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 factors of 13x if (7-$\\frac{4x}{3}$)($\\frac{3}{2}$)=$\\frac{x}{6}$.<\/p>\n a)\u00a012<\/p>\n b)\u00a04<\/p>\n c)\u00a08<\/p>\n d)\u00a06<\/p>\n Question 2:\u00a0<\/b>Find the value of $\\ \\sqrt{30+\\sqrt{30}+…}$<\/p>\n a)\u00a05<\/p>\n b)\u00a03<\/p>\n c)\u00a06<\/p>\n d)\u00a07<\/p>\n Question 3:\u00a0<\/b>$2\\frac{1}{5}x^{2}\\ $= 2750, find the value of x ?<\/p>\n a)\u00a025<\/p>\n b)\u00a0$25\\sqrt{3}$<\/p>\n c)\u00a0$25\\sqrt{2}$<\/p>\n d)\u00a020<\/p>\n Question 4:\u00a0<\/b>Sin A + Sin$^{2}\\ $A = 1, then the value of cos$^{2}\\ $A + cos$^{4}\\ $A is<\/p>\n a)\u00a02<\/p>\n b)\u00a0$\\frac{2}{3}$<\/p>\n c)\u00a0$1\\frac{1}{2}$<\/p>\n d)\u00a01<\/p>\n Question 5:\u00a0<\/b>If X = $\\ \\sqrt[3]{5}\\ $+ 2, then the value of $\\ x^{3}-6x^{2}\\ $+ 12x – 13 is<\/p>\n a)\u00a0-1<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a00<\/p>\n Download SSC GD FREE PREVIOUS PAPERS<\/a><\/p>\n LATEST FREE SSC GD MOCK 2019<\/a><\/p>\n Question 6:\u00a0<\/b>If 2x + $\\ \\frac{2}{x}\\ $= 3 then the value of $\\ x^{3}+\\frac{1}{x^{3}}$+ 2 is<\/p>\n a)\u00a0-$\\frac{9}{8}$<\/p>\n b)\u00a0-$\\frac{25}{8}$<\/p>\n c)\u00a0$\\frac{7}{8}$<\/p>\n d)\u00a011<\/p>\n Question 7:\u00a0<\/b>If$\\ a^{2}+b^{2}+c^{2}$= 2(a-b-c)-3, then the value of 4a – 3b + 5c is<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a05<\/p>\n d)\u00a06<\/p>\n Question 8:\u00a0<\/b>The value of$\\ \\frac{3\\sqrt{2}}{(\\sqrt{3}+\\sqrt{6})}-\\frac{4\\sqrt{3}}{(\\sqrt{6}+\\sqrt{2})}+\\frac{\\sqrt{6}}{(\\sqrt{2}+\\sqrt{3})}\\ $is<\/p>\n a)\u00a0$\\sqrt{2}$<\/p>\n b)\u00a00<\/p>\n c)\u00a0$\\sqrt{3}$<\/p>\n d)\u00a0$\\sqrt{6}$<\/p>\n Question 9:\u00a0<\/b>The value of $\\sqrt{2\\sqrt[3]{4}\\sqrt{2\\sqrt[3]{4}}\\sqrt[4]{2\\sqrt[3]{4}}…..}$ is<\/p>\n a)\u00a0$2\\sqrt[3]4$<\/p>\n b)\u00a0$\\sqrt{2\\sqrt[3]4}$<\/p>\n c)\u00a0$2\\sqrt4$<\/p>\n d)\u00a0$\\sqrt[3]4$<\/p>\n Question 10:\u00a0<\/b>The value of cosec$^{2}\\ 18^\\circ\\ – \\frac{1}{cot^{2}72^\\circ}\\ $is<\/p>\n a)\u00a0$\\frac{1}{\\sqrt{3}}$<\/p>\n b)\u00a0$\\frac{\\sqrt{2}}{3}$<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a01<\/p>\n DOWNLOAD APP TO ACESS DIRECTLY ON MOBILE<\/a><\/p>\n Daily Free SSC Practice Set<\/a><\/p>\n Question 