1. Mathematics Intermediate Tier Paper 2 Summer 2001 (2 hours)

2. 2 (a) Simplify: 3x - 2y - x +5y 1. Write down the next two terms of the following sequence. 21, 19, 15, 9,,, (b) Expand 5(x-2) (c) Find the value of 4c - 3d when c=-2 and d=6 1-9 = 2x + 3y = 5x - 10 = 4 x -2 – 3 x 6 = -8 - 18 = - 26

3. 3. Fifty people were asked how many pets they owned. The results were as follows. Number of pets owned012345 Number of people101215742 (a)What is the probability that a randomly chosen person from this group has exactly 3 pets? (b) How many pets have these people got altogether? = 7 50 = 0 x 10 + 1 x 12 + 2 x 15 + 3 x 7 + 4 x 4 + 5 x 2 = 0 + 12 + 30 + 21 + 16 + 10 = 89 anifail animals

4. 4. The Williams family go on holiday to Mallorca, when the exchange rate is £1 = 286 pesetas. (a)They exchange £350 into pesetas. How many pesetas did they get? (b) Whilst on holiday they bought 30 postcards at 85 pesetas each and stamps for the postcards at 70 pesetas each postcard. Calculate how much in £s, correct to the nearest penny, this cost them. = 350 x 286 = 100100 pesetas = 30 x( 85 + 70) = 4650 peseta = 4650 ÷ 286 = £16.26

5. 5. Draw, on the grid below, an enlargement of the given shape, using a scale factor of 3.

6. 6. In a survey, the type of central heating used by 240 households was as shown in the table. Type of central heatingNumber of households Solid Fuel46 Gas54 Electricity30 Oil90 None20 Draw a pie chart to illustrate the results. You should show how you calculate the angles of your pie chart. Total 240 Solid fuel = 46 x 360 240 = 23 x 3 = 69° Gas = 54 x 360 240 = 27 x 3 = 81° Electricity = 45° Oil = 135° None = 30°

7. 7. A water tank, in the shape of a cuboid, contains 56 000cm³ of water. The base of the tank measures 62cm by 35cm. (a)Calculate the depth, in cm, of the water in the tank. (b) Given that 1 gallon = 4.54 litres, calculate the number of gallons of water in the tank. Volume of a cuboid = length x width x height 56000 = 62 x 35 x d d = 56000 62 x 35 d = 56000 2170 d =25.8065cmd = 25.8cm Number of gallons = 56 4.54 56000cm³ = 56000 ÷ 1000 = 56 litres 1 litre = 1000cm³ = 12.3348gal= 12.3 gal (1 decimal place)

8. 8. Colin’s gas bill for the period April 1 st – June 30 th is calculated from the following information. Number of units used198 Charge per unit43.8p Number of says in this period91 Service charge per day 13.39p VAT5% Showing all your working, find the total cost of the gas including VAT. Cost of units = 198 x 43.8 = 8672.4p Cost of service charge = 91 x 13.39= 1218.49p Total = 8672.4 + 1218.49 = 9890.89p= £98.9089 TAW VAT = 5 x 98.9089 100 = 4.945445 Total cost of bill = £98.9089 + £4.945445 = £103.854345= £103.85

9. £26 + VAT at 17 ½ % £30 Including VAT Which price offer is the cheaper and by how much? 9. = 17 ½ x 26 100 = £4.55 Total cost = 26 + 4.55 = £30.55 £30 including VAT is the cheapest by 55pence

10. 10. The graph opposite shows Gary’s journey by car from his home to a services area, where he stops for a while before returning home. (a)How far is the services area from Gary’s home? (b) How long did Gary stop at this services area? (c) Use the graph to find Gary’s average speed, in m.p.h., for his return journey home. = 57 miles = 46 minutes Speed = Distance Time = 57 miles 72 minutes = 57 miles 72÷60 hour = 57 1.2 = 47.5 miles / hour

11. 10.00am11.00am12.00pm 1.00pm2.00pm Time 0 10 20 30 40 50 60 70 Distance from Gary’home (miles)

12. 11. The assessment for a mathematics exmaination consists of two parts, namely, coursework marked out of 50, and written papers, marked out of 200. The marks for ten pupils are given in the table. Coursework mark530154492239263327 Written papers mark2212064186177614311214892 (a)On the graph paper, draw a scatter diagram to display these results. (b) What type of correlation does your scatter diagram show? Positive (c) The mean coursework mark for the pupils is 25 and the mean mark for the written paper is 98. Draw a line of best fit on your scatter diagram.

13. 10203040 Coursework 20 40 60 80 100 120 140 160 180 0 Written Papers

14. (d) Another pupil completed the coursework and was given a mark of 19, but was absent from the written papers examination. Use your line of best fit to estimate the mark of the written papers for this pupil. = 76

15. 12. The speeds of 120 cars on a stretch of motorway were measured and the following results were obtained. Speed, s (m.p.h.)Number of cars 30 ≤ b < 406 40 ≤ b < 5024 50 ≤ b < 6030 60 ≤ b < 7045 70 ≤ b < 8012 80 ≤ b < 903 (a)Write down the modal class. (b) On the graph paper, draw a grouped frequency diagram for the data. = 60 ≤ b < 70

16. (c) find an estimate for the mean speed of the cars. 0102030405060708090 10 20 30 40 50 Speed (m.p.h.) Number of cars = (35 x 6 + 45 x 24 + 30 x 55 + 45 x 65 + 12 x 75 + 3 x 85) ÷ 120 = (210 + 1080 + 1650 + 2925 + 900 + 255) ÷ 120 = 7020 ÷ 120 = 58.5 m.p.h.

