# CAT Questions on Square Roots

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## CAT Questions on Square Roots

Download important Square Roots Questions for CAT PDF based on previously asked questions in CAT exam. Practice Square Roots Questions PDF for CAT exam.

Question 1: Let $x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$. Then x equals

a) 3

b) $(\sqrt{13} – 1)/2$

c) $(\sqrt{13} + 1)/2$

d) $\sqrt{13}$

Question 2: $(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})\div\sqrt{\frac{16}{81}}=?$

a) $\frac{5}{16}$

b) $\frac{7}{12}$

c) $\frac{3}{8}$

d) None of these

Question 3: What is the value of $\sqrt{\frac{a}{b}}$, If $\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+1$

a) -5/3

b) 2

c) 5/3

d) 1

Question 4: $2-\frac{\sqrt{6407522209}}{\sqrt{3600840049}}=$

a) 0.666039

b) 0.666029

c) 0.666009

d) None of the above

Question 5: The value of $\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$ is equal to

a) 1/3

b) 2/3

c) 5/6

d) 7/6

Question 6: The highest number amongst $\sqrt{2}, \sqrt[3]{3},$ and $\sqrt[4]{4}$ is

a) $\sqrt{2}$

b)  $\sqrt[3]{3}$

c) $\sqrt[4]{4}$

d) All are equal

Question 7: The simplest value of the expression $(\frac{4^{p+\frac{1}{4}}\times \sqrt{2 \times 2^{p}}}{2\times \sqrt{2^-p}})^\frac{1}{p}$

a) 4

b) 8

c) 4p

d) 8p

Question 8: The value of x for which the equation $\sqrt{4x – 9}$ + $\sqrt{4x + 9}$ = 5 + $\sqrt{7}$ will be satisfied, is:

a) 1

b) 2

c) 3

d) 4

Question 9: The value of $\text{log}_{7} \text{log}_{7} \sqrt{7(\sqrt{7\sqrt{7}})}$

a) 7

b) $\text{log}_7$ 2

c) $1-3 \text{log}_2$ 7

d) $1-3 \text{log}_7$ 2

Question 10: $log_{13} log_{21} (\sqrt{x+21}+ \sqrt{x} ) =0$ then the value of x is

a) 21

b) 13

c) 81

d) None of the above

$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$

=> $x = \sqrt{4+\sqrt{4-x}}$

=> $x^2 = 4 + \sqrt{4-x}$

=>$x^4 + 16 – 8x^2 = 4 – x$

=> $x^4 – 8x^2 + x +12 = 0$

On substituting options, we can see that option C satisfies the equation.

$(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})\div\sqrt{\frac{16}{81}}=x$ This can be simplified as

$(\frac{15}{27}-\frac{5}{12})\div\frac{4}{9}=x$

$(\frac{5}{36})*\frac{9}{4}=x$

x=$\frac{5}{16}$. Hence, option A is the correct answer.

$\sqrt{\frac{a}{b}}$, If $\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+\log_{4}{4}$

i.e. $\log_{4}\log_{4}4^{a-b}=\log_{4}((\sqrt{a}-\sqrt{b})^2)*4$

i.e. $\log_{4}4^{a-b}=((\sqrt{a}-\sqrt{b})^2)*4$

i.e. (a-b)*$\log_{4}4=((\sqrt{a}-\sqrt{b})^2)*4$

i.e. a-b = 4a+4b-8$\sqrt{ab}$

i.e. 3a + 5b – 8$\sqrt{ab}$ = 0

i.e. $3\sqrt\frac{a}{b}^2$ – 8$\sqrt\frac{a}{b}$+5 = 0

put $\sqrt\frac{a}{b}$ = t

therefore 3$t^2$ – 8t + 5 = 0

solving we get t = 1 or t = $\frac{5}{3}$

i.e. $\sqrt\frac{a}{b}$ = 1 or $\frac{5}{3}$

but if $\sqrt\frac{a}{b}$ = 1 then a=b then $\log_{4}(\sqrt{a}-\sqrt{b})$ will become indefinite

Therefore  $\sqrt\frac{a}{b}$ = $\frac{5}{3}$

Therefore our answer is option ‘C’

$2-\frac{\sqrt{6407522209}}{\sqrt{3600840049}}=2-\frac{80047}{60007}$
=$2-1.3339610$
$=0.666039$
Therefore, option A is the right answer.

