Download Surds and Indices Questions for XAT PDF \u2013 XAT Surds and Indices questions PDF by Cracku. Practice XAT solved Surds and Indices Questions paper tests, and these are the practice question to have a firm grasp on the Surds and Indices topic in the XAT exam. Top 20 very Important Surds and Indices Questions for XAT based on asked questions in previous exam papers. \u00a0The XAT question papers contain actual questions asked with answers and solutions.<\/p>\n
Download Surds and Indices Questions for XAT<\/a><\/p>\n Enroll to XAT 2023 Mocks & Sectionals <\/a><\/p>\n Question 1:\u00a0<\/b>If x = 1 + $\\surd{2}$+ $\\surd{3}$, then the value of 2x^{4}- 8x^{3}- 5x^{2} + 26x – 28 is<\/p>\n a)\u00a02$\\surd{2}$<\/p>\n b)\u00a03$\\surd{3}$<\/p>\n c)\u00a05$\\surd{5}$<\/p>\n d)\u00a06\u00a0$\\surd{6}$<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n x = 1 + $\\surd{2}$+ $\\surd{3}$<\/p>\n => $ (x-1)^2 = ( \\surd{2}+ \\surd{3})^{2} $<\/p>\n =>\u00a0$ x^2 + 1\u00a0-2x\u00a0= 5 + 2\\surd{6} $<\/p>\n =>\u00a0$ x^2-2x =4\u00a0+ 2\\surd{6} $ ———– (1)<\/p>\n Squaring on both sides<\/p>\n => $ (x^2-2x)^2 = x^4 + 4x^2 – 4x^3 = 40 + 16\\surd{6}\u00a0 $\u00a0—— (2)<\/p>\n Now,<\/p>\n $2x^{4}- 8x^{3}- 5x^{2} + 26x – 28 = 2(x^{4}-4x^{3})- 5x^{2} + 26x – 28 $ —- (3)<\/p>\n Substituting values in (1) & (2) in equation (3), we get value as $ 6 \\surd {6} $<\/p>\n Question 2:\u00a0<\/b>if x+ $\\frac{1}{x}$=$\\surd{3}$ then the value of $x^{18}+x^{12}+x^{6}+1$<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a03<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given that\u00a0x+ $\\frac{1}{x}$=$\\surd{3}$<\/p>\n Squaring on both sides, we get<\/p>\n $(x+ \\frac{1}{x})^{3}=(\\surd{3})^{3}$<\/p>\n => $x^{3}+\\frac{1}{x^3}+3\\surd{3}=3\\surd{3}$<\/p>\n =>\u00a0 $x^{3}+\\frac{1}{x^3} = 0 $<\/p>\n =>\u00a0 $x^{3}= –\u00a0\\frac{1}{x^3} $<\/p>\n =>\u00a0$x^{6}= -1 $<\/p>\n Squaring on both sides<\/p>\n =>\u00a0$x^{12}= 1 $<\/p>\n $ (x^{6})^{3} = (-1)^{3} = -1 $<\/p>\n Therefore,<\/p>\n $x^{18}+x^{12}+x^{6}+1$ = $ -1 + 1 -1 + 1 = 0 $<\/p>\n Question 3:\u00a0<\/b>Which of the following statement(s) is\/are TRUE? a)\u00a0only I<\/p>\n b)\u00a0only II<\/p>\n c)\u00a0Niether I nor II<\/p>\n d)\u00a0Both I and II<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\sqrt{1}+\\sqrt{2}+\\sqrt{3}+\\sqrt{4}+\\sqrt{\\ 5}+\\sqrt{6}=10.83.$<\/p>\n it means that (I) is correct.