= log x. log a
a
log y = log x [∵loga .logb = loga By properties]
b c c
Taking Exponent on both sides.
elogy =elogx
yloge =xlogx
y = x
Ans. (a)
158. Let y = 32−log 3 6
= 32.3 −log 3 6
[ ]
= 9.3 −log 3 (3x2)
[ ]
= 9.3 −log 3 3+log 3 2
[ ]
= 9.3 −1+log 3 2
= 9.3
−1.3 −log32
= 3.3log 3 (1/2)
= 3.(1/2) [∵ a log x = x properties]
a
∴ y = 3/2
Ans. (b)
159. log 30 = log (2×3×5)
= log 2 + log 3 + log 5
= 0.3010 + 0.4771 + 0.6990
= 1.4771
Ans. (c)
160. log 124.5 + log 379 = log (12.45 × 10)
10 10 10
+ log (3.79 × 100)
10
= log 12.45 + log 10
10 10
+ log 3.79 + log 100
10 10
= 1.0952 + 0.5786 +2
∴ log 124.5 + log 379 = 4.6738
10 10
Ans. (b)
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ANSWERS
n! n! 2
161. n :n =2:1 ⇒ ÷ =
P 5 P 3 (n−5)! (n−3)! 1
(n−3)! 2
⇒ = ⇒ (n – 3) (n – 4) = 2.1
(n−5)! 1
∴ n – 3 = 2 ⇒ n = 5
Ans. (b) 5
162. No. of ways to enter into room = 10
No. of ways to came out from a different door = 9
∴ Total Ways = 10.9 = 90 ways
Ans. (a) 90
163. Five digit Nos. by using digit (1, 2, 3, 4, 6) = 5! = 120
Four digit nos. by using digit (1, 2, 3, 4, 6) = 5 P
4
= 120
∴ Total Nos. greater than 1000 = 120 + 120
= 240
Ans. (c) 240
164. Total Nos. of 6 digit (greater than 1 lakh)
6!
by using (1, 1, 1, 2, 2, 3) are =
3! 2!
= 60
Ans. (a) 60
165. No. of ways in which 17 billiard can be arranged.
If 7 are black, 6 red and 4 white are
17!
= = 4084080
7! 6! 4!
Ans. (b) 4084080
∫
xex
166. Let I = dx
( )
x+12
( )
∫
x+1−1 ex
= dx
( )
x+12
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( )
= ∫
x+1ex
dx − ∫
ex
dx
( ) ( )
x+12 x+12
= ∫ ( x e + x 1 )dx − ∫ ( x e + x 1 ) 2 dx
= I –I
1 2
∫ex dx
I =
(x−1)
Integrating by parts.
⎛ ⎞
1 1
I 1 = 1+x ex − ∫ ⎜ ⎜ ⎝ − (x+1)2 ⎟ ⎟ ⎠ ex dx
1 1 ex
∴ I = ex + ∫ ex dx − ∫ dx
(1+x) (x+1)2 (x+1)2
1
I = ex
(1+x)
Ans. (b)
(x−1) x+1−2
167. ∫ ex dx = ∫ ex dx
(x+1)3 (x+1)3
= ∫ ex ⎪ ⎨ ⎧ 1 − 2 ⎪ ⎬ ⎫ dx
⎪ ⎩(x+1)2 (x+1)3⎪ ⎭
= ∫ ex { f ( x ) +f ′ ( x )} dx
1
Where f(x) =
(x+1)2
= ex f(x) + c
∫ex(x−1)
dx =
ex
+c
( )
(x+1)3 x+1 2
Ans. (c)
168. See the text book example page No. 9.28
Ans. (a)
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ANSWERS
∫ dx ∫ dx
169. = ( )( )
x2 −a2 x−a x+a
∫ 1 ⎜ ⎛ 1 − 1 ⎟ ⎞ dx
2a ⎝x−a x+a ⎠
=
1
⎢
⎡ ∫ dx −∫ dx
⎥
⎤
2a ⎣ x−a x+a⎦
1 [ ( ) ( )]
= log x−a −log x+a
2a
∫ dy = 1 log ⎢ ⎡x−a ⎥ ⎤
x2 −a2 2a ⎣x+a⎦
170.
∫ 1
dx
=∫
(
d
)
x
( )
a2 −x2 a+x a−x
= ∫ ⎜ ⎛−1 ⎟ ⎞ ⎢ ⎡ 1 − 1 ⎥ ⎤ dx
⎝2a ⎠⎣a+x a−x⎦
= − 1 ⎢ ⎡ ∫ 1 dx − ∫ 1 dx⎥ ⎤
2a ⎣ a+x a−x ⎦
1 [ ( ) ( )]
= − log a+x −log a−x
2a
−1 ⎡a+x⎤
= log ⎢ ⎥
2a ⎣a−x⎦
171. ex−y +log xy+xy=0
d.wr.t.u.
