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SMC2012_web_solutions

# Initially each player has one attempt at hitting a

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Tom and Geri have a competition. Initially, each player has one attempt at hitting a target. If one player hits the target and the other does not then the successful player wins. If both players hit the target, or if both players miss the target, then each has another attempt, with the same rules applying. If the probability of Tom hitting the target is always 5 4 and the probability of Geri hitting the target is always 3 2 , what is the probability that Tom wins the competition? A 15 4 B 15 8 C 3 2 D 5 4 E 15 13 Solution: C Let the probability that Tom wins the competition be p. The probability that initially Tom hits and Geri misses is 15 4 3 1 5 4 = × . The probability that initially they both hit is 15 8 3 2 5 4 = × and that they both miss is 15 1 3 1 5 1 = × . So the probability that either they both hit or both miss is 5 3 15 9 15 1 15 8 = = + . If they both hit or both miss the competition is in the same position as it was initially. So Tom’s probability of winning is then p . Therefore, p p 5 3 15 4 + = . So 15 4 5 2 = p and hence 3 2 = p . 24. have length 1. It may be divided into three congruent quadrilaterals as shown. What is the area of the tile? A 2 3 2 1 + B 3 3 4 C 6 D 4 3 4 3 + E 2 3 3 Solution: B The tile is made up of three congruent quadrilaterals one of which, PQRS shown in the diagram on the right. We let T be the foot of the perpendicular from Q to SR and we let U be the point shown. Then 1 = = UR SR . Also PQ=QU. Let their common length be x. Hence x QR + = 1 and x TR = 1 . Because the three congruent quadrilaterals fit together at Q , we have 0 120 = PQR . As PQ and SR are both perpendicular to PS , they are parallel and hence 0 60 = QRT . Hence, from the right-angled triangle QRT we have 0 60 sin = QR TR , that is, 2 1 1 1 = + x x , and therefore . 1 2 2 x x + = Hence 1 3 = x and so 3 1 = x . Therefore 3 2 = TR and 3 4 = QR . Now 0 60 cos = QR QT and hence 0 60 cos QR QT = 3 3 2 2 3 3 4 0 3 4 ) ( 60 cos = = = . [We could also use Pythagoras’ Theorem applied to the triangle QRT to deduce this length.] We can now work out the area of the quadrilateral PQRS in two different ways. Most straightforwardly, area( PQRS ) = area( PQTS ) + area( QRT ) 9 3 4 3 2 3 3 2 2 1 3 1 3 3 2 ) ( = × + × = . Alternatively, as PQ is parallel to SR , PQRS is a trapezium. We now use the fact that the area of a trapezium is its height multiplied by the average length of its parallel sides. Hence, PQRS has area. ) 1 ( 3 1 2 1 3 3 2 + × 9 3 4 3 2 3 3 2 = × = . The tile is made up of three copies of PQRS. So its area is . 3 3 3 4 9 3 4 = × 1 1 1 1 1 1

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25. How many distinct pairs ) , ( y x of real numbers satisfy the equation ) 4 )( 4 ( ) ( 2 + = + y x y x ? A 0 B 1 C 2 D 3 E 4 Solution: B Method 1. The most straightforward method is to rewrite the given equation as a quadratic in x , whose coefficients involve y , and then use the “ 0 4 2 ac b ” condition for a quadratic, 0 2 = + + c bx ax , to have real number solutions. Now , 16 4 4 2 ) 4 )( 4 ( ) ( 2 2 2 + = + + + = + y x xy y xy x y x y x 0 16 4 4 2 2 = + + + + y y x xy x 0 ) 16 4 ( ) 4 ( 2 2 = + + + + y y x y x .
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