2023 CCHY 4NA Prelim EM P2 Sol

1 @CCHY 2023 4045/2 [Turn Over Solution to 2023 CCHY Preliminary Examination 4NA Maths 4045/2 S/N Solution 1(a) 1.8056 8.1349 13.8321+ = 0.082196021 1b(i) 0.0822 1b(ii) 0.08 2(a) 2 459 457 – 62 671 = 2.396786 x 106 = 2.40 x 106 (3sf) # 2(b) 2459457 62671 = 39.24394058 = 40 (nearest 10) # 3(a) P(–4,3), Q(2,15) gradient PQ = ( ) 15 3 24 − −− = 2 equation of PQ : y = 2x + c At (–4,3), 3 = 2(–4) + c c = 3 + 8 = 11 ∴ equation of PQ is y = 2x + 11 # 3(b) gradient of CD = 2 equation of CD : y = 2x + c At R(7,11), 11 = 2(7) + c c = 11 – 14 = –3 ∴ equation of CD is y = 2x –3 # 4(a) 3x + y = 12 (1) x – 2y = 11 (2) from (1), y = 12 – 3x (3) subst (3) into (2), x – 2(12 – 3x) = 11 x – 24 + 6x = 11 7x = 35 ÷7, x = 5

2 @CCHY 2023 4045/2 [Turn Over subst x = 5 into (3), y = 12 – 3(5) = –3 Hence, x = 5, y = –3 # 4(b) interior angle = ( ) o8 2 180 8 −× = 135° x° + x° + 135° = 180° (∠ sum of ∆) x° = oo180 135 2 − = 22.5° ∴ x = 22.5 # 5(a) 12m – 3m3 = 3m(4 – m2) = 3m(2 + m)(2 – m) # 5(b) 63 27xx+ = = 33 x x + 6 = 3x 6 = 2x ÷2, x = 3 # 5(c) 23 4 90xx+ −= x = ( ) ( )( ) ( ) 2 4 4 43 9 23 −± − − − = 4 124 6 −± x = 4 124 6 −+ or x = 4 124 6 −− = 1.18925 = –2.52259 = 1.19 (2dp) = –2.52 (2dp) ∴ x = –2.52 or x = 1.19 # 6a(i) y = k x 54 = k 9 = 3k k = 54 3 = 18 ∴ y = 18 x # 6a(ii) 72 = 18 x

3 @CCHY 2023 4045/2 [Turn Over x = 72 18 = 4 x = 42 = 16 # 6(b) Pattern for 1st no. of 2nd column 2nd no. of 2nd column 1st no. of 3rd column 1 x 2 – 1 = 1 1 + 2 = 3 13 2 + = 2, 22 2 x 2 – 1 = 3 3 + 2 = 5 35 2 + = 4, 42 3 x 2 – 1 = 5 5 + 2 = 7 57 2 + = 6, 62 4 x 2 – 1 = 7 7 + 2 = 9 79 2 + = 8, 82 (2n – 1) x (2n – 1 + 2) = (2n – 1)(2n +1) 2 12 1 2 nn−+ + = 2n ∴ the nth pattern = (2n – 1)(2n +1) = 4n2 – 1 # 1st pattern = 1 x 3 = 22 – 1 2nd pattern = 3 x 5 = 42 – 1 3rd pattern = 5 x 7 = 62 – 1 4th pattern = 7 x 9 = 82 – 1 7(a) x°= 42°(alt ∠s, AB // CD) ∴ x = 42 # Method 1 65° + ∠AFE + 42° = 180° (int ∠s, AB // CD) ∠AFE = 180° – 65° – 42° = 73° y° = ∠AFE (vert opp ∠s) = 73° # ∴ y = 73

4 @CCHY 2023 4045/2 [Turn Over Method 2 to find y 65° + ∠AFE + 42° = 180° (int ∠s, AB // CD) ∠AFE = 180° – 65° – 42° = 73° 73° + ∠EFH = 180° (adj ∠s on a st line) ∠EFH = 180° – 73° = 107° y° + 107° = 180° (adj ∠s on a st line) y° = 180° – 107° = 73° ∴ y = 73 7(b) Price after 1st year Price after 2nd year 90% x $168000 = 90 100 x $168000 = $151200 90% x $151200 = 90 100 x $151200 = $136080 price decrease = $(168000 – 136080) = $31920 percentage decrease = 31920 168000 x 100% = 19% # 8(a) 1 cm 25000 cm = 0.25 km 10 cm2 10 x 1 cm x 1 cm = 10 x 0.25 km x 0.25 km = 0.625 km2 ∴ the actual area is 0.625 km2 # 8(b) volume of fuel per second : πr2h = π x 2 8 2 cm  x (2 x 100 cm) = 3200π cm3 Total volume of fuel = 3200π cm3 x 24 x 60 = 4608000 π cm3 = 4608π litres = 14476.45895 litres = 14476 litres (nearest litres) #