⎛ dy⎞ 1 ⎡ dy ⎤ ⎡ dy ⎤
ex−y⎜1− ⎟+ ⎢x +y.1⎥ + ⎢x +y.1⎥ =0
⎝ dx⎠ xy⎣ dx ⎦ ⎣ dx ⎦
dy 1 dy 1 dy
ex−y −ex−y. + + +x +y=0
dx y dx x dx
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⎜ ⎜ ⎛ −ex−y + 1 +x⎟ ⎟ ⎞ dy = −ex−y − 1 −y
⎝ y ⎠ dx x
1 (xy+1−y.ex.y) dy = − 1 (x.ex−y +1+xy)
y dx x
∴ dy = −y . ⎜ ⎛ x.ex−y +1+xy⎟ ⎞
dx x
⎜ ⎝1+xy−y.ex−y ⎟
⎠
Ans. (d) None of these
172. y = xlog(logx)
log y = log (log x) . log x
1 dy 1 1 1
. = log (log x). + (log x). .
y dx x (log x) x
dy y
∴ = [log (log x) + 1]
dx x
y
Ans. (a) [log (log x) + 1]
x
1
173. y = x +
1
x+
x
x3 +x+x x3 +2x
y = =
x2 +1 x2 +1
( )( ) ( )
( )
dy x2 +1 3x2 +2 − x3 +2x 2x
=
( )
dx x2 +1 2
3x4 +3x2 +2x2 +2−2x4 −4x2
= ( )
2
x2 +1
dy x4 +x2 +2
=
dx (x2 +1)2
x4 +x2 +2
Ans. (a)
(x2 +1)2
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ANSWERS
y x
174. + = 6
x y
y+x
= 6 ⇒ x+y = 6 xy
xy
dy 6 ⎛ dy ⎞
∴ 1+ = ⎜x +y.1⎟
dx 2 xy ⎝ dx ⎠
⎛ ⎞
⎜ 1−3 x ⎟dy =3 y −1
⎜ ⎟
⎝ y ⎠dx x
y ( )
3 −1
dy = x ⇒ ( 3 y − x ) y
dx x y −3 x x
1−3
y
⎛x+y⎞
3y−⎜ ⎟
dy 3y− xy ⎝ 6 ⎠
= =
dx xy −3x ⎜ ⎛x+y ⎟ ⎞ −3x
⎝ 6 ⎠
dy 17y−x x−17y
= =
dx y−17x 17x−y
x−17y
Ans. (c)
17x−y
3
∑
175. 47C + 50−i
4 C 3
i=0
⇒ 47 +50 +49 +48 +47
C4 C3 C3 C3 C3
⇒ 178365 + 19600 + 18424 + 17296 + 16215
⇒ 249900
Ans. (a) 249900
176. a = 100
S = 5. S
6 6
6 6
[2a+(6−1)d]=5. [2(a+6d)+(6−1)d]
2 2
3 [200 + 5d] = 15 [200+12d + 5d]
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200+5d = 1000 + 85d
80d = – 800 ⇒ d = –10
Ans. (a) – 10
177. S = 3n2 + n
n
∴ S = 4 ∴ a = 4
1 1
S = 14 a = 14 – 4 = 10 ∴ d = 6
2 2
S = 30 a = 30 – 14=16
3 3
∴ Tp = a + (p – 1) d
= 4 + (p – 1) 6
Tp = (6p – 2)
Ans. (b) ( 6 p – 2)
178. S = S
m n
m n
[2a+(m−1)d] = [2a+(n−1)d]
2 2
2ma – 2na = (n2 −n)d−(m2 −m)d
2a (m – n) = (n2 – n – m 2 – m)d
2a (m – n) = –(m – n) (m+n – 1)d
∴ 2a = –(m+n – 1)d
m+n
S = [2a + (m+n – 1)d]
m+n
2
m+n
[(–m+n – 1)d + (m+n – 1)d]
2
S = 0
m+n
Ans. (a) 0
9 7
179. − ,−2,− .....0
4 4
9 9 1
a = – d = –2 + =
4 4 4
9 1 9 1
0 = – +(n−1) ⇒ =(n−1)
4 4 4 4
⇒ 9 = n – 1 ⇒ n=10
Ans. (b) 10th term
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ANSWERS
180. 6.T = 15.T
6 15
∴ 6 (a+5d) = 15 (a+14d)
2a + 10d = 5a + 70d
3a = –60d ∴ a = –20d
T = a + 20d
21
= –20 d + 20d
T = 0
21
Ans. (c) 0
181. The average of n numbers is x.
If any no.is multiplied to each of datas. Then average will also multiplied by such no.
∴ New average = (n + 1) x
Ans. (c) (n + 1) x.
∑ ∑
x x
182. Av = ⇒ 1.5 =
n δ
∑
= x = 12 kg. (increased weight)
∴ Weight of New person = 65 + 12 = 77 kg.
