Tag: square root

Questions Related to square root

$\sqrt { \sqrt { 169 } +\sqrt [ 3 ]{ 1728 }  } $ equals

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. 12

  2. 5

  3. 3

  4. 9

Show answer
Correct Option: B
Explanation:

$\sqrt{169}$ $\rightarrow$ $13$

$\sqrt[3]{1728}$ $\rightarrow$ $12$
$12$ $+$ $13$ $\rightarrow$ $25$
$\sqrt{25}$ $\rightarrow$ $5$
Hence, Option B is correct.

169+17283

What is the square root of ${0.000441}$ ?

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. $0.021$

  2. $0.0021$

  3. $0.21$

  4. $2.1$

Show answer
Correct Option: A
Explanation:

$\sqrt{0.000441}$


$\sqrt{441}$ $\rightarrow$ $21$

$\sqrt{1000000}$ $\rightarrow$ $1000$

$\sqrt{0.000441}$ $\rightarrow$ $\sqrt{441}$ $/$ $\sqrt{1000000}$

$\sqrt{0.000441}$ $\rightarrow$ $\dfrac{21}{1000}$ $\rightarrow$ $0.021$

Hence, Option A is correct.

If x is a positive integer less than 100, then the number of x which make $\displaystyle \sqrt{1+2+3+4+x}$ an integer is

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. 6

  2. 7

  3. 8

  4. 9

Show answer
Correct Option: B
Explanation:

Let, $\sqrt{1+2+3+4+5+x}=l$, where $ l $ is an positive integer.
$\Rightarrow \sqrt{10+x}=l$
$\Rightarrow 10+x=l^2$
$\Rightarrow x=l^2-10$
Given, $1<x<100$
$\Rightarrow 1<l^2-10<100$
$\Rightarrow 11<l^2<110$
$\Rightarrow \sqrt{11}<l<\sqrt{110}$
$\Rightarrow 3.32<l<10.49$
$\Rightarrow l=4,5,6,7,8,9,10$              ($\because l\text{ is an integer.}$)
Therefore, there are 7 values of $x$.

The least number by which 176 be multiplied to make the result a perfect square, is:

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. 8

  2. 9

  3. 10

  4. 11

Show answer
Correct Option: D
Explanation:
$176 = \underline {2 \times 2} \times \underline {2 \times 2} \times 11.$
So, in order to make it a perfect square, it must be multiplied by 11.

Find the square roots of $100\;and\;169$ by method of repeated substraction.

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. $10; 13$

  2. $10; 17$

  3. $20; 13$

  4. $20; 17$

Show answer
Correct Option: A
Explanation:

For $\sqrt{100}$
$100-1=99$
$99-3=96$
$96-5=91$
$91-7=84$
$84-9=75$
$75-11=64$
$64-13=51$
$51-15=36$
$36-17=29$
$19-19=0$

For $\sqrt{169}$
$169-1=168$
$168-3=165$
$165-5=160$
$160-7=153$
$153-9=144$
$144-11=133$
$133-13=120$
$120-15=105$
$105-17=88$
$88-19=69$
$69-21=48$
$48-23=25$
$25-25=0$
From $100\;and\;169$ we have substracted successive odd numbers starting from $1$ and obtained $0$ at $10th\;and\;13th$ steps, respectively.
So, $\sqrt{100}=10\;and\;\sqrt{169}=13$. 

The least number which must be subtracted from 6,156 to make it a perfect square is:

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. 62

  2. 72

  3. 52

  4. 82

Show answer
Correct Option: B
Explanation:
First we will find the square root of $6156$:

$\sqrt { 6156 } =\sqrt { 2×2×3×3×3×3×19 } =2×3×3×\sqrt { 19 } =18×4.358=78.444...$

Now, we find the square of $78$ as follows:

${ 78 }^{ 2 }=78×78=6084$

Let us now subtract $6156$ from $6084$ as shown below:

$6156 - 6084 = 72$

Hence, $72$ will be subtracted from $6156$ to make it a perfect square.

