SSC GD Constable Algebra Question paper with answers download PDF based on SSC GD exam previous papers. 40 Very important Algebra questions for GD Constable.<\/p>\n
ALGEBRA QUESTIONS FOR SSC GD PDF<\/a><\/p>\n GET 20 SSC GD MOCK FOR JUST RS. 117<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Download SSC GD Important Questions PDF<\/a><\/p>\n 1500<\/strong>+ Must Solve Questions for SSC Exams<\/a> (Question bank)<\/p>\n Question 1:\u00a0<\/b>Find the number of even factors of 15680.<\/p>\n a)\u00a042<\/p>\n b)\u00a021<\/p>\n c)\u00a036<\/p>\n d)\u00a024<\/p>\n Question 2:\u00a0<\/b>Find the number of prime factors of 14560<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a05<\/p>\n d)\u00a06<\/p>\n Question 3:\u00a0<\/b>What is the square root of 97-16$\\sqrt{3}$<\/p>\n a)\u00a09-4$\\sqrt{3}$<\/p>\n b)\u00a09+4$\\sqrt{3}$<\/p>\n c)\u00a07-4$\\sqrt{3}$<\/p>\n d)\u00a07+4$\\sqrt{3}$<\/p>\n Question 4:\u00a0<\/b>Find the value of $\\Large\\frac{\\frac{1}{3}.\\frac{1}{3}.\\frac{1}{3}+\\frac{1}{4}.\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{9}.\\frac{1}{9}.\\frac{1}{9}-3.\\frac{1}{3}.\\frac{1}{4}.\\frac{1}{9}}{\\frac{1}{3}.\\frac{1}{3}+\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{9}.\\frac{1}{9}-(\\frac{1}{3}.\\frac{1}{4}+\\frac{1}{4}.\\frac{1}{9}+\\frac{1}{3}.\\frac{1}{9})}$<\/p>\n a)\u00a0$\\large\\frac{25}{36}$<\/p>\n b)\u00a0$\\large\\frac{19}{36}$<\/p>\n c)\u00a0$\\large\\frac{24}{35}$<\/p>\n d)\u00a0$\\large\\frac{17}{26}$<\/p>\n Question 5:\u00a0<\/b>Find the value of $\\Large\\frac{\\frac{1}{2}.\\frac{1}{2}.\\frac{1}{2}+\\frac{1}{4}.\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{5}.\\frac{1}{5}.\\frac{1}{5}-3.\\frac{1}{2}.\\frac{1}{4}.\\frac{1}{5}}{\\frac{1}{2}.\\frac{1}{2}+\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{5}.\\frac{1}{5}-(\\frac{1}{2}.\\frac{1}{4}+\\frac{1}{4}.\\frac{1}{5}+\\frac{1}{2}.\\frac{1}{5})}$<\/p>\n a)\u00a0$\\large\\frac{17}{20}$<\/p>\n b)\u00a0$\\large\\frac{19}{20}$<\/p>\n c)\u00a0$\\large\\frac{13}{20}$<\/p>\n d)\u00a0$\\large\\frac{11}{20}$<\/p>\n Download SSC GD FREE PREVIOUS PAPERS<\/a><\/p>\n LATEST FREE SSC GD MOCK 2019<\/a><\/p>\n Question 6:\u00a0<\/b>The lines 2x+y = 3 and x+2y = 3 intersect at points<\/p>\n a)\u00a0(1,1)<\/p>\n b)\u00a0(-1,5)<\/p>\n c)\u00a0(0,3)<\/p>\n d)\u00a0(3,-3)<\/p>\n Question 7:\u00a0<\/b>If If $(3^{x})(3^y) = 9$ and $(5^{x})(125^y) = 625$, then find (x,y)<\/p>\n a)\u00a0(4,-2)<\/p>\n b)\u00a0(0,2)<\/p>\n c)\u00a0(1,1)<\/p>\n d)\u00a0(6,-4)<\/p>\n Question 8:\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 Question 9:\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 10:\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 DOWNLOAD APP TO ACESS DIRECTLY ON MOBILE<\/a><\/p>\n Daily Free SSC Practice Set<\/a><\/p>\n Question 11:\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 12:\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 13:\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 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 $3X+\\large\\frac{3}{X}$ $= 6$, 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 SSC CGL Free Mock Test<\/a><\/p>\n Download SSC GD Constable Syllabus