SMC2012_web_solutions

Xy we are given that so we can deduce that the area

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XY We are given that so we can deduce that . The area of trapezium . Therefore SR = PQ = 25 cm ( a + d ) + ( b + c ) = 25 + 25 = 50 PQRS = 1 2 ( SP + QR ) × XY = 600 cm 2 P Q R S C X Y a a b b c c d d r r . So , i.e. . 1 2 ( a + b + c + d ) × 2 r = 600 1 2 × 50 × 2 r = 600 r = 12 21. D . Note that all of the alternatives given are of the form so we need . The only ordered pairs of positive integers which satisfy this are (1, 6), (2, 3), (3, 2), (6, 1). For these, the values of are 73, 22, 17, 38 respectively. So the required number is . ( x + y 2 ) 2 = x 2 + 2 xy 2 + 2 y 2 a + 12 2 xy = 6 ( x , y ) x 2 + 2 y 2 54 + 12 2 22. B Let the perpendicular from meet at and let . Note that as a tangent to a circle is perpendicular to the radius at the point of contact. Therefore . Consider triangle : . So . So the area of triangle . Y UV T ZXV = α VZX = 90 ° sin α = r 3 r = 1 3 YTX sin α = YT YX YT = YX sin α = 4 r 3 YVW = 1 2 × VW × YT = 1 2 × r × 4 r 3 = 2 r 2 3 U V W X Z Y r r a r 2 T 23. C Tom wins after one attempt each if he hits the target and Geri misses. The probability of this happening is . Similarly the probability that Geri wins after one attempt is . So the probability that both competitors will have at least one more attempt is . 4 5 × 1 3 = 4 15 2 3 × 1 5 = 2 15 1 - 4 15 - 2 15 = 3 5 Therefore the probability that Tom wins after two attempts each is . The probability that neither Tom nor Geri wins after two attempts each is . So the probability that Tom wins after three attempts each is and, more generally, the probability that he wins after attempts each is . 3 5 × 4 15 3 5 × 3 5 ( 3 5 ) 2 × 4 15 n ( 3 5 ) n - 1 × 4 15 Therefore the probability that Tom wins is . 4 15 + ( 3 5 ) × 4 15 + ( 3 5 ) 2 × 4 15 + ( 3 5 ) 3 × 4 15 + … This is the sum to infinity of a geometric series with first term and common ratio . Its value is . 4 15 3 5 4 15 ÷ ( 1 - 3 5 ) = 2 3 24. B The diagram shows one of the three quadrilaterals making up the tile, labelled and with a line inserted. Note that it is a trapezium. As three quadrilaterals fit together, it may be deduced that , so . It may also be deduced that the length of is , where is the length of . BE ABC = 360 ° ÷ 3 = 120 ° BAD = 60 ° AB 1 + x x BC Now . So , i.e. . cos BAD = cos60 ° = 1 2 = 1 - x 1 + x 1 + x = 2 - 2 x x = 1 3 The area of is . So the area of the tile is . ABCD 1 2 ( AD + BC ) × CD = 1 2 ( 1 + 1 3 ) × 4 3 sin60 ° = 2 3 × 4 3 × 3 2 = 4 3 9 3 × 4 3 9 = 4 3 3 1 D E B C A x x x 1 - x 25. B Starting with and expanding both sides gives , i.e. . ( x + y ) 2 = ( x + 4 )( y - 4 ) x 2 + 2 xy + y 2 = xy - 4 x + 4 y - 16 x 2 + ( y + 4 ) x + y 2 - 4 y + 16 = 0 To eliminate the term we let and then replace by . The equation above becomes . However, xy z = x + 1 2 y x z - 1 2 y z 2 + 4 ( z - 1 2 y ) + 3 4 y 2 - 4 y + 16 = 0 z 2 + 4 ( z - 1 2 y ) + 3 4 y 2 - 4 y + 16 = ( z + 2 ) 2 + 3 4 y 2 - 6 y + 12 = ( z + 2 ) 2 + 3 4 ( y 2 - 8 y + 16 ) = ( z + 2 ) 2 + 3 4 ( y + 4 ) 2 . So the only real solution is when and ; i.e. and . z = - 2 y = 4 x = - 4 y = 4 This solutions pamphlet outlines a solution for each problem on this year's paper. We have tried to give the most straightforward approach, but the solutions presented here are not the only possible solutions.
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