2010 JJC Prelim P2 Solutions

2010 JJC Prelim P2 Solutions
1(i)
x
Let the length and the width of the rectangle be 2x and y
The perimeter S = 2y + 4x (1)
y 2 = a 2 x 2 (2)
Subst (2) into (1)
S = 2 a2 x2 + 4 x
dS
2x
=
+4=0
2
dx
a x2
2x
+4=0
a2 x2
x
=2
a2 x2
4 2
a
5
1
y2 = a 2
5
4
1
x=
a
a
and y =
5
5
d 2s
2a 2
=
3 < 0
dx 2
(a 2 x 2 ) 2
x2 =
Length : width = 2x : y = 4:1
1(ii)
2
+ 3 (1)
x
Differentiate w.r.t x
dy
dy
2
2 y + x + y = 2x + 2
dx
dx
x
dy
2
(2 y + x) = 2 x + 2 y
dx
x
2
2x + 2 y
dy
x
(2)
=
2y + x
dx
Subst. x = 1 into
(1): y 2 + y = 2 y= 1 and y 2
dy
(2):
=1
dx
y 2 + xy = x 2
dy dy dx
= .
dt dx dt
1
= (1)( )
5
1
= units s 1
5
2
( x + y ) 2 = 2 xy y 3
3
2
( x + y ) 2 2 xy = y 3
3
2
y 2 + x2 = y3
3
dy
dy
2 y + 2 x = 2 y 2
dx
dx
dy
( y + y 2 ) = x (shown)
dx
Alternative:
dy
dy
dy
2( x + y ) 1 + = 2 x + 2 y 2 y 2
dx
dx
dx
dy
( y + y 2 ) = x (shown)
dx
dy
= y2 + y
dx
use y = ux
dy du
x+u
=
dx dx
2
du
x x + u = ( ux ) + ux
dx
du
x + u = u2 x + u
dx
1 du
=1
u 2 dx
1
u 2 du = 1 dx
1
= x + c , where c is an arbitrary constant
u
x
= x+c
y
x
c
y=
= 1 +
x+c
x+c
x
2 Family of solution curves:
c = 1, y = 1
c = 0, y = 1
c = 1, y = 1 +
1
x 1
c =1
1
x +1
y
c = 1
x
y = 1
c=0
x = 1
x =1
3(i) Using ratio theorem,

2OA + OB
OM =
3

3OM OB
OA =
2
0 2
1
= 3 2 0
2
2 2
1

= 3
4

= 1
3(ii)
4 2

BC = 0
1 2

2

=
1

3
Length of projection of BC onto the line OM =
2

BC OM
3
=
2
OM
2 0

1
1 1

2
3
2
1 +1
+1 = 3
2 + 2 8 = 0
= 2 or 4 (rejected)
= 2
3(iii)
2 1
1

AB = 0 3 = 3 1
2 4
2

4 1 5

AC = 2 3 = 1
1 4 5

1 5

Normal of plane ABC = 1 1
2 5

3

= 5
4

a 3 a 3

0 5 = 0 5 sin 6
1 4 1 4

1
3a + 4 = a 2 + 1 50
2
( 6a + 8 )2 = 50 ( a 2 + 1)
7 a 2 48a 7 = 0
a = 7 or -
1
(reject)
7
a = 7
4(i)
z2 = 2 z = 2
arg ( iz ) =
arg ( i ) + arg( z ) =

4 2
3
=
4
arg( z ) =
wz = 2 2
w z =2 2
w =2
z2
5
arg =
w
6

5
2 arg( z ) arg( w) =
6
3 5
arg( w) = 2 +
4 6
7
=
3
=
( pv)
3

w = 2 cos + i sin
3
3
= 1 + 3i
4(ii)
4
z + 1 3i = 0
4
z = 1 + 3i
2
+ 2 k i
= 2e
1 + k
= 2 4 e 6 2
1
(1+ 3k )i
= 24 e 6
, k = 0, 1, 2
1
1 2
1
1 5
i
i
i
i
= 24 e 6 , 24 e 3 , 24 e 3 , 24 e 6
z3
z2
z4
z1
The points form a square since the diagonals are perpendicular and of equal
length.
5(i)
The different letters left are D S T R B U O N

4!
No of different code words = 8C1 = 32
3!
5(ii)
To permute the remaining 9 letters 1st,
9!
ways.
2
To slot in the 3Is, 10C3 ways.

Thus number of code words that can be formed is
=
9! 10
C3
2
= 21772800
6(i)
6(ii)
P(1 left-handed player) =
C1 6C3 4
=
8
C4
7
P(G R G L BR BR )+P(G R G R BL BR )
P(exactly 1 left-handed)
2
C1 4 C 2 + 2 C2 4 C1
8
C4
=
4
7
8
2
= 35 =
4 5
7
P(2 G given exactly 1 left-handed)=
6(iii) P(Boy A or Girl B is chosen) = 1 P (Both Boy A and Girl B are not chosen)
6
C 11
= 1 8 4 =
C4 14
Alternative:
P(Boy A or Girl B is chosen) = P(Boy A is chosen) + P(Girl B is chosen)
P(Boy A and Girl B are chosen)
7
7
C
C 6C
= 8 3+8 38 2
C4
C4
C4
1 1 3 11
= + =
2 2 14 14
Let X denote the number of guitars sold by a music shop in a day.

