Cambridge IGCSE Mathematics (US) 0444 — 2017 May/June Paper 2 · Variant 1

1 Simplify. (x2) 5 .............................................. [1]

Mark scheme: Question Answer Marks Part marks 1 x10 1

2 The thickness of one sheet of paper is 8 # 10-3 cm. Work out the thickness of 500 sheets of paper. ........................................ cm [1]

Mark scheme: 2 4 1

3 Write 23.4571 correct to (a) 4 significant digits, .............................................. [1] (b) the nearest 10. .............................................. [1]

Mark scheme: 3(a) 23.46 cao 1 3(b) 20 cao 1

4 The table shows the temperatures in five places at 10 am one day in January. Place Temperature (°C) Helsinki -7 Chicago -10 London 3 Moscow -4 Bangkok 26 (a) Which place was the coldest? .............................................. [1] (b) At 2 pm the temperature in Helsinki had increased by 4 °C. Write down the temperature in Helsinki at 2 pm. .........................................°C [1]

Mark scheme: 4(a) Chicago 1 4(b) – 3 1

5 Factor completely. 12n2 - 4mn .............................................. [2]

Mark scheme: 5 4n(3n – m) final answer 2 B1 for 4(3n2 – mn) or n(12n – 4m) or 2n(6n – 2m) or 2(6n2 – 2mn)

6 (a) 2 r = 16 Find the value of r. r = ....................................... [1] (b) 3 t = 5 3 Find the value of t. t = ........................................ [1]

Mark scheme: 6(a) − 4 1 6(b) 1 1 or 0.2 5

7 Work out 1 + . 3 7 Give your answer as a mixed number in its simplest form. .............................................. [3]

Mark scheme: 7 8 3 50 8 29 29 2 cao M2 for or or or 21 21 121 21 121 14 k ( or35k ) 15k or M1 for + 21k 21k

8 Simon has two boxes of cards. In one box, each card has one shape drawn on it that is either a triangle or a square. In the other box, each card is colored either red or blue. Simon picks a card from each box at random. The probability of picking a triangle card is t. The probability of picking a red card is r. Complete the table for the cards that Simon picks, writing each probability in terms of r and t. Event Probability Triangle and red Square and red (1 - t) r Triangle and blue Square and blue [3]

Mark scheme: 8 3 B1 for each rt (1 – t) r (1 − r)t oe (1 – r)(1 – t) oe

9 h varies directly as the square root of p. h = 6 when p = 4. 1 Find h when p = . 4 h = ....................................... [3]

Mark scheme: 9 1.5 oe 3 M1 for h = k p oe M1 for h = their k p 6 h or M2 for = oe 4 1 4

10 y 5 4 3 2 1 x 0 1 2 3 4 5 6 –1 By shading the unwanted regions of the grid, find and label the region R that satisfies the following four inequalities. y G 2 y H 1 y G 2x - 1 y G 5 - x [3]

Mark scheme: 10 Correct region identified 3 0 1 1 2 2 3 2 1 2 1 R SC1 for

11 The two barrels in the diagram are mathematically similar. NOT TO SCALE 75 cm h cm The smaller barrel has a height of h cm and a capacity of 64 liters. The larger barrel has a height of 75 cm and a capacity of 125 liters. Work out the value of h. h = ....................................... [3]

Mark scheme: 11 60 3 125 64 M2 for 75 ÷÷ 3 or 75 × 3 64 125 125 64 or M1 for 3 soi or 3 soi 64 125  h 3 64 oe or  =  75  125

12 A line has slope 5. M and N are two points on this line. M is the point (x, 8) and N is the point (k, 23). Find an expression for x in terms of k. x = ....................................... [3]

Mark scheme: 12 kk − 3 or −+3 k 3 23 − 8 M1 for 5 = oe k − x M1 for 5 ( k − x ) = 23 − 88 or better 23 − 8 e.g. [ x = ] k −− 5

13 D C NOT TO 5 cm SCALE 3 cm A 4 cm B Angle BAD = angle DBC. Work out BC. BC = .............................. cm [3]

Mark scheme: 13 3 15 3 33 3.75 or 34 or 4 M2 for 5 × 44 44 5 or M1 for = oe 33 BC

14 The diagram shows a regular octagon joined to an equilateral triangle. NOT TO SCALE x° Work out the value of x. x = ....................................... [3]

Mark scheme: 14 165 3 3600 360 M2 for + oe 8 3 360 or M1 for [eexterior anglle of octagonn =] or 8 360 [exterior anggle of trianglle =] oee 3

15 Simplify. (a) 20 + 125 ............................................. [2] (b) 2 + 2 3 2 ^ h ............................................. [2]

Mark scheme: 15(a) 77 5 2 B1 for 2 5 or 5 5 15(b) 114 + 4 6 oe final answer 2 B1 for 3 corrrect from 2 2 ( 2 ) + 2 × 2 3 + 2 × 2 3 + (2 3 ) or better

