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MATHEMATICS SET 3 Form 4 Topic: Standard Form 1. (7.435 x 6.752)/2.314 = [correct to 3 significant figures] A B C D E 19.7 21.6 21.7 21.695 2.59 Answer: C Solution: (7.435 x 6.752)/2.314 = 50.20112/2.314 = 21.69452 = 21.7 Topic: Standard Form 2. If x = yz2, find y if x = 17.34 and z = 13.35. Round off the value ok y correct to 2 significant figures. A B C D E 0.1 0.09 0.097 0.0972 10.3 Answer: C Solution: x = yz2 y = x/z2 y = 17.34/ (13.35)2 y = 17.34/178.2225 y= 0.0972941 y = 0.097 Topic: Quadratic Expressions and Equations 3. Factorise x2 – 3x – 28. A B C D E (x + 3) (x + 8) (x – 8) (x + 3) (x – 3) (x + 8) (x – 4) (x + 7) (x – 7) (x + 4) Answer: E Solution: x2 || -3x -28 || = (x – 7) (x + 4) x(4) + x(-7) = -3x Topic: Quadratic Expressions and Equations 4. Simplify x2 – 16 / (x2 – 2x – 8). A B C D E x+2 x+4 x + 4/x + 2 4x x + 2/x + 4 Answer: C Solution: x2 – 16 _________ x2 – 2x – 8 = (x – 4) (x + 4) ___________ (x - 4) (x + 2) x+4 = _____ x+2 Topic: Quadratic Expressions and Equations 5. 0.5 m 0.5 m DIAGRAM 1 Diagram 1 shows a square mini field that is surrounded by a path 0.5 m width. The area of the field is 23 m2 bigger than the area of the path. Find the length of a side of the field. A B C D E 2m 3m 4m 5m 6m Answer: E Solution: Assume the side of the field = k. The area of the path = (2)(k + 0.5 + 0.5) + (2)(k)(0.5) = 2(k + 1)(0.5) + 1k =k+1+k = 2k + 1 Given k2 – (2k + 1) = 23 k2 – 2k – 1 – 23 = 0 k2 – 2k – 24 = 0 (k + 4)(k – 6) = 0 k = -4 (ignored) k=6 Therefore, the length of a side of the field = 6 m Topic: Sets 6. Given the universal set ε = {x: 10 ≤ x < 20, x are integers} and K = {x: x is a number where the sum of its digits is more than 3}. Find n (K’). A B C D E 3 4 7 8 10 Answer: A Solution: K = {13, 14, 15, 16, 17, 18, 19} K’ = {10, 11, 12} Therefore, n(K’) = 3 Topic: Sets 7. If the universal set, ε = {all the alphabet from a to z}, G ={vowels}, H = {consonants}, J = {e, x, c, e, l, l, e, n, t}. Find n (H J)’. A B C D E 5 6 9 20 21 Answer: D Solution: H n(H J = {x, c, l, l, n, t} J)’ = 26 – 6 = 20 Topic: Sets 8. ε •2 •10 X •1 •6 •3 •13 •9 •8 •7 •12 •11 •5 •16 •4 •14 •p •15 •17 Y W DIAGRAM 1 If ε = {x: x are integers where 1 ≤ x ≤ 18}, find X A B C D E {4, 14, 15, 17} {4} {4, 14} {14} {4, 11, 14, 16, p} Answer: D Solution: Set W = {1, 2, 3, 4, 5, 6, 7, 8, 9} Set X = {3, 4, 7, 8, 10, 12, 13, 14} Set Y = {4, 14, 15, 17} Set W’ = {10, 11, 12, 13, 14, 15, 17} X Y = {4, 14} X Y W’ = {14} Y W’. Topic: Straight Line 9. DIAGRAM 3 Find the equation of the straight line in Diagram 3. A B C D E y = 3/4x + 13/4 y = 3/8x + 13/4 y = 3/8x + 4 y = -3/8x + 13/4 y = 3/4x - 5 Answer: B Solution: m = (4 – 1) / [2 – (-6)] = 3/8 Therefore, y = 3/8x + c ----- 1 Substitute (2, 4) into 1 y = 3/8x + c 4 = (3/8) x 2 + c c=4–¾ = (16 – 3)/4 = 13/4 Hence, y = 3/8x + 13/4 Topic: Straight Line 10. TU is a straight line which is parallel to 2y = 2x – 5. If TU passes through (6, 2), find the straight line TU. A B C D E y=x-4 y = 2x - 4 y = 3x - 5 y = 1/2x + 4 y=x+4 