Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
Qn
1(a)
Page 1
Solutions
Using ratio theorem,
  8   0   6 
3OA  OB 1        
OD 
 3  0    4     1 
4
4      
  0   0    0 
1(b)
8 0  0 
     
OA  OB   0    4    0 
 0   0   32 
     
0
 
Thus a normal vector of plane OAB is  0   k . [Note: A normal vector of the plane
1
 
can also be determined by observing OA and OB are on the xy -plane. Since OAB is
on the xy-plane, k is a normal vector to OAB.]
 6  2  4 
     
CD   1    4    3 
 0   6   6 
     
Angle between CD and OAB:
cos  90    
k  CD
61
 90    39.806
ALTERNATIVELY,
sin  
1(c)

6
61    50.2 (to 1 d.p)
k  CD
6

   50.2
61
61
Realise that the z-coordinate of C is the height of the tetrahedron. Realise also that
OA  OB .
Thus, volume OABC =
1 1
 OA OB    6   32 units 2

3 2

Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
2a(i)
f ( x)  x 2   x  2
f '( x)  2 x    0
 x

2
is the x-coodinate of the min. point
Inverse of f exists when 

 2    4
2
ALTERNATIVELY,
 

f ( x)  x 2   x  2   x   
2
2
4

2
Inverse of f exists when 
(ii)

2
2
 2    4.
Let y  x 2  4 x  2   x  2   2
2
x  2 y2
f 1 : x  2  x  2,
x  2
2
2b(i)
k
k2

g( x)  x  kx  1   x    1 
2
4

2
 k

R g  1  ,  
4


hg exists if R g  D h
2
k2
0
4
 4  k2  0
Thus, 1 
   k  
(ii)
3 
For k  1, R g   ,  
4 
3 
(, )   ,    [4, )
4 
 R hg  [4, )
ALTERNATIVELY,
R hg  [4, )
Page 2
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
3(i)
xu
1
2
1
 2u
u2
dy
2
 3 2
du
u
y
dx
1

du 2 u
5
dy dy du  2

 
   3  2   2 u  4u 2 1  u 3 
dx du dx  u

3(ii)
Since x  1, u  1 .
When u  1 , y  1 and
dy
 8 .
dx
Equation of the required tangent:
y  1  8  x 1  y  8x  7 .
When y  0, x 
7
and when x  0, y  7 .
8
7 
 A  , 0  and C  0, 7  .
8 
Equation of the required normal:
1
1
9
y  1   x  1  y  x  .
8
8
8
9
When y  0, x  9 and when x  0, y   .
8
9

 B  9, 0  and D  0,   .
8

7 65
9 65
Finally AB  9  
and CD  7  
8 8
8 8
 AB  CD
(iii)
Let

dy
be G.
dx
7
 5 7 1 1 
dG
 4   u 2  u 2   2u 2  u 3  5
du
2
 2

When u  2 ,

 7 
3
dG dG du


 2  2 2   23  5   0.5  
dt du dt
8 2


dy
3
is decreasing at
units per second.
dx
8 2
Page 3
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
4(i)

6/5
0
1/ 2
2 x  1 dx   1  2 x dx  
0
1/ 2
1/ 2
  x  x 2 

(ii)
6/5
0
2 x  1 dx
6/5
  x 2  x 
1/ 2
37
50
dx
 2 cos 
d
x  0   0
x  2sin  ,
6
3
   sin 1
5
5
2
6/5
x
0 1  4 dx
x

a

a

a
0
0
0
4sin 2 
3
1
2cos d where a  sin 1
4
5
4  4sin 2 
2cos d
4
1  sin 2  2cos d
a
a
  2 cos 2  d   cos 2  1 d
0
0
a
a
1

  sin 2     sin  cos    0
2
0
 sin a cos a  a  sin a 1  sin 2 a  a
3
3
 3  4 
     sin 1 (Since sin a  )
5
5
 5  5 
12
3

 sin 1 (shown)
25
5
Page 4
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
(iii)
(a)
Area of R  
6/5
0
 
6/5
0
1
2 x  1 dx  
x2 
3
  2 x  1   dx
4 
5
6/5
0
3
x2
 1  dx
5
4
37  3 x 
 12
 3 
        sin 1   
50  5  0
 5 
 25
37 18 12
3
     sin 1    from (i) and (ii) 
50 25 25
5
23
3

 sin 1   units 2
50
5
6/5
(iii)
y
(b)
y  2x 1 
1
3
5
4/5
2/5
x
3 / 5
x2
 y2  1
4
Volume generated
Idea:
1

