SJI 2025 EMATH PRELIM P2 MS

1 ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) MA4052 Paper 2 Solutions Q/N Solutions 1(a)(i) 3 2 5 na n k  = 13 3 1 22 53 3               = 1 2 1(a)(ii) 3 2 5 na n k  2 3 2 5 na n k  2 22 5 3a n a k n  2 22 3 5a n n a k  2 2(2 3) 5n a a k  2 2 5 2 3 a kn a   or 2 2 5 3 2 a kn a 

2 1(b) 4 5 5 16 4 x x   4 5 4 ( 5) 6 4 x x   4 5 4 5 6 4 x x   4 5 1 6 4 x x   6( 1)4 5 4 xx    4(4 5 ) 6( 1)x x    16 20 6 6x x    14 22x   22 14x  11 7x or 417x 1(c) 2 3 51 1 x x x x     ( 2)( 1) ( 3)( 1) 5( 1)( 1) x x x x x x        2 2 2 ( 2 2) ( 3 3) 51 x x x x x x x         2 3 2 4 3 51 x x x      2 7 1 51 x x    27 1 5( 1)x x    25 5 7 1 0x x    25 7 4 0x x   27 (7) 4(5)( 4) 2(5)x     7 129 10x   0.4357816692x  or 1.835781669 x  0.44 or 1.84 (2dp)

3 2(a)(i) 114.9 million 6114.9 10  2 61.149 10 10    81.15 10 2(a)(ii) Population density in 2025 = 6116.79 10 300000  = 389.3 Since 389.3 > 350 The Philippines had exceeded the government’s population density. 2(b) % change = (2.8 1.5 5 ) 10 100%10 x x x x x     = 9.3 10 100%10 x x x   = – 7% 2(c) Total cost = 12 1.1 1.09x   = 14.388x Total bill = 30 11 29.7 14.388 x   25x  3(a) V olume of the solid       21 4 2 4 42 x x x x x    

4   2 216 8 (16 8 )x x x x x x      3 2 3 28 16 8 16x x x x x x      3 22 16 32x x x   3(b) The dimensions of the figure must be greater than 0. Therefore, 4 0 x  4x and 0x 0 4 x   3(c) 1.75 3(d) 3(e) 0.65 3(f) From the graph the maximum value is 18.9, the volume cannot be greater than 18.9 cm 3 . Hence the volume of the solid cannot be 20 cm 3 .

5 4(a) Using similarity, 24 10 18 h h  18 24 240h h  40h V olume of frustum =     2 21 1 12 40 9 303 3    = 1110 cm3 = 1.11 litres 4(b)   23 20 11104 r  2 74r  74 8.6023r  Therefore, area not in contact of water =    12 74 20 4      270.250 = 270 cm2 (to 3 sf) 5(a) 3 2 6x y  --- (1) 5x y  --- (2) From (2) 5x y  --- (2*) Sub (2*) into (1)  3 5 2 6y y   9 5y Sub