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Practice Aptitude Questions With Solutions for SSC CGL / Railway Exam

Practice Aptitude Questions With Solutions Railway / SSC Exams Set-10:
Dear Readers, Important Aptitude Questions for Railway / SSC / FCI Exam was given here with solutions, candidates those who are preparing for those exams can use this material.

1). In a division sum, the ivisor is 3 times the quotient and 6 times the remainder. If the remainder is 2, then the dividend is
a)   36
b)   28
c)   50
d)   48
2). 1/7 + {999 [692/693] }is equal to
a)   99800
b)   99900
c)   1
d)   99000
3). Let ∛a = ∛26 +∛7 + ∛63. Then
a)   a > 729
b)   a = 729
c)   a < 729 but a > 216
d)   a < 216
4). The number of prime factors in 6333 ×7222 × 8111 is
a)   1111
b)   1211
c)   1221
d)   1222
5).   
Is equal to







a)   3
b)   2
c)   1
d)   4
6). If 120is 20% of a number, then 120% of that number will be
a)   20
b)   120
c)   360
d)   720
7). A reduction of 10% in the price of a commodity enables a person to buy 25 kg more for Rs.225. The original price of the commodity per kg was
a)   Rs. 2.50
b)   Rs. 1.50
c)   Rs. 2
d)   Rs. 1
8). One type of liquid contains 20% water and the second type of liquid contains 35% of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. The water in the new mixture in the glass is
a)   12 1/7 %
b)   24 2/7%
c)   37%
d)   46%
9). If x:y=3:4 and y:z=3:4, then  (x+y+z)/3z is equal to
a)   73/84
b)   37/48
c)   13/27
d)   ½
10). The value of

is



a)   45/28
b)   55/28
c)   55/42
d)   45/56

Answers:
1). c)   2). d)   3). c)   4). c)   5). c)   6). d)   7). d)   8). b)   9). b)   10). b)  

Solutions:
1). Let d, q, r be the divisor quotient and remainder respectively
Then d=3q and d = 6r
Also r=2
Therefore d = 6×2 = 12
Now q = d/3 = 12/3=4
Therefore dividend = qd + r
                   = 4 ×12+2
                   = 50
Answer:c)
2). 1/7 + { 999 [692/693] } × 99
= 1/7 + { (999×693+692)/693 } × 99
= 1/7 + 692999/7
=693000/7
= 99000
Answer:d)
3). ∛26 < ∛27 = 3
∛7 < ∛8 = 2
∛63 < ∛64 = 4
∛a = ∛26 + ∛7 + √63
< 3+ 2 + 4
= 9
a < 93 = 729
a < 729 ………….. (1)
Also
∛26 > ∛8
∛7 > ∛1
∛63 > ∛27 = 3
∛a = ∛26 + ∛7 + ∛63
>   2+1+3 = 6
a > 62 = 216 ………………… (2)
From (1) and (2)
A<729 but a>216
Answer:c)
4). 6333 × 7222 × 8111
= (2×3)333  × 7222 ×(23)111
= 2333 × 3333 ×7222 ×2333
Total no of prime factors
= 333 + 333 + 222 + 333
= 1221
 Answer:c)

5).



 Answer:c)
6). Let the number be x
Then 20% of x = 120
20/100 × x = 120
X = (120×100)/20  = 600
Now 120% of x
= 120% of 600
= 120/100 × 600 = 720
Answer:d)
7). Let the original price per kg be Rs.x
Quantity of rice bought for Rs.225 = 225/x kg
Reduced price = 90% of x
= 90/100 x = 9x/10
Quantity of rice bought for Rs.225 in reduced price = 225/ (9x/10) = 2250/9x
According to the problem,
2250/9x – 225/x = 25
(2250-2025) / 9x = 25
225/9x = 25
X = 225 /(9 × 25)
= Rs.1
Original price per kg = Rs.1
 Answer:d)
8).  Water in 10 parts of first liquid = 20% of 10
20/100 × 10 = 2
Water in 4 parts of second liquid
= 35/100 × 4 = 7/5
% of water in the new mixture
= [2+(7/5)] / (10+4)  × 100
=(17/5) / 14  × 100
= 17 / 5×14  × 100 = 170/7
= 24  2/7%
 Answer:b)
9). x:y = 3:4 = 4=9:12
y:z = 3 : 4 = 12 : 16
(x : y : z ) / 3z = (9+12+16)/(3×16)
=37/48
 Answer:b)
10).
√72 = √(2×36) = 6√2
√363 = √(3×121) = 11√3
√175 = √(7×25) = 5√7
√32 = √(2×16) = 4√2
√147 = √(3×49) = 7(√3)
√252 = √(36×7) =6√7
{(√72)× √363×√175) / √32 ×√147 ×√252}
= {(6√2×11√3× 5√7) / 4√2 × 7√3 ×6√7}
=55/28
Answer:b)
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