{"id":37175,"date":"2019-11-07T18:17:47","date_gmt":"2019-11-07T12:47:47","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=37175"},"modified":"2019-11-07T18:17:47","modified_gmt":"2019-11-07T12:47:47","slug":"cat-questions-on-square-roots","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/cat-questions-on-square-roots\/","title":{"rendered":"CAT Questions on Square Roots"},"content":{"rendered":"

CAT Questions on Square Roots<\/strong><\/span><\/h2>\n

Download important Square Roots Questions for CAT PDF based on previously asked questions in CAT exam. Practice Square Roots Questions PDF for CAT exam.<\/p>\n

Download CAT Questions on Square Roots<\/a><\/p>\n

CAT Crash Couse – Sufficient To Crack The Exam<\/a><\/p>\n

Download CAT Quant Questions PDF<\/a><\/p>\n

Take 3 Free Mock Tests for CAT<\/a><\/p>\n

Question 1:\u00a0<\/b>Let $x = \\sqrt{4+\\sqrt{4-\\sqrt{4+\\sqrt{4- \\ to \\ infinity}}}}$. Then x equals<\/p>\n

a)\u00a03<\/p>\n

b)\u00a0$(\\sqrt{13} – 1)\/2$<\/p>\n

c)\u00a0$(\\sqrt{13} + 1)\/2$<\/p>\n

d)\u00a0$\\sqrt{13}$<\/p>\n

Question 2:\u00a0<\/b>$(\\sqrt{\\frac{225}{729}}-\\sqrt{\\frac{25}{144}})\\div\\sqrt{\\frac{16}{81}}=?$<\/p>\n

a)\u00a0$\\frac{5}{16}$<\/p>\n

b)\u00a0$\\frac{7}{12}$<\/p>\n

c)\u00a0$\\frac{3}{8}$<\/p>\n

d)\u00a0None of these<\/p>\n

Question 3:\u00a0<\/b>What is the value of $\\sqrt{\\frac{a}{b}}$, If $\\log_{4}\\log_{4}4^{a-b}=2\\log_{4}(\\sqrt{a}-\\sqrt{b})+1$<\/p>\n

a)\u00a0-5\/3<\/p>\n

b)\u00a02<\/p>\n

c)\u00a05\/3<\/p>\n

d)\u00a01<\/p>\n

Question 4:\u00a0<\/b>$2-\\frac{\\sqrt{6407522209}}{\\sqrt{3600840049}}=$<\/p>\n

a)\u00a00.666039<\/p>\n

b)\u00a00.666029<\/p>\n

c)\u00a00.666009<\/p>\n

d)\u00a0None of the above<\/p>\n

Question 5:\u00a0<\/b>The value of $\\log_{0.008}\\sqrt{5}+\\log_{\\sqrt{3}}81-7$ is equal to<\/p>\n

a)\u00a01\/3<\/p>\n

b)\u00a02\/3<\/p>\n

c)\u00a05\/6<\/p>\n

d)\u00a07\/6<\/p>\n

Take 3 free mock tests for CAT<\/a><\/p>\n

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Question 6:\u00a0<\/b>The highest number amongst $\\sqrt{2}, \\sqrt[3]{3},$ and $\\sqrt[4]{4}$ is<\/p>\n

a)\u00a0$\\sqrt{2}$<\/p>\n

b)\u00a0\u00a0$\\sqrt[3]{3}$<\/p>\n

c)\u00a0$\\sqrt[4]{4}$<\/p>\n

d)\u00a0All are equal<\/p>\n

Question 7:\u00a0<\/b>The simplest value of the expression\u00a0$(\\frac{4^{p+\\frac{1}{4}}\\times \\sqrt{2 \\times 2^{p}}}{2\\times \\sqrt{2^-p}})^\\frac{1}{p}$<\/p>\n

a)\u00a04<\/p>\n

b)\u00a08<\/p>\n

c)\u00a04p<\/p>\n

d)\u00a08p<\/p>\n

Question 8:\u00a0<\/b>The value of x for which the equation $\\sqrt{4x – 9}$ + $\\sqrt{4x + 9}$ = 5 + $\\sqrt{7}$ will be satisfied, is:<\/p>\n

a)\u00a01<\/p>\n

b)\u00a02<\/p>\n

c)\u00a03<\/p>\n

d)\u00a04<\/p>\n

Question 9:\u00a0<\/b>The value of $\\text{log}_{7} \\text{log}_{7} \\sqrt{7(\\sqrt{7\\sqrt{7}})}$<\/p>\n

a)\u00a07<\/p>\n

b)\u00a0$\\text{log}_7$ 2<\/p>\n

c)\u00a0$1-3 \\text{log}_2$ 7<\/p>\n

d)\u00a0$1-3 \\text{log}_7$ 2<\/p>\n

Question 10:\u00a0<\/b>$log_{13} log_{21} (\\sqrt{x+21}+ \\sqrt{x} ) =0 $ then the value of x is<\/p>\n

a)\u00a021<\/p>\n

b)\u00a013<\/p>\n

c)\u00a081<\/p>\n

d)\u00a0None of the above<\/p>\n

Download CAT Quant Questions PDF<\/a><\/p>\n

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Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$x = \\sqrt{4+\\sqrt{4-\\sqrt{4+\\sqrt{4- \\ to \\ infinity}}}}$<\/p>\n