PDF<\/a><\/p>\n Question 16:\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 17:\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 18:\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 Question 19:\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 20:\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 21:\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 22:\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 23:\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 Question 24:\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 25:\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 26:\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 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n Question 27:\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 28:\u00a0<\/b>If $x+\\large\\frac{1}{x}$ $= 3$, then $x^{5}+\\large\\frac{1}{x^{5}}$ $=?$<\/p>\n a)\u00a0123<\/p>\n b)\u00a0121<\/p>\n c)\u00a0116<\/p>\n d)\u00a0107<\/p>\n Question 29:\u00a0<\/b>If $x+\\large\\frac{1}{x}$ $=2$, then find the value of $x^{6}+\\large\\frac{1}{x^{6}}$.<\/p>\n a)\u00a02<\/p>\n b)\u00a05<\/p>\n c)\u00a08<\/p>\n d)\u00a06<\/p>\n Question 30:\u00a0<\/b>If $x+\\large\\frac{1}{x}$ $=4$, then find the value of $x^{3}+\\large\\frac{1}{x^{3}}$.<\/p>\n a)\u00a048<\/p>\n b)\u00a056<\/p>\n c)\u00a052<\/p>\n d)\u00a064<\/p>\n Question 31:\u00a0<\/b>If $x+\\large\\frac{1}{x}$ $=3$, then find the value of $x^{2}+\\large\\frac{1}{x^{2}}$.<\/p>\n a)\u00a06<\/p>\n b)\u00a07<\/p>\n c)\u00a09<\/p>\n d)\u00a08<\/p>\n Question 32:\u00a0<\/b>What is the units digit of $17^{17}$ *$33^{33}$<\/p>\n a)\u00a07<\/p>\n b)\u00a09<\/p>\n c)\u00a03<\/p>\n d)\u00a01<\/p>\n Question 33:\u00a0<\/b>What is the units digit of $27^{27}$ ?<\/p>\n a)\u00a07<\/p>\n b)\u00a09<\/p>\n c)\u00a03<\/p>\n d)\u00a01<\/p>\n Question 34:\u00a0<\/b>If $x+\\frac{1}{x}=-2$ then the value of $x^{p}+x^{q}$ is: (Where p is an even number and q is an odd number)<\/p>\n a)\u00a0-2<\/p>\n b)\u00a02<\/p>\n c)\u00a01<\/p>\n d)\u00a00<\/p>\n Question 35:\u00a0<\/b>If $p(x+y)^{2}=5$ and $q(x-y)^{2}=3$, then the simplified value of $p^{2}(x+y)^{2}+4\\ pq\\ xy – q^{2}(x-y)^{2}$ is:<\/p>\n a)\u00a0$- (p + q)$<\/p>\n b)\u00a0$2 (p + q)$<\/p>\n c)\u00a0$p + q$<\/p>\n d)\u00a0$-2 (p + q)$<\/p>\n Question 36:\u00a0<\/b>The simplified value of the following expression is: $\\frac{1}{\\sqrt{11-2\\sqrt{30}}}-\\frac{3}{\\sqrt{7-2\\sqrt{10}}}-\\frac{4}{\\sqrt{8+4\\sqrt{3}}}$<\/p>\n a)\u00a0$0$<\/p>\n b)\u00a0$1$<\/p>\n c)\u00a0$\\sqrt{2}$<\/p>\n d)\u00a0$\\sqrt{3}$<\/p>\n Question 37:\u00a0<\/b>The value of the following is: $\\sqrt{12+\\sqrt{12+\\sqrt{12+…..}}}$<\/p>\n a)\u00a0$2\\sqrt{2}$<\/p>\n b)\u00a0$2\\sqrt{3}$<\/p>\n c)\u00a0$2$<\/p>\n d)\u00a0$4$<\/p>\n Question 38:\u00a0<\/b>The value of x in the following equation is: a)\u00a0$5.3$<\/p>\n b)\u00a0$2\\frac{3}{10}$<\/p>\n c)\u00a0$2\\frac{2}{3}$<\/p>\n d)\u00a0$2.\\dot{35}$<\/p>\n Question 39:\u00a0<\/b>If $1^{2}+2^{2}+3^{2}+……..+p^{2}$ = $\\frac{p(p+1)(2p+1)}{6}$, then $1^{2}+3^{2}+5^{2}+……..+17^{2}$ is equal to:<\/p>\n a)\u00a01785<\/p>\n b)\u00a01700<\/p>\n c)\u00a0980<\/p>\n d)\u00a0969<\/p>\n Question 40:\u00a0<\/b>Given $2^{2}+4^{2}+6^{2}+……..