Answer: A Solution: 2y = 2x – 5 y = 1x – 5/2 m1 = 1 Since TU is parallel to 2y = 2x – 5 Therefore, mLM = 1 y = mx + c y = 1x + c ----- 1 Substitute (6, 2) into 1 y = 1x + c 2 = 1(6) + c c=2–6 = -4 Therefore, y = 1x – 4 y=x–4 Topic: Straight Line 11. In the diagram below, KLMN is a parallelogram. Determine the coordinates of S. DIAGRAM 4 A B C D E (3, 0) (0, -3) (0, 3) (3, -3) (0, 4) Answer: C Solution: mkl = mmn mkl = (y2 – y1) / (x2 – x1) = (6 – 4) / (5-(-2)) = 2/7 S(0, y) (because S is on y-axis) msm = (5 – y) / (7 – 0) = 2/7 5 – y = 2/7 x 7 y=5–2 =3 Therefore, the coordinates of S = (0, 3) Topic: Statistics Question 12 – 14 are based on the information below. DIAGRAM 5 Diagram 5 shows a frequency polygon representing the length of cockroach in science experiment. 12. How many cockroaches are used in the experiment? A B C D E 10 20 30 50 70 Answer: E Solution: Number of cockroaches = 10 + 20 + 10 + 22 = 70 13. What is the percentage of cockroaches longer than 1.5 cm? (correct to 3 significant figures). A B C D E 14.3% 42.9% 57.1% 85.7% 97.1% Answer: C Solution: Number of cockroaches > 1.5 cm = 40 Percentage = 40/70 x 100% = 57.14286 = 57.1% 14. Calculate the mean. A B C D E 2.0 1.896 1.986 1.8 1.1 Answer: C Solution: Mean = [(1 x 10) + (1.5 x 20) + (2 x 10) + (2.5 x 22) + (3.0 x 8)] _____________________________________________ 70 = 10 + 30 + 20 + 55 + 24 __________________ 70 = 1.9857 = 1.986 Topic: Probability I 15. The probability to pick a 50 cents coin from a coin box is 4/8. If there are 80 fifty cents coins and t 1 cents coins, find t. A B C D E 20 40 50 80 90 Answer: D Solution: 80/ (80 + t) = 4/8 8(80) = 4(80 + t) 640 = (320 + 4t) 640 - 320 = 4t t = 320/4 = 80 Topic: Probability I 16. In a dart competition, the probability of hitting the shaded region is 2/6 for a particular participant. Find the area of the unshaded region if the shaded region has an area of 13π cm2. A B C D E 13π 26π 39π 65π 78π Answer: B Solution: P (hit the shaded region) = area of shaded region / total area 2/6 = 13π/total area 2 x total area = 6(13π) Total area = 78π/2 = 39π The area of unshaded region = 39π - 13π =26π Topic: Probability I 17. The probability that a washing machine produced by a particular manufacturer is of an acceptable standard is 0.93. How many washing machines are acceptable if the manufacturer produced 1300 washing machines? A B C D E 1209 1211 1309 1311 1409 Answer: A Solution: P (washing machine of acceptable standard) = n (washing machine of acceptable standard) / n(S) = n (washing machine of acceptable standard) / 1300 = 0.93 n (washing machine of acceptable standard) = 0.93 x 1300 = 1209 Topic: Probability I 18. In a shooting competition, the probability of shooting the target by a shooter is 6/15. If the particular shooter shoots 120 times, how many times does he achieve shooting the target? A B C D E 9 40 44 48 52 Answer: D Solution: 6/15 x 120 =6x8 = 48 Topic: Circle III 19. DIAGRAM 6 From the diagram 6, EF and EG are tangents to the circle. Calculate the value of s. A B C D E 30° 