The Surds and Indices is an important topic in the Quant section of the SNAP Exam. Quant is a scoring section in SNAP, so it is advised to practice as much as questions from quant. This article provides some of the most important Surds and Indices Questions for SNAP. One can also download this Free Surds and Indices Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Surds and Indices questions in the SNAP exam. Utilize this PDF practice set, <\/strong>which is one of the best sources for practising.<\/p>\n Download Surds and Indices Questions for SNAP<\/a><\/p>\n Enroll to SNAP 2022 Crash Course<\/a><\/p>\n Question 1:\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 1)\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 Question 2:\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 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n x =\u00a0$\\sqrt{552+\\sqrt{552+}\\sqrt{552+…}}$ Question 3:\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 3)\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 4:\u00a0<\/b>The value is? a)\u00a05<\/p>\n b)\u00a025<\/p>\n c)\u00a050<\/p>\n d)\u00a010<\/p>\n 4)\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 5:\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 5)\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 Take\u00a0 SNAP mock tests here<\/strong><\/span><\/a><\/p>\n Enrol to 10 SNAP Latest Mocks For Just Rs. 499<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>If $(5.55)^x = (0.555)^y = 1000$, then the value of $\\frac{1}{x} – \\frac{1}{y}$ is<\/p>\n a)\u00a0$\\frac{1}{3}$<\/p>\n b)\u00a03<\/p>\n c)\u00a01<\/p>\n d)\u00a0$\\frac{2}{3}$<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have,\u00a0$(5.55)^x = (0.555)^y = 1000$<\/p>\n Taking log in base 10 on both sides,<\/p>\n x($\\log_{10}555$-2) = y($\\log_{10}555$-3) = 3<\/p>\n Then,\u00a0x($\\log_{10}555$-2) = 3…..(1)<\/p>\n y($\\log_{10}555$-3) = 3\u00a0…..(2)<\/p>\n From (1) and (2)<\/p>\n => $\\log_{10}555$=$\\ \\frac{\\ 3}{x}$+2=$\\ \\frac{\\ 3}{y}+3$<\/p>\n =>\u00a0$\\frac{1}{x} – \\frac{1}{y}$ =\u00a0$\\frac{1}{3}$<\/p>\n Question 7:\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 7)\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 Question 8:\u00a0<\/b>$\\left\\{\\frac{2^{\\frac{1}{2}} \\times 3^{\\frac{1}{3}} \\times 4^{\\frac{1}{4}}}{10^{\\frac{-1}{5}} \\times 5^{\\frac{3}{5}}} \\div \\frac{3^{\\frac{4}{3}} \\times 5^{\\frac{-7}{5}}}{4^{\\frac{-3}{5}} \\times 6}\\right\\} \\times 2 =$<\/p>\n a)\u00a010<\/p>\n b)\u00a020<\/p>\n c)\u00a030<\/p>\n d)\u00a040<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n we have : Question 9:\u00a0<\/b>If $x=8-\\sqrt{32}$ and $y=2+\\sqrt{2}$, then $\\left(x+\\frac{1}{y}\\right)^2$ is given by:<\/p>\n a)\u00a0$\\frac{16}{25}x^2$<\/p>\n b)\u00a0$\\frac{64}{81}y^2$<\/p>\n c)\u00a0$\\frac{25}{16}y^2$<\/p>\n d)\u00a0$\\frac{81}{64}x^2$<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $x=8-\\sqrt{32}$ and $y=2+\\sqrt{2}$,<\/p>\n We have to find the value of\u00a0$\\left(x+\\frac{1}{y}\\right)^2$<\/p>\n $\\left(8-\\sqrt{\\ 32}+\\ \\frac{\\ 1}{2+\\sqrt{\\ 2}}\\right)^2$<\/p>\n $(\\ \\frac{\\ 8\\left(2+\\sqrt{\\ 2}\\right)-\\sqrt{\\ 32}\\left(2+\\sqrt{\\ 2}\\right)+\\ \\ 1}{2+\\sqrt{\\ 2}})^2$<\/p>\n $\\left(\\ \\frac{\\ 9}{2+\\sqrt{\\ 2}}\\right)^2$<\/p>\n $\\ \\frac{\\ 81}{6+2\\sqrt{\\ 2}}$<\/p>\n $\\left(\\ \\frac{\\ 1}{y}\\right)^2=\\ \\left(\\frac{\\ 1}{2+\\sqrt{\\ 2}}\\right)^2$ =\u00a0$6+2\\sqrt{\\ 2}$<\/p>\n =$\\left(\\ \\frac{\\ 2-\\sqrt{\\ 2}}{2}\\right)^{^2}$<\/p>\n =$\\ \\frac{\\ 6-4\\sqrt{2\\ }}{4}$<\/p>\n =$\\ \\frac{\\ 3-2\\sqrt{2\\ }}{2}$<\/p>\n $x^2=64+32-64\\sqrt{\\ 2}$<\/p>\n =$96-64\\sqrt{\\ 2}$<\/p>\n =32($3-2\\sqrt{\\ 2}$) = 32*$2y^2$<\/p>\n we get,\u00a0$x^{2\\ }=\\ 64y^2$<\/p>\n $\\ \\frac{\\ 81}{y^2}=\\ \\frac{\\ 81}{64}x^2$<\/p>\n D is the correct answer.<\/p>\n Alternative solution,<\/p>\n $xy=\\left(8-\\sqrt{\\ 32}\\right)\\left(2+\\sqrt{\\ 2}\\right)=4\\sqrt{\\ 2}\\left(\\sqrt{\\ 2}-1\\right)\\times\\ \\sqrt{\\ 2}\\left(\\sqrt{\\ 2}+1\\right)=8\\left(2-1\\right)=8$<\/p>\n (As $xy=8\\ \\longrightarrow\\ y=\\dfrac{x}{8}$)<\/p>\n $\\left(x+\\frac{1}{y}\\right)^2=\\left(\\frac{\\left(xy+1\\right)}{y}\\right)^2=\\left(\\frac{\\left(xy+1\\right)\\times\\ x}{8}\\right)^2=\\left(\\frac{\\left(8+1\\right)\\times\\ x}{8}\\right)^2=\\frac{81}{64}\\times\\ x^2$<\/p>\n Question 10:\u00a0<\/b>If $2^x + 2^{x + 1} = 48$, then the value of $x^x$ is<\/p>\n a)\u00a04<\/p>\n b)\u00a064<\/p>\n c)\u00a0256<\/p>\n d)\u00a016<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $2^x\\ +\\ 2^x.2\\ =48$<\/p>\n $2^x\\ \\left(1+2\\right)\\ =48$<\/p>\n $2^x\\ =16$<\/p>\n x=4<\/p>\n $4^4=256$<\/p>\n Question 11:\u00a0<\/b>Find the value of $\\sqrt{\\frac{2 + \\sqrt3}{2 – \\sqrt3}}$ a)\u00a03.141<\/p>\n b)\u00a02.732<\/p>\n c)\u00a03.124<\/p>\n d)\u00a03.732<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let us consider $\\frac{2 + \\sqrt3}{2 – \\sqrt3}$<\/p>\n Rationalising the denominator by multiplying and diving with 2+$\\sqrt3$ we get,<\/p>\n $\\frac{(2 + \\sqrt3)\\times (2 + \\sqrt3) }{(2 – \\sqrt3)\\times (2 + \\sqrt3) } = \\frac {(2 + \\sqrt3)^2}{4 – 3} = (2 + \\sqrt3)^2$<\/p>\n Now,<\/p>\n $\\sqrt{\\frac{2 + \\sqrt3}{2 – \\sqrt3}} = \\sqrt{(2 + \\sqrt3)^2} = 2 + \\sqrt3 = 2 + 1.732 = 3.732$<\/p>\n Question 12:\u00a0<\/b>If a and b are positive real numbers and $a * b$ denotes $\\sqrt{a \\times b}$, what is the value of 8 * (4 * 16)?<\/p>\n a)\u00a0$4^{1\/3}$<\/p>\n b)\u00a016<\/p>\n c)\u00a08<\/p>\n d)\u00a0$4\\sqrt2$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given, If a and b are positive real numbers then a * b denotes \u221aab<\/p>\n Consider 4 * 16 = $\\sqrt{4\\times 16} = \\sqrt64 = 8$<\/p>\n then 8 * 8 = $\\sqrt{8\\times 8} = \\sqrt64 = 8$<\/p>\n Hence the value of\u00a0 8 * (4 * 16) = 8<\/p>\n Question 13:\u00a0<\/b>If x = $\\sqrt[6]{5}$\u00a0 and y = $\\sqrt[5]{4}$, Which of the following is true?