2012 Year 6 H1 Mathematics Preliminary Examination Solutions

Solutions 1 For 2 22 4 0 2 4 0kx kx x k kx k x k

2 2

2 2

2 4 4 0 and coefficient of 0

2 4 0

2 4 2 4 0

2 3 2 5 0

k k k x k

k k

k k k k

k k

2 25 3

k and k > 0

The solution set is 2: 0 .3

k k 2 Let length of side AC be y units and the length of side BC be x

units. Given that

Perimeter of = 36

2 362 34

34 2 ......(1)

ABCx x y

x yy x

and 22 22 2

is a right-angled triangle

2

2 4 4 ......(2)

ABC

x x y

x x y

Substitute (1) into (2): 22

2 2

2

2

2 4 4 34 2

2 4 4 1156 136 42 140 1152 0

70 576 0By GC, 60.475... (reject, since 36) or 9.5245...

x x x

x x x xx x

x xx x x

For x = 9.5245, 34 2 9.5245 14.951...y The length of side AC is 15.0 units (3sf).

2/5 2/3

Alternative

2 3 2 5 0

3 2 2 5 0

k k

k k

2/5 2/3

Solutions

3

ln 2 3e 2 3

e 32

y

y

y x

x

x

Area of shaded region

ln5

0

ln5

0

ln5

0

ln5 0

2

e 34ln 5 d2

14ln 5 e 3 d214ln 5 e 3214ln 5 e 3ln 5 e214ln 5 5 3ln 5 12

5 ln 5 2 units2

y

y

y

y

y

y

x = 3/2

(2,0)

y

x

x = 3/2

(2,0)

y

x (4,0)

(0, ln5)

ln 2 3y x

ln 2 3y x

Solutions

4(i)

2f '3

2f ' 2 22 3

4 6 21

xbx

bb

b

2f '3

2f d 2ln 33

Given that f 1 0,

2 ln 4 02ln 4

2ln 3 2ln 4

xx

x x x Cx

CC

y x

4(ii)

f 2ln 3 2ln 4

For f 2 f f 2 ,

2ln 3 ( 2) 2ln 4 2ln 3 2ln 4 2ln 3 2 2ln 4

2ln 1 2ln 3 2ln 4

ln 3 ln 1 ln 4

3ln ln 41

3 413 4 43 1

13

x x

x x

x x

x x

x x

xx

xxx x

x

x

5(i) 2 328 20 2y x x x

5(ii) The range of values of k is k < 0 or k > 49.3.

(2, 0)

(1.67, 49.3)

(3.5, 0) x

y

(0, 28)

Solutions 5(iii) 2 3 2 3For ln 28 20 2 to exist, 28 20 2 0.

From the graph, 2 or 2 3.5

y x x x x x x

x x

2 3 2 3

20 1 2 20 1 2For ln 28 to exist, 28 0.e e e e e e

Replace by e ,e 2 or 2 e 3.5(no solution 0 e 3.5

e 0 for all )

x x x x x x

x

x x

x

x

y

x

x

From the graph, ln 3.5x

6(ai) Let the height of the cone be h cm.

Using similar triangles,

22

x r xrhh

2

3

13 21 (shown)

12

xV x

x

6(aii) 2d 1 dand 3

d 4 dV Vxx t

d d dd d dV V xt x t

2d 1When 6, (6 ) 9

d 4Vxx

d3 9dxt

d 1d 3xt

Hence, rate of decrease of depth of water is 11 cm s .3

6(bi) 4 3 2

3 2

28 30 36 13

d 4 28 60 36d

h t t t t

h t t tt

For stationary points, d 0dht

3 24 28 60 36 0t t t

x h

2h

r

y = ex y

x

y = 3.5

ln 3.5

Solutions From GC, 1 or 3t t . For t = 1, 1- 1 1+

ddht

>0 0

Solutions

7 Given 2~ N ,X 2 21 50~ N , and ... ~ N 50 ,5050X X X

Given 1 50P 128 0.15 and P ... 7500 0.25X X X

Standardizing, Z~N(0,1)

128P 0.1550

Z

128 1.0364350

50 1.03643 128 50 ...(1)

Solving equation (1) and (2), using GC,

141.327... 141 (nearest whole number)90.923... 91 (nearest whole number)

8(a) Sampling interval, k = 810 2730

List the applications in order of the applicants names (or loan amount), and randomly select the first application to process. Subsequently, select every 27th application until a sample of 30 applications is selected, going back to the front of the list if necessary.

7(b) A systematic random sample may not ensure that all categories of loan applications were processed within that day if the sampling interval coincides with a cyclic pattern in the list of applications.

7(c) Stratified random sampling ensures that each category of loan is proportionately represented in the sample.

7500 50P 0.2550

Z

7500 50 0.67448950

50 0.674489 50 7500...(2)

Solutions

9(i) Unbiased estimate of population mean is 123 15 16.025120

t 9(ii) The term unbiased estimate means that the expected value of

the sample statistic is equal to the population parameter that is being measured.

9(iii) Test 0 : 15H Against 1 : 15H Conduct a 2-tailed test at 5% significance level. Since 120n is large, by Central Limit Theorem and under 0,H

24.47~ N 15,120

T

approximately.

Since 16.025,t p-value = 0.0120 (3 s.f) Since p-value = 0.0120 < 0.05, we reject 0H and conclude that there is sufficient evidence at 5% significance level that the mean waiting time differs from 15 minutes.

9(iv) Since n is large, by the Central Limit Theorem, 24.47~ N 15,T

n

approximately

Since 0H is rejected, p-value < 2P( 15.5) 0.06T P( 15.5) 0.03T Standardizing, Z~N(0,1)

15.5 15P 0.034.47

0.5 1.8807...4.47

282.69...