5 @CCHY 2023 4045/2 [Turn Over 9(a) 2.3 1.4 9(b) P1 : plot the points correctly C1 : join the points to show a cubic curve C1 : smooth curve 9(c) draw tangent line gradient = 4.5 1.8 = 2.5 acceptable range : 2.14 to 2.75

6 @CCHY 2023 4045/2 [Turn Over 9(d) 3211 232y x xx= − −+ (1) 322 12 3 6x xx+= + ÷6, 3211 232x xx+= + 3211 2032x xx− −+= (2) subst (1) into (2), y = 0 322 12 3 6x xx+= + and 3211 2032x xx− −+= are equivalent equations. The graph of 3211 232y x xx= − −+ cuts the x-axis at only 1 point, hence, the equation 322 12 3 6x xx+= + has only 1 solution. 10a(i) number of working hours per day = (2 + 6 – 1) h = 7 h number of days working in a month = 4 x 2 days = 8 days monthly salary = 8 x 7 x $11 = $616 # 10a(ii) 20% of $(616 + 1600) = 20 $2 216100× = $443.20 = $443 (nearest dollar) ∴ Amy’s monthly contribution to CPF is $443 # 10(b) take home pay = $(1600 + 616) – $443.20 = $1772.80 savings per month = $(1772.80 – 10 – 54 – 400) = $1308.80 savings for 4 months = 4 x $1308.80 = $5235.20 savings of $5235.20 > $3086.50, hence Amy is correct. 10 a.m. 12 noon 6 p.m. time 2 h 6 h

7 @CCHY 2023 4045/2 [Turn Over Yes because her savings from the 4 months is more than the course fees. 11a(i) ∠EBD = ∠EAD (∠s in the same segment) = 59° ∠AED + 50° + 59° = 180° (∠s in opp segment) ∠AED = 180° – 50° – 59° = 71° # 11a(ii) Method 1 In ∆BCD, 87° + ∠BCD = 50° + 59° (ext ∠ of ∆) ∠BCD = 50° + 59° – 87° = 22° # Method 2 50° + 59° + ∠DBC = 180° (adj ∠s on a st line) ∠DBC = 180° – 50° – 59° = 71° In ∆BCD, 87° + 71° + ∠BCD = 180° (∠ sum of ∆) ∠BCD = 180° – 87° – 71° = 22° 11b(i) ∠BAC = 180° – 33.1° – 51.4° = 95.5° Using Sine Rule, 150.8 sin 51.4 sin 95.5 AB = AB = 150.8sin 51.4 sin 95.5   = 118.39837 = 118 (3sf) ∴ AB = 118 m # 11b(ii) tan 25° = 118.39837 h h = 118.39837 tan 25° = 55.21007 = 55 (nearest whole number) ∴ height of building is 55 m # h m 118.39837 m 25° B A

8 @CCHY 2023 4045/2 [Turn Over 12a(i) 5 8 8 11 , 8 11 , 3 11 12a(ii) P(same gender) = P(both male) + P(both female) = 38 53 8 11 8 11× +× = 39 88 # 12b(i) 24.5 23 12b(ii) Students in class A take longer time than class B to complete the problem since median of class A (24.5 min) is greater than the median of class B (23 min). Interquartile range : Class A : (27 – 16) min = 11 min Class B (32.5 – 17.5) min = 15 min The spread of time taken is lesser in class A (11 min) compared to class B (15 min) as the interquartile range in class A is smaller than that in class B. Or The time taken by students in class A are more consistent than that in class B because the interquartile range for class A (11 min) is less than that for class B (15 min). Or Students in class A takes a shorter time to complete the problem than students in class B since 75% of them completed in 27 min in class A compared to 32.5 min in class B. Or Students in class A takes shorter time to complete the problem than class B since the longest time taken by students in class A is 35 min while t