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  (40 13)(40 13)    (53) (27) 2 factors are 27 and 53. 19(a) 6.25 cm2 rep 1.44 km2 2.5 cm rep 1.44 km 1: 1.2 100000 2.5  1: 48 000 19(b) Length (on the map) = 100 48000 x = 480 x cm 20(a) 3120 2 3 5   284 2 3 7   596 2 3  Largest possible length = 22 3 = 12 20(b) No. of cubes = 120 84 96 12 12 12  = 560

5 21(a) 22 4 2x xy x y   =    2 2 2x x y x y   =   2 2 1x y x  21(b) 22 8 7 0x x   2 74 0 2x x     2 2 72 2 0 2x      2 152 2x  152 2x  4.74 or 0.74x x   22(a) 8ADO   (tangents from ext point where line OD bisects ADB ) 8tan 8 AD   8 tan 8 19.3137 AD   = 19.3 cm (to 3 sf) 22 (b) Shaded area =   21 12 19.3137 8 8 22 2 2 2 4             79.111 = 79.1 cm 2 (to 3 sf) 23(a) 4 2 x        N

6 23(b) 2 1 3 4 3 2 1 2 1 4 2 x             L = 10 3 16 12 2 x x x        23(c) Matrix L represents the total litres of red, blue, and yellow base colors, respectively, needed by the shop to create the three shades, A, B, and C. OR 10 3 x represents the total litres of red base colour needed by the shop to create the 3 shades, A, B and C. 16 x represents the total litres of blue base colour needed by the shop to create the 3 shades, A, B and C. 12 2 x represents the total litres of yellow base colour needed by the shop to create the 3 shades, A, B and C. 23(d) Each colour must use at most 25 litres. For Red: 10 3 25 5 x x    For Blue: 16 25 9 x x    For Yellow: 12 2 25 6.5 x x    Largest integer of x = 5

7 24 (a) 24 (b) 5 1 5 8 3 12 x x   1 2 3 3 x x   3 3 2 6x x   3x  24 (c) As x increases, the number of red balls in Bag B increases (or decreases), while the number of blue balls remains the same. This makes it less likely to draw a blue ball from B. The probability of drawing 2 blue balls decreases (or increases) as x increases (decreases). Red Blue Red Blue Red Blue

8 25(a) 1x  25(b) 6 0 0 6m   = –1 25(c) The number of real solutions to the question is equal to the number of points where the horizontal line y m meets the graph. 25(d) Draw 0.5 2.5y x  From the graph, 1 or 5x   26(a) Lily stopped (or rest / not moving / at Town B). 26(b) Average speed = 30 14 48 60  = 20 km/h 26(c) Time taken = 20 216 3 = 1h 12 min Correct straight line drawn.

9 26(d) 26(e) Distance from Town D = 9 km