16 Six students revise for a test. The scatter diagram shows the time, in hours, each student spent revising and their mark in the test. 50 45 40 Mark 35 30 25 0 1 2 3 4 5 6 7 8 9 10 Time (hours) (a) The data for two more students is shown in the table. Time (hours) 4.5 6.5 Mark 33 35 Plot these two points on the scatter diagram. [1] (b) What type of correlation is shown on the scatter diagram? .............................................. [1] (c) Draw a line of best fit on the scatter diagram. [1] (d) Another student spent 5.5 hours revising. Estimate a mark for this student. .............................................. [1]

Mark scheme: 16(a) Points plotted at (4.5, 33) and 1 (6.5, 35) 16(b) Positive 1 16(c) Correct ruled line 1 16(d) 33.5 to 37.5 1FT FT from their line provided positive gradient

17 (a) Write down the amplitude and period of the function f (x) = cos KK OO. 2 3 L P Amplitude = ........................................... Period = ........................................... [2] J N 2 2 2 (b) The graph of y = x + x + 3 is mapped onto the graph of y = x + ux + v by the translation KK OO. 0 L P Find the value of u and the value of v. u = ....................................... v = ....................................... [2]

Mark scheme: 17(a) 1 2 B1 for each [amplitude = ] or SC1 for answers reversed 2 [period = ] 1080 17(b) [u = ] – 3 2 2 + ( x − 2) + 3 or better M1 for ( x − 2 ) [v = ] 5 If zero scored, SC1 for u = 5 and v = 9

18 The diagram shows a parallelogram OCEG. C D E NOT TO SCALE B F H b O A G a O is the origin, OA = a and OB = b . BHF and AHD are straight lines parallel to the sides of the parallelogram. OG = 3 OA and OC = 2 OB . (a) Write the vector HE in terms of a and b. HE = .................................. [1] (b) Complete this statement. a + 2b is the position vector of point ......................... [1] (c) Write down two vectors that can be written as 3a - b. ............................... and ............................... [2]

Mark scheme: 18(a) 2a + b 1 18(b) D 1 18(c) 2 B1 for each CF and BG

19 ABCD is a rhombus with side length 10 cm. A NOT TO 10 cm SCALE D 60° B C Angle ADC = 60°. DAC is a sector of a circle with center D. BAC is a sector of a circle with center B. The area shaded is pr+ q 3 cm2 . ^ h Find the value of p and the value of q. p = ....................................... q = ....................................... [4]

Mark scheme: 19 100 4  60 2   1 2   [p = ] oe M3 for 2 ×  × π × 10  −  × 10 × sin60   3  360   2   [q = ] − 50  1  2 60 2 or M2 for × 10 × sin60 and [ 2 ×] × π × 10    2  360  1  2 60 2 or M1 for × 10 × sin60 or [ 2 ×] × π × 10    2  360 3 or sin60 = 2

20 The diagram shows a fair spinner. 1 6 3 4 3 Anna spins it twice and adds the scores. (a) Complete the table for the total scores. Score on first spin 1 3 3 4 6 1 2 4 4 5 7 3 4 6 6 7 9 Score on 3 4 6 6 7 9 second spin 4 6 [1] (b) Write down the most likely total score. .............................................. [1] (c) Find the probability that Anna scores (i) a total less than 6, .............................................. [2] (ii) a total of 3. .............................................. [1]

Mark scheme: 20(a) 1 5 7 7 8 10 7 9 9 10 12 20(b) 7 1 20(c)(i) 7 2FT their 7 or 0.28 or 28% FT 25 25 k B1 for 25 2 6 If zero scored, SC1 for or if no values in the 5 15 bottom two rows of the table 20(c)(ii) 0 1FT their 0 FT 25

21 (a) D NOT TO SCALE v° C 35° A X u° B A, B, C and D are points on the circle. AD is parallel to BC. The chords AC and BD intersect at X. Find the value of u and the value of v. u = ....................................... v = ....................................... [3] (b) NOT TO SCALE 210° H p° O F G F, G and H are points on the circle, center O. Find the value of p. p = ....................................... [2]

Mark scheme: 21(a) [u =] 35 1 [v =] 110 2 B1 for ACB or ADB = 35 21(b) 75 2 B1 for 150 360 − 210 or M1 for 2

22 Write as a single fraction in its simplest form. x 2 - 3x (a) 2 x - 9 .............................................. [3] 3 2 (b) + x - 4 2x + 5 .............................................. [3]

Mark scheme: 22(a) x 3 B1 for x(x – 3) final answer x + 3 B1 for (x – 3)(x + 3) 22(b) 8 x + 7 3 B1 for common denominator of (x – 4)(2x + 5) final answer M1 for 3(2x + 5) + 2(x – 4) oe with an attempt to ( x − 4)( 2 x + 5) expand the brackets