NASS 2026 MA-O S4&5 Prelim P1 (MR)

1 Ngee Ann Secondary School Secondary 4&5 Elementary Mathematics-O 2026 Prelim paper 1 Marking Scheme Qn No. Qn Part Solutions Marks (Remarks) Total 1 2(3 2)(8 5 1)x x x   = 3 2 224 15 3 16 10 2x x x x x     = 3 224 31 13 2x x x   M1: Expansion (all terms must be correct) A1 [2] 2 (a) 50 mins B1 [1] (b) She is incorrect as the number of patients who have waiting time less than 43 minutes is the same/equal as the number of patients who have waiting time more than 66 minutes. B1 [1] 3 81 = 34 108 = 22 × 33 LCM: 22 × 34 = 324 M1: either prime factorisation A1 (to give 2 marks if no method) [2] 4 3(2 ) 6(2 ) 5 4 5 k k k   30k = 20k – 48k +20 58k = 20 10 29k  M1: substitute x with 2k A1 [2] 5 (a) 2 B1 [1] (b) 9 B1 [1] 6 (a) 15π – 24πb + 3π2 = 3π (5 – 8b + π) B1 [1] (b) 4x2 + 9y2 – 12xy – 64 = (2x – 3y)2 – 64 = (2x – 3y – 8) (2x – 3y + 8) M1: (2x – 3y)2 A1 [2]

2 7 5 3( 1) 2 7 6 4 12 x x x    = 10 9( 1) 2 7 12 12 12 x x x    = 10 9 9 2 7 12 x x x    = 3 16 12 x M1: common denominator (same form as here) A1 [2] 8 (a) 1.57 (3 sf) B1 [1] (b) 11 and 12 B1, B1 [2] 9 (a) 547.39 billion = 547.39 × 109 = 5.4739 × 1011 B1 3sf not accepted [1] (b) 11 66.037 10 5.4739 10  = USD 90 672.52/90673/90700 M1 A1: accept 3sf, whole number or 2dp Don’t accept standard form [2] 10 3 5 4 7 4 22 x y x y     12 20 16 35 20 110 x y x y     47x = – 94 x = – 2 y = 2 M1: Substitution/ Elimination (must show method) A1 A1 [3] 11 (a) Angle OAB is 90° because tangent is perpendicular to the radius. B1 (accept abbreviation) [1] (b) Area of quadrilateral – Area of sector = 6 ×11 – 2126 (6)360  = 26.4 cm2 M1:either area A1 [2]

3 12 (a) 1 cm is to 60 km 24 cm is to 1440 km B1 [1] (b) 1 cm2 is to 3600 km2 1574722 ÷ 3600 = 437.4227778 = 437 cm2 M1: area scale A1 [2]

4 13 (a) BD2 = 202 = 400 BC2 + CD2 = 122 +162 = 400 Since BD2 = BC2 +CD2 , by Converse of Pythagoras’ Theorem, it is a right angled triangle where BD is the hypotenuse. Hence BC is perpendicular to CD. OR 2 2 212 16 20cos 2(12)(16)BCD   Angle BCD = 90° Hence BC is perpendicular to CD. (must be 2 separate lines) B1 B1 [1] (b) cos sinx x = cos sinCDB CDB  = 16 12 20 20  = 1 5 M1: either value A1 [2] 14 2 3 3 7 40 18 128 x x x x    = 2 (3 8)( 5) 2 (9 64) x x x x    = (3 8)( 5) 2 (3 8)(3 8) x x x x x     = 5 2 (3 8) x x x   or 2 5 6 16 x x x   M1: factorise numerator M1: factorise denominator A1 [3] 15 Angle CDA = 132° ÷ 2 = 66° (angle at centre = 2 angles at circumference) Angle CBA = 180° - 66° = 114° (angles in opposite segment) Angle CBP = 180° - 114° = 66° (adjacent angles on a straight line) B1 with reason B1 with reason B1: value of 66° [3]

