SSC CHSL Algebra Practice Questions download PDF based on previous year question paper of SSC exams. 20 Very important Algebra Practice questions for SSC CHSL Exam.<\/p>\n
Download Algebra Practice Questions For SSC CHSL PDF<\/a><\/p>\n\n Take a free mock test for SSC CHSL<\/a><\/p>\n Download SSC CHSL Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Find the value of $1+\\Large\\frac{1}{1+\\frac{1}{1+\\frac{1}{2}}}$<\/p>\n a)\u00a0$\\large\\frac{6}{5}$<\/p>\n b)\u00a0$\\large\\frac{8}{5}$<\/p>\n c)\u00a0$\\large\\frac{8}{7}$<\/p>\n d)\u00a0$\\frac{7}{6}$<\/p>\n Question 2:\u00a0<\/b>Find the value of $1+\\Large\\frac{1}{1+\\frac{1}{1+\\frac{1}{6}}}$<\/p>\n a)\u00a0$\\large\\frac{17}{5}$<\/p>\n b)\u00a0$\\large\\frac{19}{6}$<\/p>\n c)\u00a0$\\large\\frac{20}{13}$<\/p>\n d)\u00a0$\\frac{17}{13}$<\/p>\n Question 3:\u00a0<\/b>Find the value of $1+\\Large\\frac{1}{1+\\frac{1}{1+\\frac{3}{2}}}$<\/p>\n a)\u00a0$\\large\\frac{13}{5}$<\/p>\n b)\u00a0$\\large\\frac{17}{6}$<\/p>\n c)\u00a0$\\large\\frac{17}{5}$<\/p>\n d)\u00a0$\\frac{12}{7}$<\/p>\n Question 4:\u00a0<\/b>Find the value of $\\sqrt{3\\sqrt{3\\sqrt{3……}}}$<\/p>\n a)\u00a09<\/p>\n b)\u00a03<\/p>\n c)\u00a027<\/p>\n d)\u00a01.2<\/p>\n Question 5:\u00a0<\/b>Find the value of $\\sqrt{4\\sqrt{4\\sqrt{4……}}}$<\/p>\n a)\u00a04<\/p>\n b)\u00a02<\/p>\n c)\u00a016<\/p>\n d)\u00a08<\/p>\n SSC CHSL PREVIOUS PAPERS<\/a><\/p>\n SSC CHSL Study Material<\/a> (FREE Tests)<\/p>\n Question 6:\u00a0<\/b>Find the value of $\\sqrt{7\\sqrt{7\\sqrt{7……}}}$<\/p>\n a)\u00a0$\\sqrt{7}$<\/p>\n b)\u00a049<\/p>\n c)\u00a07<\/p>\n d)\u00a02.64<\/p>\n Question 7:\u00a0<\/b>Find the value of $\\sqrt{6+\\sqrt{6+\\sqrt{6+……}}}$<\/p>\n a)\u00a09<\/p>\n b)\u00a03<\/p>\n c)\u00a027<\/p>\n d)\u00a016<\/p>\n Question 8:\u00a0<\/b>Find the value of $\\sqrt{30+\\sqrt{30+\\sqrt{30+……}}}$<\/p>\n a)\u00a012<\/p>\n b)\u00a015<\/p>\n c)\u00a06<\/p>\n d)\u00a018<\/p>\n Question 9:\u00a0<\/b>Find the value of $\\sqrt{42+\\sqrt{42+\\sqrt{42+……}}}$<\/p>\n a)\u00a011<\/p>\n b)\u00a07<\/p>\n c)\u00a06<\/p>\n d)\u00a010<\/p>\n Question 10:\u00a0<\/b>Find the value of $\\sqrt{20-\\sqrt{20-\\sqrt{20-……}}}$<\/p>\n a)\u00a08<\/p>\n b)\u00a04<\/p>\n c)\u00a06<\/p>\n d)\u00a010<\/p>\n SSC CHSL FREE MOCK TEST<\/a><\/p>\n Question 11:\u00a0<\/b>Find the value of $\\sqrt{56-\\sqrt{56-\\sqrt{56-……}}}$<\/p>\n a)\u00a09<\/p>\n b)\u00a08<\/p>\n c)\u00a011<\/p>\n d)\u00a014<\/p>\n Question 12:\u00a0<\/b>If $2X+\\large\\frac{2}{X}$ $= 6$, then