=> $x = \\sqrt{4+\\sqrt{4-x}}$<\/p>\n

=> $x^2 = 4 + \\sqrt{4-x}$<\/p>\n

=>$x^4 + 16 – 8x^2 = 4 – x$<\/p>\n

=> $x^4 – 8x^2 + x +12 = 0$<\/p>\n

On substituting options, we can see that option C satisfies the equation.<\/p>\n

2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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

$(\\frac{15}{27}-\\frac{5}{12})\\div\\frac{4}{9}=x$<\/p>\n

$(\\frac{5}{36})*\\frac{9}{4}=x$<\/p>\n

x=$\\frac{5}{16}$. Hence, option A is the correct answer.<\/p>\n

3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$\\sqrt{\\frac{a}{b}}$, If $\\log_{4}\\log_{4}4^{a-b}=2\\log_{4}(\\sqrt{a}-\\sqrt{b})+\\log_{4}{4}$<\/p>\n

i.e. $\\log_{4}\\log_{4}4^{a-b}=\\log_{4}((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n

i.e. $\\log_{4}4^{a-b}=((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n

i.e. (a-b)*$\\log_{4}4=((\\sqrt{a}-\\sqrt{b})^2)*4$<\/p>\n

i.e. a-b = 4a+4b-8$\\sqrt{ab}$<\/p>\n

i.e. 3a + 5b – 8$\\sqrt{ab}$ = 0<\/p>\n

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

put\u00a0$\\sqrt\\frac{a}{b}$ = t<\/p>\n

therefore 3$t^2$ – 8t + 5 = 0<\/p>\n

solving we get t = 1 or t = $\\frac{5}{3}$<\/p>\n

i.e.\u00a0$\\sqrt\\frac{a}{b}$ = 1 or\u00a0$\\frac{5}{3}$<\/p>\n

but if\u00a0$\\sqrt\\frac{a}{b}$ = 1 then a=b then $\\log_{4}(\\sqrt{a}-\\sqrt{b})$ will become indefinite<\/p>\n

Therefore\u00a0\u00a0$\\sqrt\\frac{a}{b}$ =\u00a0$\\frac{5}{3}$<\/p>\n

Therefore our answer is option ‘C’<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$2-\\frac{\\sqrt{6407522209}}{\\sqrt{3600840049}}=2-\\frac{80047}{60007}$
\n=$2-1.3339610$
\n$=0.666039$
\nTherefore, option A is the right answer.<\/p>\n

5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

$\\log_{0.008}\\sqrt{5}+\\log_{\\sqrt{3}}81-7$<\/p>\n

$81 = 3^4$ and $0.008 = \\frac{8}{1000} = \\frac{2^{3}}{10^{3}} = \\frac{1}{5^{3}} = 5^{-3} $<\/p>\n

Hence,<\/p>\n

$\\log_{0.008}\\sqrt{5}+ 8 -7 $<\/p>\n

$ \\log_{5^{-3}}5^{\\frac{1}{2}}+ 8 -7 $<\/p>\n

$\\frac{log 5^{0.5}}{log 5^{-3}} + 1$<\/p>\n

$ – \\frac{1}{6} + 1$<\/p>\n

= $\\frac{5}{6}$<\/p>\n

Free CAT Practice – Study Material\u00a0<\/a><\/p>\n

Download CAT Quant Formulas PDF<\/a><\/p>\n

6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Given that<\/p>\n

a = $\\sqrt{2}, b = \\sqrt[3]{3},$ and c = $\\sqrt[4]{4}$<\/p>\n

a = $(2)^{\\frac{1}{2}}$, b = $(3)^{\\frac{1}{3}}$ and c = $(4)^{\\frac{1}{4}}$<\/p>\n

Taking log both sides\u00a0 \u00a0$\\log_{}{a}$ = $\\log_{}{(2)^{1\/2}}$ , $\\log_{}{b} = \\log_{}{(3)^{1\/3}}$ , $\\log_{}{c} = \\log_{}{(4)^{1\/4}}$<\/p>\n

we know that ($\\log_{}{2}$ =\u00a00.3010 , $\\log_{}{3}$ = 0.4771 , $\\log_{}{4}$ = $2\\log_{}{2}$ =0.6020)<\/p>\n

Substituting these values<\/p>\n

$\\log_{}{a}$ = $\\frac{1}{2}$*0.3010 , $\\log_{}{b}$ =$\\frac{1}{3}$*0.4771 , $\\log_{}{c}$= $\\frac{1}{4}$*0.6020<\/p>\n

$\\log_{}{a}$ = 0.1505 , $\\log_{}{b}$ = 0.1590\u00a0, $\\log_{}{c}$= 0.1505<\/p>\n

