Kranji 4048 01 AS

2022 4E5N E Math Prelim Paper 1 (Marking Scheme) Qn Solution Remarks 1 69.5 g 2 The inconsistent scale on the vertical axis exaggerates the differences in crude birth rate between the years. OR It is not clear whether the height or the area of the baby picture should be used to compare the crude birth rate. OR The area of each baby picture is not directly proportional to its height. Do not accept ‘size’ of baby….. Do not accept the crude birth rate is proportional to the ……… 3 0.000 905 74 ÷ 10 ÷ 100 × 2 = 0.000 00 181 148 = 1.81148 × 10−6 m Radius × 2 s.o.i Do not accept 1.81× 10−6 m (3sf) 4 Distance travelled by wiper = 120 360 × 2 × 𝜋 × 48 = 32𝜋/ 100.5309649 Speed = 32𝜋 0.8 = 126 𝑐𝑚/𝑠 (3𝑠𝑓) Find arc length Do not accept answers in terms of π 5 𝐿 = 𝑘𝐴2 15 = 𝑘(3)2 𝑘 = 15 9 = 5 3 → Hence 𝐿 = 5 3 𝐴2 6 Correct shape Correct position of turning point (p, q) where p<0 and q>0 7 Oliver’s 78 marks is within top 25% of the class for the history test but below the top 25% of the class for the science test. Hence, Oliver did better for the history test. OR Oliver’s 78 marks is above the 75 th percentile for the history test but below the 75th percentile of the class for the science test. Hence, Oliver did better for the history test. Comparing 78 with upper quartile of both tests No mark will be given for stating ‘history test’ if reason is wrong Do not accept upper interquartile range x y 0

2 8 Total amount paid = 18732 100 × 1.79 = 335.1417 gallons 335.1417 x 3.785 = 1268.5113345 litres 1268.5113345 x 1.21 = USD 1535 (nearest dollar) Total petrol consumption: 18732 100 × 1.79 s.o.i Correct conversion between litres and gallons s.o.i 9(a) 1115−1081 1081 × 100% = 3.145% (3dp) 9(b) Number of boys= 9 9+16 × 50 = 18 boys After some girls join, 6 units represent 18 boys 1 unit represent 18 6 = 3 boys 19 units represent 3× 19 = 57 students New total = 57 students 10(a) Length = 100 20 × 38.2 = 191 𝑐𝑚 10(b) 1 cm : 0.2 m → 1 cm2 : 0.04 m2 Actual area = 0.04 × 80 = 3.2 m2 area scale s.o.i 11(a) 5832 = 23 × 36 11(b) The powers of the prime factors of 5832, 3 and 6, are multiples of 3. Hence, 5832 is a perfect cube. OR 5832 = 23 × 36 = (𝟐 × 𝟑𝟐)𝟑 . Hence, 5832 is a perfect cube. Do not accept 5832 has integer as a cube root (because does not use part (a)) 11(c) k = 2 12(a) 9, 15, 21, 27 12(b) 6n + 3= 3(2n + 1) OR Both 6n and 3 are multiples of 3. (Must mention specifically that 6n and 3) Do not accept “the terms can be divided by 3” / Do not accept 6 is a multiple of 3. 13(a) 3 − 𝑥 < 7 − 3𝑥 2 ≤ 5 ⇒ 3 − 𝑥 < 7 − 3𝑥 2 and 7 − 3𝑥 2 ≤ 5 6 − 2𝑥 < 7 − 3𝑥 7 − 3𝑥 ≤ 10 𝑥 < 1 − 3𝑥 ≤ 3 𝑥 ≥ −1 ∴ −1 ≤ 𝑥 < 1 for 6 − 2𝑥 < 7 − 3𝑥 o.e. OR 7 − 3𝑥 ≤ 10 o.e 13(b) – 1, 0 Allow ecf from (a)

