ACJC 8865 2022 Prelim Solution vetted

2022 JC2 H1 Math Prelim Marking Scheme Qn Solution 1 Let the amount invested in Banks A, B and C be x, y and z respectively. We have 25000 (1) 0.06 0.07 0.08 1620 (2) 6000 (3) x y z x y z yz + + = −−− + + = −−−− − = −− By GC, 15000, 8000, 2000x y z= = = .  the amount invested in Banks A, B and C are $ 15000, $ 8000 and $ 2000 respectively. 2(a) 2 31 22 5 2 31 d 2 31 d 22 3 5 x x x x x x x x c − + =+ = + +   where c is an arbitrary constant 2(b) ( ) ( ) 2 2 2 d 4 3lnd 1 d1 ln 4 3 ln 1d2 22 43 1 x x x xxx x x x  − − = − − − =− − − 3i ( )( ) 2 2 2 2 2 41 410 41 4 40 2 2 0 2 or 2 (rejected since 0) dy dx x x x x x xx xx x =− −   − − +  −  

y=x2+5 3ii 3iii ( ) 32 22 2 2 54 54 since is positive, 4 5 Add 5 to , intersection pt is (0.842,5.6 8) 0 x 0.824 x x x x x x x xx x y x C +  + +  + +  + =+    4(i) 1.5x=− y (0,5) (-0.75,0) x 4(ii) ( )( ) ( ) 2 1510 5 23 20 30 15 523 20 15 5 2 3 2 3 10 0 mxx x mxx x mx x mx m x − = ++ +− =++ + = + + + − = discriminant > 0 ( ) 2 3 10 0 10, ,0 3 m mm −  y=x x y o o 1510 23y x=− + 10y=

4(iii) 0 3 4 0 3 4 15 1 510 d 52 3 2 2 15 2510 ln | 2 3 |24 15 15 15 3 250 ln 3 ln2 2 2 2 4 15 15 15 15 25ln 3 ln 3 ln 22 2 2 2 4 15 55ln 224 1 55 30ln 24 55, 30 xx xx pq 5i 220 5 0.01 500C x x x= − + + 5ii ( ), (0, 420) and (45,60) 420 60 80 45 sub (0, 420) and 8 into straight line: 420 8(0) 420 8 420 xS m m S mx c c c Sx = −= =−− =− =+ =− + =  =− + 5iii Revenue = ( )8 420xx−+ Profit = revenue - cost 22 2 = 8 420 20 5 0.01 500 = 8.01 400 500 5 500 x x x x x x x x a − + − + − − − + − +  =− 5iv 32 2 2 d5 400 16.02d 2 d 0d 5 400 16.02 0 2 using GC graph, find -intercept 25 8(25) 420 220 million dollars =$220000,000 Method 1: d5 16.02 16.03 0 is maximum.d4 P xx x P x x x xx S P xPx − = + − = + − = =  =− + = =− − =−   Method 2:

x 24.99 25 25.01 d d P x 0.160 0 -0.160 shape is maximum.P 5v when 40, d 240.40 = 240 million dollars/plane d When the production level increased from 40 to 41 planes, the profit decreases by approximately $240 million dollars. The production level should be x P x = =− − decreased in this case. 6(i) The probability of an employee is working from home is constant at 0.37. Whether an employee is working from home is independent of any other employee. (ii) Let X be the number of employees from the 30 who are working from home. ( )30,0.37XB ( ) ( )P 18 1 P 18 = − XX = 0.0030511= 0.00305 (3sf) (iii) ( )30,0.37XB Mean = 30 0.37 11.1 (exact)= Variance = ( )30 0.37 1 0.37 6.993 (exact)  − = 6(iv) Method 1: Probability of a company getting the monetary incentive = ( )P 18 0.0030511=X Answer: 0.0030511  (1 – 0.003511)  2 = 0.00608 Method 2: Let Y be the no. of companies that will get the monetary incentive out of 2. ( )2,0.0030511YB ( )P1Y == 0.00608 7a (i) Number of ways = 8! 3!= 241 920 7a (ii) Considering only the 8 north zone and 2 south zone schools Method 1: There are 3 cases: Case 1: S _ _ _ _ _ _ S_ _ 2! 8! 80640= Case 2: _ S_ _ _ _ _ _ S_ 2! 8! 80640= N1N2N3N4N5N6N7 N8 S1 S2

Case 3: _ _ S _ _ _ _ _ _ S 2! 8! 80640= Number of ways = 3 80640 241920= Method 2: Number of ways = 8 6 6! 2! 3! 241 920   =C 7b East (out of 14) West (out of 6) Case 1 1 6 Case 2 2 5 Case 3 3 4 Number of ways = 14 6 14 6 14 6 1 6 2 5 3 4 6020 +  +  =C C C C C C S1_ _ _ _ _ _ S2 N1 N2

