YCKSS EM P2 PRELIM MS 2025

1 Yio Chu Kang Secondary School 4E/5N Prelim Mathematics P2 2025 Mark Scheme Qn Solutions Marks Marker’s Comments 1(a) 5 2 3 7 5 2 3 2 3 7 1 5 15 Prime numbers are 2, 3, 5. x xx xx x −  −  −  − −  −   −   M1 M1 A1 Many students did not combine final inequality to 15 x−   . Avoid doing by guess-and-check. Form simultaneous inequalities to solve. Those who simply wrote 2, 3, 5 get a maximum of 2 marks. Some still think that negative numbers and ‘1’ are prime. 1(b) 3 3 5 2 3 33 3 5 2 22 18 3 5 10 18 10 53 12 c d c d ff c d f f c d d cf  = = B2 (deduct 1 m for 1 error) No mark awarded for ‘keep, change, flip’. 1(c) 3 8 4 12 3 4 12 4 48 9 6 16 81 3 2 27 8 y x x y x y −   = = M1 A1 Cannot have a fraction within another fraction. That is not simplified.

2 1(d) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 32 2121 3 2 2 1 21 3 4 2 21 2 21 x xx xx x xx x x x − −− −−= − −+= − −+= − M1 A1 2(2 1) (2 1)(2 1)x x x−  − + Many still get this Sec 2 identity mixed up. 2 (a)(i) 120 32 15 1 5 5 67 68 335 3 15 3 4 5 1 1 0 8      =    = =  T T T M1 A1 1 mark awarded for matrix multiplication without 1 1 1    matrix. (a)(ii) The elements represent the total number of mangoes, apples and oranges sold by seller A and B respectively in 5 days. B1 “Weekdays” not allowed. It is ‘5 days’, “weekday” or “weekend” not mentioned. Total number, not average number. (b)(i) 80 70 0.80 0.70 50 40 or 0.50 0.40 60 50 0.60 0.50        ==          CC B1 (b)(ii) 80 7020 32 15 50 4015 35 18 60 50 4100 3430 41.00 34.30 or 4030 3350 40.30 33.50  =      =       P P A1 for 2 correct elements A2 for 4 correct elements

3 (b)(iii ) Seller A receives a total of $41 while seller B receives $33.50. Seller A made a higher profit. B1 “Total sales/total revenue” not “total selling price” (c) 0.8 0 0 80 70 0 0.9 0 50 40 0 0 1 60 50 64 56 45 36 60 50      =        = N N M1 A1 Many still do not know that the orders of matrices are important for matrix multiplication. 3(a) 17 ( 1)Gradient of 7 ( 2) = 2 1Gradient of 2 1Equation of : 2 Sub (20,3) 13 (20)2 13 1Equation of : 13 2 PR SQ SQ y x c S c c SQ y x −−= −− =− =− + =− + = =− + M1 M1 M1 A1 Error carried forward (ecf) maximum of 2 marks.

4 (b) 22 22 2 [5 ( 4)] (3 ) 15 9 (3 ) 225 (3 ) 144 3 12 or 3 12 9 or 15 Since 0, 15 h h h hh hh hh − − + − = + − = −= − = − =− =− = = M1 M1 A1 281 ( 3) 9 ( 3)!!!hh+ −  + − For example, 81 25 10.3+= correct 81 25 9 5+ = + 14= 3.74= incorrect!!! (c) 2 6 0 26 6 2 3 x ky y kk k k + − = =− + =− =− M1 A1 Many wrote 2 6 2x ky+ − =− !! ‘ 2− ’ is the y- intercept, so you should substitute (0, 2)− into the original equation 2 6 0x ky+ − = .

5 4(a) 0.66 B1 Quite a few did not follow 2 d.p. required!! (b) P2 – pts plotted correctly C1 – smooth curve Some curved were not even S-shaped! A cubic graph is S-shaped. Usually a result of one or two pointed incorrectly plotted. Do not extend your graph beyond the first and last points in the table. It is a curve. No part should be linear/straight- line. Many curves were not smooth; some parts had “double lines”. Some were using pencils not sharpened or not dark enough. Some lost marks as curve did not go through the middle of the plotted X. (c) ( ) ( ) 32 2 2 6 8 12 6 8 12 8 6 34 Draw the line 3 From the graph, 4.9 0.1 x x x x x x x xx y x − + = − + = − + = = = M1 A1 Poorly done.

