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Is equal to
Answer:c)
Answer:c)
Answer:d)
Answer:b)
Answer:b)
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).
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
5).
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
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%
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
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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