8(i) unbiased estimate of the population mean = 231 3.8560 60== x (exact) unbiased estimates of the population variance = ( ) 2 60 12.733 0.215813 0.21659 60 59  − = = =  xx (3sf) (ii) Let  cm be the average body length of the cockroach. Let X be the body length (in cm) of a randomly chosen cockroach. Test :4oH = (claim) Against 1 :4H   (researcher’s suspicion) at 1% sig. level Under H0, 0.21581~ N 4, 60   X approx. by Central limit theorem since 60=n large, Test Stats value = – 2.501 p – value = 0.0124 > 0.01 (do not reject H0) There is insufficient evidence at 1% significance level to conclude that the mean body length of the cockroach is not 4 cm. (iii) Test :4oH = (claim) Against 1 :4H   (researcher’s suspicion) at 5% sig. level Under H0, 0.22~ N 4, 60   X approx. by Central limit theorem since 60=n large, Test statistics 4 0.2260 −= kz The given conclusion is to reject H0, therefore the test stats z lies in the critical/rejection region. 44 1.96 or 1.96 0.22 0.2260 60 −− − kk 3.88(3sf) or 4.12 (3sf)  kk 0

9(i) (ii) r = 0.986 (3sf) There is a strong and positive linear correlation between the marks for the written component and the oral component. As the marks for the written component increase, the oral component marks also increases. (iii) y = 0.702 x + 7.17 a = 7.17 (2sf) and b = 0.702 (3sf) Significance of gradient b: For each increase of 1 mark in the written component (x), the oral component (y) increases by 0.702 marks. Significance of y-intercept a: The oral component (y) mark is expected to be 7.17 (x = 0, y = 7.17) if a student scores zero mark for the written component (x). (iv) Estimated oral component mark y = 0.70168 (17) + 7.1747 = 19.1 marks r = 0.986 is very close to +1, therefore there is a strong positive linear correlation between the written component marks and the oral component marks. x = 17 is within the data range of x (i.e. 2 30x ). This is an interpolation. Hence the estimate for the oral component mark is reliable. 27.8 2 x (written component mk) y (oral component mk) 8.8 30 o

10(i) no. of pupils who study Mathematics only = 393 ( 20) 20 85 18 290 no. of pupils who study Physics but not Chemistry ( 20) 290 8( 20) 50 xx x xx x − − − − − = − =− − = − = 10(ii) 105 35( ) 0.267(3 )393 131P M C sf = = = 10(iii) 375 125P( ) 393 131 123P( ) 393 5125P( )P( ) 0.299(3 ) 0.267 P( )17161 and are not independent events. M C M C sf M C MC == = = =  =   10(iv) The probability of a student taking Mathematics given that the student also takes Chemistry. 105P( ) 35 393P( | ) 123P( ) 41 393 MCMC C = = = 10(v) Number of students taking exactly two subjects=30+85=115 Method 1: 115 278 21 393 3 0.182(3 . ) Method 2: 115 114 278 3 0.182(3 . )393 392 391 CC sfC sf =    = M F C M F C

11 (i) Let M be mass of a medium size egg, ( ) 2~ N 49.6 , M  P( 55) 0.17173 55 49.6P 0.17173 By GC, 55 49.6 0.94735 5.70 = − = − = → = M Z   ~ N(0,1)Z Variance = ( ) 22 5.7001 32.5== (ii) Let L be the mass of a large size egg, ( ) 2~ N 56.8 , 6.1L ( ) ( )1 2 3 4 ~ N 4 56.8 , 4 37.21 i.e. N 227.2 , 148.84+ + +  L L L L ( )1 2 3 4 250 0.030822 0.0308+ + +  = =P L L L L (3sf) (iii) Let X be no. of packs of large egg with mass > 250grams Distribution ( )~ B , 0.030822Xn ( ) ( ) ( ) ( ) ( ) ( ) 0 0 1 0.9 1 0 0.9 0 0.1 0.030822 1 0.030822 0.1 0.96918 0 .1  − =  = −  nn n PX PX PX C Method 1: By GC, let ( )0.96918= n y n = 72, y = 0.105 n = 73, y = 0.1017 > 0.1 n = 74, y = 0.0986 Method 2: ( )0.96918 0.1 ln(0.1) ln(0.96918) 73.55    n n n Greatest value of n is 73. (iv) Let ( ) ( )1 2 3 4 5 6 1 2 3 4 1.5= + + + + + − + + +T M M M M M M L L L L where ( )T ~ N 42.2 , 167.21− ( )( ) 49.6 6 1.5 56.8 4 43.2=  −  =−ET ( ) 2Var( ) 6 32.5 (1.5) 37.21 4 529.89=  +