Ans. (c) 77 Kg.
183. If passes students = x
39x+15(120−n)
∴ 35 =
120
4200 = 39x + 1800 – 15x
2400 = 24x
∴ x = 100
Passed Students = 100
Ans. (a) 100
∑
x
184. =45
17
∑
x = 765
Total of first 9 numbers = 9 × 51 = 459
Total of last 9 numbers = 9 × 36 = 324
∴ Value of 9th number = (459 + 324) – 765
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= 18
Ans. (c) 18
∑
185. x = 11 × 30 = 330
Total of first five numbers = 5 × 25 = 125
Total of last five numbers = 5 × 28 = 140
∴ Value of 6th number = (125 + 140) ~ 330 = 65
Ans. (b) 65
186. Ans. (a) Refer properties
187. Ans. (b) Refer Properties.
188. Let A be the hearts playing cards in a part. Let B be the club playing cards in a part.
13
∴ P(A) =
52
13
P(B) =
52
Here A and B are mutually exclusive.
( ) 13 13 26 1
∴ P A∪B =P(A) +P(B) = + = =
52 52 52 2
189. Total number of balls in the bag = 6 + 4 = 10
Since three balls are drawn out of 10 balls in 10C ways
3
∴ Exhaustive number of Cases = 10 C = 120
3
The number of favourable cases two balls are blue and balls is red
= 6C x 4C
2 1
= 60
6c ×4c
∴ Probability of 2 balls are blue and 1 is red = 2 1
10c
3
60
= =½
120
Ans. (c)
190. There are 366 days in a leap year.
Now 366 = 7 x 52 + 2
∴ The leap year will contain at least 52 Mondays. The possible combination for the
remaining two days are:
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ANSWERS
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Let A be the event of getting 53 Mondays in the leap year. Therefore, only those
combinations will be favourable to the event A which contain "Monday"
∴ The combination (i) and (ii) are favourable to the happening of A
∴ P(A) = 2/7
Ans. (a)
191. Given P(A) = 1/3
P(B) = 3/4
and A and B are independent events
( ) ( )
P A∪B =1−P A∩B 1
[ ]
= 1− P(A1).P(B1)
{[ ] [ ]}
= 1– 1−P(A) 1−P(B) [A and B are independent]
⎡ ⎛ 1⎞⎛ 3⎞⎤
= 1 – ⎢ ⎜1− ⎟⎜1− ⎟⎥
⎣ ⎝ 3⎠⎝ 4⎠⎦
⎧⎛2⎞⎛1⎞⎫
= 1 – ⎨⎜ ⎟⎜ ⎟⎬
⎩⎝3⎠⎝4⎠⎭
( ) ⎡1⎤ 5
P P A∪B =1− ⎢ ⎥ =
⎣6⎦ 6
Ans. (b)
192. Out of given 4 letters, there are two letters are vowel (O,E). Let A be the first letter is
vowel.
P(A) = 2/4
Let B be the second letter is vowel
P(B) = 1/3
Here A and B are independent
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P(AB) = P(A).P(B)
2 1
= . = 1/6
4 3
Ans. (a)
193. Out of given 4 letters, there are two letters are vowel (O, E)
Let A be the first letter is vowel.
i.e. P(A) = 2/4
Let B be the second letters is vowel.
P(B) = 1/3
Here A and B are Mutually executive
( )
∴ P A∪B = P(A) + P (B)
2 1 6+4 10 5
= + = = =
4 3 12 12 6
Ans. (a)
194. Let A be the first letter selected M from the 'HOME'.
B be the second letter selected M from the 'HOME'
P(A) = 1/4, P(B) = 1/4
A and B are Mutually exclusive
∴ P(AUB)= P(A) + P(B)
1 1 2 1
= + = =
4 4 4 2
Ans. (b)
195. By addition thereon
P(A or B) = P(A) + P(B)
0.65 = [1 – P('not A)] + p
= [1 –0.65]+ p
∴ p = 0.65 – 0.35
p = 0.30
Ans. (c)
197. Since f(x) is a Polynomial.
& a , a , a are in AP
1 2 3
∴ f(a ), f(a ), f(a ) also in AP
1 2 3
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ANSWERS
∴ f ′ (a ), f ′ (a ), f ′ (a ) also in AP
1 2 3
Ans. (a) AP
⎡( 1+i ) n ⎤
198. A = P ⎢ −1⎥
⎢⎣ i ⎥⎦
⎡( 1.04 ) 10 ⎤
20,000 = P ⎢ −1⎥
⎢⎣ 0.04 ⎥⎦
After solving we get
P = 2470 (Approx)
Ans. (a) 2470
199. byx = 1.2 & bxy = – 0.5
This is wrong because bxy and byx have same sign.
Ans. (b) false.
200. The mean of poison distribution is 1.6 and variance is 2. This is wrong because P – d will
greater than 2
Ans. (b) false.
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