11:\u00a0<\/b>If$\\ x^{2}$- 3x + 1 = 0, then the value of $\\ x^{5}+\\frac{1}{x^{5}}\\ $is equal to<\/p>\n a)\u00a087<\/p>\n b)\u00a0123<\/p>\n c)\u00a0135<\/p>\n d)\u00a0201<\/p>\n Question 12:\u00a0<\/b>The value of 0.65 x 0.65 + 0.35 x 0.35+0.70 x 0.65 is<\/p>\n a)\u00a01.75<\/p>\n b)\u00a01.00<\/p>\n c)\u00a01.65<\/p>\n d)\u00a01.55<\/p>\n Question 13:\u00a0<\/b>The value of $\\frac{(75.8)^{2}-(35.8)^{2}}{40}$ is<\/p>\n a)\u00a0121.6<\/p>\n b)\u00a040<\/p>\n c)\u00a0160<\/p>\n d)\u00a0111.6<\/p>\n Question 14:\u00a0<\/b>If $tan(\\theta)tan(5\\theta)=1$, then what is the value of $sin 2\\theta$ ?<\/p>\n a)\u00a0$0$<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a0$1\\sqrt2$<\/p>\n d)\u00a0$\\frac{\\sqrt3}{2}$<\/p>\n Question 15:\u00a0<\/b>What is the simplified value of $\\ \\frac{2sin^{3}\\ \\theta\\ – sin\\theta}{cos\\theta\\ -\\ 2\\ cos^{3}\\ \\theta}$?<\/p>\n a)\u00a0tan \u03b8<\/p>\n b)\u00a0sin \u03b8<\/p>\n c)\u00a0cos \u03b8<\/p>\n d)\u00a0cot \u03b8<\/p>\n SSC<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n Download SSC GD Constable Syllabus PDF<\/a><\/p>\n Question 16:\u00a0<\/b>If$\\ \\frac{x-xtan^{2}15^\\circ}{1+tan^{2}15^\\circ}$= sin $60^\\circ\\ $+ cos 30$^\\circ\\ $, then what is then what is the value of x?<\/p>\n a)\u00a02<\/p>\n b)\u00a0-1<\/p>\n c)\u00a0-2<\/p>\n d)\u00a01<\/p>\n Question 17:\u00a0<\/b>What is the simplified value of$\\ \\sqrt{\\frac{sec^{2} \\theta+cosec^{2} \\theta}{4}}$?<\/p>\n a)\u00a0cosec 2\u03b8<\/p>\n b)\u00a0sec 2\u03b8<\/p>\n c)\u00a0cosec \u03b8 sec \u03b8<\/p>\n d)\u00a0tan \u03b8<\/p>\n Question 18:\u00a0<\/b>The side QR of \u0394PQR is produced to S. If \u2220PRS = 105\u00b0 and \u2220Q = (1\/2)\u2220P, then what is the value of \u2220P?<\/p>\n a)\u00a045<\/p>\n b)\u00a060<\/p>\n c)\u00a070<\/p>\n d)\u00a075<\/p>\n Question 19:\u00a0<\/b>In the given figure, O is the center of the circle, $\\angle$CAO = 35$^\\circ$. What is the value (in degrees) of $\\ \\angle$AOB?<\/p>\n a)\u00a090<\/p>\n b)\u00a0110<\/p>\n c)\u00a0160<\/p>\n d)\u00a0130<\/p>\n Question 20:\u00a0<\/b>In \u0394ABC, \u2220A : \u2220B : \u2220C = 3 : 3 : 4. A line parallel to BC is drawn which touches AB and AC at P and Q respectively. What is the value of \u2220AQP – \u2220APQ?