According to the second condition of the problem
x−5 4
=
y−5 5
5(x – 5) = 4 (y – 5)
5x – 25 = 4y – 20
5x – 4y = 5 → (2)
(1) × 5 ⇒ 30x – 25y = 0 →(3)
(2) × 5 ⇒ 30x – 24y = 30
Subtracting (3) from (4), we get
y = 30
(1) ⇒ 6 x – 5 (30) = 0
6x = 150
150
x = = 25
6
Hence the required numbers are 25 and 30
Ans. (c)
164. Let the given numbers be x and y. Then according to the given conditions of the problem.
x+1 1
=
y+1 2
⇒ 2x + 2 = y + 1
2x – y =–1 → (1)
x−5 5
and =
y−5 11
11x – 55 = 5y – 25
11x – 5y = 30 → (2)
∴ (1) × 11 ⇒ 22x – 11y = –11 → (3)
(2) × 2 ⇒ 22x – 10y = 60 → (4)
(3) – (4) ⇒ – y = – 71
y = 71
(1) ⇒2x – 71 = – 1
2x = 70
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x = 35
Hence, the required numbers are 35 and 71
Ans. (b)
165. Let the number to be subtracted be x.
Then according to the problem
27−x 7
=
43−x 15
⇒ 15 (27– x) = 7 (43 – x)
⇒ 405 – 15x = 301 – 7x
15x – 7x = 405 – 301
8x = 104
104
x = = 13
8
Hence the required number is 13
Ans. (a)
166. Let the unit digit = x
and ten digits = y
∴ x+y = 3 → (1)
and the number = 10y + x
Reversing the order of digits
Units digit = y
and ten's digit = x
∴ Number = 10x + y
According to the given condition of the problem
7(10y + x) = 4(10x+y)
70y + 7x = 40x + 4y
70x – 40x + 70y – 4y = 0
– 33x + 66y = 0
– x+2y = 0 → 92)
Adding (1) and (2) we get
3y = 3
y = 1
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(1) ⇒ x + 1 = 3
x = 3 –1 = 2
Hence, the required number is 12
167. The committee of six must include atleast 2 ladies.
i.e. two or more ladies. As there are only 3 ladies, the following possibilities arise:
The committee of 6 consists of
(i) 4 men and 2 ladies, (ii) 3 men and 3 ladies.
The number of ways for (i) = 7C + 3C
4 2
= 35 × 3 = 105
The number of ways for (ii) = 7C × 3C
3 3
= 35 × 1
= 35
Hence the total number of ways of forming a committee so as to include atleast two
ladies = 105 + 35 = 140
Ans. (a)
n!
168. We have nCr =
r!(n−4)!
Now substituting for n and r, we get
28!
28C =
2r (2r)!(28−2r)!
24!
24C 2r – 4 = (2r−4)! { 24−(2r−4) } !
24!
=
(2r−4)!(28−2r)!
Given 28 : 24 = 225:11
C 2r C 2r−4
28 28! (2r−4)!(28−2r)!
⇒ C 2r = ÷
24 (2r)!(28−2r)! 24!
C 2r−4
28×27×26×25×24! (2r−4)!(28−2r)!
= ×
2r(2r−1)(2r−2)(2r−3)(2r−4)!(28−2r)! 24!
28×27×26×25 225
= =
2r(2r−1)(2r−2)(2r−3) 11
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11×28×27×26×25
⇒ 2r (2r – 1) (2r – 2) (2r – 3) =
225
= 11 × 28 × 3 × 26
= 11 × 7 × 4 × 3 × 13 × 2
= 11 × 12 × 13 × 14
= 14 × 13 × 12 × 11
∴ 2r = 14
r = 7
Ans. (b)
169. Let in the number unit's digit = x
and ten's digit = y
∴ Number = 10y + x
According to the given conditions of the problem,
8 (x+y) + 1 = 10y + x
(or) 8x + 8y + 1 = 10y + x
⇒ 8x – x + 8y – 10y + 1 = 0
7x – 2y + 1 = 0 → (1)
and 13 (y – x) + 2 = 10y + x
13y – 13x + 2 = 10y + x
⇒ x + 13x + 10y – 13y – 2 = 0
⇒ 14x – 3y – 2 = 0 → (2)
(1) × 2 ⇒ 14x – 4y + 2 = 0 → (3)
(2) – (3) ⇒ y – 4 = 0
y = 4
Put y = 4 in (1), we get
7x – 8 + 1 = 0
7x = 7
x = 1
Hence, the required number is 41
Ans. (b)
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170. We want to find out the number of combination of 12 things taken 3 at a time and this is
by:
12!
12 =
C 3 3!(12−3)!
12! 12×11×10×9!
= =
3!9! 3!9!
12×11×10
=
3×2
= 220
Ans. (c)
x−3 x −3 1
172. Lt = Lt
x→9 x−9 x→9 x −3 x +3
1 1
App lt = =
3+3 6
1
Ans. (a)
6
( )
x+0 − 2a x+a − 2a x+a − 2a
173. Lt ⇒ Lt × ( )
x→9 x−0 x→9 x−0 x+a + 2a
x+a−2a (x−a)
⇒ Lt ( ) = Lt ( )
x→9 (x−a) x+a + 2a x→a (x−a) x+a + 2a
App Lt
1 1
( ) =
a+a + 2a 2 2a
1
Ans. (b)
2 2a
⎛ 6 ⎞
⎜5+ ⎟
174. Lt
6+5x2
⇒ Lt x2
⎝ x2 ⎠
x→∞ 4x+15x2 x→∞
x2⎜
⎛
15+
4
⎟
⎞
⎝ x⎠
App Lt.
5+0 1
⇒ =
15+0 3
1
Ans. (c)
3
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a−bx ⎛ a b⎞
175. Lt ⇒ Lt ⎜ − ⎟
x→∞ x2 x→∞ ⎝x2 x⎠
App Lt.
⇒ (0 … 0) = 0
Ans.(a) 0
d d
176. y= ex −e −x ⇒ dy =
(ex +e −x)
dx
(ex −e −x)−(ex −e −x).
dx
(ex +e −x)
ex +e −x dx (ex +e −x)2
(ex +e −x)(ex +e −x)−(ex −e −x)(ex −e −x)
=
(ex +e −x)2
e2x +e −2x +2−e2x −e −2x +2 4
= =
(ex +e −x)2 (ex +e −x)2
4
Ans. (b)
(ex +e −x)2
d d
(1+x)2 x−x (1+x)2
177. y = x ⇒ dy = dx dx
(1+x)2 dx (1+x)4
(1+x)2.1−x.2(1+x) 1+x2 +2x−2x−2x2
⇒ =
(1+x)4 (1+x)4
1−x2 1−x
= =
(1+x)4 (1+x)3
1−x
Ans. (b)
(1+x)3
178. y = x+ x
dy d d
= t1/2. (x+ x)
dx dt dx
1 ⎛ 1 ⎞
= .⎜ ⎜1+ ⎟ ⎟
2⎜ ⎛ x+ x ⎟ ⎞ ⎝ 2 x ⎠
⎝ ⎠
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2 x+1
=
4 x x+ x
2 x+1
Ans. (a)
4 x x+ x
179. y =
7x2+2x
dy d d
= 7t. (x2 +2x)
dx dt dx
=
7x2+2x.log
7. (2x+2)
dy =2(x+1).7x2+2.log7
dx
Ans. (b)
2(x+1).7x2+2.log7
⎛ ⎞
180. y = log ⎜x+ x2 +a2 ⎟
⎝ ⎠
dy d d ⎛ ⎞
= logt. ⎜x+ x2 +a2 ⎟
dx dt dx⎝ ⎠
⎡ ⎤
1 1
= . ⎢1+ .2x⎥
x+ x2 +a2 ⎢⎣ 2 x2 +a2 ⎥⎦
⎛ ⎞
⎜ x2 +a2 +x⎟
⎝ ⎠
=
⎛ ⎞
x2 +a2⎜x+ x2 +a2 ⎟
⎝ ⎠
dy 1
=
dx x2 +a2
1
Ans. (a)
x2 +a2
x
181. Given ( x−y ) ex−y =a
Differentiate on both sides.