X P0 ( )
2
7
2
e =
7
2
= ln = 1.253
7
Given that P( X = 0) =
7(i)
P( X < 4) = P( X 3)
= 0.96145 0.961
7(ii) Let Y be the number of days out of 100 in which at least 4 guitars were sold per day.
Y B(100,1 0.96145) = B(100, 0.03855)

Since n is large and np < 5 , we use a Poisson approximation.
Y P0 (3.855)
P(more than 95 days in which less than 4 guitars were sold per day)
= P( at most 4 days in which at least 4 guitars were sold per day )
= P(Y 4)
= 0.657 (to 3 sf)
7(iii)
X P0 (1.253)
For a large sample of size 90, by Central Limit theorem,
1.253
X N 1.253,
.
90
P X > 1.5
= 0.0182
).
Alternative : X 1 + + X 90 P0 (1.253 90 ) = P0 (112.77 )
P X > 1.5 = P ( X 1 + + X 90 > 135 )
= 1 P ( X 1 + + X 90 135 )
= 0.0183
X N (15,5)
8(i)
Let
T = X1 + X 2
E (T 3 X ) = E ( X 1 + X 2 ) 3E ( X ) = 15
Var (T 3 X ) = Var ( X 1 + X 2 ) + 9Var ( X ) = 11(5) = 55
T 3 X N (15,55)
P(T > 2 + 3 X ) = P(T 3 X > 2)

= 0.0109
8(ii)
P( X < 20) = 0.9873 0.987

Probability = 3 [ P ( X < 20) ] P ( X > 20)
2
= 3(0.9873) 2 (1 0.9873)
= 0.0371
8(iii)
Y N ( , 22.52 )
15.1 + 29.9 45
=
= 22.5
2
2
Greatest P(15.1 < Y < 29.9) = 0.258
For greatest probability, =
8(iv) (iv)
X N (15,5)
Y N (10, 2 )
X + Y N (15 + 10, 5 + 2 ) = N (25, 5 + 2 )
P ( X + Y > 27) = 0.25
27 25
PZ >
= 0.25
5+ 2
2
PZ >
= 0.25
5+ 2
2
PZ <
= 0.75
2
5+
2
= 0.6745
5 + 2
= 1.95
Assumption : The random variables X and Y are independent of each other.
10
9(i) To obtain a quota sampling of 60, divide the diners into two subgroups : male and
female. Select the first 30 males and 30 females who leaves the restaurant.
Or any other relevant answers.
9(ii) By drawing random samples according to the proportion in each stratum, lunch and
dinner, the sample will be a better representation of the population.
Draw random samples from each stratum with sample size proportional to the size of the
strata as follows :
Lunch
Number of
customers to be
sampled
10(i)
2
60 = 24
5
Dinner
3
60 = 36
5
e y = axb
y = ln a + b ln x
The scatter diagram is plotted with y against ln x .
y
ln x
From the scatter diagram, the points lie close to a straight line, so the linear model is
appropriate.
10(ii) From GC, since r 0.982 which is very close to -1, it supports the claim in part (i).
10(iii)
y = 73.3 18.4 ln x
a = 6.89 1031 (3sf )
b = 18.4 (3sf)
10(iv) 86.0% (3sf)
Since x = 0.5 is out of the given data range of 1 x 40 , the prediction is unreliable.
10(v) For large values of x , the model gives y < 0 . So the model is not valid for large values
of x .
11
11 Assume that the amount spent per customer follows a normal distribution.
Let be the actual mean amount spent per customer per visit.
H 0 : = 59
H1 : < 59
Under H 0 , X ~ N( ,
2
n
) and test statistic Z =
~ N(0, 1)
n
432
= 54, = 59, = 8, n = 8.
8
At 5% level of significance, we use a left-tailed z- test and reject H0 if p-value < 0.05.
where x =
From the GC, p value = 0.0385 .

Since p value = 0.0385 < 0.05 , we reject H 0 and conclude that at 5% level of
significance there is sufficient evidence to suggest that the mean amount spent per
customer per visit has decreased in recent months.
11
H 0 : = 0
H1 : 0
( y 70)
+ 70 = 62
9
2
72 )
(
1
2
s =
1234
= 82.25
9 1
9
y=
Under H 0 , Y ~ N(0 ,
Y 0
s2
~ t(n 1)
) and test statistic T =
s
n
n
where y = 62, = 0 , s = 82.25, n = 9
At 5% level of significance, we use a 2-tailed t- test and reject H0 if T 2.306

Since H 0 is rejected at 5% level of significance,
Y 0
Y 0
2.306 or
2.306
s
s
n
n
82.25
82.25
or 0 62 2.306
9
9
0 55.03 or 0 68.97
0 62 + 2.306
{0 : 0 55.0 or 0 69.0 } or
{0 : 0 < 55.0 or 0 >69.0 }

12

2010 JJC Prelim P2 Solutions

Uploaded by

Copyright:

Available Formats

2010 JJC Prelim P2 Solutions

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

2010 JJC Prelim P2 Solutions

Uploaded by

Copyright:

Available Formats

2010 JJC Prelim P2 Solutions

Length : width = 2x : y = 4:1

2 Family of solution curves:

3(i) Using ratio theorem,

The different letters left are D S T R B U O N

To permute the remaining 9 letters 1st,

To slot in the 3Is, 10C3 ways.

P(1 left-handed player) =

P(2 G given exactly 1 left-handed)=

Let X denote the number of guitars sold by a music shop in a day.

Y B(100,1 0.96145) = B(100, 0.03855)

Alternative : X 1 + + X 90 P0 (1.253 90 ) = P0 (112.77 )

P X > 1.5 = P ( X 1 + + X 90 > 135 )

P(T > 2 + 3 X ) = P(T 3 X > 2)

P( X < 20) = 0.9873 0.987

For greatest probability, =

Assumption : The random variables X and Y are independent of each other.

) and test statistic Z =

From the GC, p value = 0.0385 .

At 5% level of significance, we use a 2-tailed t- test and reject H0 if T 2.306

{0 : 0 < 55.0 or 0 >69.0 }

You might also like