91.
If ÃÂÃÂ the square root of ÃÂÃÂ is nearest to = ?
- (A) 0.172
- (B) 0.414
- (C) 0.586
- (D) 1.414
Solution:
$$\eqalign{ & = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr & = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr & = {\left( {\sqrt 2 - 1} \right)^2} \cr & \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr & = \left( {\sqrt 2 - 1} \right) \cr & = \left( {1.414 - 1} \right) \cr & = 0.414 \cr} $$
92.
How many two-digit numbers satisfy this property. : The last digit (unit's digit) of the square of the two-digit number is 8 ?
- (A) 1
- (B) 2
- (C) 3
- (D) None of these
Solution:
A number ending in 8 can never be a perfect square.
93.
If ÃÂÃÂ then the value of ÃÂÃÂ is equal to :
- (A) 5.59
- (B) 7.826
- (C) 8.944
- (D) 10.062
Solution:
$$\eqalign{ & \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \, \cr & = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr & = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr & = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{35\sqrt 5 }}{{10}} \cr & = \frac{{7 \times 2.236}}{2} \cr & = 7 \times 1.118 \cr & = 7.826 \cr} $$
94.
Given that ÃÂÃÂ the value of ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ is ?
- (A) 1.414
- (B) 1.732
- (C) 2.551
- (D) 4.899
Solution:
$$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 2\sqrt {12} - \sqrt {32} + \sqrt {50} }} \cr & = \frac{{3 + \sqrt 6 }}{{5\sqrt 3 - 4\sqrt 3 - 4\sqrt 2 + 5\sqrt 2 }} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \cr & = \frac{{\left( {3 + \sqrt 6 } \right)}}{{\left( {\sqrt 3 + \sqrt 2 } \right)}} \times \frac{{\left( {\sqrt 3 - \sqrt 2 } \right)}}{{\left( {\sqrt 3 - \sqrt 2 } \right)}} \cr & = \frac{{3\sqrt 3 - 3\sqrt 2 + 3\sqrt 2 - 2\sqrt 3 }}{{\left( {3 - 2} \right)}} \cr & = \sqrt 3 \cr & = 1.732 \cr} $$
95.
The value of ÃÂÃÂ is = ?
- (A) 0.00021
- (B) 0.0021
- (C) 0.021
- (D) 0.21
Solution:
$$\eqalign{ & = \sqrt {0.000441} \cr & = \sqrt {\frac{{441}}{{{{10}^6}}}} \cr & = \frac{{\sqrt {441} }}{{\sqrt {{{10}^6}} }} \cr & = \frac{{21}}{{{{10}^3}}} \cr & = \frac{{21}}{{1000}} \cr & = 0.021 \cr} $$
96.
What is the smallest number by which 3600 be divided to make it a perfect cube ?
- (A) 9
- (B) 50
- (C) 300
- (D) 450
Solution:
$$3600 = {2^3} \times {5^2} \times {3^2} \times 2$$ To make it a perfect cube, it must be divided by $${5^2} \times {3^2} \times 2,i.e.,450$$
97.
Solved
- (A)
- (B) 25
- (C) 125
- (D) None of these
Solution:
$$\eqalign{ & {\text{Given,}} \cr & = \root 4 \of {{{\left( {625} \right)}^3}} \cr & = {\left( {625} \right)^{\frac{3}{4}}} \cr & = {\left( {5 \times 5 \times 5 \times 5} \right)^{\frac{3}{4}}} \cr & = {\left( {{5^4}} \right)^{\frac{3}{4}}} \cr & = {5^3} \cr & = 125 \cr} $$
98.
If ÃÂÃÂ , then the value of n is ?
- (A) 6
- (B) 8
- (C) 10
- (D) 12
Solution:
$$\eqalign{ & \Leftrightarrow \sqrt {{3^n}} = 729 \cr & \Leftrightarrow \sqrt {{3^n}} = {3^6} \cr & \Leftrightarrow {\left( {\sqrt {{3^n}} } \right)^2} = {\left( {{3^6}} \right)^2} \cr & \Leftrightarrow {3^n} = {3^{12}} \cr & \Leftrightarrow n = 12 \cr} $$
99.
If ÃÂÃÂ ÃÂÃÂ ÃÂÃÂ = 37.25, the x is equal to ?
- (A)
- (B)
- (C)
- (D) None of these
Solution:
$$\eqalign{ & \Leftrightarrow 37 + \sqrt {0.0615 + x} = 37.25 \cr & \Leftrightarrow \sqrt {0.0615 + x} = 0.25 \cr & \Leftrightarrow 0.0615 + x = {\left( {0.25} \right)^2} = 0.0625 \cr & \Leftrightarrow x = 0.001 \cr & \Leftrightarrow x = \frac{1}{{{{10}^3}}} \cr & \Leftrightarrow x = {10^{ - 3}} \cr} $$
100.
The cube root of .000216 is:
- (A) .6
- (B) .06
- (C) 77
- (D) 87
Solution:
$$\eqalign{ & {\left( {.000216} \right)^{\frac{1}{3}}} = {\left( {\frac{{216}}{{{{10}^6}}}} \right)^{\frac{1}{3}}} \cr & = {\left( {\frac{{6 \times 6 \times 6}}{{{{10}^2} \times {{10}^2} \times {{10}^2}}}} \right)^{\frac{1}{3}}} \cr & = \frac{6}{{{{10}^2}}} \cr & = \frac{6}{{100}} \cr & = 0.06 \cr} $$