Z

n

n

n

283n , where n . Least value of n = 283.

Solutions

10(i) From GC, 0.92152... 0.922 (3 s.f)r

10 (iia)

From GC, regression line of y on x is 1.8308 35.0301.83 35.0 (3 s.f)

y xy x

10 (iib)

From GC, regression line of x on y is 0.46384 14.4380.464 14.4 (3 s.f)

x yx y

10(iii) The gradient of the regression line of y on x represents the estimated increase in test score per hour spent on studying.

10(iv) Using the regression line of y on x, 60 1.8308 35.030

Estimated value of 13.638...13.6 (3 s.f)

xx

The regression line of x on y is not suitable as the test score depends on the number of hours spent studying.

10(v) Let the mean number of hours spent studying and mean score of the 53 students be p and q respectively.

1.6 35 0.5 14 2( 14)

y xx y y x

Solving for point of intersection from GC, ( , ) (17.5,63)p q Mean number of hours studied

1 2 3 6 12 24 36 53(17.5)60

16.858...16.9 (3 s.f)

Mean score of 60 students

20 33 49 52 70 80 95 53(63)60

62.3

10 20 30 40

2040

6080100

y

x

Qn Suggested Solution 11(i)

20 19

P( 0) P( 19) 0.120(1 ) (1 ) 0.119

X X

p p p

By GC, p = 0.10874 = 0.109 (3 s.f.)

11(ii) Reasons should address either one of these behaviour: 1: The probability of "success" p is the same for each outcome. 2: The responses are independent of each other.

Behaviour of X Behaviour of Y 1. Probability of responding

positively to the email (success) is unlikely to be the same for each student since - not everyone is a fan of action thrillers, or - not everyone reads email.

Probability of picking a girl (success) is the same for each of the 5 weekdays since all names are in the bag.

OR 2.

Responses may not be independent since decision of a student is likely to influence or be influenced by other students in his/her clique.

The trials of picking a student on each of the five day are independent since the names are drawn from a bag at random.

11 (iii) Y~B(5,

1221 ) i.e. Y~B(5,

47

)

P(Y 3) = 1 P(Y 2) = 1 0.36788 = 0.63212 = 0.632 (3 s.f.)

11 (iv)

Let W be the number of weeks in which a girl is on duty more often than a boy, out of 20. W ~ B(20, 0.63212) Since n=20 is sufficiently large such that np = 12.642 > 5 and nq = 7.3576 > 5, W ~ N(12.642 , 4.6508) approximately.

P(Y < 11) continuity correction P(Y < 10.5) = 0.16029 = 0.160 (3 s.f.)

Solutions

12(ai)

2

2

2

P( ) P( | )P( )5 5

16 165 70.25

16 165 3 0

16 160.2 or 3 reject since 0 < 1

A B A B B

x x x

x x

x x

x x x

12(aii) 25P( ) (0.2) 0.012516

P( ) P( ) (0.25)(0.2) 0.05

A B

A B

Since P( ) P( ) P( ),A B A B thus A and B are not independent.

12 (b) Let A, B & C be the events a drawer containing: 2 gold coins; 1 gold coin and 1 silver coin; and 2 silver coins is selected, respectively. Let G & S be the events: a gold coin is selected; and a silver coin is selected, respectively

12 (bi)

1P( ) P( ) P( ) 3A B C 3 1P( ) P( ) 6 2G S or

1 1 1 1P( ) 3 3 2 2G

12 (bii)

P(Drawer containing 2 gold coins is selected given that the coin selected is gold) =P( | )A G

P( )P( )

13

12

23

A GG

12 (biii)

Coins in the drawer after the older son picks a gold coin from the top

drawer:

Gold

Gold, Silver

Silver, Silver

Required probability = 1 1 1 13 3 2 2

Solutions

12 (biv)

P(both sons will pick a gold coin each)

P(elder son picks gold from top & younger picks gold)+ P(elder picks gold from middle drawer & younger picks gold)

1 1 1 1 13 2 3 2 329

13(i) Let A be the random variable that denotes the score of a prospective

student for Kingdom University. Let B be the random variable that denotes the score of a prospective student for Island University. Given 2~ N 500,100A and 2~ N 480,60B . P 0.3A a

552.44...a The minimum score a randomly chosen student must achieve is 553 (3 s.f).

13(ii) P 420 540 0.68268...0.683 (3sf )

B

Island University has admitted 68.3% of its prospective students. 13(iii) 2100~ N 500,

5A

and

260~ N 480,5

B

2 2100 60~ N 500 480,5

~ N 20,2720

A B

A B

Method 1 Required Probability P 50 P 50

0.089767... 0.28256...0.37232...0.372 (3s.f )

A B A B

500 552.44

0.3

Alternative

P 0.7A a

Solutions Method 2 Required Probability

P 50 P 50

1 P 50 50

1 0.62766...0.37234...0.372 (3s.f )

A B A B

A B

13(iv)

2

1 2 5

1 2 5

... ~ N 5 480,5 60

... ~ N 2400,18000

B B B

B B B

and

2 2

2

5 ~ N 5 500,5 100

5 ~ N 2500,500

A

A

1 2 5

2

Let ... 5

~ N 2400 2500,18000 500

~ N 100,268000

P 0 0.42341...0.423 (3sf)

L B B B A

L

L

L

13(v) Required Probability

5 5

5 5

4

4

P 500 P 500

0.5 0.36944...

2.1506... 102.15 10 (3sf )

A B

This probability found has considered fewer cases than the probability where the total score of these 10 prospective students is at least 5000.