17. 13. The shape shown below is made up of two semicircles. The diameter of the smaller semicircle is 8.6cm. C is the mid-point of the diameter of the larger semicircle. Stating clearly the units of your answers, calculate the perimeter of the shape, giving your answer to an appropriate degree of accuracy. 8. 6cm (b) the area of the shape, giving your answer to the nearest whole number. C = π d Small circle C = 3.142 x 8.6 = 27.0212 Large circle C = 3.142 x 17.2 = 54.0424 ½ circle = 13.5106 ½ circle = 27.0212 Perimeter = 8.6 + 13.5 + 27.0 = 49.1cm A = π r² Small circle = 3.142 x 4.3x4.3 Large circle = 3.142 x 8.6x8.6 = 58.09558 = 232.38232 ½ circle = 29.04779cm² ½ circle = 116.19116cm² Total area = 29.05 + 116.19= 145.24cm² C

18. 14. On April 1 st Marcus owed £250 on his credit card account. The credit card company requires Marcus to pay at least 10% of the balance on the 20 th of each month. The company charges interest at 2% on what the balance is on the 28 th of every month. Marcus pays the minimum payment on time every month. Write down full details of his account up to May 31 st. April 1 st £250 April 20 th May 31 st £210.68 10% = £25 250 -25£225 April 28 th 2% of £225 = £4.50£229.50 May 1st£229.50 May 20 th 10% = £22.95 £229.50 - £22.95£206.55 May 28 th 2% of £206.55= £4.13£210.68

19. 15. (a) Expand 2x(x² + 3). (b) Expand and simplify 4(3x – 1) + 3(x – 5). = 2x³ + 6x = 12x – 4 + 3x - 15 = 15x - 19

20. 16. Use your calculator to find the value of √845.6 253.9 – 46.74 correct to 2 decimal places. = √845.6 207.16 = 0.14037 = 0.14 (2.d.p.)

21. 17. (a) The following numbers have been written in standard form. Write each in decimal form. (i)(3.7 x 10 6 ) (ii) (8.2 x 10 -4 ) (b) Find in standard form, the value of: (4.2 x 10 8 ) x (9.1 x 10 4 ) (ii) (6.2 x 10 -9 ) ÷ (8.3 x 10 6 ) = 3700000 = 0.00082 =38220000000000= 3.822 x 10 13 =7.4698 x 10 -16 =7.47 x 10 -16

22. 18. 53.1cm 16.3cm 24.7cm S P Q R Diagram not drawn to scale. The diagram shows a cuboid of length 53.1cm. The cross-section, PQRS, is such that PR=24.7cm and QR=16.3cm. (a)Calculate the length PQ. (b) The density of the material from which the cuboid is made is 4.3g/cm³. Calculate the mass of the cuboid in kilograms. PQ² + QR² = PR² PQ² + 16.3² = 24.7² PQ² = 24.7² - 16.3² PQ² = 344.4PQ = √344.4PQ = 18.558 PQ = 18.6cm Mass = Density x Volume = 4.3 x 16.3 x 53.1 x 18.6= 69225 g= 69.225 kg

23. 19. A solution to the equation x³ - 6x – 3 = 0 lies between 2.6 and 2.7. Use the method of trial and improvement to find this solution correct to 2 decimal places. Try x = 2.652.65³ - 6x2.65 – 3= -0.290Too low Try x = 2.672.67³ - 6x2.67 – 3 = 0.014Too high Try x = 2.66 2.66³ - 6x2.66 – 3 = -0.139Too low Try x =2.665 2.665³ - 6x2.665 – 3= -0.0626Too low 2.67 is solution to 2.d.p.

24. 20. A bag contains 7 blue balls and 5 green balls. Another bag contains 4 blue balls and 6 red balls. A ball is drawn at random from the first bag and its colour is noted. A ball is then drawn at random from the second bag and its colour is noted. (a)Complete the following tree diagram. 1 st ball2 nd ball Blue Green Blue Red 7/12 (b) Calculate the probability that both balls are blue. (c) Calculate the probability that at least one ball is blue 5/12 4/10 6/10 4/10 6/10 = 7 x 4 12 10 = 7 30 = P(BB)+P(BR)+P(GB) = 7 x 4 + 7 x 6 + 5 x 4 12 10 12 10 12 10 = 28 + 42 + 20 120 120 120 = 90 120 = 3 4

25. 21. (a) Simplify (5x³)². (b) Expand the following expression, simplifying your answer as far as possible. (x + 7)(x – 3) (c) Make d the subject of the following formula. 4(d – 2e) = 7 + 3e = 5x 3 x 5x 3 = x² - 3x + 7x - 21 = x² + 4x - 21 4d – 8e = 7 + 3e 4d = 7 + 3e + 8e 4d = 7 + 11e d = 7 + 11e 4 = 25x 6 First Outside Inside Last

26. 22. In the diagram below, ABC = 90°, BED = 90°, AB = 17.8m, CD = 23.6m, BE = 21.4m and BÂC= 37°. Diagram not drawn to scale Calculate the size of angle BDE. 23.6m 17.8m 21.4m E D C A B 37° Tan 37° = BC 17.8 BC = 17.8 Tan37° BC = 13.41m BD = 13.41 + 23.6 = 37.0m Sin d = 21.4 37.0 Sin d = 0.5783 d = Sin -1 0.5783d = 35.3°

27. 23. Solve the following equation. 3 x + 1 2 x + 1 = 3 4 2 4 4 (3 x + 1) 4( 2 x + 1) = 4 x 3 4 2 4 3 x + 1 - 2(2 x + 1) = 3 3x + 1 – 4x – 2 = 3 - x – 1 = 3 -x = 4 x = -4