YCKSS EM P1 PRELIM MS 2025

1 Yio Chu Kang Secondary School 4E/5N Prelim Mathematics P1 2025 Marking Scheme Qn Solutions Marks Marker’s Comments 1 ( )5 4 3 2 6 5 4 3 2 6 4 13 p p p p p p p − − + + − = − − − + − =− M1 A1 Well done 2a 03x or x A1 Poorly done 2b 3 B1 3 ( ) ( ) ( ) ( ) 713 2 5 3 12 8 6 2 3 2 3 5 3 4 7 1 24 6 4 15 9 28 28 24 37 23 24 xxx x x x x x x x +−− −− − − − −  += − − + − −= −−= M1 M1 A1 Many did not use LCM of denominators, hence some final answers were not in lowest term 4a 35 B1 4b Incorrect. Median mark means 50% of students scored both more than or less than the mark. B1 Poor 5a ( ) 2224 15 9 3 8 5 3 pq pq p q pq q p +− = + − B1 5b ( ) ( ) ( )( ) ( )( ) 2 2 5 2 5 2 5 5 2 2 5 1 2 1 1 5 2 accept 1 2 5 a bc ac ab ac a bc ab ac ab ac ac ab ac ab + − − = − + − = − − − = − − − − M1 A1 6a 1.25 : 1 11 : 1.25 1010 : 8 1.25 km cm km cm km cm = M1 A1 Poor conversion of units 6b 2 2 2 2 2 2 2 2 1 :1.25 1.5625 6.56 : 6.56 1.25 10.25 cm km km cm km km = = M1 A1 Some did not take square of the map scale 7 ( ) ( ) ( ) ( )23 3 23 2 23 23 23 80 5 115 80 7 2, 9, 16, 30 d d d d d d w x y z − + − + − + + + = − + = = = = = = B2 Deduct 1 mark for each error Poorly done despite being a common past yr’s O lvl qn

2 8a ( )2 44 28 28 44 72 76 72 72 76 32 72 108 72 1.5angle AOB       − + + = − = − − −= =− M1 A1 Poorly done as many used the formula for degree 8b ( ) ( ) ( ) 22 2 11 44 1.5 28 1.522 946 3 area cm sf =  − −  − = M1 A1 (ecf) Poorly done as many used the formula for degree 9a 11 55cos or cos12 12 55cos cos 65.37512 12 114.624 180 65.375 114.6 ACD ACB ACD ACB ACD −−  =−  =  = −  = =  =   = −  = M1 A1 Some did not leave answer in 1dp 9b ( )360 114.624 90 335.4− −  =  B1 Some did not leave answer in 1dp 10ai 60 – 30 = 30 M1A1 10aii 36 marks B1 10b 390 180 21 400 40 − = (accept 0.525 & 52.5%) M1A1 11a 168.0 (4sf) B1 Generally ok 11b 2 2 22 2 2 2 2 2 2 2 2 xyPw w xy wPw x y w Pw w Pwx y w Pwx y =− =− =− −= −= M1 M1 A1 Some wrote: 2 2 2 2 xyPw w Pw x y =− =− Quite a number still weak in algebraic manipulation . Many did not include the  12 22 2 2 2 2 2 26 676 24 10 676 AB AC BC AB == + = + = = By the converse of Pythagoras’ Theorem, ∆ABC is a right-angled triangle. Hence, by the converse of right-angle in semi-circle, AB is the diameter of the circle. M1 M1 A1 Poorly done as students are weak in explanation. Did not use the word ”converse” when explaining using Pythagoras’ Thm

3 13 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 8 3 25 1 12 5 9 2 1 5, 40 15 125 3 2 3, 36 15 27 4 3 4 , 76 152 2 3 xy xy xy xy x x y − = −−−−− + = −−−−−  − = −−−−−  + = −−−−− += = =− M1 A1 A1 Generally well done. 14 S1 – correct shape MP1 –