find the value of $X^{5}+\\large\\frac{1}{X^{5}}$<\/p>\n a)\u00a0123<\/p>\n b)\u00a0121<\/p>\n c)\u00a0116<\/p>\n d)\u00a0107<\/p>\n Question 13:\u00a0<\/b>If $3X+\\large\\frac{3}{X}$ $= 6$, then find the value of $X^{6}+\\large\\frac{1}{X^{6}}$<\/p>\n a)\u00a04<\/p>\n b)\u00a03<\/p>\n c)\u00a09<\/p>\n d)\u00a02<\/p>\n Question 14:\u00a0<\/b>If a+b = 5 and a-b = 1, Then find the value of ab<\/p>\n a)\u00a04<\/p>\n b)\u00a06<\/p>\n c)\u00a08<\/p>\n d)\u00a012<\/p>\n Question 15:\u00a0<\/b>If $(a-b)^{2} = 16$ and $(a+b)^{2} = 36$, then find the value of $\\frac{ab}{a+b}$<\/p>\n a)\u00a0$\\frac{5}{6}$<\/p>\n b)\u00a0$\\frac{8}{11}$<\/p>\n c)\u00a0$\\frac{6}{7}$<\/p>\n d)\u00a0$\\frac{7}{6}$<\/p>\n FREE SSC MATERIAL – 18000 FREE QUESTIONS<\/a><\/p>\n Question 16:\u00a0<\/b>If $x-\\large\\frac{1}{x}$ $= 3$, then $x^{3}-\\large\\frac{1}{x^{3}}$ $=$ ?<\/p>\n a)\u00a024<\/p>\n b)\u00a028<\/p>\n c)\u00a036<\/p>\n d)\u00a042<\/p>\n Question 17:\u00a0<\/b>Find the value of $1+\\Large\\frac{1}{1-\\frac{1}{1+\\Large\\frac{1}{1-\\frac{1}{7}}}}$<\/p>\n a)\u00a0$\\Large\\frac{15}{7}$<\/p>\n b)\u00a0$\\Large\\frac{19}{8}$<\/p>\n c)\u00a0$\\Large\\frac{20}{7}$<\/p>\n d)\u00a0$\\Large\\frac{17}{8}$<\/p>\n Question 18:\u00a0<\/b>If a = 48, b = 16, c = -64, then find the value of $\\large\\frac{a^{3}+b^{3}+c^{3}}{abc}$<\/p>\n a)\u00a0176<\/p>\n b)\u00a064<\/p>\n c)\u00a03<\/p>\n d)\u00a012<\/p>\n Question 19:\u00a0<\/b>If a = 17, b = -4, c = -13, then find the value of $\\large\\frac{3a^{3}+3b^{3}+3c^{3}}{4abc}$<\/p>\n a)\u00a03<\/p>\n b)\u00a0$\\frac{3}{4}$<\/p>\n c)\u00a01<\/p>\n d)\u00a0$\\frac{9}{4}$<\/p>\n Question 20:\u00a0<\/b>If $(2^{x})(2^y) = 16$ and $(3^{x})(9^y) = 27$, then find (x,y)<\/p>\n a)\u00a0(4,0)<\/p>\n b)\u00a0(3,2)<\/p>\n c)\u00a0(5,-1)<\/p>\n d)\u00a0(6,-2)<\/p>\n DOWNLOAD APP TO ACESSES DIRECTLY ON MOBILE<\/a><\/p>\n SSC CHSL Important Q&A PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $1+\\Large\\frac{1}{1+\\frac{1}{1+\\frac{1}{2}}}$ = $1+\\Large\\frac{1}{1+\\frac{1}{\\frac{3}{2}}}$<\/p>\n = $1+\\Large\\frac{1}{1+\\frac{2}{3}}$<\/p>\n = $1+\\Large\\frac{1}{\\frac{5}{3}}$<\/p>\n = $1+\\Large\\frac{3}{5}$<\/p>\n = $\\large\\frac{8}{5}$<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $1+\\Large\\frac{1}{1+\\frac{1}{1+\\frac{1}{6}}}$ = $1+\\Large\\frac{1}{1+\\frac{1}{\\frac{7}{6}}}$<\/p>\n = $1+\\Large\\frac{1}{1+\\frac{6}{7}}$<\/p>\n = $1+\\Large\\frac{1}{\\frac{13}{7}}$<\/p>\n = $1+\\Large\\frac{7}{13}$<\/p>\n = $\\large\\frac{20}{13}$<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $1+\\Large\\frac{1}{1+\\frac{1}{1+\\frac{3}{2}}}$ = $1+\\Large\\frac{1}{1+\\frac{1}{\\frac{5}{2}}}$<\/p>\n = $1+\\Large\\frac{1}{1+\\frac{2}{5}}$<\/p>\n = $1+\\Large\\frac{1}{\\frac{7}{5}}$<\/p>\n = $1+\\Large\\frac{5}{7}$<\/p>\n = $\\large\\frac{12}{7}$<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let $\\sqrt{3\\sqrt{3\\sqrt{3……}}}$ = X<\/p>\n Then, $\\sqrt{3X} = X$<\/p>\n Squaring on both sides, 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let $\\sqrt{4\\sqrt{4\\sqrt{4……}}}$ = X<\/p>\n Then, $\\sqrt{4X} = X$<\/p>\n Squaring on both sides, 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let $\\sqrt{7\\sqrt{7\\sqrt{7……}}}$ = X<\/p>\n Then, $\\sqrt{7X} = X$<\/p>\n Squaring on both sides, 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let $\\sqrt{6+\\sqrt{6+\\sqrt{6+……}}}$ = X<\/p>\n Then, $\\sqrt{6+X} = X$<\/p>\n Squaring on both sides, X cannot be negative when all the terms are positive. 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let $\\sqrt{30+\\sqrt{30+\\sqrt{30+……}}}$ = X<\/p>\n Then, $\\sqrt{30+X} = X$<\/p>\n Squaring on both sides, X cannot be negative when all the terms are positive. 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let $\\sqrt{42+\\sqrt{42+\\sqrt{42+……}}}$ = X<\/p>\n Then, $\\sqrt{42+X} = X$<\/p>\n Squaring on both sides, X cannot be negative when all the terms are positive. 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let $\\sqrt{20-\\sqrt{20-\\sqrt{20-……}}}$ = X<\/p>\n Then, $\\sqrt{20-X} = X$<\/p>\n Squaring on both sides, Hence, Option B is correct answer.<\/p>\n GK Questions And Answers PDF<\/a><\/p>\n 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let $\\sqrt{56-\\sqrt{56-\\sqrt{56-……}}}$ = X<\/p>\n Then, $\\sqrt{56-X} = X$<\/p>\n Squaring on both sides, 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given $2X+\\large\\frac{2}{X}$ $= 6$<\/p>\n $\\Rightarrow 2(X+\\large\\frac{1}{X})$ $= 6$<\/p>\n $\\Rightarrow X+\\large\\frac{1}{X}$ $= 3$ –> (1)<\/p>\n Squaring (1) on both sides<\/p>\n $(x+\\large\\frac{1}{x})^{2}$ $= 9$<\/p>\n $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$ $+2\\times x\\times\\large\\frac{1}{x}$ $= 9$<\/p>\n $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$ $+2 = 9$<\/p>\n $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$ $= 7$ –> (2)<\/p>\n Cubing (1) on both sides<\/p>\n $(x+\\large\\frac{1}{x})^{3}$ $= 27$<\/p>\n $\\Rightarrow