AISS Prelim EMath Paper 2 Marking Scheme

1 2025 Sec 4E Math Prelim Paper 2 Marking Scheme Qn Working Mark Awarded Sub- total Remarks 1 (a) 5 2 1 3xx−  −  + 5 2 1x−  − and 2 1 3xx−  + 42 x− 4x 2 x− 24 x−   Primes that satisfy the inequality: 2, 3 M2 A1 3 M1 for 2 x− M1 for 4x (b) 22 330 11 25 22 xz y xy z  = 22 3 30 22 11 25 x y y z xz = 3 4 12 5 xy z B3 2 4 terms to simplify: 12 5 , x, y3, z4. B1 for one correct term in the correct position (numerator or denominator). B2 for two or three correct terms. B3 for all four correct terms. (c) 3 12 4 1681 p q −   = 3 16 4 12 81q p   = 12 9 27q p B2 2 B1 for numerator, B1 for denominator. (d) 4212 x xx +=−− 2( 1) 4 12 xx xx +− =−− 3 2 4 12 x xx − =−− (3 2)(2 ) 4( 1)x x x− − = − 6x – 3x2 – 4 + 2x = 4x – 4 –3x2 +4x = 0 x(–3x + 4) = 0 x = 0 or –3x + 4 = 0 x = 4 3 M1 M1 A1 3 M1 for correct quadratic equation. M1 for correct factorisation or correct substitution into quadratic formula A1 for both correct answers.

2 Qn Working Mark Awarded Sub- total Remarks 2 (a)(i) ( –4, 9) B1 1 (a)(ii) riseGradient = run 5 = 3 5 = 3 − − y = mx + c y = 5 3 xc−+ When x = –1 and y = 4, 54 ( 1)3 c=− − + 7 3c= 57 33yx=− + M1 A1 2 M1 for correct gradient. (b) D(3, y) Length of CD = 3 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 22 2 2 2 2 22 2 3 ( 2) 5 13 3 ( 2) 5 13 5 5 13 5 13 5 5 144 5 12 5 12 y y y y y y y − − + − = − − + − = + − = − = − −= − = = y = –7 or y = 17 D(3, –7) D(3, 17) M1 M1 A1 3 M1 for identifying that the x-coordinate of D is 3. M1 for correct formula for length of line segment. A1 for both correct answers.

3 Qn Working Mark Awarded Sub- total Remarks 3 (a) Cosine Rule 2 2 2 2 2 2 2 2 2 2 2 2 1 62 37 50 2(37)(50)cos 62 37 50 2(37)(50)cos 62 37 50 cos2(37)(50) 62 37 50cos 2(37)(50) 89.61286341 89.6 (to 1 d.p.) ABC ABC ABC ABC ABC ABC − = + −  − − =−  −− =−  −−=  −  =   =  M1 M1 A1 3 M1 for correct substitution of values into Cosine Rule. M1 for rearranging and making ABC the subject of the equation. (b) Method 1: Interior angles 180o – 124o = 56o (interior angles, parallel lines) 360o – 56o = 304o (angles at a point) Method 2: Alternate angles 180o + 124o = 304o B1 [B1] 1 B1 for 304o Angle properties not required. (c) Sine Rule 1 sin sin 52 73 62 73sin 52sin 62 73sin 52sin 62 CAD CAD CAD −  = = =  Let the North of A be N. Since A is due west of C, 90NAC =  M1 M1 M1 for substituting the values correctly into Sine Rule. M1 for rearranging the equation and finding the value of CAD .