$\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$

$81 = 3^4$ and $0.008 = \frac{8}{1000} = \frac{2^{3}}{10^{3}} = \frac{1}{5^{3}} = 5^{-3}$

Hence,

$\log_{0.008}\sqrt{5}+ 8 -7$

$\log_{5^{-3}}5^{\frac{1}{2}}+ 8 -7$

$\frac{log 5^{0.5}}{log 5^{-3}} + 1$

$– \frac{1}{6} + 1$

= $\frac{5}{6}$

Given that

a = $\sqrt{2}, b = \sqrt[3]{3},$ and c = $\sqrt[4]{4}$

a = $(2)^{\frac{1}{2}}$, b = $(3)^{\frac{1}{3}}$ and c = $(4)^{\frac{1}{4}}$

Taking log both sides   $\log_{}{a}$ = $\log_{}{(2)^{1/2}}$ , $\log_{}{b} = \log_{}{(3)^{1/3}}$ , $\log_{}{c} = \log_{}{(4)^{1/4}}$

we know that ($\log_{}{2}$ = 0.3010 , $\log_{}{3}$ = 0.4771 , $\log_{}{4}$ = $2\log_{}{2}$ =0.6020)

Substituting these values

$\log_{}{a}$ = $\frac{1}{2}$*0.3010 , $\log_{}{b}$ =$\frac{1}{3}$*0.4771 , $\log_{}{c}$= $\frac{1}{4}$*0.6020

$\log_{}{a}$ = 0.1505 , $\log_{}{b}$ = 0.1590 , $\log_{}{c}$= 0.1505

We know that if $A_{1}$ > $A_{2}$ > $A_{3}$ > 1   Then  $log_{}{A_{1}}$ > $log_{}{A_{2}}$ > $log_{}{A_{3}}$

Here clearly  $log_{}{b}$ > $log_{}{a}$ = $log_{}{c}$  hence we can say that b = $\sqrt[3]{3}$ is the highest number among all.

Alternate method:

a = $\sqrt{2}, b = \sqrt[3]{3},$ and c = $\sqrt[4]{4}$

$a^{12} = 2^6, b^{12} = 3^4, c^{12} = 4^3$

$a^{12} = 64, b^{12} = 81, c^{12} = 81$

We can see that $b^{12}$ > $c^{12}$ = $a^{12}$

Also, a, b, c > 1. Hence, we can say that b > a = c.

***  Point to remember —— $\sqrt[N]{N}$ is highest for N=3 (N being natural number) ***

Simplifying the surds, and writing everything on numerator we get:

= ${(2^{2p + 1/2 + 1/2 + p/2 – 1 + p/2}})^{1/p}$

= ${(2^{3p}})^{1/p}$

= $2^{3}$ = 8

This question can be solved with the help of options easily.
We can say that 4x – 9 $\geq$ 0. Hence x $\geq$ 2.25. Now we can check option C and D.

Option C: $\sqrt{4x – 9}$ + $\sqrt{4x + 9}$ = $\sqrt{3}$ + $\sqrt{21}$ Which is not same as what we have in the question. Hence, this is not the correct answer.

Option D: $\sqrt{4x – 9}$ + $\sqrt{4x + 9}$ = $\sqrt{7}$ + $\sqrt{25}$ = 5 + $\sqrt{7}$. Which is the same as what we have in the question. Hence, we can say that option D the correct answer.

$\text{log}_{7} \text{log}_{7} \sqrt{7(\sqrt{7\sqrt{7}})}$
= $\text{log}_{7} [\frac{1}{2} (\text{log}_{7} 7+\text{log}_{7} \sqrt{7(\sqrt{7})})]$
= $\text{log}_{7} [\frac{1}{2} (1 +\frac{1}{2}\text{log}_{7}{7(\sqrt{7})})]$
= $\text{log}_{7} [\frac{1}{2} (1 +\frac{1}{2}(1+1/2))]$
= $\text{log}_{7}\frac{7}{8}$
= $\text{log}_{7}7 – \text{log}_{7}8$
= $1-3 \text{log}_{7}2$
Hence, option D is the correct answer.

$log_{13} log_{21} (\sqrt{x+21}+ \sqrt{x} ) =0$
Thus, $log_{21} (\sqrt{x+21}+ \sqrt{x} ) = 1$
Thus, $(\sqrt{x+21}+ \sqrt{x} ) = 21$
Let, $\sqrt{x} = t$
Thus, $x = t^2$
Thus, $x+21 = t^2+21$
Thus, $\sqrt{t^2+21}+t = 21$
Thus, $(t^2+21) = (21-t)^2$
=> $t^2 + 21 = 441 – 42t + t^2$
=> $42t = 420$
Hence, $t = 10$
Hence, option D is the correct answer.

We hope this Square Roots Questions PDF for CAT with Solutions will be helpful to you.