June 2009 Paper 4

Tania went to Italy. She changed £325 into euros Tania went to Italy. She changed £325 into euros. The exchange rate was £1 = 1.68 euros a) Change £325 into euros 325 x 1.68 = 546

1) When she came home she changed 117 euros into pounds. The new exchange rate was £1 = 1.5 euros. Change 117 euros into pounds 117 ÷ 1.5 = 78

2)a)

2)b) Reflection in the y axis (1 mark) (1 mark)

3) The nth term of a number sequence. is n2 + 1 3) The nth term of a number sequence is n2 + 1. Write down the first three terms of the sequence. n=1 12 + 1 = 2 n=2 22 + 1 = 5 n=3 32 + 1 = 10

4)a)

4)b) Positive correlation or the taller the sheep, the longer it is

4)c) 108cm

5) Julie buys 19 identical calculators. The total cost is £143.64 Work out the total cost of 31 of these calculators £143.64 ÷ 19 = £7.56 £7.56 x 31 = £234.36

6) F = 1.8C + 32 Work out the value of F when C = −8 F = 1.8 x -8 + 32 (1 mark) F = -14.4 + 32 F = 17.6 (1 mark)

6) F = 1.8C + 32 b) Work out the value of C when F = 68 68 = 1.8C + 32 (1 mark) 36 = 1.8C C = 36 ÷ 1.8 C = 20 (1 mark)

7) The bearing of a boat R from boat P is 060°. The bearing of boat R from boat Q is 310°. R 310° 60°

8)a) There are some sweets in a bag. 18 of the sweets are toffees. 12 of the sweets are mints. Write down the ratio of the number of toffees to the number of mints. Give your ratio in its simplest form. 18 : 12 3 : 2

8)b) There are some oranges and apples in a box 8)b) There are some oranges and apples in a box. The total number of oranges and apples is 54. The ratio of the number of oranges to the number of apples is 1 : 5 Work out the number of apples in the box. 1 + 5 = 6 54 ÷ 6 = 9 5 x 9 = 45 45 apples in the box

9). The equation x3 + 20x = 71 has a solution between 2 and 3 9) The equation x3 + 20x = 71 has a solution between 2 and 3. Use a trial and improvement method to find this solution. Give your answer correct to 1 decimal place. Answer = 2.6 (1 mark) x x3 + 20x 2.5 65.625 too low (1 mark) 2.6 69.576 2.7 73.683 too high 2.65 71.609625

10) Leave in your construction lines!

11) Tarish says, ‘The sum of two prime numbers is always an even number’. He is wrong. Explain why. Give a counter example e.g. 2 + 3 = 5

0<t<6 15 6<t<12 25 12<t<18 20 18<t<24 12 12) Sethina recorded the times, in minutes, taken to repair 80 car tyres. Information about these times is shown in the table. Calculate an estimate for the mean time taken to repair each car tyre. Time Frequency 0<t<6 15 6<t<12 25 12<t<18 20 18<t<24 12 24<t<30 8

0<t<6 15 3 45 6<t<12 25 9 225 12<t<18 20 300 12) Sethina recorded the times, in minutes, taken to repair 80 car tyres. Estimated Mean = 1038 ÷ 80 = 12.975 Time Frequency Midpoint 0<t<6 15 3 45 6<t<12 25 9 225 12<t<18 20 300 18<t<24 12 21 252 24<t<30 8 27 216 80 1038

13) 8cm Perimeter = (π x 8 ÷ 2) + 8 = 12.568 + 8 = 20.57cm (2dp)

14) a) Simplify a × a × a a3 b) Expand 5(3x – 2) 15x – 10 c) Expand 3y(y + 4) 3y2 + 12y d) Expand and simplify 2(x – 4) + 3(x + 2) 2x – 8 + 3x + 6 5x – 2 e) Expand and simplify (x + 4)(x – 3) x2 + 4x – 3x – 12 x2 + x – 12