4 Qn Working Mark Awarded Sub- total Remarks Bearing of D from A 1 73sin 5290 sin 62 158.0973671 158.1 (to 1 d.p.) NAC CAD − = + = +  = = A1 3 (d) Using TOA CAH SOH, tan 8 50 50 tan 8 CF CF = = 1 tan 73 50 tan 8tan 73 50 tan 8tan 73 5.498398955 5.5 (to 1 d.p.) CFFDC FDC FDC FDC FDC − = = =   =   =  M1 M1 A1 3 M1 to find correct height of flagpole. M1 for correct equation involving FDC and CF. 4 (a) Town A to Town B Town B Town B to Town C Speed x km/h Speed (x – 20) km/h Time 1 h 30 min x min Total distance ( )(1) ( 20) 60 ( 20) 60 xxx xxx = + −  −=+ 2 2 ( 20) 12060 60 ( 20) 7200 60 20 7200 40 7200 0 (shown) xxx x x x x x x xx −+= +−= + − = + − = M1 M1 A1 3 M1 for correct distance from Town B to Town C ( 20) 60 xx − . M1 for equating total distance to 120. Correct expansion and simplication of equation

5 Qn Working Mark Awarded Sub- total Remarks (b) 2 2 2 40 7200 0 4 2 40 40 4(1)( 7200) 2(1) 40 30400 2 xx b b acx a x x + − = −  −= −  − −= −= x = 67.17797887 or x = –107.1779789 x = 67.18 (to 2 d.p.) or x = –107.18 (to 2 d.p.) M1 A2 3 M1 for substitution of values into quadratic formula correctly. A1 for each correct x. Deduct one mark if not 2 d.p. (c) x = –107.18 is rejected as time/speed cannot be negative. B1 1 (d) Difference in time = 67.17797887 – 60 minutes = 7.17797887 minutes = 7 minutes 0.17797887  60)(  seconds = 7 minutes 10.67873225 seconds = 7 minutes 11 seconds B1 1 (e) Total distance = 120 km Total time = 1 h + 0.5 h + 67.17797887 60 h = 2.619632981 h Average speed total distance total time 120 2.619632981 45.8 km/h = = = M1 M1 A1 3 M1 for correct calculation of total time. M1 for substitution of total distance / total time. 5 (a)(i) 150 160x B1 1 (a)(ii) Estimate of Mean Height 3(145) 10(155) 7(165) 4(175) 1(185) 25 161 cm + + + += = B1 1

6 Qn Working Mark Awarded Sub- total Remarks (a)(iii) 10.19803903 = 10.2 (to 3 s.f.) B1 1 (a)(iv) We do not know the actual heights of the students. B1 1 (a)(v) The mean height of the five students is less that 161 cm. B1 1 5 (b)(i) 31 15 5= B1 1 (b)(ii) P(both different colours) = 1 – P(both same colours) = 1 – [P(red, red) + P(blue, blue) + P(Y, Y)] = 1 – 5 4 7 6 3 2 15 14 15 14 15 14       +  +            = 341 105− = 71 105 Alternative Method P(red, blue) + P( red, yellow) + P(blue, red) + P(blue, yellow) + P(yellow, red) + P(yellow blue) = 5 7 5 3 7 5 15 14 15 14 15 14       +  +            7 3 3 5 3 7 15 14 15 14 15 14      +  +  +            = 71 105 M1 A1 [M1] [A1] 2 [2] M1 for correct calculation. M1 for correct calculation. (b)(iii) P(two red, one blue) = P(R, R, B) + P(R, B, R) + P(B, R, R) = 5 5 7 315 15 15 = 7 45 M2 A1 3 M1 for 5 5 7 15 15 15 M1 for 3 6 (a) 30 310 = m/s2 B1 1

7 Qn Working Mark Awarded Sub- total Remarks (b) Duration of deceleration = 30 152 = s Total time = 45 s OR Method 2: Use gradient 30 0 230 30 2( 30) 30 2 60 45 t t t t − =− =− =− = -------------------------------------- Total distance = area of trapezium = 1 (20 45) 302 +  = 975 km M1 [M1] M1 A1 3 M1 for finding the correct total time of the journey, 45 s. M1 for area of trapezium. 7 (a)(i) 15 10 25 15 1025 6 cm d d d = = = M1 A1 2 M1 for correct equation showing the relationship between the diameter and height of the two similar bottles. (a)(ii) smaller bottle (new) larger bottle 400 ml 1.2 litres = 1200 ml