SMC2012_web_solutions

# 15 a let professor rosseforps usual journey take

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15. A Let Professor Rosseforp's usual journey take minutes at an average speed of metres/ minute. Then the distance to work is metres. On Thursday his speed increased by 10%, i.e. it was metres/minute. The time taken was minutes. t v vt 11 v / 10 ( t - x ) Therefore . So , i.e. . 11 v / 10 × ( t - x ) = vt 11 ( t - x ) = 10 t t = 11 x 16. A At points and , . So , i.e. , i.e. or . A C x = 0 y 2 - 4 y = 12 ( y - 6 )( y + 2 ) = 0 y = 6 y = - 2 So is and is . At points and , . So , i.e. , i.e. or . So is and is (3, 0). C ( 0 , - 2 ) A ( 0 , 6 ) B D y = 0 x 2 + x = 12 ( x - 3 )( x + 4 ) = 0 x = 3 x = - 4 D (- 4 , 0 ) B Therefore the areas of triangles and are and . DAB DBC 1 2 × 7 × 6 = 21 1 2 × 7 × 2 = 7 So has area 28. { It is left to the reader to prove that area .} ABCD ABCD = 1 2 BD × AC 17. B In the diagram, is the centre of the quarter-circle arc ; is the point where the central square touches arc ; is the point where the central square touches arc ; is the centre of the circle. B AC D AC F CE O As both the circle and arc have radius 1, is a square of side 1. AC OABC By Pythagoras' Theorem: . So . Therefore . By a similar argument, . Now since . So the side of the square is OB 2 = 1 2 + 1 2 OB = 2 OD = OB - DB = 2 - 1 OF = 2 - 1 DF 2 = OD 2 + OF 2 = 2 × OD 2 OD = OF 2 × OD = 2 ( 2 - 1 ) = 2 - 2 . A B C D E F O 18. A In the diagram, is the midpoint of . Triangle is isosceles since . Therefore, bisects and is perpendicular to . The angles at a point total , so . Therefore . So . D AC ABC AB = BC = 1 2 BD ABC BD AC 360 ° ABC = 360 ° - 2 × 90 ° - 2 α = 180 ° - 2 α ABD = ∠ CBD = 90 ° - α BAD = ∠ BCD = α 1 2 1 2 2 α x A B C D Therefore . x = AC = 2 × AD = 2 × AB cos α = 2 × 1 2 cos α = cos α 19. E Note that the number represented by appears in both the horizontal row and the vertical column. Note also that 2 + 3 + 4 + 5 + 6 + 7 + 8 = 35. Since the numbers in the row and those in the column have sum 21, we deduce that . x x = 2 × 21 - 35 = 7 We now need two disjoint sets of three numbers chosen from 2, 3, 4, 5, 6, 8 x so that the numbers in both sets total 14. The only possibilities are {2, 4, 8} and {3, 5, 6}. We have six choices of which number to put in the top space in the vertical line, then two for the next space down and one for the bottom space. That leaves three choices for the first space in the horizontal line, two for the next space and one for the final space. So the total number of ways is . 6 × 2 × 1 × 3 × 2 × 1 = 72
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