Tag: approximation of square roots

Questions Related to approximation of square roots

Estimate: $\sqrt { 60 } $

  1. $7.7$

  2. $7$

  3. $7.2$

  4. $7.5$


Correct Option: A
Explanation:

$60$ is in between two perfect squares: $49$, which is ${7}^{2}$ and $64$ which is ${8}^{2}$. The difference between $64$ and $49$ is $15$ so $60$ is little more than $\cfrac{2}{3}$ of the way toward $64$ from $49$. A reasonable estimate for $\sqrt {60}$, then would be about $7.7$ which is a little more than $\cfrac{2}{3}$ toward $8$ from $7$.

If $\sqrt{0.01+\sqrt{0.0064}}=x$, then the value of $x$ is ____________.

  1. $0.3$

  2. $0.03$

  3. $\sqrt{0.18}$

  4. None of these


Correct Option: A
Explanation:

Given, $\sqrt{0.01+\sqrt{0.0064}}=x$

Taking square of both sides.
$0.01+\sqrt{0.0064}=x^2$
We know that $\sqrt{0.0064}=0.08$
So, $x^2=0.08+0.01$
$x^2=0.09$
$x=0.3$

The square root of $\displaystyle\frac{36}{5}$ correct to two decimal places is _____________.

  1. $2.68$

  2. $2.69$

  3. $2.67$

  4. $2.66$


Correct Option: A
Explanation:

We know $\sqrt{\dfrac{36}{5}}=\dfrac {6}{\sqrt 5}$

$=\dfrac{6}{2.236}$
$=2.6833$
Upto two decimal places:
$=2.68$

Which of the following is not a perfect square?

  1. $16384$

  2. $23857$

  3. $18496$

  4. $11025$


Correct Option: B
Explanation:

$16384= 128\times 128$         (perfect square)

$18496=136\times 136$           (perfect square)
$11025=105\times 105$         (perfect square)
$23857 = 1\times  23857$        (prime no. so not a perfect square)
Hence, option B is correct.

The number that must be subtracted from $16161$ to get a perfect square is ________.

  1. $31$

  2. $32$

  3. $33$

  4. $34$


Correct Option: B
Explanation:

Let the number is $x$

Finding the square root of $16161$
$\sqrt{16161}=127.1259$

Finding the square of $127$
$127^2=16129$
$x=16161-16129$
$x=32$

Determine
(a) $\sqrt{147} \times \sqrt{243} \$

  1. $189$

  2. $209$

  3. $197$

  4. $138$


Correct Option: A
Explanation:

$√143 = √3  × √49 = √3  × 7$
$√243 = √3  × √81 = √3  × 9$
$√143  ×√243  = 3  × 9  × 7 = 189$

If $x=\sqrt {12}-\sqrt {9},y=\sqrt {13}-\sqrt {10}$ and $z=\sqrt {11}-\sqrt {8}$, then which of the following is true?

  1. $z > x > y$

  2. $z > y > x$

  3. $y > x > z$

  4. $y > z> x$


Correct Option: A
Explanation:

$x=\sqrt{12}-\sqrt{9}$

$x=2\sqrt{3}-3$
$x=2(1.73)-3$
$x=3.46-3$
$\therefore$   $x=0.46$                    ----- ( 1 )
Now,
$y=\sqrt{13}-\sqrt{10}$
$y=3.60-3.16$
$\therefore$  $y=0.44$                ---- ( 2 )
Now,
$z=\sqrt{11}-\sqrt{8}$
$z=3.32-2.83$
$\therefore$  $z=0.49$             ---- ( 3 )
From ( 1 ), ( 2 ) and ( 3 )
$\Rightarrow$  $0.49>0.46>0.44$
$\therefore$  $z>x>y$

Find the atleast number which must  be added to each of the following numbers to get a perfect square. Also find the square root of the perfect square numbers.


$a)525$

  1. 4

  2. 3

  3. 1

  4. 6

  5. 12


Correct Option: A
Explanation:
$i)\ 23^2=529$
$\therefore \ $ it will add $4$ to $525$ we get $529$
which is payout square
$\therefore \ 525+4=529=23^2$

$\therefore 4$ should be added


The value of $\sqrt { \sqrt [ a ]{ { 4 }^{ { a }^{ { a }^{ 2 } } }\sqrt { { 6 }^{ { a }^{ 3 } }\sqrt [ { a }^{ 3 } ]{ { 12 }^{ { a }^{ 6 } }\sqrt [ { a }^{ 4 } ]{ { 18 }^{ { a }^{ 10 } } }  }  }  }  } $ is equal to

  1. $\sqrt {216}$

  2. $\sqrt {72}$

  3. $72$

  4. $216$


Correct Option: C
Explanation:

$\begin{array}{l} =\sqrt { \sqrt [ a ]{ { { 4^{ { a^{ { a^{ 2 } } } } } }\sqrt { { 6^{ { a^{ 3 } } } }\sqrt [ { { a^{ 3 } } } ]{ { { { 12 }^{ { a^{ 6 } } } }\sqrt [ { { a^{ 4 } } } ]{ { { { 18 }^{ { a^{ 10 } } } } } }  } }  }  } }  }  \ =\sqrt { 4\times 6\times 18\times 12 }  \ =72 \ Hence, \ option\, \, C\, \, is\, correct\, \, answer. \end{array}$

The approximate value of $\sqrt{(1.97)^2+(4.02)^2+(3.98)^2}$.

  1. $5.99$

  2. $5.099$

  3. $5.009$

  4. $5.734$


Correct Option: A
Explanation:

According to Question

$\sqrt{(2-0.03)^2+(4+0.02)^2+(4-0.02)^2}$
Opening Brackets
$\sqrt{(4+0.0009-0.12)+(16+0.0004+0.16)+(16+0.0004-0.16)}$
$\sqrt{36-0.1183}$
$\sqrt{35.8817} \approx 5.99$