177. Ans. (a) Refer Properties
178. Ans. (b) Refer Properties
179. Ans. (a) Refer Properties
180. Ans. (a) Refer Properties
181. Let A be the number which is multiply 3 with in 1 to 20
A = {3, 6, 9, 12, 15, 18}
6 3
Probability of A =P(A) = =
20 10
Let B be the number which is multiply 7 with in 1 to 20
B = {7, 14}
2 1
P(B) = =
20 10
∴ Probability of number which is multiple of 3 or 7
P(AUB) = P (A) + P (B) = 3/10 + 1/10 = 4/10 = 2/5
Ans. (b)
182. Let A be the Card drawn King from the pack
P(A) = 4/52
Let B be the card drawn heart from the pack
P(B) = 13/52
P(King and Heart) = P (A ∩B) = 1/5
Here, A and B are non − mutually exclusive
∴ P (King or Heart) = P(A∪B) = P(A) + P(B) − P(A ∩B)
4 13 1
= + −
52 52 52
16
=
52
4
=
13
But P (neither a king nor a heart) = 1 − P (A∪B)
13−4 9
−
= 1 4/13 = =
13 13
Ans. (b)
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ANSWERS
183. Total number of balls = 3 Red + 5 yellow + 4 green. Since 3 balls are drawn at Random,
total number of possible outcomes = 12 C
3
Probability of balls drawn contain exactly two green balls.
4C .8C
= 2 1
12C
3
(Since out of Four green balls two green exactly taken 4C and the remaining one balls
2
from total number of other two colours).
6×8 48 12
= = =
220 220 55
Ans. (a)
184. Let A = event that Husband is selected.
B = event that wife is selected
∴ P(A) = 3/5 and P(B) = 1/5
⇒ P( A) = 1 − P(A) = 1 − 3/5 = 2/5
P(B) = 1 − P(B) = 1 − 1/5 = 4/5
Now AB = The event that only Husband is selected.
AB = the event that only wife is selected.
∴ AB ∪ AB = the event that only one of them is selected.
Now AB and AB are mutually exclusive events.
( )
∴ By Addition thereon P AB∪AB = P( AB) + P( AB)
Also, the interviews of husband and wife are independent experiments.
∴ P( AB) = P(A) . P(B) = 3/5 × 4/5 = 12/25
and P(AB) = P(A) P(B)
= 2/5 * 1/5 = 2/25
∴ P (only one is selected)
( )
= P AB∪AB
= P( AB) + P( AB)
12 2 14
= + =
25 25 25
Ans. (c)
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185. Balls in first bag = 4 White + 2 Black
Balls in Second bag = 3 White + 5 Black.
The draws from bags are independent.
∴ Required probability = (w1B2 or B1W2)
= (PW ). P(B ) + P(B ) P(W )
1 2 1 2
4 5 2 3
= . + .
6 8 6 8
26
=
48
13
=
24
Ans. (b)
186. Ans. (b) … Refer Properties
187. Ans. (a) … Refer Properties
188. Ans. (c) … Refer Properties
189. Ans. (a) … Refer Properties
190. Ans. (b) … Refer Properties
192. Probability to get red ball case I
⎛5⎞ ⎛ 4 ⎞
= ⎜ ⎟×⎜ ⎟ (Red ball from 1st bag and also from 2nd)
⎝9⎠ ⎝11⎠
20
=
99
Case II If Red ball from 1st bag no t drawn but from 2nd bag Red ball drawn
⎛4⎞ ⎛ 3 ⎞ 12
= ⎜ ⎟×⎜ ⎟ =
⎝9⎠ ⎝11⎠ 99
20 12 32
∴ Total probability = + =
99 99 99
32
Ans. (a)
99
196. Six boys & five girls may sit in such manner
(B) G (B) G (B) G (B)G (B)G (B)
∴ Total No. of ways to sit the girls = 5 !
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ANSWERS
Total No. of ways to sit the boys = 6P = 6 !
6
∴ Total No. of way that they sit
(No. 2 Girls and Boys sit together = 5! × 6! = 120 × 720
= 86400
Ans. (a) 86400
197. If sum of two dice throw is odd
= { (1,2), (1,4) (1,6), (2,1), (2,3) (2,5), (3,2)
(3,4) (4,1), (4,3), (4,5), (5,2), (5,4), (5,6)
(6,1), (6,3), (6,5)}
∴ Probability to get sum as odd
16 16 4
P = = =
(6)2 36 9
∴ Probability to get sum as even nos.