5 16 Size of interior angle : (360° – 60°)÷ 2 = 150° Size of exterior angle: 180° – 150° = 30° Number of sides: 360° ÷ 30° = 12 sides M1: interior angle M1: exterior angle A1 [3] 17 (a) The principal value increases every month, hence the appreciation value is not 1.2% of $56700. B1 (must see the word ‘principal’) [1] (b) 36 0.158900 1 100     = $61057.93/61058/61100 M1 A1 [2] 18 Let x be the cash price of the sofa bed. Total amount: 0.25x + 26×88 Interest: (0.75x)(0.08)(26/12) = 0.13x 1.13x = 0.25x + 2288 0.88x = 2288 x = $2600 M1: Interest M1: correct equation A1 [3] 19 (a) 1 cupboard : 96 man- hours 96 ÷ (3× 4) = 8 days 8 – 2 = 6 days M1: total number of days A1 [2] (b) 3 2 where k is a constantky x 2 2 1 1 34 8k      1 1 316 64k      3 364 k  k = 64 x = 8 or – 8 M1 M1: Find k A1 [3]

6 Generally quite well done if they managed to get the first equation. Some mistakes include not having a constant (k) in the equation. If there are 2 sets of workings, we will pick the one that results in lower marks and there would be no marks if there is no k. 20 (a)(i) 3 3 2 2x y xy     3 32 2x y xy   3 3 2 8x y x y  3 3 8 2x x y y  3 3 (8 2)x y x  3 38 2 xy x  M1: Cube on both sides M1: Terms with y on one side and factorise A1 [3] Generally well done and most can get the first method mark. However, many did not know how to make y the subject. (a)(ii) 3 3 ( 2) 8( 2) 2y    4 33y B1 [1] (b) 3 2p q  3 2pq q  ( 2) 3q p   3 2q p  It is inaccurate as q is undefined only when p = 2. OR B1: Make q the subject B1 B1: Substitute p =3 [2]

7 When p =3 , 3= ଷ ௤ + 2 1= ଷ ௤ Thus q=3 instead of undefined when p =3. B1 Well done with most using the alternative method. 21 (a) 58 – 30 = 28 M1: either value A1 [2] (b) 7 12 × 120 = 70 Score: 50 B1: 70 B1 [Allow B2] [2] Well done. Some makes the mistake of read off from 50 instead. 22 (a)(i) 12 13 B1 [1] (a)(ii) 5 12 B1 [1] (b) 1 (5)( ) 502 b  b = 20 20 – 3 = 17 R (17, 0) M1 [ Shoelace will not be given if there is any error] A1 [2] 23 (a) Since (x – h)2 is always positive or zero for all values of x, the maximum value of y occurs when (x – h)2 =0. Hence, the maximum value happens when x – h = 0 or x = h. B1 [1] (b) 241 12 0x x   2 12 41 0x x   2( 6) 36 41 0x    2( 6) 77x  ( 6) 77x  M1

8 6 77x  2.77x or – 14.77 A1, A1 [3] Not as well done as many made mistake when forming perfect squares and others left their answers in 3 sf or did double rounding off. 24 (a) 1 22 (3 1) ( 1)n nT n n      B1: all 3 terms correct B1: Any correct term In the given space. [2] (b)(ii) ܶ௡ −ܶ௡ିଵ = 1 2 2 22 (3 1) ( 1) [2 (3( 1) 1) ( ) ]n n n n n n          = 1 2 2 22 3 1 2 1 2 3 4n n n n n n n          = 1 22 2 2 4n n n    = 22 (2 1) 2 4n n    = 22 2 4n n   M1:Replace n by n – 1 M1: expansion M1 [3] (c) Since the difference of the consecutive terms could be expressed as 32(2 2)n n   and n must be a natural number/ positive integer, the difference is always an even number. This is a hence question. There is a condition that n >2 missing. B1: must take out factor of 2 [1] 25 (a) Let r be the radius of the cylinder. 24 = 2πr 12r  V olume:   2 12 17       = 2448  cm3 M1: length = circumference A1 [2] Not well done. Some assumed 17 cm to be the circumference, and others did not understand what it meant by leaving your answer in pi.

9 (b) Let the radius of circle be l and radius of the cone be r. Arc length FH: 1 (2 )4 l 1 (2 )4 l = 2πr 1 4r l Curved surface area: 1 1 724 4l l     17l  V olume of cone: 2 1 1 289 2893 4 16 l     = 311.34438 cm3 = 311 cm3 M1: arc length = circumference M1: Find