   4 1 y
4/5
–
+
2

2
 y 4
dy       dy
3/5 2
5

4/5
2
1 y
  dy
3/5  5
2

 
2/5
 3.56 units3 (to 3 s.f) by GC or
851
 units 3
750
Page 5
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
5(a)
Page 6
Number the citizens from 1 to 500, 000.
Randomly select a citizen from the first
500 000
 5000 citizens.
1000
Select every 5000th citizen thereafter.
Stratified sampling with appropriate strata such as income levels is more appropriate
because the sample is more representative of the population, as citizens’ choice of
president may vary according to their income levels.
OR
Quota sampling is more appropriate as it is generally easier and faster to conduct than
systematic sampling. Hence it will require lesser amount of resources to conduct the
survey.
OR
Quota sampling is more appropriate as it can be conducted without knowledge of the
sampling frame since the editor may not have access to the sampling frame.
(b)
Any of the following:
People may lie in the pre-election poll.
People may change their minds.
Sample may not be representative of the population even though the sample is randomly
chosen.
Sample size may not be large enough to reflect the votes of the citizens.
6( i)
No. of ways = 45 = 1024
(ii)
Method 1:
Required no. of ways = 4  3 4  4  4  768
A
Method 2:
B C D
E
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
Page 7
Required no. of ways
= Total no. of ways without restriction  No. of ways to paint block A and block B with
the same colour
= 45  4(43 ) = 768
(iii)
Method 1:
No. of ways = 5C2  4! = 240. ( 5C2 = no. of ways to choose 2 blocks from the 5 blocks to
be painted with same colour)
Method 2:
Required no. of ways =
5! 4
5!
 C1 = 240. (where
is the no. of ways to arrange 5 colours
2!
2!
with 2 same colour used and 4C1 is the no. of ways to choose one colour to be used
twice)
7(a)
Method 1:
n
 9 
   0.03
 10 
ln 0.03
n
= 33.3
ln 0.9
Hence, n is an integer and n ≥ 34.
Method 2:
Let A be the no. of accidents that occur due to driver’s negligence in City X. A ~ B(n,
0.9)
P(A = n)  0.03
Using GC,
P(A = 33) = 0.0309 > 0.03
P(A = 34) = 0.0278 < 0.03
Hence, n is an integer and n ≥ 34.
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
(b)
8 (i)
Page 8
Let X and Y be the no. of accidents that occur due to driver’s negligence in City X and
City Y on that day respectively.
9

X ~ B  60,

10 

9
Since n = 60 is large, np = 60   54 > 5,
10
1
27 

n(1 – p) = 60   6 > 5, X ~ N  54,
 approximately
10
5 

2

Y ~ B  50, 
9

2 100
Since n = 50 is large, np = 50  
> 5,
9
9
7 350
 100 700 
,
n(1 – p) = 50  
>5, Y ~ N 
 approximately.
9
9
81 
 9
 586 5687 
X Y ~ N
,
 approximately
405 
 9
P( X  Y  60) = 0.0863
(Note: need to do cc, so it should be P( X  Y  60)  P( X  Y  60.5) = 0.109 )
Unbiased estimate of the population mean, x 
 ( x  5)  5 
10
9
 5  5.9
10
2

( x  5)  

1

2

s   ( x  5) 

9
10


Unbiased estimate of the population variance,
2
1
9 
 19.96    1.317777778  1.32
9
10 
2
(ii)
Let  be the mean yield of the new variety of fruit bush.
H0 :   5
H1 :   5
n  10, x  5.9, s 2  1.31778.
Under H0, test statistic T 
X 5
2
S
10
Using t-test, t = 2.479
~ t (9)
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
Page 9
p-value = 0.0175 (critical value = 1.833)
Conclusion: Since p-value = 0.0175 < 0.05 (or t = 2.479 > 1.833), we reject H0 at 5 %
level of significance and there is sufficient evidence to conclude that the horticulturist’s
claim is correct, i.e. the new variety of fruit bush gives a higher yield.
(iii)
Assume that the yield of the new variety fruit bush follows a
normal distribution for the t-test to be valid.
If the actual value of  ( x  5)2 is less than 19.96, the actual s 2 will be smaller. This
implies that the actual test statistic will be larger.
Therefore, actual test statistic will still fall within the rejection region.
OR
The actual test statistic is larger  the actual p-value is smaller. Therefore, the actual pvalue will still be < 0.05.
Hence there is sufficient evidence to conclude that the new variety of fruit bush produces
a higher yield at 5% level of significance. Thus, the conclusion would still be the same.
9(a)
Let A be the distance travelled on a single trip.
A ~ N(8, 0.72 )
Let T  A1  A2  A3  3 A ~ N(3(8)  3(8), 3(0.72 )  32 (0.72 ))
i.e. T~N(0, 5.88)
P( T  3)  P(T  3)  P(T  3)
= 0.216.
(b)
(i)
Mean = ( 0.5  8  0.3 5 ) + 2.8 = 8.3
(b)
Let F be the total fare paid by a customer for a randomly chosen trip.
(ii)
F ~ N(8.3, 0.5194)
Variance = 0.52  0.7 2  0.32  2.12 = 0.5194
P(F < 8) = 0.33861
Let X be the no. of trips out of 10, that cost less than $8.
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
X ~ B(10, 0.33861)
P( X  6)  1  P( X  5)
= 0.0821
10
(i)
Method 1:
P(at least one blue ball is drawn)
4
C
 1  P(all red)  1  12 3
C3