4 5
= 1 − P(E) = 1− =
9 9
5
Ans. (b)
9
198. If 0 not selected then total no.of expectation to select two digits
⎛ 9 ⎞ ⎛8⎞
P(E) = ⎜ ⎟×⎜ ⎟
⎝10⎠ ⎝9⎠
72
P(E) =
90
∴ Probability to get one digit as 0 so product will be zero
72
= 1 − P(E) = 1 −
90
18 1
= =
90 5
1
Ans. (a)
5
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195 n+3P
199. If x = − 3
n 4n! (n+1)!
195 (n+1) 195 1
x = − ⇒ −
n 4n! (n+1)! 4n! n!
After solving we get 4 term will be positive
Ans. (c) 4
1 1 1
200. , , in AP
x+y 2y y+z
1 1
+
1 x+y y+z
∴ =
2y 2
1 y+2+x+y
=
2y (x+y)(y+z)2
⇒ xy + y2 +xz + yz = 2y2 + yz + xy
xz = y2
y z
∴ =
x y
∴ x, y, z in GP
Ans. (b) G.P.
Model Test Paper – BOS/CPT – 3
151. Let the two numbers are x and y
Given x + y : x − y = 7 : 1
i.e. x + y = 7 → (1)
x − y = 1 → (2)
(1) + (2) ⇒ 2x = 8
x = 4
∴ (1) ⇒ 4+y = 7
y = 7 − 4 = 3
∴ x : y = 4 : 3
Ans. (b)
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ANSWERS
152. Let the unit's digit of the number be x and the ten's digits by y.
Then the number = 10y + x
Reversing the order of digits of the given number
Unit's digit becomes y
and ten's digit becomes x
∴ New number = 10x + y
According to the given condition of the problem
( ) ( )
10x+y − 10y+x =54
9 x − 9y = 54
x − y = 6
i.e. The differences of the digit is 6.
Ans. (c)
153. Let the fraction be x/y
According to the first condition of the problem,
x = y − 4
x − y = -4 .................(i)
According to the second condition of the problem,
y+1 = 8(x −2)
y+1 = 8x −16
⇒ 8x − y=1+16
⇒ 8x-y=17 .....................(ii)
subtraction 1 from 2, we get
7x = 21
21
x = = 3
7
(i) ⇒ 3 − y = − 4
y = 3 + 4 = 7
Hence, the required fraction is 3/7
154. Let the present ages of father and his son be x and y years respectively. According to the
first condition of the problem,
x = 6y
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x − 6y = 0 ..............(i)
Four years hence
Age of father = (x+4) years
Age of son = (y+4) years
According to the second condition of the problem
x+4 = 4(y+4)
x+4 = 4y+16
x – 4y = 16 – 4
x – 4y = 12 ................ (ii)
(ii) – (i) ⇒ 2y = 12
y = 12/2 = 6
(i) ⇒ x – 6(6) = 0
x = 36
Hence, present age of father = 36 years
and present age of son = 6 years
∴Ans.(b)
n(n+1)
155. Sum of n natural number =
2
105(105+1)
=
2
= 5565
0.03 ⎛ 3 10⎞
156. log = log ⎜ × ⎟
0.7 ⎝100 7 ⎠
⎛ 3 ⎞
= log ⎜ ⎟
⎝70⎠
= log3 – log 70
= log 3 – (log 7 + log 10)
= 0.48 – (0.84 + 1)
= –1.36
Ans. (c)
157. let x = 4 0.5173
= (0.5173)1/4
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ANSWERS
Taking log on both sides
log x = log [0.5173) 1/4
log x = 1/4 log (0.5173)
= 1/4 (1.7138) (from the table)
= 1/4 (– 1 + 0.7138)
= 1/4 (–1–3+3+0.7138)
= 1/4 (–4 + 3.7138)
= –1 + 0.9284
= 1.9284
∴ x = Anti log (1.9284) = 0.8480
Ans. (a)
0.7214 ×20.37
158. Let x = 3
69.8
Taking log on both sides
log x = 1/3 (log 0.7214+log 20.37– log 69.8)
= 1/3 (1.8581+1.3090–1.8439)
= 1/3 (1.3232)
= 1/3 (3+2.3232)
= 1 + 0.7744
= 1.7744
∴ x = Antilog (1.7744) = 0.5948
Ans. (b)
159. Here P(O) = 4000
i = 0.06
P(n) = 5353
and we are required to find n.