We know that if $A_{1}$ > $A_{2}$ > $A_{3}$ > 1\u00a0 \u00a0Then\u00a0 $log_{}{A_{1}}$ > $log_{}{A_{2}}$ > $log_{}{A_{3}}$<\/p>\n

Here clearly\u00a0\u00a0$log_{}{b}$ > $log_{}{a}$ = $log_{}{c}$\u00a0 hence we can say that b = $\\sqrt[3]{3}$ is the highest number among all.<\/p>\n

 <\/p>\n

Alternate method:<\/p>\n

a = $\\sqrt{2}, b = \\sqrt[3]{3},$ and c = $\\sqrt[4]{4}$<\/p>\n

$a^{12} = 2^6, b^{12} = 3^4, c^{12} = 4^3$<\/p>\n

$a^{12} = 64, b^{12} = 81, c^{12} = 81$<\/p>\n

We can see that $b^{12}$ > $c^{12}$ = $a^{12}$<\/p>\n

Also, a, b, c > 1. Hence, we can say that b > a = c.<\/p>\n

***\u00a0 Point to remember —— $\\sqrt[N]{N}$ is highest for N=3 (N being natural number) ***<\/p>\n

7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Simplifying the surds, and writing everything on numerator we get:<\/p>\n

= ${(2^{2p + 1\/2 + 1\/2 + p\/2 – 1 + p\/2}})^{1\/p}$<\/p>\n

= ${(2^{3p}})^{1\/p}$<\/p>\n

= $2^{3}$ = 8<\/p>\n

8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

This question can be solved with the help of options easily.
\nWe can say that\u00a04x – 9 $\\geq$ 0. Hence x $\\geq$ 2.25. Now we can check option C and D.<\/p>\n

Option C:\u00a0$\\sqrt{4x – 9}$ + $\\sqrt{4x + 9}$ =\u00a0$\\sqrt{3}$ + $\\sqrt{21}$ Which is not same as what we have in the question. Hence, this is not the correct answer.<\/p>\n

Option D:\u00a0$\\sqrt{4x – 9}$ + $\\sqrt{4x + 9}$ =\u00a0$\\sqrt{7}$ + $\\sqrt{25}$ = 5 + $\\sqrt{7}$. Which is the same as what we have in the question. Hence, we can say that option D the correct answer.<\/p>\n

9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$\\text{log}_{7} \\text{log}_{7} \\sqrt{7(\\sqrt{7\\sqrt{7}})}$
\n=\u00a0$\\text{log}_{7} [\\frac{1}{2} (\\text{log}_{7} 7+\\text{log}_{7} \\sqrt{7(\\sqrt{7})})]$
\n= $\\text{log}_{7} [\\frac{1}{2} (1 +\\frac{1}{2}\\text{log}_{7}{7(\\sqrt{7})})]$
\n= $\\text{log}_{7} [\\frac{1}{2} (1 +\\frac{1}{2}(1+1\/2))]$
\n= $\\text{log}_{7}\\frac{7}{8}$
\n= $\\text{log}_{7}7 – \\text{log}_{7}8$
\n= $1-3 \\text{log}_{7}2$
\nHence, option D is the correct answer.<\/p>\n

10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$log_{13} log_{21} (\\sqrt{x+21}+ \\sqrt{x} ) =0 $
\nThus, $log_{21} (\\sqrt{x+21}+ \\sqrt{x} ) = 1 $
\nThus, $(\\sqrt{x+21}+ \\sqrt{x} ) = 21 $
\nLet, $\\sqrt{x} = t$
\nThus, $x = t^2$
\nThus, $x+21 = t^2+21$
\nThus, $\\sqrt{t^2+21}+t = 21$
\nThus, $(t^2+21) = (21-t)^2$
\n=> $t^2 + 21 = 441 – 42t + t^2$
\n=> $42t = 420$
\nHence, $t = 10$
\nHence, option D is the correct answer.<\/p>\n

Download CAT Previous Papers PDF<\/a><\/p>\n

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We hope this Square Roots Questions PDF for CAT with Solutions will be helpful to you.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. Download CAT Quant Questions PDF Take 3 Free Mock Tests for CAT Question 1:\u00a0Let $x = \\sqrt{4+\\sqrt{4-\\sqrt{4+\\sqrt{4- \\ to \\ infinity}}}}$. Then x equals a)\u00a03 b)\u00a0$(\\sqrt{13} […]<\/p>\n","protected":false},"author":42,"featured_media":37177,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3,169,125,350,362,366],"tags":[1713,2961],"class_list":{"0":"post-37175","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"category-downloads","9":"category-featured","10":"category-iift","11":"category-snap","12":"category-xat","13":"tag-cat-2019","14":"tag-square-roots-questions-for-cat"},"better_featured_image":{"id":37177,"alt_text":"CAT Questions on Square Roots","caption":"CAT Questions on Square Roots\n","description":"CAT Questions on Square 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