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