+40^{2} = 11480$, then the value of $1^{2}+2^{2}+3^{2}+……..+20^{2}$ is:<\/p>\n a)\u00a02870<\/p>\n b)\u00a02868<\/p>\n c)\u00a02867<\/p>\n d)\u00a02869<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We have to factorise the number into prime factors i.e 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We have to factorise the number into prime factors i.e 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n we have $(a-b)^{2}$=$a^{2}+b^{2}-2ab$ 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The given equation is in the form of We know that $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-(ab+bc+ca))$<\/p>\n => $\\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-(ab+bc+ca)} = a+b+c$<\/p>\n Then, $\\Large\\frac{\\frac{1}{3}.\\frac{1}{3}.\\frac{1}{3}+\\frac{1}{4}.\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{9}.\\frac{1}{9}.\\frac{1}{9}-3.\\frac{1}{3}.\\frac{1}{4}.\\frac{1}{9}}{\\frac{1}{3}.\\frac{1}{3}+\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{9}.\\frac{1}{9}-(\\frac{1}{3}.\\frac{1}{4}+\\frac{1}{4}.\\frac{1}{9}+\\frac{1}{3}.\\frac{1}{9})}$ $= \\large\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{9} = \\frac{12+9+4}{36} = \\frac{25}{36}$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The given equation is in the form of We know that $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-(ab+bc+ca))$<\/p>\n => $\\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-(ab+bc+ca)} = a+b+c$<\/p>\n Then, $\\Large\\frac{\\frac{1}{2}.\\frac{1}{2}.\\frac{1}{2}+\\frac{1}{4}.\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{5}.\\frac{1}{5}.\\frac{1}{5}-3.\\frac{1}{2}.\\frac{1}{4}.\\frac{1}{5}}{\\frac{1}{2}.\\frac{1}{2}+\\frac{1}{4}.\\frac{1}{4}+\\frac{1}{5}.\\frac{1}{5}-(\\frac{1}{2}.\\frac{1}{4}+\\frac{1}{4}.\\frac{1}{5}+\\frac{1}{2}.\\frac{1}{5})}$ $= \\large\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{5} = \\frac{10+5+4}{20} = \\frac{19}{20}$<\/p>\n SSC GD FREE STUDY MATERIAL<\/a><\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given 2x+y = 3 and x+2y = 3 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given $(3^{x})(3^y) = 9$<\/p>\n => $3^{x+y} = 3^{2}$<\/p>\n => x+y = 2 — (1)<\/p>\n $(5^{x})(125^y) = 625$ => 2y = 2 => y = 1 Therefore, (x,y) = (1,1)<\/p>\n 8)\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 9)\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 10)\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 11)\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 12)\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 13)\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 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(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 16)\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 17)\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, 18)\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 19)\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. 20)\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. 21)\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. 