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⎡ ( )( ) ⎛ dy⎞⎤
x ⎢ x−y 1 −x.⎜1− ⎟⎥ x
( x−y ) .ex−y⎢ ⎝ dx⎠ ⎥+ex−y⎜ ⎛ 1− dy ⎟ ⎞ =0
( )
⎢ x−y 2 ⎥ ⎝ dx⎠
⎢ ⎥
⎣ ⎦
⎡ dy ⎤
x ⎢x−y−x+x ⎥
ex−y⎢ ( ) dx +1− dy ⎥ = 0
⎢ x−y dx⎥
⎣ ⎦
dy
−y+x
(
d
)
x +1− dy =0
x−y dx
−y x dy dy
+ +1− = 0
x−y x−y dx dx
dy ⎡ x ⎤ y
⎢
−1⎥ = −1
dx ⎣x−y ⎦ x−y
dy ⎡x−x+y⎤ y−x+y
⎢ ⎥
=
dx ⎣ x−y ⎦ x−y
dy ⎡ y ⎤ 2y−x
⎢ ⎥
=
dx ⎣x−y⎦ x−y
dy 2y−x
=
dx y
dy ⎡2y−x⎤
∴ y +x=y⎢ ⎥ + x
dx ⎣ y ⎦
= 2y – x + x
dy
y +x =2y
dx
Ans. (c)
182. Given demand law x = 10−p2
x = (10−p2)1/2
dx 1 1 −1
= (10−p2)2 (−2p)
dp 2
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1(−2p)
2 10−p2
dx −p
=
dp 10−p2
p dx
ed = .
x dp
p −p
= .
(10−p)1/2 2 10−p2
when p = 2
2 −2
ed = .
6 6
ed = –2/3
2
ed =
3
Ans. (a)
Alternate
x3
183.
∴∫
dx
x+1
⎛ 1 ⎞
= ∫⎜x3 −x+1− ⎟dx
⎝ x+1⎠
x3 x2
= − +x – log(x+1) + c
3 2
Ans.(c)
⎛ e4x +e2x ⎞
∫⎜ ⎟
184. ⎜ ⎟dx
⎝ e3x ⎠
⎛ e4x e2x ⎞
= ∫ ⎜ ⎜ + ⎟ ⎟ dx
⎝e3x e3x ⎠
=
∫ exdx+∫
e
−xdx
= ex – e – x + c
Ans.(b)
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x4 +1 ⎛ 2 ⎞
185. ∫ dx = ∫⎜x2 −1+ ⎟ dx
x2 +1 ⎝ x2 +1⎠
1
= ∫ x2dx−∫ dx+2 ∫ dx
x2 +1
x3
= – x + 2 tan–1 x + c
3
Ans.(c)
186. Let I = ∫ log (x+1)dx
∴ I = ∫ log (x+1).1dx
Integrating by parts
[Here log (x+1) is to be taken as first function and unity as second function)
I = [log (x+1] integral of '1' – integral of
[d/dx (log (x+1)] + integral of '1']
∫ 1
= log (x+1).x – .xdx
x+1
∫x+1−1
= x log (x+1) – dx
x+1
= x log(x+1) – ∫ ⎜ ⎛ 1− 1 ⎟ ⎞ dx
⎝ x+1⎠
= x log (x+1) – [x – log (x+1)]
I = x log (x+1) – x + log (x+1) +c
1 x − 1+x
187. Consider =
x + 1+x x−(1+x)
= 1+x − x
∴ I = ∫ dx
x + 1+x
= ∫ 1+x dx −∫ x dx
I = I – I
1 2
I = ∫ 1+x dx
1
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Let z = 1 + x
dz = dx
∴ I = ∫ 1+x dx =∫ z dz
1
2
= z3/2
3
2
= (1+x)3/2
3
x3/2
∴ I = 2/3 (1+x)3/2 – + c
3/2
{ }
( )
I = 2/3 1+x 3/2 −x3/2 +c
188. Consider x3 +x2 −2x=x(x2 +x−2)
= x(x2 +2x−x−2)
= x {x(x+2) – (x+2)}
= x (x – 1) (x + 2)
x2 −x+2 x2 −x+2
∴ We may write =
x3 +x2 −2x x(x−1)(x+2)
x2 −x+2 A B C
Let = + +
x(x−1)(x+2) x x−1 x+2
(or) x2 −x+2=A(x−1)(x+2) +Bx(x+2)+Cx(x−1)
Substituting x = 1, We find 2 = 3B
B = 2/3
Substituting x = – 2, We find 8 = 6c
i.e. C = 4/3
Substituting x = 0, We find 2 = – 2A
A = – 1
∴ I = ∫ x2 −x+2 dx=−∫dx + 2∫ dx + 4 ∫ dx
x3 +x2 −2x x 3 x−1 3 x+2
I = – log x + 2/3 log (x – 1) + 4/3 log (x+2) + log c
Ans. (c)
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ANSWERS
∫ 1
189. Let I = dx
3x2 +13x−10
1 1
∫
= dx
3 x2 + 13x − 10
3 3
1∫ 1
= dx
3 ⎡
13
⎛13⎞2⎤ ⎛13⎞2
10
⎢x2 +2. x+⎜ ⎟ ⎥−⎜ ⎟ −
⎢⎣ 6 ⎝ 6 ⎠ ⎥⎦ ⎝ 6 ⎠ 3
1∫ 1
= dx
3 ( x+13/6 ) 2 − 289
36
Let t = x+13/6
∴ dt = dx
∴ I = 1/3 ∫ 1 dt
t2 −(17/6)2
1 1 ⎡t−17/6⎤
= . ( )log ⎢ ⎥
3 217/6 ⎣t+17/6⎦
1 ⎡6t−17⎤
= log ⎢ ⎥
17 ⎣6t+17⎦
⎡ ⎤
⎢ ⎥
1 6(x+13/6)−17
⎢ ⎥
= log
17 ⎢ ⎛ 13⎞ ⎥
⎢
6⎜x+ ⎟+17
⎥
⎣ ⎝ 6 ⎠ ⎦
1 ⎡ 3x−2 ⎤
= log ⎢ ⎥ +c
17 ⎣3x+15⎦
Ans. (b)
190. ∫ ex { f ( x ) + f ′ (x) } dx
By the method of integration by parts, we may write
∫ ex f ( x ) dx = f ( x ) ∫ ex dx − ∫ ⎨ ⎧ d f(x) ∫ ex dx⎬ ⎫ dx
⎩dx ⎭
= ex f ( x ) −∫ ex f ′ (x)dx
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Transposing ∫ ex f(x)dx + ∫ ex f ′ (x)dx =ex f ( x )
(or) ∫ ex { f(x) +f ′ (x) } dx = ex f ( x )
b log x
∫
191. dx Let log x = t x = a, t = log a
x
a
1 dt
= x = b, t = log b
x dx
∫
loga
t.dt. ⇒ ⎢
⎡ t2
⎥
⎤logb
logb ⎢⎣2 ⎥⎦
loga
[ ]
1 ( ) ( ) 1 ⎡ ⎛b⎞⎤
⇒ logb 2 − loga 2 ⇒ ⎢log(ab).log⎜ ⎟⎥
2 2 ⎣ ⎝a ⎠⎦
1 ⎡ ⎛b⎞⎤
Ans. (a) ⎢log(ab).log⎜ ⎟⎥
2 ⎣ ⎝a ⎠⎦
192. ∫ [ f(x)+f(−x) ][ g(x)−g(−x) ] dx
⇒ ∫ 0. [ g(x)−g(−x) ] dx ⇒ 0
Ans. (a) 0
b dx
∫
193. Let a + b – x = t x = a, t = b
(a+b−x)2/3
a
dt
– 1= x = b, t = a
dx
a b
⇒ – ∫ t −2/3 dt ⇒ ∫ t −2/3 dt
b a
[ ] [ ]
⇒ 3t1/3 b ⇒ 3b1/3 −a1/3
a
[ ]
Ans. (a) 3.b1/3 −a1/3
2 x
∫
194. I = dx ... (i)
x + 2−x
0
2 2−x
I = ∫ ... (ii) [f(x)=f(a−x)]
2−x + x
0
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ANSWERS
2
2I = ∫ dx =[x]2 = 2
0
0
1
∴ I = ×2=1
2
Ans. (a) 1
1 ⎛1 ⎞ 1 ⎛1−x⎞
195. I = ∫ log ⎜ −1⎟dx ⇒ ∫ log ⎜ ⎟dx ... (i)
⎝x ⎠ ⎝ x ⎠
0 0
1 ⎛1−1+x⎞
I = ∫ log ⎜ ⎟dx [f(x)=f(a−x)]
⎝ 1−x ⎠
0
1 x 1 ⎛1−x⎞
I = ∫ log ⇒ −∫ log ⎜ ⎟ dx ... (ii)
1−x ⎝ x ⎠
0 0
2I = 0 ∴ I = 0
Ans. (c) 0
196. No. of ways in which 7 dept distributed among 3 minister
( )
= 7 ×4 ×1)+(7 ×4 x ×3!
