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Int 2 Prelim Practice Questions

Int 2 Prelim Practice Questions. 1. Factorise a) x 2 – 3x – 18 b) 3x 2 – 12 . a) (x – 6)(x + 3) b) 3( x 2 – 4 ) = 3( x – 2 )( x + 2 ). 2. Find equation. (3, 11). (0, -1). m = c → -1 y = 4x – 1 . 11- (-1). 12. =. =. 4. 3 – 0 . 3.

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Int 2 Prelim Practice Questions

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  1. Int 2 Prelim Practice Questions

  2. 1. Factorise a) x2 – 3x – 18 b) 3x2 – 12 a) (x – 6)(x + 3) b) 3(x2 – 4) = 3(x – 2)(x + 2)

  3. 2. Find equation. (3, 11) (0, -1) m = c → -1 y = 4x – 1 11- (-1) 12 = = 4 3 – 0 3

  4. 3. Solve Graphically These Simultaneous Equations. x + 3y = 6 2x – 2y = 4 x + 3y = 6 x = 0, y = 2 (0 , 2) y = 0, x = 6 (6 , 0) (could also have (3 , 1) ) 2x – 2y = 4 x = 0, y = -2 (0 , -2) y = 0, x = 2 (2 , 0) (other points include (3 , 1) (4 , 2) (5 , 3) ) Point of Intersection (3 , 1) Solution x = 3, y = 1 x + 3y = 6 2x -2y = 4

  5. D 850 11cm 8cm E F 4. a) Calculate area of triangle DEF b) Calculate the length of FE a) Area = ½ efSinD = ½ X 8 X 11 X Sin850 = 43.8cm2 b) d2 = e2 + f2 – 2efCosD = 82 + 112 – 2 X 8 X 11 X Cos 850 = 169.661 d = √169.661 = 13.0cm

  6. 5. Break the brackets and collect like terms. a) (x + 2)(x2 + 3x – 4) b) (2x – 4)(3x + 5) a) (x+ 2)(x2 + 3x – 4) x(x2+ 3x – 4) + 2(x2 + 3x – 4) x3 + 3x2 – 4x + 2x2 + 6x – 8 x3 + 5x2 + 2x – 8 b) Using FOIL 6x2 + 10x – 12x – 20 6x2 – 2x – 20

  7. 6. a) Calculate the volume of this cylinder – Non calculator 20cm 4cm v = πr2h = 3.14 X 102 X 4 = 3.14 X 100 X 4 = 314 X 4 = 1256cm3

  8. b) This triangular prism has the same volume as the cylinder. Calculate x 8cm x Volume = Area of Face X Height/Length Area of Face = ½ X 10 X 8 = 40cm2 → 1256 = 40 X height/length x = 1256 ÷ 40 = 31.4cm 10cm

  9. Calculate the area of one of the sectors. Method 1 Sector angle = 360 ÷ 5 = 720 Area = 72/360 X π X 82 = 40.2cm2 7. The circle below has been split into 5 equal sectors. 8cm Method 2 (Handy if Non Calc.) Area = 1/5 X π X 82 = 40.2cm2

  10. 8. Calculate length of side DE. D 940 480 380 E 13cm F ?  d e f = angles add up to 1800 = sin D sin E sin F    d f = sin D sin F 13 f = sin940 sin380 f X sin940 13 X sin380 = 13 X sin380 = f sin940 8.02 cm = f

  11. 9. Delihla’s Ming(ing?) vase was worth £4 000. A year later its value had appreciated by 7.2%. Sadly following an unfortunate accident its value depreciated by 94% the following year. What was its value after 2 years? After 1 year: 1.072 X 4000 = 4288 After 2 years: 0.06 X 4288 = £257.28

  12. 10. The shoe sizes of some pupils are shown below. 5 2 4 5 2 4 5 3 2 1 6 2 a) Construct a frequency table with a cumulative frequency column and a relative frequency column to show the data.

  13. b) What is the probability of a pupil chosen at random having feet that are over size 4. c) Calculate the size of each angle if you were to show the results in a pie chart. b) 4/12 = 1/3 c) Size 1: 1/12 X 360 = 300 Size 2: 1/3 X 360 = 1200 Size 3 300 Size 4 600 Size 5 900 Size 6 300

  14. 11. Calculate the standard deviation of these 5 numbers 11 13 8 9 18 2 n = 5 (∑x)/n ∑x2 – n – 1 s = 2 59 / 759 – 5 s = x x2 4 11 13 8 9 18 121 169 64 81 324 62.8 = 4 ∑x ∑x2 59 759 √ = 15.7 = 4.0

  15. 12. In a dodgy café Percy buys 3 cakes and 2 scones. The cost is £1. Not to be outdone (or show any awareness of the merits of healthy eating) Petula purchases 5 cakes and 3 scones which sets her back £1.59 a) Draw up an equation for each situation. b) Assuming all cakes and scones cost the same, use simultaneous equations to calculate the cost of each cake and scone. a) and b) 3c + 2s = 100 5c + 3s = 159 Subtract c = 18 Substituting c = 18 54 + 2s = 100 2s = 46 s = 23 Cakes 18p each, Scones 23p each. X3 → 9c + 6s = 300 X2 → 10c + 6s = 318

  16. O C 340 A B 13. In diagram below, AB is a tangent to circle with centre O. Calculate the size of angle BCA Equal 560 620 1180 620

  17. A ship sails from A to B on a bearing of 0600 as shown above. It then alters course and sails 12km to C on a bearing of 1250. At C it is 19km from point A. a) Caclulate the size of angle ABC. 1250 B 12km 0600 A C 19km 14. 600 650 1150

  18. 1150 1250 B 12km 0600 A C 19km 14. b) Hence calculate the bearing from A to C.   a b c = = sin A sin B sin C ?  550 a b = sin A sin B 12 19 = sinA sin1150 19 X sinA 12 X sin1150 = 12 X sin1150 = sin A 19 so bearing = 60 + 35 = 0950 0.5724 = A 350 =

  19. A plan view of a right angled triangular enclosure is shown with the circles representing fence posts and the straight lines representing wire fencing. If the posts are evenly spaced at 18m apart calculate the total length of wire fencing required. 15. x2 = 722 + 542 = 8100 x = √8100 = 90 Total Fencing = 90 + 72 + 54 = 216m 72m 54m x (3 X 18)

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