<\/p>\n And,\u00a0$\\sqrt{10}+\\sqrt{12}+\\sqrt{14}=10.36.$<\/p>\n but,\u00a0$3\\sqrt{12}=10.39.$<\/p>\n So, (II) is not correct .<\/p>\n So, A is correct choice.<\/p>\n Question 4:\u00a0<\/b>Which of the following statement(s) is\/are TRUE<\/strong>? a)\u00a0Only I<\/p>\n b)\u00a0Only II<\/p>\n c)\u00a0Both I and II<\/p>\n d)\u00a0Neither I nor II<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\surd(64)+\\surd(0.0064)+\\surd(0.81)+\\surd(0.0081)=8+0.08+0.9+0.09=9.07\\ .$<\/p>\n So, I is correct .<\/p>\n $\\surd(0.010201)+\\surd(98.01)+\\surd(0.25)=10.501\\ .$<\/p>\n II is not correct .<\/p>\n A is correct choice.<\/p>\n Question 5:\u00a0<\/b>If $(\\sqrt{\\frac{7}{5}})^{3x-y}=\\frac{875}{2401}$ and $(\\frac{4a}{b})^{6x-y}=(\\frac{2a}{b})^{y-6x}$, for all non-zero real values of a and b, then the value of $x+y$ is<\/p>\n 5)\u00a0Answer:\u00a014<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $(\\sqrt{\\frac{7}{5}})^{3x-y}=\\frac{875}{2401}$<\/p>\n $\\left(\\frac{7}{5}\\right)^{\\frac{\\left(3x-y\\right)}{2}}=\\frac{125}{343}$<\/p>\n $\\left(\\frac{7}{5}\\right)^{\\frac{\\left(3x-y\\right)}{2}}=\\left(\\frac{7}{5}\\right)^{-3}$<\/p>\n 3x-y = -6<\/p>\n $(\\frac{4a}{b})^{6x-y}=(\\frac{2a}{b})^{y-6x}$<\/p>\n Therefor, y=6x as the bases are different so the power should be zero for the results to be equal.<\/p>\n 3x-y=-6<\/p>\n or, 3x – 6x = -6<\/p>\n or x= 2<\/p>\n y= 6x = 12<\/p>\n x+y = 14<\/p>\n Take XAT 2023 Mock Tests<\/a><\/p>\n Question 6:\u00a0<\/b>Find the value of $\\sqrt{552+\\sqrt{552+}\\sqrt{552+…}}$<\/p>\n a)\u00a026<\/p>\n b)\u00a0-24<\/p>\n c)\u00a024<\/p>\n d)\u00a0-26<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n x =\u00a0$\\sqrt{552+\\sqrt{552+}\\sqrt{552+…}}$ Question 7:\u00a0<\/b>If $x=(4096)^{7+4\\sqrt{3}}$, then which of the following equals to 64?<\/p>\n a)\u00a0$\\frac{x^{7}}{x^{2\\sqrt{3}}}$<\/p>\n b)\u00a0$\\frac{x^{\\frac{7}{2}}}{x^{\\frac{4}{\\sqrt{3}}}}$<\/p>\n c)\u00a0$\\frac{x^{\\frac{7}{2}}}{x^{2{\\sqrt{3}}}}$<\/p>\n d)\u00a0$\\frac{x^{7}}{x^{4\\sqrt{3}}}$<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n $x=2^{12\\left(7+4\\sqrt{\\ 3}\\right)}$.<\/p>\n $x^{\\frac{7}{2}}=2^{42\\left(7+4\\sqrt{\\ 3}\\right)}$<\/p>\n $x^{2\\sqrt{\\ 3}}=2^{24\\sqrt{\\ 3}\\left(7+4\\sqrt{\\ 3}\\right)}$<\/p>\n $\\frac{x^{\\frac{7}{2}}}{x^{2{\\sqrt{3}}}}$ =\u00a0$2^{\\left(7+4\\sqrt{\\ 3}\\right)\\left(42-24\\sqrt{\\ 3}\\right)}=2^{\\left(7+4\\sqrt{\\ 3}\\right)\\left(7-4\\sqrt{\\ 3}\\right)6}$ =$2^6$.