 3x)° NOT TO SCALE ( 70+ 5x)° Work out the value of x. x = ................................................ [3]

Mark scheme: 6 9 3 M2 for 162 + 2x = 180 or better or M1 for 92 – 3x + 70 + 5x = 180 oe

7 234 = 2 # 3 2 # 13 1872 = 2 4 # 3 2 # 13 234 # 1872 = 438048 Use this information to write 438 048 as a product of its prime factors. ................................................. [1]

Mark scheme: 7 25 × 34 × 132 1

8 Work out 2 - # . 3 8 25 Give your answer as a fraction in its simplest form. ................................................. [4]

Mark scheme: 8 7 4 M2 for correct method for common denominator 20 56 21 e.g. − 24 24 7 or B1 for 3 35 6 M1 for their × 24 25

9 Factor completely. (a) 21a 2 + 28ab ................................................. [2] (b) 20x 2 - 45y 2 ................................................. [3]

Mark scheme: 9(a) 7a(3a + 4b) final answer 2 B1 for partial factoring 7(3a2 + 4ab) or a(21a + 28b) 9(b) 5(2x + 3y)(2x – 3y) final 3 B2 for (2x + 3y)(2x – 3y) answer or (10x + 15y)(2x – 3y) or (2x + 3y)(10x – 15y) or B1 for 5(4x2 – 9y2)

10 E NOT TO SCALE D 24° F y° O 110° C x° A B Points A, B, C, D, E and F lie on the circle, center O. Find the value of x and the value of y. x = ................................................ y = ................................................. [2]

Mark scheme: 10 [x =] 55 2 B1 for each [y = ] 24

11 320 280 240 200 Distance (meters) 160 120 80 40 0 0 10 20 30 40 50 60 70 80 90 100 Time (seconds) The diagram shows the distance traveled by a cyclist during the first 100 seconds of her journey. (a) Work out her average speed. ........................................... m/s [1] (b) Find an estimate of the speed of the cyclist 60 seconds after she started. ........................................... m/s [3]

Mark scheme: 11(a) 3.2 1 11(b) 2.5 to 3.5 3 B1 for correct tangent at time = 60 M1 for rise/run dep on good attempt at tangent at time = 60

12 19 11 13 10 12 19 14 15 19 13 The list shows 10 test scores. Find (a) the mode, ................................................. [1] (b) the median. ................................................. [2]

Mark scheme: 12(a) 19 1 12(b) 13.5 2 M1 for at least first 6 values in correct order or B1 for 13 and 14 indicated

13 C B 12 cm NOT TO SCALE 9 cm 12 cm D E A C lies on a circle with diameter AD. B lies on AC and E lies on AD such that BE is parallel to CD. AB = 12 cm, CD = 12 cm and BE = 9 cm. Work out the radius of the circle. ............................................ cm [5]

Mark scheme: 13 10 5 12 M2 for 12 × = [ AC ] oe 9 12 or M1 for scale factor oe soi 9 Then Pythagoras method: and M2 for 16 2 + 12 2 [÷ 2] or ( theirAC ) 2 + 12 2 [÷ 2] or M1 for AD 2 = 1 6 2 + 12 2 2 2 2 or AD = ( theirAC ) + 12 OR alternative trigonometry method e.g. 12 M1 for tan E = 9 16 and M1 for AD = sin their 53.1

14 (a) f ( x) = 4 x + 3 g ( x) = 5 x - 4 f g ( x) = 20 x + p ` j Find the value of p. p = ................................................ [2] 5x - 1 (b) h ( x) = 3 Find h -1 ( x) . h -1 ( x) = ................................................. [3]

Mark scheme: 14(a) [p = ] –13 2 M1 for 4(5x – 4) + 3 or better 14(b) 3 x + 1 3 3 y + 1 1 3 x M2 for x = , 5y = 3x + 1 or y − = 5 5 5 5 5 y –1 1 5 x M1 for x = , 3y = 5x − 1 or y + = 3 3 3

15 A NOT TO Q SCALE x° B R S Q, R and S are points on the circle. QS is a diameter. AB is a tangent to the circle at Q. Angle BQR = x°. Show that angle QSR = x°. Give a reason for each step of your work. ............................................................................................................................................................. ............................................................................................................................................................. ............................................................................................................................................................. [3]

Mark scheme: 15 Angle SQR = 90 – x and 1 tangent – radius Angle SRQ = 90 and semi- 1 circle Angle QSR = x and angles in 1 a triangle

16 m varies inversely as the square of ( p - 1 ) . When p = 4 , m = 5 . Find m when p = 2. m = ................................................ [3]

Mark scheme: 16 45 3 45 M2 for m = ( p − 1) 2 k M1 for m = or better ( p − 1) 2

17 (a) (i) m = 7 Find 3m. [1] f p p (ii) The magnitude of the vector is 13. e12o Find the positive value of p. p = ................................................ [2] (b) E A B p NOT TO SCALE O C q OABC is a parallelogram. OA = p and OC = q . E is the midpoint of AB. Find OE in terms of p and q. OE = ................................................ [2]

Mark scheme: 17(a)(i)  15  1    21  17(a)(ii) 5 2 M1 for p2 + 122 = 132 oe 17(b) 1 2  1 p + q M1 for a correct route or for AE = q 2 2

18 Simplify 250 + 40 . ................................................. [2]

Mark scheme: 18 7 10 2 B1 for 5 10 or 2 10 seen

19 F G NOT TO SCALE E H B C 5 cm A 30 cm D The diagram shows a solid cuboid ABCDEFGH of length 30 cm and width 5 cm. The volume of the cuboid is 600 cm 3. Find the total surface area of the cuboid. ......................................... cm2 [4]

Mark scheme: 19 580 4 600 M1 for implied by 4 seen 30 × 5 M2 for 2 × (30 × 5 + 30 × their 4 + 5 × their 4) or M1 for 2 different correct rectangles

20 Simplify. x - 8 - ax + 8 a x 2 - 15x + 56 ................................................. [5]

Mark scheme: 20 1 − a 5 B2 for ( x − 8 )(1 − a ) oe oe x − 7 or B1 for x − 8 − a ( x − 8) or x (1 − a ) − 8(1 − a ) B2 for ( x − 8 )( x − 7 ) or B1 for ( x + p )( x + q ) where pq = 56 or p + q = –15 or for x ( x − 7) − 8( x − 7) or x ( x − 8) − 7( x − 8)

21 The area of a regular hexagon with side length 8 cm is k 3 cm 2. Find the value of k. k = ............................................... [3]

Mark scheme: 21 96 3 1 M1 for [6] × × 8 × 8sin60 2 3 B1 for seen 2