50° 130° 140° 170° Answer: B Solution: ΔFOG = 360° - 230° = 130° ΔEFO = ΔEGO = 90° Therefore, ΔFEG = 360° – ΔFOG – ΔEFO - ΔEGO = 360° – 130° – 90°- 90° = 50° Topic: Circle III 20. DIAGRAM 7 TUV is a tangent to the circle. Find the value of s. A B C D E 15° 38° 40° 53° 59° Answer: E Solution: ΔPRU = 78° ΔQRU = 180° – 43° = 137° (cyclical rectangle) s = 137° – 78° = 59° Topic: Circle III 21. DIAGRAM 8 In Diagram 8, FGH and HIJ are two common tangents. What is the major angle of FDJ? A B C D E 104° 114° 128° 214° 232° Answer: E Solution: ΔDFH = ΔDJH = 90° Hence the minor angle of FDJ = 360° – 90° – 90° – 52° = 128° Therefore, the major angle of FDJ = 360° – 128° = 232° Topic: Circle III 22. DIAGRAM 9 KLM is a common tangent to the two circles centered O and S. Calculate the length of OS. A B C D E 15.56 cm 14.56 cm 15.65 cm 16.34 cm 15.75 cm Answer: A Solution: OS = OL + LS sin 40° = 4/OL OL = 4/sin 40° = 6.2229 sin 40° = 6/LS LS = 6/sin 40° = 9.3343 Therefore,OP = 6.229 + 9.3343 = 15.5633 cm = 15.56 cm Topic: Trigonometry II 23. DIAGRAM 10 From Diagram 10, tan θ = 3/4. Find the length of MN. A B C D E 3 cm 4 cm 5 cm 7 cm 9 cm Answer: C Solution: tan θ= KN/JN 3/4 = 6/KJ Therefore, KJ = (4 x 6) / 3 = 8 cm Hence JN = 10 cm (Phytagoras Theorem) JM = 5 cm (Phytagoras Theorem) MN = 10 – 5 = 5 cm Topic: Trigonometry II 24. Evaluate 3 sin 30° – 4 cos 45° + tan 45°. __ A -5/2 - 2√2 __ B 5/2 + 2√2 __ C 5/2 - 2√2 __ D 3/2 - 2√2 __ E 5/2 - √2 Answer: C Solution: 3 sin 30° – 4 cos 45° + tan 45° _ = 3(0.5) + 4(√2 / 2) + 1 _ = 3/2 + 1 - 2√2 _ = 5/2 - 2√2 Topic: Trigonometry II 25. DIAGRAM 11 In Diagram 11, given that cos ΔEFD = 4/7. Calculate the perimeter of the triangle DEF. A B C D E 42.66 cm 52.66 cm 52.60 cm 55.66 cm 59.66 cm Answer: B Solution: Cos ΔEFD = EF/DF = 9/DF = 4/7 (given) DF = 9(10) / 4 = 22.5 cm __________ DE = √ (22.5)2 - 102 = 20.1556 Therefore, perimeter = 10 + 22.5 + 20.1556 = 52.66 cm Topic: Trigonometry II 26. Given tan θ = 2.156 and θ is an reflex angle. Find θ. A B C D E 245°5’ 245°7’ 244°7’ 244°5’ 245°8’ Answer: B Solution: θ = tan-1 2.156 = 65°7’ since 180° ≤ θ ≤ 360° θ = 180° + 65°7’ = 245°7’ Topic: Angles of Elevation and Depression 27. C θ A B AC is horizontal line. If A is the observer and C is the object, find the angle of elevation. A B C D E ΔACB ΔABC ΔBCA ΔCAB ΔCBA Answer: D Solution: The angle of elevation is ΔCAB. Topic: Angles of Elevation and Depression 28. In the diagram above, UV is a tree on a horizontal plane. T is a point on the horizontal plane. If ΔTUV = 90, UV = 4 m and TU = 7 m, calculate the angle of elevation of the point V from the point T. 27°15’ 55°9’ 29°45’ 60°15’ 34°60’ A B C D E Answer: C Solution: tan ΔUTV = 4/7 Therefore UTV = tan-1 4/7 = 29°45’ Topic: Angles of Elevation and Depression 29. In the diagram above, GH is a vertical pole and H is a point on the same horizontal plane. If the angle of elevation of F from H is 10 and the distance GH = 28 m, find the length of the pole. A B C D E 2.354 m 2.623 m 4.892 m 4.927 m 4.937 m Answer: E Solution: F 10° G