C3 C3 C3 C2
= (120 + 330) × 6 = 1980
Ans. (d) None of these
197. No. of selections of letters
(i) 2 like and 1 different = 3 ×2 =3×2 = 6
C C
1 1
(ii) 3 different = 5 = 10
C
3
∴ Total no. of ways of selection letters = 16
∴ Total words = 16×3! = 96 words.
Ans. (b) 96
198. No. of ways to form three digit nos. by using (1, 2, 3, 4, 3, 2) are – 42.
Ans. (b) 42
n n−2
199. S +S −2.S = [2a+(n−1)d] + [2a+(n – 2 – 1)d]
n n−2 n−1 2 2
n−1
…2. [2a + (n – 1 – 1)d]
2
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n ⎛n−2⎞
= [2a+(n – 1)d] + ⎜ ⎟[2a+(n−3)d] – (n−1)[2a+(n−2)d]
2 ⎝ 2 ⎠
(n−1)n (n−3)(n−2)
= an + d+a(n−2)+ d−2a(n−1)–(n−1)(n−2)d
2 2
⎡(n−1) (n−3)(n−2) ⎤
= d⎢ n+ −(n−1)(n−2)⎥
⎣ 2 2 ⎦
⎜
⎛ n2 −n+n2 −5n2 +6−2n2 −4+6n⎟ ⎞
d
⎜ ⎟
⎝ 2 ⎠
2d
= =d
2
Ans. (a) d
A3 1
200. =
A(n−1) 3
a+3d 1
= ⇒ (3+3 d) 3 = 3 + (n – 1)d
a+(n−1)d 3
6
∴ d =
n−10
6
∴ 7n = a + (N…1) d ⇒ 31 = 3+ (n + 2 – 1)
n−10
6
28 = (n+1) ∴ n = 13
n−10
Ans. (c) 13
Model Test Paper – BOS/CPT – 8
151. The first no. divide by 8 between 100 and 200 is 104
The last no. divide by 8 between 100 and 200 is 200
∴ The total number divide by 104 and 200 is 13.
All number divide by 8 also divide by 2 is 13
Ans. (b)
152. Sum of 1st n odd number
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ANSWERS
S = 1+3+5+ ... + (2n – 1)
n[ ]
Since S = 2a+(n−1)d
2
S = n/2 [2.1 + (n – 1)2]
= n (1 +n –1)
= n(n)
S = n2
Ans. (a)
153. Let the number be x.
According to the given condition of the problem is
36x = x + 1050
36x = 1050
1050
x =
35
x = 30
Ans. (b)
⎧n(n+1)⎫2
154. The formula is 13 +23 +33 +........ +n3 =⎨ ⎬
⎩ 2 ⎭
⎧12(12+1)⎫2
∴ 13 +23 +33 +........ +123 =⎨ ⎬
⎩ 2 ⎭
[ ]
= 6(13) 2
[ ]
= 78 2=6084
Ans. (c)
155. Let S = 1+9+24+46+75+........+t
n
shifting 1 places to the right in the RHS
S = 1+9+24+46+75+...........t +t
n-1 n
subtracting term by term,
0=1+8+15+22+29+............+(t -t ) - t
n n-1 n
Transposing
t = 1+8+15+22+29+ ............ to nth term
n
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n
= {2.1+(n – 1)7}
2
7 5
= n2 − n
2 2
Now Sn = ∑ tn= 7∑ n2 − 5∑ n
2 2
7 n(n+1)(2n+1) 5 n(n+1)
= −
2 6 2 2
n(n+1)⎧7 5⎫
= ⎨ (2n+1)− ⎬
2 ⎩6 2⎭
n(n+1)⎛7 4⎞
= ⎜ n− ⎟
2 ⎝3 3⎠
n(n+1)(7n−4)
Sn =
6
156. Let the number to be added be x then according to the problem.
83+x 1
= '
263+x 3
3(83+x) = 263 + x
249 + 3x = 263 + x
3x – x = 263 – 249
2x = 14
14
x = = 7
2
Hence the required number is 7
Ans. (c)
157. Initially, let the number of employees be 9 and wages per head be Rs. 14. Then, total
wages bill = Rs. (9×14) = Rs. 126
Further, the number of employees becomes 8 and the wages per head becomes Rs. 15.
∴ Now total wages bill = Rs. (8×15) = Rs. 120
∴ Ratio of the wages bill = 126:120
= 21:20
Thus the wages bill is decreased in the ratio 21:20
Ans. (c)
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ANSWERS
158. Let C gets Rs. = x
Given B = ¼ of C = ¼ (x)
2 1
and A = of B = 2/3 (¼x) = x
3 6
1 1
Also, given x+ x+x = 680
6 4
2x+3x+12x
= 680
12
17x
=680
12
680×12
x = = 480
17
Ans. (c)
159. Let us assume that when x is added to each of the four given numbers, they become in
proportion.
⇒ 10 + x : 18 + x = 22 + x : 38 + x
∴ Product of the means = Product of the extremes.
∴ (10+x) (38+x) = (18+x) (22+x)
⇒ 380+48x + x 2 =396 + 40x + x2
8x = 16
x = 2
Required number = 2
Ans. (a)
160. Let the required numbers be x and y. Since the mean proportional between a and c is
given by the relation b= ac
∴ Mean proportional = xy
According to the question,
xy = 24
xy = 576 → (1)
Again suppose that the third proportional to x and y is z. Then
x : y = y : z
398 Common Proficiency Test (CPT) Volume - II
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⇒ x : z = y : y
xz = y2
y2
⇒ z =
x
According to the question,
y2
= 192
x
⇒ y2 =192 x → (2)
y2
From equation (2), x=
192
Putting this value of x in equation (1)
y2
.y = 576
192
y3 = 576×192
= 24 × 24 × 24 × 8
= 24 × 24 × 24 × 2 × 2 × 2
y = (24 × 24 × 24 × 2 × 2 × 2 × 2)1/3
= 24 × 2
y = 48
(1) ⇒ xy = 576
x (48) = 576
576
x = = 8
48
Hence the required number are 12 and 48.
ex +e −x −2 ⎡ ex−1 (e −x −1 ⎤
161. Lt ⇒ ⎢ + ⎥
x→0 x ⎢⎣ x x ⎥⎦
⎡⎛ ex −1 ⎞ ⎛ e −x −1 ⎞⎤ ⎪⎧ ex −1 ⎪⎫
⇒ Lt ⎢ ⎜ ⎜ ⎟ ⎟+ ⎜ ⎜ ⎟ ⎟⎥ ⎨ Lt =1⎬
x→0 ⎢⎣⎝ x ⎠ ⎝ x ⎠⎥⎦ ⎪ ⎩x→0 x ⎪ ⎭
1 – 1 = 0
Ans. (b) 0
Common Proficiency Test (CPT) Volume - II 399
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ANSWERS
162. Lt
ex −1 −1
⇒ Lt ⎜
⎛ e1/x −1⎟ ⎞
x→0 ex −1 +1 x→0 ⎜ ⎝e1/x +1 ⎟ ⎠
⎜
⎛ e1/x −1⎟ ⎞ ∞
RHL Lt App Lt =
⎜ ⎟
x→0 + ⎝e1/x +1⎠ ∞
→ RHL does not exist
⎜
⎛ e1/x −1⎟ ⎞ ∞
LHL Lt App Lt
x→0 − ⎜ ⎝e1/x +1 ⎟ ⎠ ∞
LHL does not exist.