Since p(n) = (1+i)n × P(0)
⇒ 5353 = (1+0.06)n × 4000
5353
=(1+0.06)n
4000
(or) 1.3382 = (1.06)n
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Taking log on both sides.
log (1.3382) = n log (1.06)
0.1265 = n ( 0.0253)
0.1265
n = = 5
0.0253
Hence, the required number of years is 5
Ans. (b)
23
160. Given log x+log x+log x=
2 8 32
15
1 1 1 23
+ + =
log 2 log 8 log 32 15
x x x
1 1 1 23
+ + =
log 2 log 23 log 25 15
x x x
1 1 1 23
+ + =
log 2 3log 2 5log 2 15
x x x
⎛ 1 1⎞ 23
log 2⎜1+ + ⎟ =
x ⎝ 3 5⎠ 15
1 ⎡15+5+3⎤ 23
⎢ ⎥ =
log 2⎣ 15 ⎦ 15
x
1 ⎛23⎞ 23
⎜ ⎟=
log 2⎝15 ⎠ 15
x
1
=1
log 2
x
log x= 1
2
∴ The value of x = 2
Ans. (c)
161. The no. of ways to arrange n different books if two are always together = (n–1) ! × 2!
(Because two books taken together as 1 book)
Ans. (b) (n–1)! ×2!
162. No. of ways to arrange two books (each 3 copies) and 5 book (each 2 copies) = 7!
= 5040
Ans. (a) 5040
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ANSWERS
163. Total No. of words by letters (P, A, R, A, L, L, E, L)
8!
= = 3360
2!3!
6!
No. of words if all 'L comes together = = 360
2!
∴ Total words if 'L' does not come together
= 3360 – 360 = 3000
Ans. (b) 3000
4 !
164. Total no. of 4 digit by (1, 3, 3, 0) = = 12
2!
If 0 comes at thousandth place
3!
then total Nos. = = 3
2!
∴ Net 4 digit Nos. by (1, 3, 3, 0) = 12 – 3= 9
(1, 3, 3, 0) each comes at Unit, tenth, hundredth place 2! times.
and 1, 3, 3, each comes at thousandth place 3 times
∴ Sumof digit = 1+3+3+0 = 7
∴ Total sum = 7 × 2 [10 0 + 101 + 102) + 7×3[103]
= 14×111+21×1000
Total sum = 22554
Ans. (a) 22554
166. Let I = ∫ x3 3+5x4 dx
Put 3 + 5x4 = t
20x3 dx = dt
1
x3dx = dt
20
1
∴I = ∫ 3+5x4 x3dx = ∫ t . dt
20
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1 1 t3/2
= ∫ t1/2 dt = +c
20 20 3/2
1
= t3/2+c
30
( )
∫ x3 3+5x4dy = 1 3+5x4 3/2 +c
30
Ans.(c)
∫
2x+1
167. Let I = dx
x(x+1)
∫
2x+1
= dx
x2 +x
= log (x2 +x)+c
Ans. (b)
168. Put x =t
∴x=t2
dx = 2t dt
dx 2tdt dt
∴∫ = ∫ = 2 ∫
x+ x t2+t t+1
= 2 log (t+1) + c
( )
2 log x+1 +c
169. Let z = log x
= log(x)1/2
1
z = log x
2
1 1
dz = . dx
2 x
log x 2 log x
∴∫ dx = ∫ dx
3x 3 2x
2
∫
= z dz
3
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ANSWERS
2
⎡ z2⎤
= ⎢ ⎥+c
3⎣ z ⎦
z2
= +c
3
( )
∫ log x dx = 1 log x 2 +c
3x 3
Ans.(a)
170. Let I = ∫ x2e2xdx
Integrating by parts
x2e2x 2x
I = − ∫ e2x dx
2 2
x2e2x
I = − ∫ xe2xdx ................... (i)
2
consider ∫ xe2xdx
Integrating by parts,
xe2x e2x
= − ∫ dx
2 2
xe2x e2x
= −
2 4
(i) becomes
x2e2x ⎡ xe2x e2x⎤
I = −⎢ − ⎥
2 ⎣ 2 4 ⎦
x2e2x xe2x e2x
I = − + +c
2 2 4
Ans.(b)
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