x^{3}+\\large\\frac{1}{x^{3}}$ $+3\\times x\\times\\large\\frac{1}{x}$ $\\times(x+\\large\\frac{1}{x})$ $= 27$<\/p>\n $\\Rightarrow x^{3}+\\large\\frac{1}{x^{3}}$ $+3\\times3 = 27$<\/p>\n $\\Rightarrow x^{3}+\\large\\frac{1}{x^{3}}$ $= 27-9 = 18$ –> (3)<\/p>\n Multiplying (2) and (3)<\/p>\n $x^{2}+\\large\\frac{1}{x^{2}}$ $\\times x^{3}+\\large\\frac{1}{x^{3}}$ $= 18\\times7$<\/p>\n $\\Rightarrow x^{5}+\\large\\frac{1}{x^{5}}$ $+x^{2}\\times\\large\\frac{1}{x^{3}}$ $+x^{3}\\times\\large\\frac{1}{x^{2}}$ $= 126$<\/p>\n $\\Rightarrow x^{5}+\\large\\frac{1}{x^{5}}$ $+x+\\large\\frac{1}{x}$ $= 126$<\/p>\n Substituting $x+\\large\\frac{1}{x}$ $= 3$ in above equation<\/p>\n $\\Rightarrow x^{5}+\\large\\frac{1}{x^{5}}$ $+3 = 126$<\/p>\n $\\Rightarrow x^{5}+\\large\\frac{1}{x^{5}}$ $= 123$<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given $3X+\\large\\frac{3}{X}$ $= 6$<\/p>\n $\\Rightarrow 3(X+\\large\\frac{1}{X})$ $= 6$<\/p>\n $\\Rightarrow X+\\large\\frac{1}{X}$ $= 2$<\/p>\n Squaring on both sides $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$+$2\\times x\\times\\large\\frac{1}{x}$ $=4$<\/p>\n $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$ $+2 = 4$<\/p>\n $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$ $= 2$<\/p>\n Cubing on both sides<\/p>\n $(x^{2}+\\large\\frac{1}{x^{2}})^{3}$ $= 8$<\/p>\n $x^{6}+\\large\\frac{1}{6}$ $+3\\times x^{2}\\times\\large\\frac{1}{x^{2}}$ $\\times(x^{2}+\\large\\frac{1}{x^{2}})$ $= 8$<\/p>\n $\\Rightarrow x^{6}+\\large\\frac{1}{6}$ $+3\\times2 = 8$<\/p>\n $\\therefore x^{6}+\\large\\frac{1}{6}$ $= 8-6 = 2$<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given, a+b = 5 Then, 2a = 6 ==> a = 3<\/p>\n Substituting a = 3 in above equation Hence, ab = 3*2 = 6<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given $(a-b)^{2} = 16$ and $(a+b)^{2} = 36$ $(a+b)^{2} = 36$ Hence, $\\frac{ab}{a+b} = \\frac{5}{6}$<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given $x-\\large\\frac{1}{x}$ $= 3$<\/p>\n Cubing on both sides<\/p>\n $x^{3}-\\large\\frac{1}{x^{3}}$ $-3\\timesx\\times\\large\\frac{1}{x}$ $(x-\\large\\frac{1}{x})$ $= 27$<\/p>\n => $x^{3}-\\large\\frac{1}{x^{3}}$ $-3\\times3 = 27$<\/p>\n => $x^{3}-\\large\\frac{1}{x^{3}}$ $-9 = 27$<\/p>\n => $x^{3}-\\large\\frac{1}{x^{3}}$ $= 36$<\/p>\n 17)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $1+\\Large\\frac{1}{1-\\frac{1}{1+\\Large\\frac{1}{1-\\frac{1}{7}}}}$ = $1+\\Large\\frac{1}{1-\\frac{1}{1+\\Large\\frac{1}{\\frac{6}{7}}}}$<\/p>\n = $1+\\Large\\frac{1}{1-\\frac{1}{1+\\Large\\frac{7}{6}}}$<\/p>\n = $1+\\Large\\frac{1}{1-\\frac{1}{\\Large\\frac{13}{6}}}$<\/p>\n = $1+\\Large\\frac{1}{1-\\frac{6}{13}}$<\/p>\n = $1+\\Large\\frac{1}{\\frac{7}{13}}$<\/p>\n = $1+\\Large\\frac{13}{7}$<\/p>\n = $\\Large\\frac{20}{7}$<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given a = 48, b = 16, c = -64 We know that if a+b+c = 0, then $a^{3}+b^{3}+c^{3} = 3abc$<\/p>\n Hence, $\\large\\frac{a^{3}+b^{3}+c^{3}}{abc} = \\frac{3abc}{abc}$ $= 3$<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given a = 17, b = -4, c = -13 Then, $\\large\\frac{3a^{3}+3b^{3}+3c^{3}}{4abc}$ $= \\large\\frac{3(a^{3}+b^{3}+c^{3})}{4abc} = \\frac{3(3abc)}{4abc} = \\frac{9}{4}$<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given $(2^{x})(2^y) = 16$<\/p>\n => $2^{x+y} = 2^{4}$<\/p>\n => x+y = 4 — (1)<\/p>\n $(3^{x})(9^y) = 27$ => y = -1 Therefore, (x,y) = (5,-1)<\/p>\n\n
\n$3X = X^{2}$
\n\u21d2 X $= 3$<\/p>\n
\n$4X = X^{2}$
\n\u21d2 X $= 4$<\/p>\n
\n$7X = X^{2}$
\n\u21d2 X $= 7$<\/p>\n
\n$6+X = X^{2}$
\n\u21d2 $X^{2}-X-6 = 0$
\n\u21d2 $X^{2}-3X+2X-6 = 0$
\n\u21d2 $X(X-3)+2(X-3) = 0$
\n\u21d2 $(X-3)(X+2) = 0$
\n\u21d2 $X = 3$ or $X = -2$<\/p>\n
\nHence, $X = 3$<\/p>\n
\n$30+X = X^{2}$
\n\u21d2 $X^{2}-X-30 = 0$
\n\u21d2 $X^{2}-6X+5X-30 = 0$
\n\u21d2 $X(X-6)+5(X-6) = 0$
\n\u21d2 $(X-6)(X+5) = 0$
\n\u21d2 $X = 6$ or $X = -5$<\/p>\n
\nHence, $X = 6$<\/p>\n
\n$42+X = X^{2}$
\n\u21d2 $X^{2}-X-42 = 0$
\n\u21d2 $X^{2}-7X+6X-42 = 0$
\n\u21d2 $X(X-7)+6(X-7) = 0$
\n\u21d2 $(X-7)(X+6) = 0$
\n\u21d2 $X = 7$ or $X = -6$<\/p>\n
\nHence, $X = 7$<\/p>\n
\n$20-X = X^{2}$
\n\u21d2 $X^{2}+X-20 = 0$
\n\u21d2 $X^{2}-4X+5X-20 = 0$
\n\u21d2 $X(X-4)+5(X-4) = 0$
\n\u21d2 $(X-4)(X+5) = 0$
\n\u21d2 $X = 4$ or $X = -5$<\/p>\n
\n$56-X = X^{2}$
\n\u21d2 $X^{2}+X-56 = 0$
\n\u21d2 $X^{2}-8X+7X-56 = 0$
\n\u21d2 $X(X-8)+7(X-8) = 0$
\n\u21d2 $(X-8)(X+7) = 0$
\n\u21d2 $X = 8$ or $X = -7$
\nHence, Option B is correct answer.<\/p>\n
\n$(x+\\large\\frac{1}{x})^{2}$ $=4$<\/p>\n
\na-b = 1<\/p>\n
\n==> b = 2<\/p>\n
\n$(a+b)^{2} = (a-b)^{2}+4ab$
\n$36 = 16+4ab$
\n=> $4ab = 20$
\n$ab = 5$<\/p>\n
\n=> $a+b = 6$<\/p>\n
\nThen, a+b+c = 48+16-64 = 0<\/p>\n
\nThen a+b+c = 0.
\nWe know that if a+b+c = 0, then $a^{3}+b^{3}+c^{3} = 3abc$<\/p>\n
\n=> $(3^{x})((3^2)^y) = 3^3$
\n=> $(3^x)(3^2y) =3^3$
\n=> $3^x+2y = 3^3$
\n=> $x+2y = 3$ — (2)
\nSolving (1) and (2)<\/p>\n
\nSubstituting y = -1 in (1) –> x = 5<\/p>\n