→ Lt does not exist in f(x)
Ans. (c) does not exist
3x−|x|
163. Lt
x→0 7x−5|x|
3x−x 2x
RHL Lt = = 1
x→0 + 7x−5x 2x
3x−(−x) 4x 1
LHL Lt = =
x→0 − 7x−5(−x) 12x 3
LHL ≠ RHS
∴ f(x) does not exist at x = 0
Ans. (c) does not exist
( ) ( )
eax −ebx ⎡ eax −1 ebx −1 ⎤
164. Lt ⇒ Lt ⎢a. −b ⎥
x→0 x x→0 ⎢⎣ ax bx ⎥⎦
⎪⎧ ex −1 ⎪⎫
⇒ a.1 – b.1 ⇒ a – b ⎨ Lt =1⎬
⎪ ⎩x→0 x ⎪ ⎭
Ans. (a) a … b
165. Lt ex −1 ⇒ Lt ⎢ ⎡ ⎜ ⎛ ex −1⎟ ⎞ . x ⎥ ⎤
⎜ ⎟
x→0 log(1+x) x→0 ⎢⎣⎝ x ⎠ log(1+x)⎥⎦
⇒ Lt ⎜ ⎛ ex −1⎟ ⎞ . Lt 1
x→0
⎜
⎝ x
⎟
⎠ x→0
log(1+x)
x
⇒ 1.1 = 1
Ans. (b) 1
400 Common Proficiency Test (CPT) Volume - II
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1−x
166. Given y =
1+x
( ) ( ) ( ) ( )
d d
1+x 1−x − 1−x 1+x
dy = dx ( ) dx
dx 1+x 2
( ) ( )
1+x
⎢
⎡1(
1−x
)
2
1−1(−1)⎥ ⎤
− 1−x
⎢
⎡1(
1+x
)
2
1−1
⎥
⎤
⎣2 ⎦ ⎣2 ⎦
=
(1+x)
1 1+x 1 1−x
− −
2 1−x 2 1+x
=
(1+x)
1 ⎡ (1+x)+(1−x) (1+x) ⎤
= − ⎢ . ⎥
2⎣ 1−x 1+x 1 ⎦
dy 1⎡ 2 1+x⎤ −1
= − ⎢ . ⎥ = ( ) ( )
dx 2⎣ 1−x 1+x 1 ⎦ 1+x 3/2 1−x
Ans. (b)
x
167. Given y =
1+x2
( ) 1 ( )
1+x2 1 −x. 2x
dy = 2 1+x2
dx ⎛ ⎞2
⎜ 1+x2 ⎟
⎝ ⎠
x2
1+x2 −
1+x2
= ( )
1+x2
( )
1+x2 −x2
= ( )
⎛ ⎞
1+x2 ⎜ 1+x2 ⎟
⎝ ⎠
dy 1
=
dx ( )3
1+x2 2
Common Proficiency Test (CPT) Volume - II 401
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ANSWERS
dy x3 ⎡ x ⎤3
∴ x3 = =⎢ ⎥
dx (
1+x2
)
2
3 ⎢⎣ 1+x2 ⎥⎦
dy [ ]
x3 = y 3
dx
Ans. (c)
168. Given xy =ex−4
Taking log on both sides
( ) ( )
log xy = log ex−y
y log x = (x – y) log e
y log x = x – y
y (1 + log x) = x
x
y =
1+logx
Differentiate on both sides
⎛1⎞
(1+logx).1− x⎜ ⎟
dy ⎝x⎠
=
( )
dx 1+log x 2
1+log−1 logx
= =
( ) ( )
1+logx 2 1+logx 2
Ans. (a)
( )
169. Given y3x5 = x+y 8 → (1)
Differentiate on both sides.
( )
dy ( ) ⎡ dy⎤
y3 5x4 +x5.3y2 =8 x+y 7 ⎢1+
⎥
dx ⎣ dx⎦
dy ( ) ( ) dy
5y3x4 +3x5y2 =8 x+y 7 +8 x+y 7
dx dx
[ ]
dy ( ) ( )
3x5y2 − 8 x+y 7 = 8 x+y 7 − 5y3x4
dx
402 Common Proficiency Test (CPT) Volume - II
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( )
( )
x+y 8
dy = 8 ( x+y ) 7 −5y3x4 = 8 x+y 7 −5 x Using equation (1)
( )
dx 3x5y2−8 x+y 7 3( x+y ) 8 −8 ( x+y ) 7
y
( ) ⎡ 5( )⎤
x+y 7 ⎢8− x+y
⎥
⎣ x ⎦
=
( )
⎛3(
)
⎞
x+y 7
⎜
⎜ x+y −8⎟ ⎟
⎝y ⎠
[ ( )] [ ]
y 8x−5 x+y y 8x−5x−5y
= [ ( ) ] = [ ]
x3 x+y −8y x3x+3y−8y
[ ]
y 3x−5y y
= [ ] =
x3x−5y x
Ans. (a)
170. Given y =
xxx.....∞
i.e. y=xy
Taking log on both sides
( )
log y = log xy
log y = y log x
Differentiate on both sides
1 dy 1 dy
=y. +log x.
y dx x dx
dy ⎡1 ⎤ y
⎢ −log x⎥ =
dx ⎣y ⎦ x
dy ⎡1−y logx⎤ y
⎢ ⎥
=
dx ⎣ y ⎦ x
dy y2
=
( )
dx x1−y log x
dy y2
∴ x. =
dx 1−y log x
Ans. (b)
Common Proficiency Test (CPT) Volume - II 403
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ANSWERS
1
171. ∫ (ex −ex)dx = ∫1 (0)dx
−1
−1
= 0
Ans. (b) 0
e 1+logx
∫
172. dx
x
1
Let 1 + log x = t x = 0, t = 1
1 dt
= x = e, t = 2
x dx
2 ⎡ t2⎤2
= ∫+dt = ⎢ ⎥
⎣2 ⎦
1 1
1 3
= 2 – =
2 2
3
Ans. (a)
2
log3
ex
∫
173. dx If 1 + ex = t x = 0, t = 2
1+ex
0
dt
ex = x = log3, t = 4
dx
4 1
∫ dt = [logt]4 ⇒ log 4 – log 2
−t 2
2
= log 2
Ans. (b) log 2
1 x
∫
174. let (1 + x2) = t2 x = 0, t = 1
1+ 1+x2
0
x=1t= 2
dt t
2x = 2t ⇒ dx = dx
dx x
2 tdt 2⎛ 1 ⎞
∫ = ∫ ⎜1− ⎟dt
1+t ⎝ 1+t⎠
1 1
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[ ( )]
= t−log 1+t 2
1
( ) [ ( ) ]
= 2 −1 − log 1+ 2 −log1
( ) ( )
= 2 −1 −log 1+ 2 +0
( ) ( )
= 2 −1 −log 1+ 2
Ans. (d) None of these
1 dx 1⎛ 1 1 ⎞
175. ∫ ⇒ ∫⎜ − ⎟dx
(1+x)(2+x) ⎝1+x 2+x⎠
0 0
[ ( ) ( )] ⎡ 1+x⎤1
⇒ log1+x −log 2+x 1⇒ ⎢log ⎥
0 ⎣ 2+x⎦
0
2
⇒ log
3
2
Ans. (a) log
3
176. a + ar = 15 ⇒ a (1+ r) = 15
a = ar + ar 2 + ar3 – ∞
ar
a = ⇒ 1– r = r ⇒ r = 1/2
1−r
15×2
∴ a = = 10
3
a 10
∴ Sum of Series = = = 20
1−r 1−1/2
Ans. (a) 20
1 1
177. a = ⇒ =1 – x
1−x a
1 1
b = ⇒ = 1 – y
1−y b
1 1
∴ + =1−x+1−y⇒ 2 – (x+y) = 2.1 = 1
a b
Ans. (c) 1
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ANSWERS
R 3
178. 90 = 2000 × ×
100 4
∴ R = 6%
Ans. (b) 6%
6
179. 216 = 5400 × ×n
100
n = 4/6 yrs. = 8 months
Ans. (b) 8 months
R
180. I = 10000 ×2=200R
1
100
R
I = 6000 × ×3=1800R
2
100
I + I = 1900 ⇒ 200R + 180R = 1900
1 2
∴ 380R = 1900 ∴ R = 5%
Ans. (b) 5%
181.If all the observation are equal.