3 14 Interior angle of polygon B at O = 360° – 115° – 90° = 155° Exterior angle of polygon B = 180° – 155° = 25° Number of sides of polygon B= 360° 25° = 14.4 OR ( )2 180 155 180 360 155 25 360 14.4 n n nn n n −   = −= = = Since 14.4 is not an integer , polygon B cannot be a regular polygon. Find interior angle of polygon B at O Find number of sides of polygon B (if regular) Explain 14.4 not an integer 15(a)(i) {2,3,5,7,11,13,17,19}A= ; { 1,3,5,15}B= {3,5}AB= Accept: 3, 5 15(a)(ii) (a) 8  A (b) {15}  B 15(b) 'PQ 16(a) RS⃗⃗⃗⃗ = OS⃗⃗⃗⃗ − OR⃗⃗⃗⃗⃗ = ( 5 −2) − (4 1) = ( 1 −3) 16(b) |𝑅𝑆⃗⃗⃗⃗⃗ | = √(1)2 + (−3)2= 3.16 units (3sf) Allow ecf for answer in (a). 16(c) 𝐴𝐵⃗⃗⃗⃗⃗ = ( 0.8 −2.4) = 0.8 ( 1 −3) = 0.8𝑅𝑆⃗⃗⃗⃗⃗ 17(a) 17(b) Correct perpendicular bisector & angle bisector with construction arcs 17(c) The playground is equidistant from the points C and D and equidistant from lines AB and AD.

4 18(a) 90 (in square PQRS) (in square STXY) (Shown) RSY PSY PST  = − = 90° - ÐPSY 18(b) ÐRSY = ÐPST (Prove in (a)) ÐYRS = ÐTPS = 90° (Right angles of square PQRS and square STXY) SY = ST (sides of square STXY) DRSY congruent to DPST (AAS) All 3 correct statements with reasons AND conclusion with congruence test stated. 19(a) Reflex angle AOC = 360° – 153° = 207° Angle ABC = 207 2 = 103.5° Finding reflex angle AOC s.o.i 19(b) Angle OAW = 90° Angle OAB = 90° – 31° = 59° Angle BCO = 360° – 153° – 59° – 103.5° = 44.5° Angle OAW = 90° s.o.i 20(a) Acceleration = 293 60 5.5m/s6 − = 20(b) Let speed at t = 4s be p m/s 60 5.54 p− = → 82m/sp= 20(c) ( ) ( )1 93 9 6 3422 v +  − = v = 135 Any equivalent method of finding distance between t = 6s to t =9s Qn Solution Remarks

5 21(a) 21(b) 3Height X 960 4 Height Y 1875 5== 2 Base area X 4 Base area Y 5 = Base area 16 50 25 X = Base area of X = 16 5025 = 32 cm2 Correct ratio of areas seen 22(a) x = 1.5 22(b) Gradient 37 19 306 −= =−− Draw tangent at (3,28) (accept – 2.6 to – 3.4 inclusive) 22(c) Draw the line y = 35: There are no points of intersection between the curve 2 3 28y x x=− + + and the line y = 35m. Hence, the height of the ball will never reach 35m Section I Section II 0 Time (min) Depth 13 8 52 Top of flask Level M

6 23(a)(i) 7q – 2(q + 3) = 7q – 2q – 6 = 5q – 6 23(a)(ii) 𝑥𝑦−1 (2𝑦2)3 = 𝑥𝑦−1 8𝑦6 = 𝑥 8y7 Correct use of indices law (ab)m = ambm 23(b) 7𝑦 (𝑦 − 3)2 − 1 3 − 𝑦 = 7𝑦 (𝑦 − 3)2 + 1 𝑦 − 3 = 7𝑦 + (𝑦 − 3) (𝑦 − 3)2 = 7𝑦 + 𝑦 − 3 (𝑦 − 3)2 = 8𝑦 − 3 (𝑦 − 3)2 8𝑦 − 3 (𝑦 − 3)2 𝐎𝐑 8𝑦 − 3 (3 − 𝑦)2 24(a) 8𝑦2 + 20𝑦 − 12 = 4(2𝑦2 + 5𝑦 − 3) = 4(2𝑦 − 1)(𝑦 + 3) 24(b) 𝑥3 + 𝑥2 − 9𝑥 − 9 = 𝑥2(𝑥 + 1) − 9(𝑥 + 1) = (𝑥 + 1)(𝑥2 − 9) = (𝑥 + 1)(𝑥 + 3)(𝑥 − 3) 25(a) 3𝑥 3𝑥 + 4𝑥 = 3 7 25(b) 3𝑥 − 1 7𝑥 − 1 25(c) 3 7 (3𝑥 − 1 7𝑥 − 1) = 6 35 3𝑥 − 1 7𝑥 − 1 = 2 5 5(3𝑥 − 1) = 2(7𝑥 − 1) 15𝑥 − 5 = 14𝑥 − 2 𝑥 = 3