SJI 2025 EMATH PRELIM P1 MS

1 ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2025 (YEAR 4) MA4052 Paper 1 Solutions Q/N Solutions 1 26450 26549x  2 I = 50000 12 100 qr  125 3 rq or 2413 rq 3    2 2 29 10 8cos 2 9 10BDC    = 13 20 13cos 20BDC  4(a) P(walks to school) = 11 5 = 4 5 4(b) No. of days =  4 1905 = 152 5 (2 2) 180 ( 2) 180 202 n n n n      (2 2)(180) (2 4)(180) 40n n n    360 40 n n 9 6(a) Median mark = 34 6(b) Max mark 17 32 

2 Max mark 49 x  9 7(a) % reduction = 142 130 100%142   = 8.45% (to 3 sf) 7(b) The graph gives the impression that water consumption in 2021 doubled compared to 2018/19, when the actual increase is only about one-third. This misinterpretation arises because the vertical axis does not start from zero, exaggerating the fluctuations and making the changes appear more significant than they really are. OR It misleads us into thinking that there is a gradual downward trend in water consumption from 2024 to 2030, which might not be the case. 8 3 3 6 8 2 2 8 6 16 9 9 16 m q q m            12 9 27 64 q m 9 218 2 3    6 2 3N   = 46 656 10 2 1 3 2 2 7 10 x x x x    1 3 2 2 ( 2)( 5) x x x x      5 3 2 ( 2)( 5) x x x x       4 3 ( 2)( 5) x x x     11(a) 2 2 5 17 5T    314 11(b) 2 2(4 3)nT n n  

3 2 216 24 9n n n    217 24 9n n   12(a) x = 9 12(b) 3 8 6 2 1x     x=14 13 Let x be the total distance travelled. 0.35 0.65 20 30 x x x 1.05 1.3 60 x xx      2.35 60 xx   25.5 km/h (to 3 sf) 14(a) Set B has the larger standard deviation. All values in Set A are the same, so there is no spread and the standard deviation is 0. Set B’s values are spread out around the mean, so its standard deviation is greater than 0. 14(b) 14 Adding 10 to each score does not change the standard deviation, so it remains 7. Multiplying every score by 2 doubles the standard deviation. 15(a) 6, 9, 12, 15, 18, 21, 24 15(b) Probability 9 24 = 3 8 16(a) 90BAD    tan rad  90 in semicircleACB     90 adj s on a str lineACD    90BAD ACD    BDA ADC  (common angle) CDA and ADB are similar (AA Similarity Test) 16(b) 25 9 x x 2 225x  15x ( 0)x

4 17 2F kv 2( )new newF k v 2(0.8 )newF k v Percentage change in F = 2 2 2 (0.8 ) 100%k v kv kv   36% 18(a) 2 169x   ( 13)( 13)x x  18(b) 21431 40 169 