Then standard deviation = 0
Ans. (a) 0
182. If every item is increased by 5 then mean (x) also increased by 5, but the value of
∑
(x−x)2 remain same.
∴ Standard deviation will remain same,
Standard deviation = 10
Ans. (c) – 10
∑
d2
360
183. S.D. = = = 6
N 10
xSD
Coefficient of variation = 100
Am
100×6
= =15
40
Ans. (a) 15
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M.D
184. Coefficient of M.D. = ×100
Am
5.77×100
44 =
AM
A.M. = 13.11
Ans. 13.11
185. The S.D. of two values is equal to half their difference.
a−b
S.D. =
2
The Statement is correct
Ans. (a) True
186. Computation of Correlation Coefficient
x y xy x2 y2
50 40 2000 2500 1600
50 40 2000 2500 1600
Total 100 80 4000 5000 3200
100 80
x = =50, y = =40
2 2
4000
Cov (x, y) = – (50) (40) = 0
2
∴ r = 0
Ans. (c)
187. Given r = 2 , ∑ di2 = 55
R 3
∑
b di2
∴ r =1− ( )
R nn2 −1
( )
2 6 55
=1− ( )
3 nn2 −1
2 −330
−1= ( )
3 nn2 −1
1 330
− = − ( )
3 nn2 −1
Common Proficiency Test (CPT) Volume - II 407
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ANSWERS
( ) ( )
nn2 −1 = 990 = 10 102 −1
∴ n = 10 as n must a positive
Ans. (a)
188. Let us assume that 4x + 3y + 7 = 0 → (1)
represent the regression line of x on y and 3x + 4y + 8 = 0 → (2) represent the
regression line of y on x.
(1) 4x = –7 – 3y
7 3
x = – – y
4 4
3
∴ bxy = –
4
(2) 4y = – 8 – 3x
3
y = –2 – x
4
3
∴ byx = –
4
⎛ 3⎞ ⎛ 3⎞ 9
∴ r2 = byx. bxy = ⎜− ⎟ ⎜− ⎟ =
⎝ 4⎠ ⎝ 4⎠ 16
9 3 3
∴ r = = ± = − = −0.75
16 4 4
(We take the sign of n as negative since both the regression coefficient are negative).
Ans. (c)
189. Ans. (d) Refer Properties
190. Given byx = 1.2 → (1)
x−100
U=
2
⇒ x = 100 + 2U
⇒ x = 100 + 2 U
y−200
and v =
3
⇒ y = 200 + 3 v
⇒ y = 200 + 3 v
408 Common Proficiency Test (CPT) Volume - II
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∑
(x−x)(y−y)
byx=
E(x−x)2
[ ( ) ]
∑ ( )
2U−U.3 v−v
=
[ ]
∑ 2(U−U) 2
( )( )
∑
2×3 U−U V−V
=
∑
4 (U−U)2
= 3/2 bvu
⇒ bvu = 2/3 byx = 2/3 × 1.2 = 0.8
Ans. (b)
191. Let A, is First bag is selected
A2 is second bag is selected.
B: In a draw of 2 balls, one is red and the other is black.
The required probability
P = P(A ∩B) + P (A ∩B)
1 2
= P(A ) P (B/A ) + P (A ). P(B/A )
1 1 2 2
Since there are two bags, the selection of each being equally likely.
∴ P(A ) = P(A ) = 1/2
1 2
P(B/A ) = Probability of drawing one red and one black ball in a draw of 2 balls from the
1
1st bag
5C ×3C 15
= 1 1 =
8C 28
2
P(B/A ) = Probability of drawing one red and one black ball in a draw of 2 balls from the
2
2nd bags.
4C ×5C 5
= 1 1 =
9C 9
2
15 1 5
(1) ⇒ p = 1/2 × + ×
28 2 9
15 5 135+140 275
p. = + = =
56 18 504 504
Ans. (a)
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ANSWERS
192. Let A1 is first purse is selected.
A2 is second purse is selected.
Let B: In a draw of one coin, one coin must be silver
The required probabilities.
( ) ( )
P = P A ∩B + P A ∩B
1 2
= P(A ).P(B/A ) +P(A ).P(B/A )→ (1)
1 1 2 2
Since there are two purse, the selection of each being equally likely
1 1
∴ P(A )= ,P(A )=
1 2 2 2
P(B/A )= Probability of drawing one silver coin from the first purse.
1
3C
= 1 =3/7
7C
1
P(B/A ) = Probability of drawing one silver coin from the second purse.
2
4C 4
= 1 =
7C 7
1
⎛3⎞ 1⎛4⎞
Substituting (1) ⇒ p=1/2 ⎜ ⎟+ ⎜ ⎟
⎝7⎠ 2⎝7⎠
3 4 7 1
= + = =
14 14 14 2
Ans. (a)
193. When two tosses of unbiased dice the total sample space.
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
n(s) = 36
In the above sample space, let x be the number of sines getting from the experiment.
Let x = 0, means no sin. = number of times = 25
x = 1, means no sin. = number of times = 10
410 Common Proficiency Test (CPT) Volume - II
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x = 2, means no sin. = number of times = 01
36
∴ Expected table is:
x 0 1 2
p(x) 25 10 1
∴ The required probability = Mean = Expected Value
∑x p(x)
=
n(s)
0×25+1×10+2×1
=
36
12 1
= =
36 3
Ans. (a)
194. The experiment of throwing three dice is theoretically same as that of throwing a die
thrice.
Let E be the event of throwing six in a throw of die.
∴ P(E) = 1/6 and P ( E) = 1 – P(E)
= 1–1/6 = 5/6
Let x denotes the random variable "number of Sixes".
∴ The possible values of x are 0, 1, 2, 3
∴ P(x=2) = P(E E E or E E E or E E E )
1 2 3 1 2 3 1 2 3
= P(E ) . P(E ) P( E ) + P(E ) P( E ) P(E )+ P( E )+ P(E ) P(E )
1 2 3 1 2 3 1 2 3
1 1 5 1 5 1 5 1 1
= . . + . . + . .
6 6 6 6 6 6 6 6 6
15
P(x=2) =
216
Ans. (c)
195. let A and B denote the events that the Chartered Accountant is selected in firms X and Y
respectively. Then in the usual notations, we are given.