22)\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, 23)\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, 24)\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, 25)\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 26)\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 27)\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 28)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given $x+\\large\\frac{1}{x}$ $= 3$ –> (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 29)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given $x+\\large\\frac{1}{x}$ $=2$<\/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 30)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given $x+\\large\\frac{1}{x}$ $=4$<\/p>\n Cubing on both sides $\\Rightarrow x^{3}+\\large\\frac{1}{x^{3}}$+ $3\\times x\\times\\large\\frac{1}{x}\\times(x+\\frac{1}{x})$ $= 64$<\/p>\n $\\Rightarrow x^{3}+\\large\\frac{1}{x^{3}}$ $+3\\times4 = 64$<\/p>\n $\\Rightarrow x^{3}+\\large\\frac{1}{x^{3}}$ $+12$ $= 64$<\/p>\n $\\therefore$ $x^{3}+\\large\\frac{1}{x^{3}}$ $= 52$<\/p>\n 31)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given $x+\\large\\frac{1}{x}$ $=3$<\/p>\n Squaring on both sides $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$+ $2\\times x\\times\\large\\frac{1}{x}$ $=9$<\/p>\n $\\Rightarrow x^{2}+\\large\\frac{1}{x^{2}}$ $+2 = 9$<\/p>\n $\\Rightarrow x^{2}+\\frac{1}{x^{2}} = 7$<\/p>\n 32)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In the power cycle of 7 we get units digit for 33)\u00a0Answer\u00a0(C)<\/strong><\/p>\n In the cycle of 7 power we get 34)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given : $x+\\frac{1}{x}=-2$<\/p>\n => $\\frac{x^2+1}{x}=-2$<\/p>\n => $x^2+1+2x=0$<\/p>\n => $(x+1)^2=0$<\/p>\n => $x+1=0$<\/p>\n => $x=-1$<\/p>\n $\\therefore$ $x^{p}+x^{q}$ (let $p=2$ and $q=1$)<\/p>\n => $(-1)^2+(-1)^1=1-1=0$<\/p>\n => Ans – (D)<\/p>\n 35)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 36)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Using, $a^2+b^2+ab=(a+b)^2$<\/p>\n => $\\sqrt{11-2\\sqrt{30}}=\\sqrt{(\\sqrt6)^2+(\\sqrt5)^2-2\\sqrt6\\sqrt5}=(\\sqrt6-\\sqrt5)$<\/p>\n Similarly, $\\sqrt{7-2\\sqrt{10}}=(\\sqrt5-\\sqrt2)$<\/p>\n and $\\sqrt{8+4\\sqrt3}=\\sqrt{8+2\\sqrt{12}}=(\\sqrt6+\\sqrt2)$<\/p>\n To find : $\\frac{1}{\\sqrt{11-2\\sqrt{30}}}-\\frac{3}{\\sqrt{7-2\\sqrt{10}}}-\\frac{4}{\\sqrt{8+4\\sqrt{3}}}$<\/p>\n = $\\frac{1}{(\\sqrt6-\\sqrt5)}-\\frac{3}{(\\sqrt5-\\sqrt2)}-\\frac{4}{(\\sqrt6+\\sqrt2)}$<\/p>\n Rationalizing the denominator, we get :<\/p>\n = $[\\frac{1}{\\sqrt6-\\sqrt5}\\times\\frac{\\sqrt6+\\sqrt5}{\\sqrt6+\\sqrt5}]-[\\frac{3}{\\sqrt5-\\sqrt2}\\times\\frac{\\sqrt5+\\sqrt2}{\\sqrt5+\\sqrt2}]-[\\frac{4}{\\sqrt6+\\sqrt2}\\times\\frac{\\sqrt6-\\sqrt2}{\\sqrt6-\\sqrt2}]$<\/p>\n = $(\\sqrt6+\\sqrt5)-(\\sqrt5+\\sqrt2)-(\\sqrt6-\\sqrt2)$<\/p>\n = $0$<\/p>\n => Ans – (A)<\/p>\n 37)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let $x=\\sqrt{12+\\sqrt{12+\\sqrt{12+…..}}}$<\/p>\n => $x=\\sqrt{12+x}$<\/p>\n Squaring both sides, we get :<\/p>\n => $x^2=x+12$<\/p>\n => $x^2-x-12=0$<\/p>\n => $x^2-4x+3x-12=0$<\/p>\n => $x(x-4)+3(x-4)=0$<\/p>\n => $(x-4)(x+3)=0$<\/p>\n => $x=4,-3$<\/p>\n $\\because x$ cannot be negative, => $x=4$<\/p>\n => Ans – (D)<\/p>\n 38)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 39)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression : $1^{2}+3^{2}+5^{2}+……..