2017 HCI H1 Maths Prelims Answers

1 2017 C2 H1 Prelim Solutions 1 Let x, y and z be the original selling price per pack of organic quinoa, organic feed eggs and chia seeds in dollars. 3 0.85 2 72.28xy z  --- (1)  2 2 0.85 5 93.85xy z  --- (2)  6 3 0.85 3 145.43xy z  --- (3) $14.90, $11.49, $8.90xyz 2(a)      2d1 ln 2 ln 1d 2 = ln 1 xx xxx x xx xx      2(b)  11 22 3 2 d1eed 22 xx x x        3 2.5 1.41x   or 1.41 1.41x  4(i) Using long division 221 3 3 2844 x xxx          2 2 421OR 44 21 4 AxBx Cx xx x Ax B x C        Compare coefficient: 28 3 3A,B ,C 4(ii) 266 55 62 5 21 3 3 d2 8 d44 83 3 l n 4 84 33ln 2 65 19 33ln 2 x xx xxx xx x           32x  0.5y   2.5,0.75  1.41,0.414  1.41, 2.41

2 5(i) 5(ii)    2 2 31 = 3 3 = 2 3 y xx x xx x x      2 2 23 33 0 x kx x xx k        2 2 40 34 30 94 1 20 42 1 52 5 ba c k k k k.       { : 5.25}kk 5(iii)   31 5yx xx     2 2 23 5 32 0 21 0 21 x xx xx xx x, x           1 2 2 1 2 2 132 2 2 area 2 3 5 d = 3 2 d 3 = 2 32 13 81 2 = 2 432 3 2 1 = units6 x xxx xx x xx x                         Or x y  30,  1, 4 10, 03,   31yx x 

3 1 2 2 132 2 2 1area ( 2 3)d (3 4) 1 2 27 3 32 2 1 = units6 xx x xx x             6(i) 221 π 33Area = 3 sin 2 223 4x xy xx y      2 2 33 21 54 3321 5 4 15 3 32 4 xx y xy x y xx    Let P be the perimeter. 52 15 3 35 4 Px y x xx   2 d3 3 1 5 = 5d4 P x x For cost to be minimum, perimeter has to be minimum. 2 2 2 d3 3 1 505 0d4 15 3.700962 4.053 2.0132 2.01, 2.42 P xx xx x xy         2 23 d3 0 0d P xx  Therefore P is a minimum.

4 6(ii) Let N be the perimeter of the fence in integral value Cost from Company 90A N    Cost from Company 10 95 84 10 110 84 BN N    110 84 90 61 1 0 18.3 18 NN N NN     6(iii) When x = 2.0132, y = 2.4177 14.902P Since 14.902 18 , therefore it is cheaper to choose Company A. 7(a) No. of words that can be formed 7! 1 5039    (b) No. of words if 3 vowels are altogether 3! 5! 720    No. of words 5040 720 4320   (c) 4 2No. of words 2! 5! 1440 C   8(i) Let B be the event that the lunch box is produced by production line B. Let F be the event that the lunch box is faulty.             PP P = P P (*) P 0.03 0.05 0.4 2P 3 FB B F B FB F B B      8(ii) Let A be the event that the lunch box is produced by production line A.   P 0.05 0.6 0.03 0.03P | 0.09 1 3 AF FA    8(iii)  2P ' 0.02 0.646673BF   0.64667 0.02 2 | only 1 faulty 0.95 0.05 2 0.272 P B     9(i) Let X denote the number of diners, out of 20, who choose a burger.  ~B2 0 ,0 . 0 5X    P 3 1 P 3 0.0159XX  

5 9(ii)  P0 . 9Xn --- (1)  P1 0 . 9Xn --- (2) Using GC,  P 1 0.736X   P 2 0.925X   smallest value of n is 3. 9(iii) Let Y denote the number of diners, out of 20, buying a drink in the cafe.  ~B2 0 ,Yp  20 1 4.55pp  --- (1) 2 0.2275 0pp  0.35p  or 0.65 Since 0.5p , 0.65p 10(i) 10(ii) Product moment correlation coefficient 0 998r. which indicates a strong positive linear correlation between the number of items produced per month by the company together with the total cost of production 10(iii) 45 65x, y 12 81 No.of items (1000s) Production cost ($1000) 102 35 y x

