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Surds and Indices
1.The cube root of .000216 is:

Answer:Option 1

Explanation:

Solution:
(.000216)13=(216106)13=(6×6×6102×102×102)13=6102=6100=0.06

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2.What should come in place of both x in the equation
x128=162x

Answer:Option 1

Explanation:

Solution:
Letx128=162xThenx2=128×162=64×2×18×9=82×62×32=8×6×3=144x=144=12

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3.The least perfect square, which is divisible by each of 21, 36 and 66 is:

Answer:Option 1

Explanation:

Solution:
L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 22 x 32 x 72 x 112 = 213444

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4.If
35+125
  = 17.88, then what will be the value of
80+65?

Answer:Option 1

Explanation:

Solution:
35+125=17.8835+25×5=17.8835+55=17.8885=17.885=2.23580+65=16×5+65=45+65=105=(10×2.235)=22.35

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5.If a = 0.1039, then the value of
4a24a+1+3a
    is:

Answer:Option 1

Explanation:

Solution:
4a24a+1+3a=(1)2+(2a)22×1×2a+3a=(12a)2+3a=(12a)+3a=(1+a)=(1+0.1039)=1.1039

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6.If
x=3+131
  and
y=313+1,
  then the value of
(x2+y2)
  is?

Answer:Option 1

Explanation:

Solution:
x=(3+1)(31)×(3+1)(3+1)=(3+1)2(31)=3+1+232=2+3y=(31)(3+1)×(31)(31)=(31)2(31)=3+1232=23x2+y2=(2+3)2+(23)2=2(4+3)=14

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7.A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

Answer:Option 1

Explanation:

Solution:
Money collected = (59.29 x 100) paise = 5929 paise.
∴ Number of members =
5929
  = 77

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8.If
5=2.236,
  then the value of
52
105
+
125
  is equal to :

Answer:Option 1

Explanation:

Solution:
52105+125=(5)220+25×5525=520+5025=3525×55=35510=7×2.2362=7×1.118=7.826

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9.
(62511×1425×11196)
    is equal to :

Answer:Option 1

Explanation:

Solution:
GiveExpression=2511×145×1114=5

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10.
0.0169×?=1.3

Answer:Option 1

Explanation:

Solution:
Let0.0169×x=1.3Then,0.0169x=(1.3)2=1.69x=1.690.0169=100

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