1 / 32

Mathematics Intermediate Tier Paper 2 Summer 2002 (2 hours)

Mathematics Intermediate Tier Paper 2 Summer 2002 (2 hours). (a) Calculate 26% of 34. = 26 x 34 100. =£8.84. (b) The total cost of 8 loaves and 12 baguettes is £15.48 . One loaves cost 93p. Find the cost of one baguette. Cost of 8 loaves = 8 x 93 = £7.44.

ssabrina
Télécharger la présentation

Mathematics Intermediate Tier Paper 2 Summer 2002 (2 hours)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Mathematics Intermediate Tier Paper 2 Summer 2002 (2 hours)

  2. (a) Calculate 26% of 34. = 26 x 34 100 =£8.84

  3. (b) The total cost of 8 loaves and 12 baguettes is £15.48 . One loaves cost 93p. Find the cost of one baguette. Cost of 8 loaves = 8 x 93 = £7.44 Cost of baguettes = 15.48 – 7.44 = £8.04 Cost of 1 baguette = 8.04 ÷ 12 = £0.67 = 67c

  4. 2. The diagram below represents a number machine. Divide 5 Input Add 7 Output If the input is x, write down the output in terms of x. = x + 7 5

  5. 3. (a) Simplify 7a – 5 – 3a – 4. = 7a – 3a – 5 – 4 = 4a - 9

  6. (b) Find the value of 4x + 3y when x = - 6 and y = 5. = 4x -6 + 3 x 5 = -24 + 15 = -9

  7. 4. Draw, on the grid, an enlargement of the given shape, using a scale factor of 4.

  8. 5. (a) A solid cuboid measures 6.4 cm by 4.8,cm by 3.5cm, as shown in the diagram. Calculate its volume, clearly stating the units of your answer. 3.5cm 4.8cm 6.4cm Volume = length x width x height = 6.4 x 4.8 x 3.5 = 107.52cm³

  9. Calculate the total surface area of the cuboid. 3.5cm 4.8cm 6.4cm = 44.8cm² Front and back= 2 x 6.4 x 3.5 Top and bottom = 2 x 6.4 x 4.8 = 61.44cm² Sides= 2 x 4.8 x 3.5 = 33.6cm² Total = 44.8 + 61.44 + 33.6 = 139.84cm²

  10. 6. Eighty pupils were asked what they drank with their breakfast. Of these pupils, 36 drank tea, 18 drank coffee, 16 drank milk and 10 drank other drinks. • What is the probability that a randomly chosen pupil (i) drank coffee at breakfast? = 18 80 (ii) did not drink tea at breakfast? = 44 80 = 1 - 36 80

  11. (b) Draw a pie chart to illustrate the different drink that the pupils had with their breakfast.You should show how you calculate the angles of your pie chart. 1 drink = 360 ÷ 80 = 4.5° Tea = 36 x 4.5 = 162° Coffee = 18 x 4.5 = 81° Milk = 16 x 4.5 = 72° Others = 10 x 4.5 = 45°

  12. 7. Mr and Mrs Gann received their electricity bill.The details were as follows. Present meter reading 54261 Previous meter reading 52815 Charge per unit 6.52 pence per unit VAT 5% Service charge £10.56 Showing all your working, find the total cost of the electricity including VAT. = 1446 Units used = 54261 - 52815 Cost of units = 1446 x 6.52 = 9427.92c = £94.28 Cost including service charge = £94.28 + 10.56 = £104.84 VAT = 5 x 104.84 100 = £5.24 Total = £104.84 + 5.24 = £110.08

  13. A 130° B F D Diagram not drawn to scale. 60° C E 8. ABCD and ADEF are two parallelograms in which ADC=130 ° and DEF=60 °. Find BAF BAD = 50° DAF = 60° 50° 60° BAF = 50 + 60 = 110°

  14. 9. The weight of eighty eggs were measured and the results are sumarised in the following table. (a) On the graph paper, draw a grouped frequency diagram for the data.

  15. 30 25 20 15 Frequency 10 5 70 80 100 90 60 50 Weight (grams) (b) Write down the modal class. 70 - 80

  16. 10. Write down, in terms of n, the nth term of each of the following sequences. n (n + 2) • 1 x 3 2 x 4 3 x 5 4 x 6 ….. x ….. 7n - 4 (b) 3 10 17 24 ……… +7 +7 +7

  17. 11. A gardner is making a circular lawn of radius 6m • Calculate the area of the lawn. = 3.142 x 6 x 6 A = πr² = 113.097 = 113.1 m² (b) The gardener wishes to put an edging around the circumference of the lawn. Calculate the length of edging needed. C = 2 x 3.142 x 6 C = π d or C = 2 π r C = 37.699 C= 37.7 m

  18. 12. Beverley leaves home at 11.00 a.m. to go for a drive in her car. She travels a certain distance then stops for three quarters of an hour before starting back for home at a speed of 40 m.p.h. (a) Calculate the speed for the first part of her journey. = 68 2 = 34 m.p.h.

