2024 TPSS EMATH P1 PRELIM MS

Tampines Secondary School Sec 4E/5NA/4NA OOS Math Prelim Exam Paper 1 2024 Marking Scheme Total Marks: 90 √ = follow through No. Answers Marks 1 3 8 56x−< 3 64 21.3 x x < < The largest prime number is 19 M1 A1 2 2 15ab−+ B1 3 334 ( 5) ( 5)x yx y−+ B1 for 334 ( 5) ( 5)x yx y−+ or 33(2 10 )(2 10 )x yx y−+ seen 4(a) B1 (b) 3p=− B1 5 Total number of rhinos 4 327000 1 100 = +  30388.73 30000= ≈ M1 A1

No. Answers Marks 6 90° for French --- B1 Thai and Japanese with correct angles measured --- B1 7 B1 --- perpendicular line of GH drawn. B1--- for the circle drawn passing through F, G and H and has the centre at the intersection of the two lines

No. Answers Marks 8(a) 222420 2 5 11= ×× Not all the powers of the prime factors are multiples of 3, hence 2420 is not a perfect cube. B1 B1 (b) LCM 222420 2 5 11= = ×× HCF 110 2 5 11= =×× The two numbers are 22 5 11 220×× = and 22 5 11 1210×× = B1 , B1 9(a) 11 15 1233 12 15 27 8 8 27 xy yx −   =   4 5 2 3 y x= B1 for 42y B1 for 53x (b)(i) ( 2)(4 1)yx−+ B1 , B1 (ii) ( 2)(4 1) 0yx− += 12 or 4yx= =− √ √B1 for both correct values 10(a) For the vertical bars, the area of each bar is not directly proportional to the height, hence readers might be misled that the amount spent on hotel stays in 2021 is about 4 times that in 2020 instead of only 2 times as shown by the height. [Accept other reasonable responses] B1 for misleading fact B1 for explanation of why this misleading fact cause misinterpretation. (b) The chart does not support her claim because between 2021 and 2022, the amount spent on flight increases but the amount spent on hotel stays decreases. [Accept other reasonable responses] B1

No. Answers Marks 11(a) (i) ⊂ B1 (ii) ∉ B1 (b) B1 (c)(i) 21x= B1 (ii) 21 20 74 74 21 20 21 74 20 120 119 118 120 119 118 120 119 118   × ×+× ×+× ×     111 2006= or 0.0553 Alternative Method 21 20 74 111 3120 119 118 2006  × × ×= or 0.0553 M1 for 21 20 74 120 119 118×× seen M1 for addition A1 M2 + A1 12 3, 5, 5, 7, 10 or 4 , 5, 5, 5, 11 B1 13(a) 10 : 4 : 7 B1 (b) Flour : Sugar : Butter 1500 : 500 : 1000 1500 250 6÷= 500 100 5÷= 1000 175 5÷= Maximum number of biscuits made 5 25= × 125= M1 A1

No. Answers Marks 14(a) ( ) 22 255 9 2.5 9 4xx x− += − − + ( ) 211 2.54 x=+− 11 4p= or 2.75p= B1 (b) 5 or 2.52xx= = B1 15(a) 2 45yx= −+ B1 for the correct gradient B1 for the correct y-intercept (b) Let the shortest distance from P to line L be h . Let θ be the angle made between the line and the x-axis. 04tan 21.801410θθ= ⇒= sin21.80147 h = 7 0.713 2.60h= ×≈ Alternative Method 7 4 116 h = 28 2.60 116 h= ≈ M1 M1 A1 M1 , B1 for 116 seen A1 16 Interior angle of polygon A 0 00 0360 135 60 165= − −= Let n be the number of sides of polygon A. ( 2) 180 165n n −× = 180 360 165nn−= 360 2415n= = B1 M1 A1

No. Answers Marks 17 kP Q = 21 9 16 kk+= 2134 kk+= 7 2112 k = 7 252k = 36k = 36 100 Q = 2 36 0.1296100Q = = [Accept 81 625] M1 M1 A1 18(a) Perimeter 216 2 10 20360 π= ××× + 12 20π= + M1 for 216 2 10360 π××× (b) 2 12rππ = 6r = Height of the cone = 2210 6 8−= Volume of the cone 21 683 π=×× × 301.59 302= ≈ M1 M1 M1 A1 19 AD BC= (opposite length of parallelogram) EAD ADB DBC∠= ∠= ∠ (alternate angle) EDA EDP ADP∠ =∠ −∠ DPC DBC=∠ −∠ (alternate angle) PCB=∠ (exterior angle) BCP ADE∴∆ ≡∆ (ASA) B1 B1 B1 for using alternate angle B1 for using exterior angle or other equivalent reason to conclude that EDA PCB∠= ∠ A1 for ASA shown

No. Answers Marks 20(a) (i) 6 , 12, 22, 36 B1 (ii) 222 4 2( 2)nn+= + is a multiple of 2 for all values of n , hence it is an even number and not an odd number. B1 (b) 223k − B1 21 4 5 4 11 2 xx xx ++ = + ( )( )4 5 2 (4 11)x x xx+ += + 224 13 10 4 11xx xx+ += + 2 10x=− 5x=− M1 M1 A1 22(a) 47 B1 (b) 20.5 B1 (c) 1. On average Team B scored more points than Team A as their median score was 53 which was higher than the median score of 47 achieved by Team A. 2. The interquartile range of Team B was higher than that of Team A which was 17.5. Hence the scores for Team B was more widely spread out. B1 B1 (d) There is an outlier in the distribution scores of Team A which is significantly greater than the rest of the scores. B1

No. Answers Marks 23(a) 560 420 140 490 280 280   B1 (b) ( ) 560 420 1401.25 2.50 490 280 280T =  ( )1925 1225 875= B1 (c) The total cost of the small and large vegetarian pies B1 24(a) 068 2 136DOB∠ = ×= (angle at centre = twice angle on circumference) 0180 68 112BCD∠ = −= (angles in opposite segment) 0360 52 136 112 60ODC∠ = −− − = (sum of angles in a quadrilateral is 360°) B1 for correct reasoning B1 for correct reasoning B1 (b) Area of the shaded segment = Area of sector OCB − Area of triangle OCB 22 076 1 5 5 sin 76360 2π= ×× −×× 2 4.451932 4.45 cm = = M1 , M1 A1

No. Answers Marks 25 Curved surface of the small hemisphere ( ) 2 2 3.6 25.92ππ= = cm2 Curved surface area of the cylinder ( )( )2 3.6 6.5 46.8ππ= = cm2 Total surface area of the large hemisphere ( ) ( ) 2 222 5.4 5.4 3.6 74.52ππ π= + −= cm2 Total surface area of the solid 25.92 46.8 74.52 462.56 463ππ π++ = ≈ cm2 M1 M1 M1 , M1 A1 26(a) 40 60.5 2 v− =−− 31v= M1 A1 (b) B1 for correct curve drawn from t = 0 to t=0.5h B1 for correct curve drawn from t = 0.5h to t = 2h B1 for values 10 and 63.25 labelled correctly on the vertical axis