6 10(iv) 09 8 2 09 9y .x . - 0.98 gives the rate at which the production costs are increasing i.e. for every additional item produced, the production cost increases by $0.98 - OR For every increase in 1000 items produced, there is an increase in the total production cost by 980 dollars. - $20,990 is the fixed cost of production. 10(v) 0 9781 20 985y .x. When 70x , 0 9781 70 20 985 89 452y .. . ,  an estimate for the production cost of 70 thousand items is $89452. Since 70x lies within the data range 12 81x  and r is close to 1, therefore this estimate is reliable. 10(vi) If x items(1000s) are produced, Total income = $2.20 x The total cost for producing x items is 0 9781 20 985y. x .  If there is no loss, 2 20 0 9781 20 985 12 2 1 9 2 09 8 5 17 17407 .x . x . .x. x.    Therefore the min number of items to be produced per month is 17175 items. 11(i) Let X be the weight of a packet of Calhwa potato chips and  denotes the population mean weight of a packet of potato chips in grams 0 1 H: 8 4 H: 8 4     No.of items (1000s) Production cost ($1000) 12 81 35 102 x y X x ,y

7 11(ii) At 1% level of significance, under H0, since 100n is large, by Central limit theorem, 25N8 4 100X,   approximately Test statistic 2 0,1XZN n    0 0139 0 01p .. , we do not reject H0 and conclude that at the 1% level of significance, there is insufficient evidence to say that the average weight of a packet of potato chips is less than 84 grams 11(iii) When the level of significance is set at 1%, there is 1% chance that we wrongly conclude the mean weight of a packet of potato chips is less than 84 grams when in fact the mean weight of a packet of potato chips is at least 84 grams. 11(iv) Since the sample size 100n is sufficiently large, the sample mean weight of the packets of potato chips will be normally distributed by the Central Limit Theorem. Therefore it is not necessary to assume the weight of packets of potato chips follow a normal distribution. 11(v) 0 1 H: 8 4 H: 8 4     Level of significance: 5% Under H0, since n =100 is large, by Central limit theorem, 25N8 4 100X,   approximately. Test statistic N0 1XZ , n    Rejection region: Reject H0 if 1 95996z. or 1 95996z. Since there is sufficient evidence, at 5% level significance to conclude that the average weight of the potato chip has changed, 84 1 959965 100 t .  or 84 1 959965 100 t .  2 84 1 95996t.  or 2 84 1 95996t.  83 020t. or 84 979t. Range of t : 83 0t. or 85 0t.

8 12(a)  15P 15 0.841 P 0.841XZ        --(1) 15 0.99858    91 5P 9 15 0.682 P 0.682XZ         9P 0.841 0.682 0.159Z      ---(2) 9 0.99858    --- (2) Solving (1) & (2), 12  and 3.00  Alternatively By observation , 12  .  P1 5 0 . 8 4 1X  Using GC, 3.00  12(bi)  2N, 3S  P2 . 5S --- (1) 2.5=P 33 S    2.5=P 3Z  2.52P 3Z   --- (2) 0.405 b(ii) P1 1 0 . 7 5S  --- (1) 11P0 . 2 5 3Z   11 0.674493   , 13.0 C   b(iii) P 17.5 23 0.786T  b(iv) Find 12P0 1 0 2 SST    Let 12 2 SSWT  E8W  Var 9.34W  N8 , 9 . 3 4W 12P 0 10 0.73915 0.739 3s.f2 SST      (v) Assume that the minimum and maximum temperatures are independent of each other. It is unrealistic because the weather, e.g. wind direction, rainy weather, etc, will affect both the minimum and maximum temperature of the city.