  19. 70 60 50 Distance (miles) from home 40 30 20 10 Time 14:00 11:00 13:00 15:00 12:00 (b) On the graph paper, draw lines to represent (ii) Her ¾ hour stop and (ii) her return journey home.

  20. 13. (a) The population of a country increased from 56 000 000 to 59 500 000. What percentage increase is this? Increase = 59 500 000 – 56 000 000 = 3 500 000 Percentage increase= 3 500 000 x 100 56 000 000 = 6.25%

  21. (b) What will be the amount if £5000 is invested for 3 years at the rate of 4% compound interest per annum? = £200 Interest 1st year = 4 x 5000 100 Amount in the bank = 5000 + 200 = £5200 = £208 Interest 2nd year = 4 x 5200 100 Amount in the bank = 5000 + 208 = £5408 Interest 3rd year = 4 x 5408 100 = £216.32 Amount in the bank = 5408 + 216.32 = £5624.32

  22. 14. Solve the following equation. 15-4x = 3 7 15 – 4x = 3 x 7 15 – 4x = 21 15 – 21 = 4x -6 = 4x - 6 = x 4 - 3 = x 2 -1 ½ = x x = -1 ½

  23. 15. A solution to the equation x³ + 5x – 30 = 0 lies between 2 and 3. Use the method of trial and improvement to find this solution correct to one decimal place. = -1.875 Try x = 2.5 2.5³ + 5 x 2.5 – 30 Too small Too big Try x = 2.6 2.6³ + 5 x 2.6 – 30 = 0.576 Too small = -0.6686 Try x = 2.55 2.55³ + 5 x 2.55 – 30 Felly x = 2.6 (1 ll.d.)

  24. 16. (a) The batting scores of 100 cricketers were recorded and the results are summarised in the following table. On the graph paper, draw a frequency polygon for the data.

  25. Frequency 50 40 30 20 10 Batting Score 60 80 100 0 20 40 (b) Find an estimate for the mean of the batting scores. =35 = 3510 100 = Σfx Σf = 20x9.5+45x29.5+24x49.5+9x69.5+2x89.5 20+45+24+9+2

  26. C 5.4cm B A 8.7cm 17. The diameter of a circle, AB is of length 8.7cm, BC has length 5.4cm and ACB = 90 °. Calculate the length of AC. Diagram not drawn to scale. AB² = AC² + CB² Neu / or AC² = AB² - BC² 8.7² = AC² + 5.4² AC² = 8.7² - 5.4² 8.7² - 5.4² = AC² 46.53 = AC² AC = √ 46.53 AC = 6.82cm

  27. 18. ABCD is a rectangle. B C D A • Draw the locus of all the points inside the rectangle whose distance from AB is the same as their distance from AD. (b) Draw the locus of all the points inside the rectangle which are 6cm from DC. (c) Draw the locus of all the points inside the rectangle whose distance from A is the same as the length of AB.

  28. 19. Find, in standard form, the value of (a) (7.4 x 10 -5) x (3.9 x 10 -4) = 2.889 x 10 -8 (b) 59639 0.087 = 6.86 x 10 5 (3 sig.fig)

  29. 20 (a) Simplify the expression (4x3y2) x (2x4y5) = 8x7y7 (b) Expand and simplify (x + 5) (x – 6) First Outside Inside Last = x² -6x +5x - 30 = x² -x - 30 (c) Make d the subject of the following formula: h = √ t - d h² = t - d d = t - h²

  30. 21. Solve the following equation: 4x – 8 – x = 2 3 6 6 (4x – 8) – 6 x = 6 x 2 multiply every term by 6 3 6 2(4x – 8) – x = 12 expand brackets 8x – 16 – x = 12 8x - x = 12 + 16 collect terms 7x = 28 x = 28 7 x = 4

  31. C 4.7m D 13.5m 62° A 8.4m E B 22. A vertical flagpole, BDC, stands on horizontal ground ABE. It is supported by two ropes AC and DE. The length of AC is 13.5m, and the distance CD is 4.7m. The rope AC makes an angle of 62° with the ground and the rope DE is fixed to the ground at E such that BE is 8.4m. Diagram not drawn to scale. Sin 62° = BC 13.5 BD = 11.9 – 4.7 = 7.2m 13.5 Sin 62° = BC Tan BDE = 8.4 7.2 BC = 11.9m Tan BDE = 1.16667 BDE = Tan-1 1.16667 Calculate the size of BDE. BDE = 49.4°

  32. First bag Second bag Red Red ⅔ Not red Red Not red Not red 23. Two bags contain some coloured balls, which are identical except for their colour. One ball is taken at random from each bag and their colours noted. The probability of the selected ball from the first bag being red is ¼ . The probability of the selected ball from the second bag NOT being red is ⅔. (a) Complete the following tree diagram. (b) Calculate the probability that both balls are red. ⅓ P(red, red) ¼ = 1 x 1 4 3 = 1 12 ⅓ ¾ (c) Calculate the probability that only one ball is red. ⅔ = P (red, not red) + P(not red, red) = 1 x 2 + 3 x 1 4 3 4 3 = 2 + 3 12 12 = 5 12

More Related