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 2 3 5 Since 54 150 2 3 5 ,54 150 is perfect square. Since the indices of all prime factors are multiples of 2 (or even), 2 3 5 is a perfect square. OR                         M1 – show prime factorised expression AG1 N1 AO3 13b 2 2 2 150 2 3 5 2 3 5 180 the smallest possible integer 180. k k k k          B1 N1 AO2 14a 5 16 5 4 5 3 16 4 5 3 x y y x x y x y y x x y            B1, B1 for forming any 2 equations N7 AO2

5 Q Solution Marks AO 14b 2 2 5 16 6 15 (1) 5 4 5 3 4 3 (2) sub (2)into(1) 6(4 3) 15 24 18 15 9 18 2 Sub y = 2into(2) 4(2) 3 5 5, 2 4(5) 5(2) 3 27 27Perpendicular ht from to 27 2 23.382 1 2 x y y x x y x y x y x y y y y y y y x x x y AC cm B AC cm Area of ABC                                      2 27 23.382 315.657 316 (3 )cm sf    OR 2 60 ( ) 1 27 27 sin 602 315.666 316 (3 ) ABC angles of an equilateral Area of ABC cm sf            M1 - Using Substitution or Elimination method A1, A1 A1 A1 N7 AO2

6 Q Solution Marks AO 15a 1 2 2 1 3(2 ) 4( 1) 6 3 4 4 1 (4 4) 6 3 4 816 2 22 2 22 2 2 1 (4 4) 6 3 1 4 4 6 3 1 4 6 9 y y y y y y y y y y y y y y                          M1 - 1 2(4 2) or 4( 1)2 y or 3(2 )2 y or 1 (4 4)2 y  M1 - 1 4 4 6 3y y    A1 N1 AO1 15b 500 2 250 250 250 250 250 250 2 (2 ) 4 4 5 I disagree with her claim because 4 5 .     M1 - 2 250 250(2 ) or 4 AG1 – must state 250 2504 5  o.e. N1 AO3 16a 5: 7 B1 G2 AO1 16b 3 3 5 7 125 36 343 125 36343 4500 343 450036 343 22.8804 22.9 (3 ) Mass of X Mass of C Mass of X Mass of X kg mass of Y kg sf          M1 – finding 3 3 5 7 A1 G2 AO2

7 Q Solution Marks AO 17ai   2 2 2 2 2 2 12 1212 5 12 5 2 2 12 36 36 5 ( 6) 31 6, 31 x x x x x x x a b                             B1, B1 N7 AO1 17aii 6x B1 N7 AO1 17bi B1 – correct shape of grph B1 – x and y intercepts shown N6 AO1 17bii  1,16 B1 N6 AO1 18a  3,0Q B1 G6 AO2 18b 0 4 3 ( 3) 4 6 2 3 6 4 2 ( 3) 3 2 2 3 3 3 3 6 x x x x             M1 – finding gradient A1 G6 AO2 Q Solution Marks AO O y x 15 3 -5

8 18c 2 3 2 3 ( 3, 4) 24 ( 3)3 4 2 2 2 2 . .3 m y x c Sub c c c y x o e             B1 G6 AO1 19ai (2 1.8) rad  B1 G5 AO1 19aii (2 1.8) 5 (10 9) cm     B1 G5 AO1 19b 2 2 2 2 2 2 2 1sec 5 1.8 2 22.5 1 5 5 sin1.82 12.173 (5 . .) se 22.5 12.173 10.327 10.3(3 ) Area tor OADB cm Area of AOB rad cm s f Area of shaded gment cm cm cm sf cm               M1 M1 A1 G5 AO2 20ai Integers that are perfect squares. B1 N8 AO1 20aii {1,9} B1 N8 AO1 20aiii 11 B1 N8 AO1 20bi B1 N8 AO1 Q Solution Marks AO P Q

9 20bii      ' ' ' 'P Q or P Q or P Q P Q     B1 N8 AO1 21ai 2 2 2 2 2 2 2 2 2 8 6 100 10 100 Since By the converse of Pythagoras' thereom, 90 . AD DC AC AD DC AC and ADC            OR 2 2 2 1 Using Cosine rule, 10 8 6cos 2(8)(6) 90 ( ) ADC Shown            M1 – show Pyth Thm AG1 M1 – show Cosine Rule AG1 G4 AO3 21aii 6cos 10 3 5 ACB   B1 – lowest term G4 AO1 21b 1 sin 0.8929 sin 0.8829 63.239 180 63.239 63.2 116.8(1 ) x x x or x or dp       B1 B1 G4 AO1 22ai (5 2)180 5 108 BAE     B1 G1 AO1 22aii 180 108 ( . )2 36 AEB base s of isos      B1 G1 AO1 22aiii are congruent to . 108 36 36 36 AEB DEC BEC BAE AEB DEC BEC             B1 G1 AO2

10 22b 180 108Sin 36 ( ), 2 36 , , . , . ce ECD base of isosceles BEC ECD by the property of converseof alternate angles BE is parallel toCD OR They form a pair of alternate angles BE is parallel toCD           B1 – converse of alternate angles G1 AO3 23a 5p B1 S1 AO1 23b 19 27 27 19 8 LQ UQ IQR      M1, A1 S1 AO1 23c is any positive integer 22x  B1 S1 AO1 24i 44 11 64 16 B1 S2 AO1 24ii 20 3 64 14 14(20 ) 3(64 ) 280 14 192 3 11 88 8 x x x x x x x x           M1 A1 S2 AO2 25a 2.530 30 35 42 40 5.5 387.5 2.5 4 8 220 387.5 6.5 228 R x x R x x R x                         B1 N9 AO1 25b The total amount of money collected from Outlet A selling the three types of drinks on a particular day. B1 N9 AO3