15) Work out 4.6 + 3.85 3.22 – 6.51 Write down all the numbers on your calculator display. 8.45 = 2.26541555 3.73

16)a) Simplify t6 × t2 t8 b) Simplify m8 m5 m3 c) Simplify (2x)3 8x3 d) Simplify 3a2h × 4a5h4 12a7h5

17) AB2 = 92 – 62 AB2 = 81 – 36 AB2 = 45 AB = 45 AB = 6.71(3sf)

18)a) The box plot gives information about the distribution of the weights of bags on a plane. Jean says the heaviest bag weighs 23 kg. She is wrong. Explain why. The heaviest bag weighs 29kg

18) b) Write down the median weight. 17 c) Work out the interquartile range of the weights. 23 – 10 = 13

18) There are 240 bags on the plane. d) Work out the number of bags with a weight of 10 kg or less. 25% x 240 = 60

19)a) Toby invested £4500 for 2 years in a savings account 19)a) Toby invested £4500 for 2 years in a savings account. He was paid 4% per annum compound interest. How much did Toby have in his savings account after 2 years? 4% of 4500 = 180 After 1 year = 4500 + 180 = 4680 4% of 4680 = 187.20 After 2 years = 4680 + 187.20 = £4867.20

19)b) Jaspir invested £2400 for n years in a savings account 19)b) Jaspir invested £2400 for n years in a savings account. He was paid 7.5% per annum compound interest. At the end of the n years he had £3445.51 in the savings account. Work out the value of n. 2400 x (1.075)n = 3445.51 n = 5

20)a) cos x = 5 8 x = cos-1(5/8) x = 51.3178... x = 51.3°

20)b) tan 40 = y 12.5 y = tan 40 x 12.5 = 10.4887... = 10.5cm (1dp)

21)a) A sample, stratified by the language studied and by gender, of 50 of the 258 students is taken. Work out the number of male students studying Spanish in the sample. 26 x 50 = 5.03... 258 Therefore 5 students in the sample Language studied German French Spanish Male 45 52 26 Female 25 48 62

21)b) A sample, stratified by the language studied and by gender, of 50 of the 258 students is taken. Work out the number of female students in the sample. 25+48+62 x 50 = 26.16... 258 26 (or 27) students in the sample Language studied German French Spanish Male 45 52 26 Female 25 48 62

Prove that (3n + 1)2 – (3n –1)2 is a multiple of 4, for all positive integer values of n. = (9n2 + 3n + 3n + 1) – (9n2 – 3n – 3n + 1) = 9n2 + 6n + 1 – 9n2 + 6n – 1 = 12n 12n is a multiple of 4 or 12n = 4n x 3

23)a) AB = -a + b

23)b) P is the point on AB such that AP:PB = 3:2. Show that OP = 1/5(2a + 3b) OP = OA + AP = a + 3/5(-a+b) = a – 3/5a + 3/5b = 2/5a + 3/5b = 1/5(2a + 3b)

24) Area of triangle ABC = ½ x 6 x 6 x sin 60 = 15.588... Area of sector APQ = 60/360 x π x 32 = 4.712... Shaded Area = 15.588... – 4.712... = 10.9cm2 (3sf)

25) x2 – 8x + 15 2x2 – 7x – 15 = (x – 3)(x – 5) (2x + 3)(x – 5) = (x – 3) (2x + 3)

26). Phil has 20 sweets in a bag. 5 of the sweets are orange 26) Phil has 20 sweets in a bag. 5 of the sweets are orange. 7 of the sweets are red. 8 of the sweets are yellow. Phil takes at random two sweets from the bag. Work out the probability that the sweets will not be the same colour. P(not same colour) = 1 – P(same colour) P(same colour) = (5/20 x 4/19) + (7/20 x 6/19) + (8/20 x 7/19) = 0.311 P(not same colour) = 1 – 0.3105 = 0.689