tan 10° = GH/28 GH = 28 tan 10° = 4.937 m H Topic: Angles of Elevation and Depression 30. DIAGRAM 12 Diagram 12 shows a pyramid on a square base of side 8 cm. If MD = MG = MF = ME = 14 cm, find the height MN of the pyramid. A B C D E 6.432 cm 12.8063 cm 13.1432 cm 14.2442 cm 15.3543 cm Answer: B Solution: _______ DF = √82 + 82 = 11.3137 DN = 11.3137/2 = 5.6569 By using the Pythagoras Theorem, ____________ 2 MN = √142 – 5.65692 = 164 ____ Hence, MN = √164 = 12.8063 cm Form 5 Topic: Number Bases 31. The value of digit 5, in base ten, in the number 952610 is A B C D E 5 50 10 25 500 Answer: E Solution: 5 x 102 = 500 Topic: Number Bases 32. Express 2210 as a number in base two. A B C D E 1012 10102 101102 110002 100002 Answer: C Solution: Therefore, 2210 = 101102 Topic: Number Bases 33. It is given that s8 = 10110102, find the value of s. A B C D E 113 131 121 112 132 Answer: E Solution: 001 1 011 3 010 2 Therefore, s8 = 1328 Topic: Graphs of Functions II Question 34 – 35 are based on the following table which shows the corresponding values of x and y of the function y = 2x2 + x + 5. x y -3 k -2 11 34. The value of k is A B C D E 6 8 14 20 21 Answer: D Solution: Replace x = -3 in the equation, y = 2x2 + x + 5 y = 2(-3)2 + (-3) + 5 y = 2(9) – 3 + 5 y = 20 35. Find the value of m. A B C D E 5 10 15 17 19 Answer: C -1 6 0 5 1 8 2 m Solution: Replace x = 2 in the equation, y = 2x2 + x + 5 y = 2(2) + 2 + 5 y = 2(4) + 2 + 5 y = 15 Topic: Graphs of Functions II 36. DIAGRAM 13 Diagram 13 shows the graph of y = c + mx – x2. Find the value of m. A B C D E 5/2 3/2 1 11/2 -5/2 Answer: A Solution: y = c + mx – x2 y = 6 + mx – x2 (c = 6 because the curve cuts the y-axis at (0, 6)) Using point (3, 0), 0 = 6 + m(4) – 42 4m = 10 m = 10/4 m = 5/2 Topic: Transformation III 37. DIAGRAM 14 Diagram 14 shows six trapezoids on a Cartesian plane. Given that G is the translation 3 -2 which of the trapezoids A, B, C, D and E is the image of trapezoid H under the combined transformation G2? Answer: E Solution: E is the image of trapezoid H under the combined transformation G2. Topic: Transformation III 38. DIAGRAM 15 Diagram 15 shows a point F on a Cartesian plane. Transformation T is a reflection about the line y = 2 and transformation U is a clockwise rotation of 90° about the centre (5, -2). Find the coordinates of the image of point F under transformation TU. A B C D E (11, 4) (7, 4) (11, 2) (11, -4) (11, 2) Answer: A Solution: The coordinates of the image of point F under transformation TU is (11, 4). Topic: Transformation III 39. DIAGRAM 16 Diagram 16 shows two regular pentagons JKLMN and PQRST. PQRST is the image of JKLMN under transformation V. Given that the area of JKLMN is 25 cm2, calculate the area of PQRST. A B C D E 50 cm2 100 cm2 225 cm2 75 cm2 115 cm2 Answer: B Solution: Transformation V is an enlargement about the centre (6, 8) with scale factor 2. Therefore, the area of PQRST = 25 x 22 = 100 cm2 Topic: Matrices 40. Given that s 6 = 3 3 10 3 A B C D E 6 2s + t , find the value of t. 1 2 4 -4 16 Answer: C Solution: s=3 2s + t = 10 2(3) + t = 10 t = 10 – 6 =4 Topic: Matrices 41. It is given that p + 2 p – 4 = 16 . Find the value of p and q. q q–2 11 A B C D E p = 8 and q = 5 p = -8 and q = -5 p = 5 and q = 8 p = -5 and q = 8 p = 5 and q = -8 Answer: A Solution: p + 2 p – 4 = 16 q q–2 11 p + 2p – 8 = 16 q 2q – 4 11 Therefore, p + 2p – 8 = 16 3p = 24 p=8 and q + 2q – 4 = 11 3q = 15 q=5 Topic: Matrices 42. It is given that matrix K = u -1 6 v KL = 1 0 . 