6 (d)(i) 1Line : 2 Sub (3,1), 11 (3) 2 12 2 15 22 L y x c c c yx =− + =− + = =− + B1 for eq B1 for drawing line or B2 for drawing line Poorly done. (ii) From the graph, 4.2 0.1x= B1 Poorly done. Some wrote unreadable answers. You can only read to the accuracy of half the smallest box. (iii) ( ) ( ) 2 2 32 15864 2 2 8 6 2 10 6 10 10 0 6, 10, 10 x x x x x x x x x x x A B C − + =− + − + =− + − + − = =− = =− M1 A2 for 3 answers Poorly done. 5(a) 2 2 2 5.4 3.1 2(5.4)(3.1) cos128 59.382346 7.705994 7.71 km (to 3 s. f.) HB HB HB HB = + −  = = = M1 A1

7 (b) 180 133 47 5.4 6.5 sin 47 sin 6.5sin 47sin 5.4 61.68256 133 61.68256 71.31744 71.3 (to 1 d.p.) (shown) HCA CHA CHA CHA CAH CAH CAH  = − =  = =  =   = −  =  =  M1 M1 A1 When question asks for a number to “show”, you need to show the unrounded number before the rounded number that you are aiming to get. (c) Draw North line at 180 61.7 118.3 (int ) Bearing of from 360 128 118.3 =113.7 A NAH s BA  = − =   = − −   M1 A1 Poor presentation in finding bearings. Many still rounded to 3 s.f. (d) tan14 3.1 0.772917 km 0.772917tan 6.5 6.8 (to 1 d.p.) angle of elevation 6.8 h h   = = = = = M1 M1 A1 A few students have ecf mark(s) here.

8 6(a) Time taken to travel from A to B = 12 h3x− Time taken to travel from B to C = 9 h3x+ ( ) ( ) ( )( ) ( ) 2 2 2 12 9 1 13 3 4 12 3 9 3 5 3 3 4 48 144 36 108 5 9 84 36 5 45 5 84 81 0 (shown) xx xx xx x x x xx xx +=−+ + + − =−+ + + − = − + = − − − = M1 A1 M1 M1 A1 Many students were not able to connect speed, distance and time to form an equation. (b) 284 ( 84) 4(5)( 81) 2(5) 84 8676 10 17.7145 or 0.91450 17.71 or 0.91 x x x x  − − −= = =− =− M1 A1, A1 Some students failed to give the answers correct to 2 d.p. (c) The solution x = 0.91− must be rejected, as it results in the speed x km/h, to be a negative value. B1 Some students took x to be time rather than speed. (d) 12 9Difference h 17.7145 3 17.7145 3 0.381044 h = 22.8626 min = 22 min 52 sec =− −+ = M1 A1 Some used the 3 s.f. value of 17.7 to find the difference in time. This resulted in inaccurate answer. Some students do not know how to convert minutes to minutes and seconds.

9 7(a(i) 100 120p B1 Most were able to get the correct interval. (ii) 125.6 mmHg B1 Well done (iii) 20.1 mmHg B1 Well done (iv) The blood pressures of the second group of patients is less consistent than that of the first group of patients, since it has a higher SD compared to the first group. B1 Good. Many correctly used the terms ‘less consistent’ or ‘more consistent’. (b)(i) Group B: 142, 153, y, 157, 158, 160, 162, 164, 178 3 1 1 1 162 164 1632 163 10 153 153 1532 153 3 (shown) Q Q Q yQ y x +== −= = +== = = M1 M1 A1 Poorly done. A number of students gave wrong presentations: 15x means 15 multiplied by x. 1 153 15 1532 15 306 153 15 153 3 xQ x x x +== =− = = Wrong interpretation: x is not the height 1 153 1532 306 153 153 3 xQ x x x +== =− = = (ii) 51Prob 10 2== B1 7 10 was a common answer, not realising that lengths 152 cm and 164 cm are not included. (iii) 5 3 2 3Prob 10 9 10 9 7 = 30 =  +  M1 A1 Many had only one possible outcome, 53 ,10 9 not realizing that there was another possible outcome.

10 8(a(i) BAC = DAB (common) ACB = 90 (angle in semicircle) ABD = 90 (angle in semicircle) ACB = ABD Triangle ABC is similar to triangle ADB (AA Similarity Test) M1 A1 (only if both proofs are given) Poorly done. Instead of proving 2 pairs of equal angles, answers such as AB is common were seen. Some proved 1 pair of equal angles only and concluded that the 2 triangles were similar. (ii) 2 18 6 108 10.4 cm (to 3 s. f.) AB AD AC AB