+17^{2}$<\/p>\n = $[1^{2}+2^{2}+3^{2}+4^{2}……..+16^{2}+17^{2}]$ $-[2^2+4^2+………+16^2]$<\/p>\n = $[1^{2}+2^{2}+3^{2}+4^{2}……..+16^{2}+17^{2}]$ $-(2^2)[1^2+2^2+3^2………+8^2]$<\/p>\n = $[\\frac{17(17+1)+(34+1)}{6}]-[4\\times\\frac{8(8+1)(16+1)}{6}]$<\/p>\n = $[\\frac{17(17+1)+(34+1)}{6}]-[4\\times\\frac{8(8+1)(16+1)}{6}]$<\/p>\n = $[51\\times35]-[48\\times17]$<\/p>\n = $17\\times(105-48)=969$<\/p>\n => Ans – (D)<\/p>\n 40)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given : $2^{2}+4^{2}+6^{2}+……..+40^{2} = 11480$<\/p>\n => $2^2[1+2^2+3^2+…..+20^2]=11480$<\/p>\n => $1^{2}+2^{2}+3^{2}+……..+20^{2}=\\frac{11480}{4}$<\/p>\n => $1^{2}+2^{2}+3^{2}+……..+20^{2}=2870$<\/p>\n => Ans – (A)<\/p>\n
\n$0.\\dot{3}+0.\\dot{6}+0.\\dot{7}+0.\\dot{8}=x$<\/p>\n
\n15680=$2^{6}*5*7^{2}$
\nNo of even factors =6*(1+1)*(2+1)
\n=36<\/p>\n
\n14560=$2^{5}*5*13*7$
\nThere are 4 different prime factors namely 2,5,7 and 13.<\/p>\n
\nComparing this with 97-56$\\sqrt{3}$=$a^{2}+b^{2}-2ab$
\nWe have 97=$a^{2}+b^{2}$
\nFor a=7 and b=4$\\sqrt{3}$ it gets satisfied and also 2ab=2*7*4$\\sqrt{3}$
\nSo the $(7-4\\sqrt{3})^{2}$=$7^{2}+(4\\sqrt{3})^{2}-2*7*4*\\sqrt{3}$
\nAnd so required answer is 7-4$\\sqrt{3}$<\/p>\n
\n$\\large\\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-(ab+bc+ca)}$<\/p>\n
\n$\\large\\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-(ab+bc+ca)}$<\/p>\n
\nSolving above equations,
\nWe get x = 1 and y = 1.
\nHence, the lines intersect at (1,1)<\/p>\n
\n=> $(5^{x})((5^3)^y) = 5^4$
\n=> $(5^x)(5^3y) =5^4$
\n=> $5^x+3y = 5^4$
\n=> $x+3y = 4$ — (2)
\nSolving (1) and (2)<\/p>\n
\nSubstituting y = 1 in (1) –> x = 1<\/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
\nThen a+b+c = 0.
\nWe know that if a+b+c = 0, then $a^{3}+b^{3}+c^{3} = 3abc$<\/p>\n
\nThen, a+b+c = 48+16-64 = 0<\/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
\na-b = 1<\/p>\n
\n==> b = 2<\/p>\n
\n$(x+\\large\\frac{1}{x})^{2}$ $=4$<\/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$20-X = X^{2}$
\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$42+X = X^{2}$
\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$30+X = X^{2}$
\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$6+X = X^{2}$
\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$7X = X^{2}$
\n\u21d2 X $= 7$<\/p>\n
\n$4X = X^{2}$
\n\u21d2 X $= 4$<\/p>\n
\n$3X = X^{2}$
\n\u21d2 X $= 3$<\/p>\n
\n$(x+\\large\\frac{1}{x})^{2}$ $=4$<\/p>\n
\n$(x+\\large\\frac{1}{x})^{3}$ $=64$<\/p>\n
\n$(x+\\large\\frac{1}{x})^{2}$ $=9$<\/p>\n
\n$7^{1}=7$
\n$7^{2}=49$
\n$7^{3}=243$
\n$7^{4}=1701$
\nAnd this cycle repeats. Cyclicity =4
\n$17^{4(4)+1}$ has 7 as its unit digit.
\nIn the power cycle of 3 we get units digit for
\n$3^{1}=3$
\n$3^{2}=9$
\n$3^{3}=27$
\n$3^{4}=81$
\nAnd this cycle repeats. Cyclicity =4
\n$33^{8(4)+1}$ has 3 as its unit digit.
\nSo the product of 7 and 3 is 21 and so the units digit is 1.<\/p>\n
\n$7^{1}=7$
\n$7^{2}=49$
\n$7^{3}=243$
\n$7^{4}=1701$
\nAnd this cycle repeats. Cyclicity =4
\n$27^{4(6)+3}$ has 3 as its unit digit.<\/p>\n