2017 SNGS Sec 4 EM Prelim P2 Solutions

3 CHIJ SNGS Preliminary Examinations 2017 - Mathematics 4048/02 1 (a) Solve the simultaneous equations .1943 ,82   yx xy [3] (b) Simplify 22 22 65 4 yxyx yx   . [3] (c) (i) Express 372  xx in the form .)( 2 bax  [2] (ii) Hence solve the equation ,0372  xx giving your answers correct to two decimal places. [2] 1(a) 82  yx 194)82(3  yy 52 y 5.2y 3x 1(b) 22 22 65 4 yxyx yx        yxyx yxyx 23 22      yx yx 3 2   1(ci) 3)2 7()2 7(737 2222  xxxx = 4 1152 7 2      x 1(cii) 04 1152 7 2      x 4 1152 7 2      x 905.32 7      x 41.041.7  xx

4 CHIJ SNGS Preliminary Examinations 2017 - Mathematics 4048/02 2 The diagram shows a road junction at O with PO perpendicular to OQ and OP = 60 km. A cyclist, X, starts from P and travels towards O at a constant speed of 10 km/h. At the same time, another cyclist, Y, starts from O and travels towards Q at a constant speed of 15 km/h. (a) Cyclist X leaves from P and cyclist Y leaves from O at noon. Given that t is the time in hours, write down, in terms of t, (i) the distance of X from O after t hours, [1] (ii) the distance of Y from O after t hours. [1] (b) After t hours, the cyclists are 80 km apart. Write down an equation in t to represent this information and show that it reduces to .01124813 2  tt [2] (c) Solve the equation .01124813 2  tt [3] (d) Find the distance of cyclist X from O when the cyclists are 80 km apart[1] 2(ai) t1060 cm 2(aii) 15t cm 2(b) 222 80)15()1060(  tt 640022510012003600 22  ttt 028001200325 2  tt 01124813 2  tt 2(c) 01124813 2  tt 26 )112)(13(44848 2 t 62.131.5  tt 2(d) 60 ̶ 10(5.313) = 6.87 km Q 10 km/h P X O 15 km/h Y

5 CHIJ SNGS Preliminary Examinations 2017 - Mathematics 4048/02 3 The diagram shows a circle ABCDE, with centre O. AEF is a straight line. Angle BAE = 100 and angle CEF = 120. AE = ED. EC is a diameter. (a) Show that triangle ACE and triangle DCE are congruent. [2] (b) Find, giving reasons for each answer, (i) angle ECD, [2] (ii) angle ABD, [1] (iii) angle ECB. [1] (c) Given that AC = 8 cm, find the diameter of the circle. [2] 3(a) In  EDC and  EAC EDAE (given) ECEC (common side) 90 EACEDC (rt  in semi-circle)  EDC ≡  EAC (RHS) 3(bi)  60120180 AEC (s in a st line)  306090  ECDACE (part (i) ) 3b(ii)  60302  ACDABD (s in the same segment) 3b(iii)  80100180 ECB (opt  in cyclic quad) 3(c) EC 860sin  cm 24.9AC B C 120 O D F E A 100

6 CHIJ SNGS Preliminary Examinations 2017 - Mathematics 4048/02 4. The first four terms in a sequence of numbers, ..,.........,,, 4321 uuuu are given below. 1030 1 u 5231 2 u 13432 3 u 33633 4 u (a) Write down an expression for 5u and show that .895 u [1] (b) Write down an expression for 8u and evaluate it. [1] (c) Find an expression, in terms of ,n for the nth term, ,nu of the sequence. [2] (d) Explain why the value of nu must be odd for all values of n. [2] (e) (i) Show that   221 3)2()3(3   nnn [2] (ii) Find , and simplify, an expression, in terms of n , for 1 nn uu . [2] 4(a) 89834 5 u 4(b) 2201143 7 8 u 4(c) )1(23 1   nu n n 4(d) 13 n is an odd number for all values of n, )1(2 n is always an even number for all values of n , The sum of odd no and even no gives odd number 4e(i)   2221 )3(2)13(3)3(3   nnnn 4e(ii) )2(23 2 1    nu n n )1(23 1   nu n n 423223 21 1    nnuu nn nn 233 21   nn   232 2  n

7 CHIJ SNGS Preliminary Examinations 2017 - Mathematics 4048/02 5 (a) The diagram shows an open container A, made from a cylinder and a cone. The cylinder has a radius of 8 cm and a height of 16 cm. The cone has a height of 12 cm. The container is completely filled with water. Find the volume of water in the container. [3] (b) The diagram shows a pot which is part of a right circular cone of height h cm. The open end of the pot is a circle of radius 6 cm. The base of the pot is a circle of radius 4 cm. The height of the pot