<\/p>\n Hence C is correct answer.<\/p>\n Question 8:\u00a0<\/b>The value is? a)\u00a05<\/p>\n b)\u00a025<\/p>\n c)\u00a050<\/p>\n d)\u00a010<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $5^{\\frac{1}{4}}$ *\u00a0$125^{0.25}$<\/p>\n =\u00a0$5^{0.25}$ *\u00a0$5^{3*0.25}$<\/p>\n = $5^{0.25+0.75}$<\/p>\n =$5^1$<\/p>\n =5<\/p>\n Question 9:\u00a0<\/b>If m and n are integers such that $(\\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\\sqrt[4]{64})$ then m is<\/p>\n a)\u00a0-20<\/p>\n b)\u00a0-24<\/p>\n c)\u00a0-12<\/p>\n d)\u00a0-16<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have,\u00a0$(\\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\\sqrt[4]{64})$<\/p>\n Converting both sides in powers of 2 and 3, we get<\/p>\n $2^{\\ \\frac{19\\ }{2}}3^42^43^{2m}2^{3n}$ =\u00a0$3^n2^{4m}2^{\\frac{\\ 6}{4}}$<\/p>\n Comparing the power of 2 we get,\u00a0$\\ \\frac{\\ 19}{2}+4+3n\\ =4m+\\frac{\\ 6}{4}\\ $<\/p>\n => 4m=3n+12 …..(1)<\/p>\n Comparing the power of 3 we get,\u00a0$4+2m=n$<\/p>\n Substituting the value of n in (1), we get<\/p>\n 4m=3(4+2m)+12<\/p>\n => m=-12<\/p>\n Question 10:\u00a0<\/b>If $R_c = mln \\left(1 + \\frac{R_m}{m}\\right)$ then $R_m$ is equal to<\/p>\n a)\u00a0$R_m = ln\\left(1 + \\frac{R_c}{m}\\right)$<\/p>\n b)\u00a0$R_m = ln\\left(1 + \\frac{R_c}{e}\\right)$<\/p>\n c)\u00a0$R_m = m\\left(e^{\\frac{R_c}{m}} – 1\u00a0\\right)$<\/p>\n d)\u00a0Cannot be determined<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $R_c = mln \\left(1 + \\frac{R_m}{m}\\right)$<\/p>\n $\\ \\frac{\\ R_c}{m}$ = $ln \\left(1 + \\frac{R_m}{m}\\right)$<\/p>\n $\\ \\frac{\\ R_m}{m}+\\ 1\\ =\\ e^{\\ \\frac{\\ R_c}{m}}$<\/p>\n $\\ \\frac{\\ R_m}{m}\\ =\\ e^{\\ \\frac{\\ R_c}{m}-1}$<\/p>\n $R_m = m\\left(e^{\\frac{R_c}{m}} – 1 \\right)$<\/p>\n C is the correct answer.<\/p>\n
\nI. $\\surd1+\\surd2+\\surd3+\\surd4+\\surd5+\\surd6>10$
\nII. $\\surd(10)+\\surd(12)+\\surd(14)>3\\surd(12)$<\/p>\n
\nI. $\\surd(64) + \\surd(0.0064) + \\surd(0.81) + \\surd(0.0081) = 9.07$
\nII. $\\surd(0.010201) + \\surd(98.01) + \\surd(0.25) = 11.51$<\/p>\n
\n$x^2-552$ =\u00a0$\\sqrt{552+\\sqrt{552+}\\sqrt{552+…}}$
\n$x^2 – x –\u00a0552 = 0$
\n(x-24)(x+23) = 0
\nx cannot be negative. Therefore,\u00a0$\\sqrt{552+\\sqrt{552+}\\sqrt{552+…}}$ = 24
\nAnswer is option C.<\/p>\n
\n$5^{\\frac{1}{4}} \\times (125)^{0.25}$<\/p>\n