2024 FMSS EMATH P1 PRELIM MS

1 Sec 4 Exp/5 NA Prelim Paper 1 2024 Q Solution Marks AO 1 :16 3: 20 3 16 20 3 1620 122.4 5 p p p or      B1 N3 AO1 2a 22 2 (2 ) (2 ) ( )(2 ) ( )( 2 ) a c ad abc bd a ac d b ac d a b ac d or b a d ac             B1, B1 / B2 N5 AO1 2b    2 2 2 2 7 4 3 7 21 4 12 7 17 12 x y x y x xy xy y x xy y          M1 A1 N5 AO1 3 7, 7, 8, 15, 28 B1 – 7, 7, 8 B1 – 15, 28 S1 AO2 4a ( ) ( ) Since 2 pairs of corresponding angles are equal, triangle and are similar. OR by AA CAB BAD common angle ABC ADB given ABC ADB SimilarityTest     M1 – show two pairs of corresponding angles are equal AG1 – correct reason G2 AO3 4b 8 8 5 8 85 412.8 12 . . 5 AC AB AB AD AC AC AC m or m o e      B1 G2 AO1 5 It may mislead readers thinking the number of EVs manufactured in 2023 is at least twice the number manufactured in 2022 based on the height or size of the picture. B1 – Accept any similar answers on comparing size of the pictures S1 AO3 6 2 2 2 2 2 2 2 1 : 2000000 1 : 20 1 : 20 1 : 400 400 0.55 220 cm cm cm km cm km cm km km  M1 either 202 or 400 A1 N2 AO1

2 Q Solution Marks AO 7a 4+7(n-1) or 7n-3 B1 N5 AO1 7b If 121 is a term, 7n-3=121 7n=124 n= 17.714 (5sf) or 124 5 177 7or o.e. Since n is not a positive integer, 121 is not a term B1 with working of showing n = 17.714 N5 AO3 8 2 2 2 2 2 4.5 (4.5 ) 20.25 4 81 4 181% 100 1 7700 5% 95 % 95.1%(3 . )81 81 kF d k Fd new d d knew F d Fdnew F d new F F change or or s f             M1 – show 1 4 20.25 81or A1 N2 AO2 9a n = any negative odd integer (-1, -3, -5 etc) B1 N6 AO1 9bi B1 – line must cut y = 1 N6 AO1 . O y x 1 1

3 Q Solution Marks AO 9bii M1 - The graph must be above (1,1). N6 AO1 10 2.5 8 100 10 100 5 4 100 2 5 100 4 1005 5 1002 5 4 100 2 5 100 5 4(100 ) 2 500 2500 4(100 )2 2500 8(100 ) 2500 800 8 8 1700 212.5% newArea oldArea oldArea AB h xnewArea AB h xoldArea AB h xAB holdArea oldArea AB h x x x x x x x                        M1 – form the equation for new Area A1 N3 AO2 11a SD = 1.23 (3s.f.) B2 S1 AO1 . O y x 1 1

4 Q Solution Marks AO 11b As all the marks are increased by 3 marks, the mean will also be increased by 3 marks but the standard deviation will remain the same hence the spread of marks remains unchanged. B1 S1 AO3 12a, b B1 – construction of perpendicular bisector B1 – construction of angle bisector G1 AO1 AO1 12c The point P is equidistant from the lines AC and BC and equidistant from the points B and C . B1 – all correct G1 AO1 13a     3 2 3 2 2 4 2 22 22 2 4 2 54 2 3 150 2 3 5 54 150 2 3 2 3 5 2 3 5