54
or 0.982
55
Method 2:
P(at least one blue ball is drawn)
 1  P(all red)  1 

(ii)
4 3 2
 
12 11 10
54
or 0.982
55
Method 1:
P(at least one ball of each colour)
 P(at least one blue ball is drwan)  P(all three balls are blue)

54 8C 3

55 12C3

8
or 0.727
11
Method 2:
P(at least one ball of each colour)
 P(at least one blue ball is drawn)  P(all balls are blue)
54 8 7 6
  
55 12 11 10
8
 or 0.727
11

Method 3:
Page 10
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
Page 11
P(at least one ball of each colour)
 1  P(all red balls)  P(all blue balls)
4
C3 8C3
8
 1  12  12  or 0.727
C3
C3 11
(iii)
P(the sum of the numbers on the balls drawn is at least 8)
 P(sum  8) + P(sum = 9)  P(2,3,3)  P(3,3,3)
C1  4C2 4C3
8 4 3 4 3 2
 12
or 3      
12
12 11 10 12 11 10
C3
C3
8
=

(iv)
12 1 13


or 0.236
55 55 55
P(sum of the numbers on the balls drwan is at least 8 at least one
blue ball)
P(sum  8 and at least one blue ball)

P(at least one blue ball)
P(sum  8)  P(sum  8 and all red balls)

P(at least one blue ball)
13 2C1 2C2
13
2 2 1
 12
 3  
55
C3
12 11 10

or 55
54
54
55
55
25

or 0.231
108
Method 1:
Since P( sum  8 at least on blue ball) 
25
13
 P(sum  8) 
, the two events are not
108
55
independent.
Method 2:
5
22
13 54
 P(sum  8)  P(at least 1 blue)   ,
55 55
the two events are not independent.
Since P(sum  8 and at least 1 blue) 
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
11
Page 12
p
(i)
0.07020
0.00021
20
(ii)
50
x
Method 1:
For p = a + bx, r = 0.834.
For p = abx , i.e. ln p  ln a  x ln b , r = 0.998.
Since the value of r for the model p = abx is closer to 1, it is the more appropriate model.
Method 2:
It can be observed from the scatter diagram that the data points fit a curve of the form p =
abx better than a linear model. Therefore the model p = abx is more appropriate.
p  ab x  ln p  ln a  x(ln b) .
Using GC, ln p  12.37899  0.195196426 x
Hence,
ln a  12.37899  a  0.000004206  0.00000421
ln b  0.195196426  b  1.2155497  1.22
(iii)
When x = 32, p  ab x  (4.21106 )(1.21555)32  0.0021727641 .
The expected no. required  5000(0.0021727641)  10.9
(iv)
p = abx  1000p = 1000abx  q = 1000abx which is in the form of q   x .
Hence,   1000a and   b .
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
Page 13
12(a) Let A and C be the no. of enquiry calls on accounts services and credit cards respectively
in 1 hour.
Let T be the total no. of enquiry calls received in 1 hour.
3
 11 
A ~ Po   C ~Po(4) T ~Po  
2
2
E(T) =
11
2
Req. prob. =

P( A  4  T  5.5) P( A  4  T  5)
=
P(T  5)
P(T  5.5)
P( A  4)P(C  1)  P( A  5)P(C  0)
= 0.00864
P(T  5)
All the calls are made independently.
OR
The mean rate of calls received is assumed to be a constant throughout the given time
interval.
(b)
Let X be the duration of a call. X ~ N 12, 3 .
(i)
3

X ~ N 12, 
n

P( X  12.3)  0.03







12.3  12
0.3 
  0.03  P  Z 
  0.97
P Z 


3 
3




n 
n


0.3
 1.88079
3
n
n  117.9 . Hence least n = 118.
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
(i)
Page 14
Alternative method (using graph or table)
Let X be the duration of a call. X ~ N 12, 3 .
3

X ~ N 12, 
n

P( X  12.3)  0.03
Solving the above inequality using GC with n as variable, n  117.9
Method 1 (Use graph):
Method 2 (Use table):
Hence least n = 118.
(ii)
Since n = 118 is sufficiently large, by Central Limit Theorem, X follows a normal
distribution approximately. Hence the answer in b(i) is still valid.
(c)
X ~ N 12, 3
P(10 < X  15)  0.8342611874
Let Y be the no. of calls with duration between 10 and 15 minutes out of 100 calls.
Y ~ B(100, 0.83426) .
Hwa Chong Institution 2011 Preliminary Examination Paper 2 Solutions
Page 15
E(Y) = 100  0.83426  83.426 . Since mode is close to mean, try values of Y close to
83.4.
P(Y  83)  0.10498
P(Y  84)  0.10694 (highest)
P(Y  85)  0.10133
Most likely number of calls = 84.