0 1 A B C D E and matrix L = 1/9 v 1 such that -6 3 u = -1 and v = -3 u = 1 and v = 3 u = 3 and v = -1 u = -3 and v = 1 u = 3 and v = 1 Answer: E Solution: u -1 -1 = 1/9 v 1 6 v -6 3 1/(uv + 6) v 1 -6 u = 1/9 v 1 -6 3 Therefore, u = 3 and uv + 6 = 9 3v = 3 v=1 Topic: Variations 43. If J varies directly as L to the power of 2 and J = 36 when L = 3, find the value of L when J = 144. A B C D E 2 4 18 6 36 Answer: D Solution: J α L2 J = kL2 k = J/L2 k = 36/32 k=4 When J = 144, 144 = 4(L2) L2 = 36 L=6 Topic: Variations __ 44. If D α 1/√E and D = 3 when E = 144, find the value of D when E = 36. A B C D E 6 8 10 12 14 Answer: A Solution: __ D α 1/√E __ D = k/√E ___ k = 3 x √144 k = 3 x 12 = 36 __ Therefore, D = 36/√E When E = 9, __ D = 36/√36 D = 36/6 D=6 Topic: Variations 45. It is known that X varies directly as Y and inversely as the square of Z. If X = 3 when Y = 15 and Z = 5, then the relation of X, Y and Z is A B C D E X = 5Z2/Y X = 5Y/Z2 X = 4Y/Z2 X = 4Z2/Y X = Y/Z2 Answer: B Solution: X α Y/Z2 X = kY/Z2 2 = k(15)/52 k = (3 x 52)/15 k=5 Therefore, X = 5Y/Z2. Topic: Gradient and Area under a Graph Question 46 – 48 are based on the following distance-time graph that shows the journey of a bus departed from a place 20 km from town M, and then returned to town M over a period of 120 minutes. DIAGRAM 17 46. Calculate the distance, in km, traveled by the bus in the first 50 minutes. A B C D E 10 km 15 km 20 km 25 km 30 km Answer: D Solution: Distance traveled = 45 – 20 = 25 km 47. Calculate the distance, in km, traveled by the bus over the period of 120 minutes. A B C D E 50 km 60 km 80 km 130 km 140 km Answer: C Solution: Distance traveled = 50 + 30 = 80 km 48. Find the average speed, in km-1, of the bus in the first 40 minutes of its journey. A B C D E 30 km h-1 20 km h-1 60 km h-1 50 km h-1 70 km h-1 Answer: A Solution: Average speed = 20/(40/60) = 20 x (60/40) = 30 km h-1 Topic: Probability II 49. A box contains 18 yellow marbles and 24 silver marbles. A marble is drawn at random from the box. What is the probability of drawing a yellow marbles? A B C D E 1 4/7 3/7 5/7 2/7 Answer: C Solution: P (yellow marbles) = n (yellow marbles) / n(S) = 18 / (18 + 24) = 18 / 42 =3/7 Topic: Probability II 50. DIAGRAM 17 Diagram 17 shows a group of letter cards in a box. Sherri took out all the cards marked with the letter S and replaced them with several cards marked with the letter T. A card is then drawn at random from the box and the probability of drawing a card with the letter I is 1/8. Calculate the number of cards with the letter T in the box. A B C D E 3 5 9 11 13 Answer: E Solution: Let the number of cards with the letter T added = p n (S) = 14 – 2 + p = 12 + p n (letter I) = 3 Hence, 3/(12 + p) = 1/8 12 + p = 24 p = 12 Therefore, number of cards with letter T = 12 + 1 = 13 Topic: Probability II 51. Naji has 60 bags of graded tomatoes. 