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41
The value of $$\sqrt {0.000441} $$   is = ?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & = \sqrt {0.000441} \cr & = \sqrt {\frac{{441}}{{{{10}^6}}}} \cr & = \frac{{\sqrt {441} }}{{\sqrt {{{10}^6}} }} \cr & = \frac{{21}}{{{{10}^3}}} \cr & = \frac{{21}}{{1000}} \cr & = 0.021 \cr} $$
42
$${1.5^2} \times \sqrt {0.0225} = ?$$
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & = {1.5^2} \times \sqrt {0.0225} \cr & = {1.5^2} \times \sqrt {\frac{{225}}{{10000}}} \cr & = 2.25 \times \frac{{15}}{{100}} \cr & = 2.25 \times 0.15 \cr & = 0.3375 \cr} $$
43
The value of $$\sqrt {0.01} {\text{ + }}$$ $$\sqrt {0.81} {\text{ + }}$$ $$\sqrt {1.21} {\text{ + }}$$ $$\sqrt {0.0009} $$   is = ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Given expression,
$$ = \sqrt {\frac{1}{{100}}} + \sqrt {\frac{{81}}{{100}}} + \sqrt {\frac{{121}}{{100}}} + $$      $$\sqrt {\frac{9}{{10000}}} $$
$$\eqalign{ & = \frac{1}{{10}} + \frac{9}{{10}} + \frac{{11}}{{10}} + \frac{3}{{100}} \cr & = 0.1 + 0.9 + 1.1 + 0.03 \cr & = 2.13 \cr} $$
44
$$\sqrt {1.5625} = ?$$
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \,\,\,\,\,\,\,1|\overline 1 \,.\,\overline {56} \,\,\overline {25} \,(1.25 \cr & \,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,22|\,\,\,\,\,\,56 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,44 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 245\,|\,\,\,\,\,\,\,12\,25 \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,12\,25 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\text{x} \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \therefore \sqrt {1.5625} = 1.25 \cr} $$
45
Given that $$\sqrt {13} = 3.605$$   and $$\sqrt {130} = 11.40{\text{,}}$$   find the value of $$\sqrt {1.30} $$  $$ + $$ $$\sqrt {1300} $$  $$ + $$ $$\sqrt {0.0130} $$   = ?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Given expression,}} \cr & \sqrt {1.30} + \sqrt {1300} + \sqrt {0.0130} \cr & = \sqrt {\frac{{130}}{{100}}} + \sqrt {13 \times 100} + \sqrt {\frac{{130}}{{10000}}} \cr & = \frac{{\sqrt {130} }}{{10}} + \sqrt {13} \times 10 + \frac{{\sqrt {130} }}{{100}} \cr & = \frac{{11.40}}{{10}} + 3.605 \times 10 + \frac{{11.40}}{{100}} \cr & = 1.14 + 36.05 + 0.114 \cr & = 37.304 \cr} $$
46
If $$\frac{{52}}{x} = \sqrt {\frac{{169}}{{289}}} {\text{,}}$$   the value of x is = ?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & \Leftrightarrow \frac{{52}}{x}{\text{ = }}\sqrt {\frac{{169}}{{289}}} \cr & \Leftrightarrow \frac{{52}}{x} = \frac{{13}}{{17}} \cr & \Leftrightarrow x = \left( {\frac{{52 \times 17}}{{13}}} \right) \cr & \Leftrightarrow x = 68 \cr} $$
47
For what value of * the statement $$\left( {\frac{*}{{15}}} \right)$$ $$\left( {\frac{*}{{135}}} \right)$$  = 1 is true ?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Method 1:}} \cr & {\text{Let the missing number be }}x \cr & {\text{Then, }} \cr & \Leftrightarrow {x^2} = 15 \times 135 \cr & \Leftrightarrow x = \sqrt {15 \times \left(15 \times 9 \right)} \cr & \Leftrightarrow x = \sqrt {{{15}^2} \times {3^2}} \cr & \Leftrightarrow x = 15 \times 3 \cr & \Leftrightarrow x = 45 \cr & \cr & {\text{Method 2:}} \cr & {\text{Let the missing number be }}x \cr & {\text{Then, }} \cr & \Leftrightarrow {x^2} = 15 \times 135 \cr & \Leftrightarrow {x^2} = 2025 \cr & \Leftrightarrow x = \sqrt {{2025} } \cr & \Leftrightarrow x = 45 \cr} $$
48
Which number should replace both the question marks in the following equation ?
$$\frac{?}{{1776}} = \frac{{111}}{?}$$
Discuss
Answer & Solution
Answer: Option E
Solution:
$$\eqalign{ & {\text{Let }}\frac{x}{{1776}} = \frac{{111}}{x} \cr & {\text{Then, }} \cr & \Leftrightarrow {x^2} = 111 \times 1776 \cr & \Leftrightarrow {x^2} = 111 \times 111 \times 16 \cr & \Leftrightarrow x = \sqrt {{{\left( {111} \right)}^2} \times {{\left( 4 \right)}^2}} \cr & \Leftrightarrow x = 111 \times 4 \cr & \Leftrightarrow x = 444 \cr} $$
49
Which number can replace both the question marks in the equation ?
$$\frac{{4\frac{1}{2}}}{?} = \frac{?}{{32}}$$
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Let,}} \cr & {\text{ }}\frac{{4\frac{1}{2}}}{x} = \frac{x}{{32}} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = 32 \times \frac{9}{2} \cr & \Leftrightarrow {x^2} = 144 \cr & \Leftrightarrow x = \sqrt {144} \cr & \Leftrightarrow x = 12 \cr} $$
50
What should come in place of both the question marks in the equation ?
$$\frac{?}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{?}$$
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let,}} \cr & {\text{ }}\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = \sqrt {128 \times 162} \cr & \Leftrightarrow {x^2} = \sqrt {64 \times 2 \times 18 \times 9} \cr & \Leftrightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & \Leftrightarrow {x^2} = 8 \times 6 \times 3 \cr & \Leftrightarrow {x^2} = 144 \cr & \Leftrightarrow x = \sqrt {144} \cr & \Leftrightarrow x = 12 \cr} $$