2001SDA

# 2001SDA - -O°.L,5 0 L U T 1 0 N 5 5 E N 1 0 R D 1V 1 51 0...

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Unformatted text preview: )-O°.L ,5 0 L U T 1 0 N 5 - 5 E N 1 0 R D 1V 1 51 0 N 7. (Also 113 and J21) Clearly, since f::.PQS is isosceles, LPSQ = LPQS = ~(180- 80) = 50 degrees. Thus LSQR = 180-50 = 130 degrees. Finally, since f::.QRS is isosceles LQRS = LQSR = ~(180 - 130) = 25 degrees, 1. (Also 12) 10 7 0.2 = 10072 = 50, hence (D). 2. Y + 2x =odd + even = odd, all the others are even, hence (D). 3. (Also J7 and 13) As shown in the diagram, x is the exterior angle of the tri- angle and is equal to the sum of the two interior angles, so x = 110 + 50 = 160, ~R S hence (D). 8. The line through (0,1) and (1,3) has equation y = 2x+ 1. So, when x = 3 then y = 6 + 1 = 7, 9. Let the radius of the circle be x. The side of the larger square is 2x, so it has area 4X2. The diagonal of the smaller square is 2x, so it has area ~(4x2) = 2x2. So, the ratio of the areas of the larger square to the smaller square is 4x2 : 2X2 = 2 : 1, hence (E). 4.-11-6w :::::-35-6w :::::-24 6w :::; 24 w :::; 4, hence (A). hence (C). hence (D).- 7 29 29+ 693 722 361 10. 0.729 = 10 + 990 = 990 = 990 = 495' 11. Alternative 1 2000 = 24 X 53. Either one factor has no fives and the other has 3 fives, or one has one five and the other has two fives. Factors with no fives are: 2,4,8,16. Factors with one five are: 5, 10, 20, 40, 80. Hence there are 9 factorisations, 5. (Also 18and J19) First note that 97 = 92+42. Suppose that either 98 or 99 = a2 +b2 with a > b. Then, since 98 = 72 + 72 we must have a = 8 or a = 9. Now 82 + 52 = 89 and 82 + 62 = 100 so a =J 8. Also, 92+ 52 = 106,so a =J 9. Hencethe largest such number is 97, hence (C). 6. 315 X 2710 = 9n 315 X (33)10 = (32t 315 X 330 = 32n 15+ 30 = 2n hence (D). hence (C). 2n = 45 45 n-- 2' Alternative 2 2000 = 24 X 53. The total number of factors is (4 + 1) x (3 + 1) = 20, so there are 10 factorisations 200 = ab, but one is 2000 = 1 x 2000, hence (A). . . hence (C). 12. Alternative 1 Join RQ. Then LP RQ = 90° (angle in a semi- circle) Then LRQP = 50°. . . LPSR = 130° (opp. angles of a cyclic quad) LSRP = 180° - 130° = 25°, 2 Alternative 2 Let the centre of the circle be O. Join SO and RD. LQPR = 40°, so LQOR = 80° (L at centre = 2 x L at circumference) LPOR = 100° and LSOP = 50°. . . LSRP = 25° (L at centre = P 2 x L at circumference), Alternative 2 Draw the perpendicuar from P to QR extended as shown in the dia- gram. Then, from the right-angled trian- gle P RS we get RS = J8 x sin 30° =)2 and PS = J8 x cos30° = 03. P s hence (D). Hence QS = 2)2 Then, from the right-angled triangle PQS, we get x2 = (2)2)2 + (03)2 = 8 + 6 = 14, x = VU, 15. Let x and y denote the numbers, then xy = x+y y(x-1) = x x y =- x-I X X2 x+y = x+- =-, x-I x-I hence (D)....
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## This note was uploaded on 09/27/2010 for the course 456 9852 taught by Professor Chaohue during the Spring '10 term at Mackenzie.

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2001SDA - -O°.L,5 0 L U T 1 0 N 5 5 E N 1 0 R D 1V 1 51 0...

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