P(A) = 0.7
P( A) = 1– P(A) = 1=0.7=0.3
P( B) = 0.5
Common Proficiency Test (CPT) Volume - II 411
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ANSWERS
∴ P(B) = 1 – P(B) = 1 – 0.5 = 0.5
( )
and P A∪B = 0.6
By De – Morgan's law
( )
A∩B = A∪B
( ) ( )
∴ P A∪B = 1– P A∩B
( )
= 1–P A∪B
= 1 – 0.6
= 0.4
The probability that the Chartered Accountant will be selected in one of the two firms X or
Y is given by:
( )
P (AUB) = P(A) + P(B) – P A∩B
= 0.7 + 0.5 – 0.4
= 0.8
Ans. (a)
⎡ ⎤
196. ∫ ⎢log(logx)+ 1 ⎥dx
⎢⎣ (logx)2⎥⎦
I = log (log x)
∫ dx−∫
⎢
⎡ d [
log(logx)
]
⎥
⎤
.xdx+
∫ 1
= dx
⎣dx ⎦ (logx)2
∫ 1 ∫ 1
= log (log x).x – .dx + dx
(logx) (logx)2
⎡ ⎛ ⎞⎤
x. log (log x) – ⎢ ⎢ ⎣(log 1 x) ∫ dx+ ⎜ ⎜ ⎝(log 1 x)2 . x x dx⎟ ⎟ ⎠⎥ ⎥ ⎦ + ∫ ( log 1 x ) 2 =dx
⇒ x log (log x) – x – ∫ 1 + ∫ 1 dx
log x (log x)2 (log x)2
x
⇒ x. log (log x) − +c
log x
x
Ans. (a) x. log (log x) − +c
log x
197. 'is equal to' Satisfies Reflexive, Symnetric and transitive Relation
∴ This is Equivalence Relation.
Ans. (d) Equivalence Relation.
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198. f(x) = x2+2
∴ f(−x)=(−x)2 +2
= x2 +2
f(– x) = f(x)
∴ f(x) is even function
Ans. (b) even function.
199. f(x) =
121+x
= 12.12 x 0 < x < 9
Range = 12×12°, 12×121 .....12×12 9
∴ Range = 12 < f(x) < 1210
Ans. (a) 12 ≤f(x)≤1210
200. 'Is greater than' over the set of real number is not satisfied Reflexive and Symmetric
relation it only satisfied transitive Relat ion
Ans. (a) Transitive relation.
Model Test Paper – BOS/CPT – 9
x 17
151. Given =
x+y 23
⇒ 23x = 17x + 17y
23x – 17x = 17y
6x = 17y
17
x= y
6
17
y+y
Now,
x+y
= 6
x−y 17
y−y
6
17y+6y 6
= ×
6 17y−6y
Common Proficiency Test (CPT) Volume - II 413
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ANSWERS
23y 23
= =
11y 11
x+y 23
∴ =
x−y 11
25 x
152. Given 1+ =1+
144 12
Squaring on both sides:
25 ⎛ x ⎞2
1+ =⎜1+ ⎟
144 ⎝ 12⎠
144 +25 x2 2x
=1+ +
144 144 12
169 144 +x2 +24x
=
144 144
∴x2 +24x−25=0
( )( )
x+25 x−1 = 0
x=−25, x=1
x = 1 (∴ negative neglected)
Ans. (a)
( )
153. Given ( 4 ) 3× 2 8 = 2n
( ) ⎛ 1 ⎞8
i.e. (2)2 3 × ⎜ (2)2 ⎟ = 2n
⎜ ⎟
⎝ ⎠
26 ×24 = 2n
210 = 2n
∴ n = 10
Ans. (a)
154. Let total number of men went to a hotel = x
Given, A man Spent Rupees = Total number of men
= x
∴ Given Data = x + x = 15625
414 Common Proficiency Test (CPT) Volume - II
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x2= 15625
x= 15625 =125
Ans. (b)
155. Given A + B + C = 1000 → (1)
A + C = 400 → (2)
B + C = 700 → (3)
(2) ⇒ A = 400 – C
(3) ⇒ B = 700 – C
(1) ⇒ 400 – C + 700 – C + C = 1000
C = 100
Ans. (a)
⎛a−b⎞
156. Given log ⎜ ⎟ = 1/2 (log a + log b)
⎝ 2 ⎠
a−b
∴ 2 log = log a + log b
2
⎛a−b⎞2
log ⎜ ⎟ = log ab
⎝ 2 ⎠
⎛a−b⎞2
⇒ ⎜ ⎟ = ab
⎝ 2 ⎠
⎛a−b⎞2
⎜ ⎟ = ab
⎝ 4 ⎠
(a – b) 2= 4ab
a2 +b2 −2ab =4ab
a2 +b2 =6ab
Ans. (a)
157. Given log x = 4
10
∴ x = 104
Ans. (c) x = 10000
158. log 225 = log (9×25)
= log 9 + log 25
Common Proficiency Test (CPT) Volume - II 415
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ANSWERS
= log 32 + log 52
= 2log 3 + 2 log 5
= 2log 3 + 2 log 10/2
= 2log 3 + 2 log 10 – 2 log 2
= 2 × 0.477 + 2 – 2 (0.301)
log 225 = 2.352
Ans. (a)
159. Let 2100 = x
Taking log on both sides.
log 2100 = log x
100 log 2 = log x
log x = 100 × 0.3010
log x = 30.1000
∴ the no. of digits in 2100 is 31
Ans. (b)
n!
160. Given nP3 =
=60
(n−3)!
∴ n(n – 1) (n – 2) = 60 = 5 × 4 × 3
∴ n = 5
Ans. (c)
ax +bx −2 (ax −1)+(bx −1)
161. Lt ⇒ Lt
x→0 x x→0 x
ax −1 bx −1 ⎡ ax −1 ⎤
⇒ Lt + Lt ⎢ Lt =loga⎥
e
x→0 x x→0 x ⎢⎣x→0 x ⎥⎦
App Lt
= log a + log b = log (ab)
Ans. (a) log (ab)
10x −5x −2x +1
162. Lt
x→0 x
⎡ (10x −1) (5x −1) (2x −1) ⎤ ⎧ ax −1 ⎫
⇒ Lt ⎢ − − ⎥ ⎨ Lt =loga⎬
x→0 ⎢⎣ x x x ⎥⎦ ⎩x→0 x e ⎭
416 Common Proficiency Test (CPT) Volume - II
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App Lt
= log 10 – log 5 – log 2
⎛ 10 ⎞
= log ⎜ ⎟ = log 1 ⇒ 0
⎝5×2⎠
Ans. (b) 0
10x −5x −2x −1 [5x×2x −5x −2x −1] 5x(2x −1)−1(2x −1)
163. Lt ⇒ Lt =
x→0 x2 x→0 x2 x2
(2x −1)(5x −1) (2x −1) (5x −1)
Lt ⇒ Lt × Lt
x→0 x2 x→0 x x→0 x
⇒ log 2 × log 5
Ans. (a) log 5 × log 2
e5x −e3x −e2x +1
164. Lt
x→0 x
⎡ (e5x −1) (e3x −1) (e2x −1) ⎤
⇒ Lt ⎢5 −3. −2 ⎥
x→0 ⎢⎣ 5x 3x 2x ⎥⎦
App Lt
5.1 – 3.1 – 2.1
⎪⎧ ex −1 ⎪⎫
= 5 – 3 – 2 = 0 ⎨ Lt =1⎬
⎪ ⎩x→0 x ⎪ ⎭
Ans. (b) 0
e5x −e3x −e2x −1 (e3x.e2x −e3x −e2x −1)
165. Lt = Lt
x→0 x2 x→0 x2
[ ] [ ]
e3x(e2x −1)−1(e2x −1) (e3x −1)(e2x −1)
Lt ⇒ Lt
x→0 x2 x→0 x2
⎜
⎛ e3x −1
⎟
⎞
⎜
⎛ e2x −1⎟ ⎞
Lt ⎜ ⎟ × Lt ⎜ ⎟
x→0 ⎝ x ⎠ x→0 ⎝ x ⎠
App Lt
3 × 2 = 6
Ans. (a) 6
Common Proficiency Test (CPT) Volume - II 417
