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 9 5y into (2*) 95 5x      = 16 5 T is 1 43 ,15 5     

6 5(b) 5x y  A is  5,0 B is  0,5 Length of AB     2 2 5 5  = 5 2 (= 7.07106) [Using area of triangle OAB]     1 17.07106 5 52 2 h 3.54h units (to 3 sf) 5(c) Let m be the distance from T  1 5 3.2 182 m   6.25m 4.45y  or 8.05y 6(a) DF AF (given F midpoint of AD) 90BFD BFA    ( bisector of chord) BF is common triangle ABF  triangle DBF (SAS) 6(b)(i) DAC DBC   45 ( s in the same segment) 6(b)(ii) 90 (ADC    in a semicircle) 180 90 45DCA     ( s sum of a triangle)  45 6(b)(iii) 45DBA DCA    ( s in the same segment) 45 (2OBA   triangle ABF  triangle DBF) 22.5  22.5OAB OBA    (base  s of an isosceles triangle) 180 22.5 22.5AOB     ( s sum of a triangle)  135

7 6(b)(iv) 135 2ACB   ( at centre = 2  s at circumference) 67.5  180 67.5 45CEB     ( s sum of a triangle)  67.5 AED CEB   67.5 (vertically opposite  s) 7(a) 200 sin 44.2 sin 95 BD   200 sin 44.2sin 95BD    139.965 140 m 7(b)   13300 48 139.965 sin2 CBD    3300 2sin 48 139.965CBD   79.230CBD  bearing of C from B  360 180 45.8 95 79.230        051.57 = 051.6o (to 1 dp) 7(c) Let h be the height of the vertical tower. sin 79.230 48 opp 48 sin 79.230opp   47.1544 80tan 47.1544 e  59.5e   (to 1 dp)

8 8(a) AC AB BC      3b  2a 8(b)   1 1 2 2XZ AZ AC CZ       1 2 (3b  2a + a) = 3 1 2 2b a 8(c) CX CZ ZX     a + 1 2 a  3 2 b = 3 3 2 2a b 8(d) CY CA AY      3b + 2a + b  2 2a b 8(e) Since 4 3CY CX   and C is a common point, C, X and Y are collinear. 8(f) area of area of area of area of area of area of ACY ACY ABC ACZ ABC ACZ       1 2 3 1  2 3

9 9(a)(i) 56 minutes 9(a)(ii) IQR = 65 51.5  13.5 minutes 9(a)(iii) Number of students who took more than 70 minutes 600 490  110 Probability that a pupil chosen at random from the school took more than 70 minutes 110 11 600 60  9(b)(i) The median time taken by students from School X to complete the task is 56 minutes, while for School Y it is 52 minutes. This means that students from School Y, on average, complete the task faster than those from School X. 9(b)(ii) For School X: Upper boundary = 85.25 For School Y: Upper boundary =89.25 88 minutes is considered an outlier for School X, but not for School Y . Therefore, 88 minutes would be considered more unexpected for School X. 9(c) Since both tasks have the same IQR, the steepness of the curve will remain the same, but the curve will be shifted to the left because of the smaller median.

10 10(a) Total Stamp Duties = 1% 180000 2% 180000 3% 640000 4% 500000 5% 500000         + 17% 2000000 = 69 600 + 340 000 = $409 600 If sold in April 2025 ( > 2 years) Cost incurred = 2 000 000 + 409 600 = 2 409 600 Proceeds after SSD = 2 300 000 – 0.04*2 300 000 = 2 208 000 Net Profit = 2 208 000 – 2 409 600 = – 201 600 ----------------------------- if Sold in March 2026 ( > 3 years) Future price   3 2000000 1.07  = 2 450 086 SSD (more than 3 years) dropped to 0% Proceeds after SSD = 2 450 086 Net Profit = 2 450 086– 2 409 600 = $40 486 If Rachel in April 2025: Loss = 201 600 Wait for another year: Net profit = $40 486 The drop in seller’s SSD to 0 and potential capital appreciation (property rise of 7%) mean Rachel stands to gain much more by waiting until April 2026.