$\displaystyle {\sqrt{\frac{36.1}{102.4}}}\, =\, ?$

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. $\displaystyle \frac{29}{32}$

  2. $\displaystyle \frac{19}{72}$

  3. $\displaystyle \frac{19}{32}$

  4. $\displaystyle \frac{29}{62}$

Show answer
Correct Option: C
Explanation:

$\displaystyle {\sqrt {\frac{36.1}{102.4}}\, =\, \sqrt {\frac{361}{1024}} = \frac{19}{32}}$

Find the square root of $100$ by the method of repeated substraction.

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. $10$.

  2. $11$

  3. $9$

  4. $12$

Show answer
Correct Option: A
Explanation:

We know that the sum of the first n odd natural numbers is $n^{2}$
From $100$, we subtract successive odd numbers starting from $1$ as under
$100-1=99\;\;\;99-3=96$
$96-5=91\;\;\;\;91-7=84$
$84-9=75\;\;\;\;75-11=64$
$64-13=51\;\;\;51-15=36$
$36-17=19\;\;\;19-19=0$
and obtain $0$ at $10th$ step.
$\therefore\;\sqrt{100}=10$.
From $169$, we subtract successive odd numbers starting from $1$ as under
$169-1=168\;\;\;168-3=165$
$165-5=160\;\;\;\;160-7=153$
$153-9=144\;\;\;\;144-11=133$
$133-13=120\;\;\;120-15=105$
$105-17=88\;\;\;88-19=69$
$69-21=48\;\;\;48-23=25$
$25-25=0$
and obtain $0$ at $13th$ step.
$\therefore\;\sqrt{169}=13$.





The value of $\sqrt{1\displaystyle\frac{1}{2}-\begin{bmatrix}1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{2}+\begin{pmatrix}1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{4}\end{pmatrix}\end{bmatrix}}$ is

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. $\displaystyle\frac{1}{2}$

  2. $\displaystyle\frac{1}{4}$

  3. $\displaystyle\frac{1}{16}$

  4. $1\displaystyle\frac{1}{5}$

Show answer
Correct Option: A
Explanation:

$\sqrt{1\displaystyle\frac{1}{2}-\begin{bmatrix}1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{2}+\begin{pmatrix}1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{2}-1\displaystyle\frac{1}{4}\end{pmatrix}\end{bmatrix}}$
$\Rightarrow \sqrt{\frac{3}{2}-\left [ \frac{3}{2}-\frac{3}{2}+\left ( \frac{3}{2}-\frac{3}{2}-\frac{5}{4} \right ) \right ]}$
$\Rightarrow \sqrt{\frac{3}{2}-\left [ 0+\left ( \frac{6-6-5}{4} \right ) \right ]}$
$\Rightarrow \sqrt{\frac{3}{2}-\frac{5}{4}}$
$\Rightarrow \sqrt{\frac{6 -5}{4}}$
$\Rightarrow \sqrt{\frac{1}{4}}$
$\Rightarrow \frac{1}{2}$




If $\sqrt{1\, +\, \displaystyle \frac{27}{169}}\, =\, 1\, +\, \displaystyle \frac{x}{13}$, then the value of $x$ is

maths square root square root of perfect square finding square root of a number square root of a perfect square

  1. 1

  2. 14

  3. Cannot be determined

  4. None of these

Show answer
Correct Option: A
Explanation:

$\sqrt{1\, +\, \displaystyle \cfrac{27}{169}}\, =\, \displaystyle \cfrac{196}{169}\, =\, \displaystyle \cfrac{14}{13}$
$\Rightarrow 1\, \displaystyle \cfrac{1}{13}\, =\, 1\, +\, \displaystyle \cfrac{1}{13}$
$\therefore\, x\, =\, 1$