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ANSWERS
∫ 1
166. Given dx
x2 +a2
Let x2 +a2 = z−x
∴ z = x + x2 +a2
dz 1 ( )
=1+ 2x dx
dx 2 x2 +a2
x2 +a2 + x z
= =
x2 +a2 x2 + a2
dz dx
∴ =
z x2 +a2
∴ ∫ dx = ∫dz =log z +c
x2 +a2 z
⎛ ⎞
= log ⎜x+ x2 +a2 ⎟ + c
⎝ ⎠
∫ 1
167. dx
x2 −a2
Let x2 −a2 =z−x
∴ z = x + x2 −a2
dz 1 ( ) x
= 1 + 2x = 1 +
dx 2 x2 −a2 x2 −a2
x2 −a2 +x z
= =
x2 −a2 x2 −a2
dz dx
=
z x2 −a2
∴ ∫ 1 dx = ∫1 dz + c
x2 −a2 2
⎛ ⎞
= log (z) = log ⎜x+ x2 −a2 ⎟ + c
⎝ ⎠
Ans. (c)
418 Common Proficiency Test (CPT) Volume - II
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168. Let t = 3x
∴ dt = 3 dx
∴ I = ∫ 1 dt = 1 ∫ 1 dt
t2 −1 3 3 t2 −12
1 1 ⎛t−1⎞
= log ⎜ ⎟ + c
3 2+1 ⎝t+1⎠
1 ⎛3x−1⎞
= log ⎜ ⎟ +c
6 ⎝3x+1⎠
Ans. (b)
∫
x−1
169. Let I = dx
x2 +1
⎡ ⎤
∴ I = ∫ ⎢ x − 1 ⎥ dx
⎢⎣ x2 +1 x2 +1 ⎥⎦
x 1
=
∫ dx−∫
dx
x2 +1 x2 +1
I = I – I (say)
1 2
∫ x
I = dx
1
x2 +1
Let t = x2 + 1
∴ dt = 2x dx
∴ I = ∫ 1 dt
1
t 2
=
1∫
t
−1/2
dt
2
1 t1/2
= = t = x2 +1
2 1/2
∫ 1
I = dx
2
x2 +1
⎛ ⎞
= log ⎜x+ x2 +1⎟
⎝ ⎠
Common Proficiency Test (CPT) Volume - II 419
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ANSWERS
∴ I = I – I
1 2
⎛ ⎞
I = x2 +1 −log⎜x+ x2 +1⎟ +c
⎝ ⎠
Ans. (a)
( )
170. Let I = ∫ 1−x2 log x dx
∴ I = ∫ log x (1−x2)dx
[Here (log x) is to be take n as first function and (1 – x2) as second function]
Integrating by parts:
⎛ x3 ⎞ 1 ⎛ x3 ⎞
I = log x ⎜ ⎜ x− ⎟ ⎟ − ∫ ⎜ ⎜ x− ⎟ ⎟ dx
⎝ 3 ⎠ x⎝ 3 ⎠
⎛ x2 ⎞ ⎛ x2 ⎞
= ⎜ ⎜ x− ⎟ ⎟ xlog x− ∫ ⎜ ⎜ x− ⎟ ⎟ dx
⎝ 3 ⎠ ⎝ 3 ⎠
⎛ x2 ⎞ ⎛ x3 ⎞
= ⎜ ⎜ 1− ⎟ ⎟ xlog x− ∫ ⎜ ⎜ x− ⎟ ⎟ +c
⎝ 3 ⎠ ⎝ 3*3⎠
I = ⎜
⎛
1−
x2
⎟
⎞
xlog x−⎜
⎛
x−
x3
⎟
⎞
+ c
⎜ ⎟ ⎜ ⎟
⎝ 3 ⎠ ⎝ 9 ⎠
Ans. (c)
171. Among 4 doctors, 4 officers and 1 doctor who is also an officer committee of 3 can be
form in such manner.
(i) 1 doctor, 1 officer, 1 doctor who is also officer = 4 ×4 ×1 = 16
C c
1 1
(ii) 2 doctor and doctor – officer = 4 ×1=6
C2
(iii) 2 officer and doctor – officer = 4C ×1= 6
2
(iv) 2 doctor and 1 officer = 4 ×4 =24
C C
2 1
(v) 1 doctor and 2 officer = 4 ×4 =24
C C
1 2
Total no. of ways = 16 + 6 + 6 + 24 + 24 = 76
Ans. (a) 76
172. Elector can vote for one or more vacancies in such manner …
(i) For 3 vacancies – 5 =10
C
3
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(ii) For 2 vacancies – 5 =10
C
2
(iii) For 1 vacancy – 5 =5
C
1
∴ Total ways = 10 + 10 + 5 = 25
Ans. (c) 25
173. No. of ways in which 12 different thing distributed in 4 groups.
12!
= = 15400
(3!)4
Ans. (a) 15,400
174. Factor of 420 is = {2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 30, 28, 35, 42, 60,
84, 105, 210, 140, 420}
No. of factor of 420 = 22
Ans. (b) 22.
175. Five balls are kept in 3 boxes as no box will empty
( )
= 5 ×4 ×3 ×3!
C c c
1 1 3
= (5 × 4 × 1) × 6 = 120 ways.
Ans. (b) 120 ways.
176. 243 + 324 + 432 + – n terms
35.1 + 34.4 + 33.42 + – n terms
∴ a = 35 r =4/3
⎡ ⎛4⎞n ⎤
35.⎢⎜ ⎟ −1⎥
⎢⎣ ⎝3⎠ ⎥⎦ ⎡ ⎛4⎞n ⎤
Sn = = 35.3 ⎢⎜ ⎟ −1⎥
⎜
⎛4
−1⎟
⎞ ⎢⎣ ⎝3⎠ ⎥⎦
⎝3 ⎠
⎡ 4n ⎤
= 36⎢ −1⎥
⎢⎣3n ⎥⎦
⎡ 4n ⎤
Ans. (a) 36⎢ −1⎥
⎢⎣3n ⎥⎦
[ ] [ ]
a r8 −1 5.a r4 −1
177. S =5.S ⇒ =
8 4 r−1 r−1
( )
⇒ r4 +1 = 5
Common Proficiency Test (CPT) Volume - II 421
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ANSWERS
( )
4
r4 = 4 = 2
∴ r = ± 2
Ans. (c) ± 2
178. 4+44+444 ... n terms
4
= [9+99+999+.....n terms]
9
4
= [(10 – 1)+(100 – 1) + (1000 – 1) + ... n terms)
9
4 ⎡ 10(10n −1) ⎤ 4 ⎡ 10(10n −1) ⎤
= ⎢ −n⎥ ⇒ ⎢ −n⎥
9 ⎢⎣ 10−1 ⎥⎦ 9 ⎢⎣ 9 ⎥⎦
4 ⎡ 10(10n −1) ⎤
Ans. (a) ⎢ −n⎥
9 ⎢⎣ 9 ⎥⎦
a+b
179. =15 and ab = 9 ∴ ab = 81
2
a = (30 – b) & (30 – b)b = 01
∴ b2 – 30b + 81 = – 0 ∴ b = 27, 3 and a = 3, 27
∴ Nos are 27, 3
Ans. (a) 27, 3
180. Product of n Gm between two No. is equal to n th Power of single Gm between two nos.
This statement is correct.
Ans. (a) True
181. The weighted arithmetic mean of first a natural numbers whose weights are equal to the
corresponding number is equal to
2n+1
3
2n+1
Ans. (a)
3
w x +w x +w x
182. x = 1 1 2 2 3 3
(x +x +x )
1 2 3
100×5 =125×5 +w ×5
110 = 3
15
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1650 = 500 + 625 + 5.w
3
525
w = = 105 kg.
3
5
Ans. (b) 105 Kgs.
∑
183. x−n×2.5=50
∑
x−2.5n=50 → (i)
∑
x−3.5n=−50 → (ii)
[Eg. (i) – Eg. (ii)]
1.0n = 100
∴ n = 100 ∴ Σx = 300
Σx 300
∴ mean = = = 3
n 100
Ans. (a) 100, 3
184. The most reliable value is mean.
Ans. (a) Mean
185. In which Central Value arranging is required – Median
Ans. (c) Median
186. There are 365 days in a normal year (without leap year)
No. 365 = 7 × 52 + 1
∴ In a year will contain at least 52 Tuesday
The possible remaining one Tuesday
Let A be the event of getting 53 Tuesday in the year.