25 bags are grade S tomatoes and the rest are grade B and grade C tomatoes. If a bag is picked at random, the probability of picking a bag of grade B tomatoes is 1/4. Calculate the probability of picking a bag of grade C tomatoes. A B C D E 2/3 1/3 1/5 2/5 1/4 Answer: B Solution: Probability of picking a grade S bag = 25/60 = 5/12 Probability of picking a grade C bag = 1 – 5/12 – 1/4 = 4/12 = 1/3 Topic: Bearing 52. North K L 31° M DIAGRAM 18 In Diagram 18, K, L and M are three points on a horizontal plane. The bearing of the point M from the point K is A B C D E 31° 45° 59° 111° 121° Answer: E Solution: ΔLKM = 180° – 90° – 31° = 59° Therefore, the bearing of the point M from the point K is = 180° – ΔLKM = 180° - 59° = 121° Topic: Bearing 53. Given that the bearing of point X from point Y is 232°, then the bearing of point Y from point X is A B C D E 52° 62° 128° 138° 148° Answer: A Solution: Therefore, the bearing of point Y from point X is 52°. Topic: Bearing 54. DIAGRAM 19 In Diagram 19, S, T and U are three points on a horizontal plane. The point S is due west of T. Find the bearing of S from U. A B C D E 45° 115° 135° 225° 240° Answer: D Solution: Therefore, the bearing of S from U = 180° + 32° + 13° = 225° Topic: Earth as a Sphere Question 55 – 56 are based on Diagram 20 where O is a centre of the earth and G, H and K are the three places on the equator such that GH is a diameter of the equator. DIAGRAM 20 55. It is given that ΔKOG = 45°, the longitude of G is A B C D E 50°W 45°W 45°E 135°E 135°W Answer: C Solution: The longitude of G is 45°E. 56. The longitude of H is A B C D E 45°W 115°W 125°W 135°W 180°W Answer: D Solution: The longitude of H is = (180 – 45) °W = 135°W Topic: Earth as a Sphere 57. What is the distance, in nautical miles, between the points Q (32°N, 115°W) and R (69°S, 115°E) along a meridian? A B C D E 6060 n.m 2220 n.m 3030 n.m 1920 n.m 4140 n.m Answer: A Solution: Q and R lie on the same meridian. Difference between latitudes 32°N and 69°S = 32° + 69° = 101° Therefore, the distance between points Q and R = 101° x 60’ = 6060’ Topic: Earth as a Sphere Question 58 – 60 are based on Diagram 21 where J, K, M, U, T and L are places on the surface of the earth. O is the centre of the earth. DIAGRAM 21 58. Which of the following points lie on the same parallel of latitude as point U? A B C D E J U K M L Answer: B Solution: U and T lie on the same parallel of latitude. 59. The location of the place J is A B C D E (53°N, 70°W) (0°, 18°W) (53°S, 18°W) (53°N, 18°W) (53°S, 70°E) Answer: A Solution: J has latitude 53°N and longitude 70°W. 60. The location of the place U is A B C D E (0°, 18°W) (36°N, 70°W) (53°N, 18°W) (53°N, 70°W) (36°S, 18°W) Answer: E Solution: U has latitude 36°S and longitude 18°W. 60 Questions