<\/p>\n a)\u00a0x > y<\/p>\n b)\u00a0y > x<\/p>\n c)\u00a0x = y<\/p>\n d)\u00a0None<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given \u00a0x = $\\sqrt[6]{5}$ and y = $\\sqrt[5]{4}$<\/p>\n which can also be written as x = $5^{\\frac{5}{30}}$ and y = $4^{\\frac{6}{30}}$<\/p>\n which can be further written as x = $\\sqrt[30]{5^5}$ and y = $\\sqrt[30]{4^6}$<\/p>\n As we know $4^6 > 5^5$<\/p>\n $\\Rightarrow\u00a0\\sqrt[30]{4^6}\u00a0 > \\sqrt[30]{5^5} $<\/p>\n $\\Rightarrow$ y > x<\/p>\n Question 14:\u00a0<\/b>If N and x are positive integers such that $N^{N}$ = $2^{160}\\ and \\ N{^2} + 2^{N}\\ $ is an integral multiple of $\\ 2^{x}$, then the largest possible x is<\/p>\n 14)\u00a0Answer:\u00a010<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n It is given that $N^{N}$ = $2^{160}$<\/p>\n We can rewrite the equation as $N^{N}$ = $(2^5)^{160\/5}$ =\u00a0$32^{32}$<\/p>\n $\\Rightarrow$ N = 32<\/p>\n $N{^2} + 2^{N}$ = $32^2+2^{32}=2^{10}+2^{32}=2^{10}*(1+2^{22})$<\/p>\n Hence, we can say that\u00a0$N{^2} + 2^{N}$ can be divided by $2^{10}$<\/p>\n Therefore, x$_{max}$ = 10<\/p>\n Question 15:\u00a0<\/b>Given that $x^{2018}y^{2017}=\\frac{1}{2}$, and $x^{2016}y^{2019}=8$, then value of $x^{2}+y^{3}$ is<\/p>\n a)\u00a0$\\dfrac{31}{4}$<\/p>\n b)\u00a0$\\dfrac{35}{4}$<\/p>\n c)\u00a0$\\dfrac{37}{4}$<\/p>\n d)\u00a0$\\dfrac{33}{4}$<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Given that $x^{2018}y^{2017}=\\frac{1}{2}$\u00a0 … (1)<\/p>\n $x^{2016}y^{2019}=8$ … (2)<\/p>\n Equation (2)\/\u00a0Equation (1)<\/p>\n $\\dfrac{y^2}{x^2} = \\dfrac{8}{1\/2}$<\/p>\n $\\dfrac{y}{x} = 4$ or $-4$<\/p>\n Case 1: When\u00a0$\\dfrac{y}{x} = 4$<\/p>\n $x^{2018}(4x)^{2017}=\\dfrac{1}{2}$<\/p>\n $x^{2018+2017}(2)^{4034}=\\dfrac{1}{2}$<\/p>\n $x^{4035}=\\dfrac{1}{(2)^{4035}}$<\/p>\n $x=\\dfrac{1}{2}$<\/p>\n Since, $\\dfrac{y}{x} = 4$, => y = 2<\/p>\n Therefore,\u00a0$x^{2}+y^{3}$ = $\\dfrac{1}{4}+8$ = $\\dfrac{33}{4}$<\/p>\n Case 2: When\u00a0$\\dfrac{y}{x} = -4$<\/p>\n $x^{2018}(-4x)^{2017}=\\dfrac{1}{2}$<\/p>\n $x^{2018+2017}(2)^{4034}=\\dfrac{-1}{2}$<\/p>\n $x^{4035}=\\dfrac{1}{(-2)^{4035}}$<\/p>\n $x=\\dfrac{-1}{2}$<\/p>\n Since, $\\dfrac{y}{x} = -4$, => y = 2<\/p>\n Therefore,\u00a0$x^{2}+y^{3}$ = $\\dfrac{1}{4}+8$ =\u00a0$\\dfrac{33}{4}$. Hence, option D is the correct answer.<\/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
\n$\\frac{\\left(2^{\\frac{1}{2}}\\times\\ 3^{\\frac{1}{3}}\\times\\ 4^{\\frac{1}{4}}\\right)}{10^{-\\frac{1}{5}}\\times\\ 5^{\\frac{3}{5}}}\\ \\ \\ \\ $
\nNow 4 = 2^2 and 10 = 2*5
\nWe get $\\ \\frac{\\left(2^{\\frac{1}{2}+\\frac{1}{2}}\\times3^{\\frac{1}{3}}\\ \\right)}{2^{-\\frac{1}{5}}\\times\\ 5^{-\\frac{1}{5}}\\times\\ 5^{\\frac{3}{5}}}=\\frac{\\left(2^{\\frac{6}{5}}\\times\\ 3^{\\frac{1}{3}}\\right)}{5^{\\frac{2}{5}}}$ \u00a0 \u00a0 (1)
\nNow the next term we have is :$\\frac{\\left(3^{\\frac{4}{3}}\\times\\ 5^{-\\frac{7}{5}}\\right)}{4^{-\\frac{3}{5}}\\times\\ 6}$
\n6= 2*3 and 4=2^2
\nWe get $\\frac{\\left(3^{\\frac{1}{3}}\\times\\ 5^{-\\frac{7}{5}}\\right)}{2^{-\\frac{1}{5}}}$ \u00a0 \u00a0 (2)
\nDividing (1) and (2) we get
\n$\\frac{\\left(2^{\\frac{6}{5}}\\times\\ 3^{\\frac{1}{3}}\\right)}{5^{\\frac{2}{5}}}\\times\\ \\frac{\\left(2^{-\\frac{1}{5}}\\right)}{3^{\\frac{1}{3}}\\times\\ 5^{-\\frac{7}{5}}}$
\n= $\\frac{2}{5^{-1}}=10$
\nNow we have to multiply by 2
\nso we get 10*2=20<\/p>\n
\nCorrect to three places of decimal.<\/p>\n