∴ P (A) = 1/7
Ans. (b)
187. Given two unbiased dice are thrown, then the simple space are:
S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Common Proficiency Test (CPT) Volume - II 423
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ANSWERS
∴ n(S) = 36
Sample space of sum of the faces is not less than 10
A = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
n(A) = 6
n(A) 6 1
∴ Required probability = = =
n(S) 36 6
Ans. (a)
188. Let A be the person travels by a plane
1
∴ P(A) =
5
Let B be the person travels by a train
2
∴ P (B) =
3
∴ Probability of his travelling neither by plane nor by train.
P(AB) = P(A). P(B) (Since A and B are mutually exclusive conditional probability)
⎛1⎞⎛2⎞ 2
= ⎜ ⎟⎜ ⎟ =
⎝5⎠⎝3⎠ 15
Ans. (b)
189. Let A denote the event of drawing a diamond and B denote the event of drawing a King
13 1
from a pack of Cards. Then we have P(A) = =
52 4
4 1
and P(B) = =
52 13
P(AUB) = P(A) + P(B) – P(A ∩B)
1 1
= + −P(A∩B) → (1)
4 13
There is only one case favourable to the event A∩B vize, king of diamond.
1
Hence, P (A∩B) =
52
1 1 1 13+4−1 16 4
∴ (1) ⇒ P(AUB) = + − = = =
4 13 52 52 52 13
Ans. (c)
424 Common Proficiency Test (CPT) Volume - II
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190. Let us define the events:
E : A solves the problem
1
E : B solves the problem
2
then we are given
6 6 2
P(E ) = = =
1 6+9 15 5
10 5
and P (E ) = =
2 10+12 11
Assuming that A and B try to solve the problem independently. E and E are
1 2
independent.
2 5 2
∴ P (E ∩E ) = P(E ).P(E ) = × =
1 2 1 2
5 11 11
The problem will be solved if at least one of the students A and B solves the problem.
Hence, the probability of the problem being solved is given by
P(E ∪E )=P(E )+P(E )−P(E ∩E )
1 2 1 2 1 2
2 5 2
= + −
5 11 11
22+25−10 37
= = =0.673
55 55
Ans. (a)
192. Given Mean of Binomial distribution = μ = np = 3 → (1)
and Variance of Binomial distribution = σ2 = npq =.2 → (2)
(2) npq 2
⇒ =
(1) np 3
∴q = 2/3
∴p + q = 1
2 1
p = 1– q = 1− =
3 3
(1) ⇒ n(1/3) = 3
= n = 9
∴ p=1/3, q = 2/3, n=9
By the Binomial distribution p(x) = nC px qn – x . The probability that the variate takes
x
values less than or equal to 2
Common Proficiency Test (CPT) Volume - II 425
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ANSWERS
i.e. p (x < 2) = p(x = 0) + P (x = 1) + P (x = 2).
= 9C (1/3)0 (2/3)9 – 0
0
+ 9C (1/3)1 (2/3)9 – 1
1
+ 9C (1/3)2 (2/3)9 – 2
2
= 9C (1) (2/3)9
0
+ 9C (1/3)1 (2/3)8 + 9C (1/3)2 (2/3)7
1 2
= (2/3)9 + 3 (2/3)8 + 4 (2/3)7
P(x < 2) = 0.3767
Ans. (a)
193. Exhaustive cases: 2 digits can be selected out of 9 digits 1 through 9 in 9C ways.
2
9×8
∴ Exhaustive number of cases = 9C = =36
2 1×2
Favoarable number of cases. Among the digits 1 through 9.
Even digits are: 2, 4, 6 and 8 i.e. 4 in all
Odd digits are 1, 3, 5, 7 and 9 i.e. 5 in all.
The sum of the two digits drawn will be even if
(i) Either both the selected digits are even (or)
(ii) both the selected digits are odd.
Two even digits can be selected out of the 4 even digits in 4C ways and two odd digits
2
can be selected out of the 5 odd digits in 5C ways.
2
Hence, the favourable number of cases that the sum of the two selected digits in even.
= 4C +5C
2 2
4×3 5×4
= +
1×2 1×2
= 6+10 = 16
∴ P (sum of the two selected digits is even)
Number of favourable cases
=
Exhaustive number of cases
16 4
= =
36 9
5C 10 5
and P(Both selected digits are odd) = 2 = =
9C 36 18
2
Ans. (e)
426 Common Proficiency Test (CPT) Volume - II
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194. Let S be the sample space of the experiment
∴ S = {(1,1), (1,2), ... (6,5), (6,6)}
Let A = event of getting sum 6
and B = event of getting 4 at least once.
∴ A = {(1,5), (2,4), (3,3), (4,2), (5,1)}
and B = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}
11
∴ P(A) = 5/36 and P(B) =
36
Also, AB = {(4,2), (2,4)}
2
∴ P(AB) =
36
∴ The required probability
= Probability of getting 4 on at least one die given that sum is 6
P(BA) P(AB)
= P(B|A) = =
P(A) P(A)
2/36 2
= =
5/36 5
Ans. (b)
σ
196. Standard Error of mean =
n
12.6 12.6
= =
36 6
Standard Error of mean = 2.1
Ans. (a) 2.1
197. Standard Error of Mean without replacement
σ N−n
=
n
N−1
12.6 101−36
= = 2.1× 0.65
36
101−1
SE = 2.1 × 0.806 = 1.69
Ans. (b) 1.69
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ANSWERS
5 1
198. P = =
25 5
1 4
Q = 1− =
5 5
n = 5
PQ
S.E. of proportion of defectives =
n
1 4 1
= × × = 0.032
5 5 5
SE = 0.1088
Ans. (b) 0.1088
199. n = 2, N = 4
Total number of possible sample of size of with replacement = 42 = 16
Ans. (a) 16
200. n = 2, N = 4
Total number of possible sample of size without replacement = N = 4 = 6
C C
n 2
Ans. (b) 6
Model Test Paper – BOS/CPT – 10
151. The value of 33 +43 +53 +.....+113
⎡11(11+1)⎤2 [ ]
= ⎢ ⎥ − 13 +23
⎣ 2 ⎦
( ) ( )
= 11×6 2 − 1+8
( ) ( )
= 66 2 − 9
= 4356 – 9 = 4347
Ans. (c)
152. Let the two numbers are x and y
Given x + y = 75 → (1)
x – y = 20 → (2)
428 Common Proficiency Test (CPT) Volume - II
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(1)+(2) ⇒ 2x = 95
95
x =
2
95
∴ (1) ⇒ = 75 –
2
55
=
2
∴ The difference of their squares = x2−y2
⎛95⎞2 ⎛55⎞2
= ⎜ ⎟ −⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
9025 3025
= −
4 4
6000
= =1500
4
Ans. (a)
153. Let the two numbers are x and y
Given x + y = 13 → (1)
and x2 +y2 = 85 → (2)
(1) ⇒ y = 13 – x
Substitute y = 13 – x in (2)
x2 +(13−x)2 = 85
x2 +169+x2 −26x−85 =0
2x2 −26x+84=0
x2 −13x+42 =0
( )( )
x−7 x−6 =0
x =7, x = 6
When x = 7 (1) ⇒ y = 13 – 7 = 6
When x = 6 (1) ⇒ y = 13 – 6 = 7
∴ The number (7, 6)
Ans. (a)
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ANSWERS
154. Let the two consecutive members are x and x – 1
Given x2 −(x−1)2 =37
x2 −(x2 +1−2x)=37
x2 −x2 −1+2x=37
2x = 38
x = 19
∴ x – 1 = 19 – 1 = 18
Ans. (a)
155. Let the number be x.
x
Given condition (1) → (1)
x+3
x+7
Given condition (2) ⇒ = 2
x+3−2
x + 7 = 2 (x+1)
= 2x + 2
x = 5
5 5
∴ (1) ⇒ =
5+3 8
1
156. Given log 3 =
x
6
1
x6 = 3
( )
6
x = 3
( ) ( ) ( )
2 2 2
= 3 3 3
=3×3×3
x = 27
Ans. (b)
157. Let y = alogax
Taking log on both sides
[ ]
log y = log alogax
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