<\/p>\n a)\u00a012<\/p>\n b)\u00a018<\/p>\n c)\u00a024<\/p>\n d)\u00a036<\/p>\n Question 21:\u00a0<\/b>If $x = 5 – \\frac{1}{x} $, then what is the value of $x^{5} + \\frac{1}{x^{5}}$?<\/p>\n a)\u00a0625<\/p>\n b)\u00a03125<\/p>\n c)\u00a02525<\/p>\n d)\u00a02500<\/p>\n Question 22:\u00a0<\/b>If $x^{3} – y^{3} = 112$ and $x – y = 4$, then what is the value of $x^{2} + y^{2}$?<\/p>\n a)\u00a016<\/p>\n b)\u00a020<\/p>\n c)\u00a024<\/p>\n d)\u00a028<\/p>\n Question 23:\u00a0<\/b>If $3x – \\frac{1}{3x} = 9$, then what is the value of $x^{2} + \\frac{x^2}{81}$?<\/p>\n a)\u00a07<\/p>\n b)\u00a083\/9<\/p>\n c)\u00a011<\/p>\n d)\u00a0121\/9<\/p>\n Question 24:\u00a0<\/b>If $x^{4}+\\frac{1}{x^{4}}= 198$ and $x>0$, then what is the value of $x^{2}-\\frac{1}{x^{2}}$?<\/p>\n a)\u00a0$14$<\/p>\n b)\u00a0$2\\sqrt{7}$<\/p>\n c)\u00a0$10\\sqrt{2}$<\/p>\n d)\u00a0$10$<\/p>\n Question 25:\u00a0<\/b>If $x + y = 4$, then what is the value of $x^{3} +y^{3} +12xy$?<\/p>\n a)\u00a016<\/p>\n b)\u00a032<\/p>\n c)\u00a064<\/p>\n d)\u00a0256<\/p>\n Question 26:\u00a0<\/b>If $N=(\\frac{\\sqrt{7}-\\sqrt{5}}{\\sqrt{7}+\\sqrt{5}})$, then what is the value of $\\frac{1}{N}$ ?<\/p>\n a)\u00a0$6-\\sqrt{35}$<\/p>\n b)\u00a0$6+\\sqrt{35}$<\/p>\n c)\u00a0$7+\\sqrt{35}$<\/p>\n d)\u00a0$7-\\sqrt{35}$<\/p>\n Question 27:\u00a0<\/b>If cosec \u03b8 + 3 sec \u03b8 = 5 cosec \u03b8, then what is the value of cot \u03b8?<\/p>\n a)\u00a04\/3<\/p>\n b)\u00a03\/4<\/p>\n c)\u00a01\/\u221a3<\/p>\n d)\u00a0\u221a3<\/p>\n Question 28:\u00a0<\/b>What is the simplified value of $\\ \\frac{7}{sec^{2} \\theta}+ \\frac{3}{1+cot^{2} \\theta}+ 4\\ sin^{2} \\theta$?<\/p>\n a)\u00a03<\/p>\n b)\u00a04<\/p>\n c)\u00a05<\/p>\n d)\u00a07<\/p>\n Question 29:\u00a0<\/b>If \u221a5 tan \u03b8 = 5 sin \u03b8, then what is the value of (sin$^{2}$ \u03b8 – cos$^{2}$\u03b8)?<\/p>\n a)\u00a03\/5<\/p>\n b)\u00a01\/5<\/p>\n c)\u00a04\/5<\/p>\n d)\u00a02\/5<\/p>\n Question 30:\u00a0<\/b>If tan $\\ \\theta\\ $= $\\ \\frac{2}{3}\\ $, then what is the value of $\\ \\frac{15sin^{2} \\theta-3cos^{2} \\theta}{5sin^{2} \\theta+3cos^{2} \\theta}$?<\/p>\n a)\u00a0$\\frac{33}{32}$<\/p>\n b)\u00a0$\\frac{11}{29}$<\/p>\n c)\u00a0$\\frac{33}{47}$<\/p>\n d)\u00a0$\\frac{11}{32}$<\/p>\n Question 31:\u00a0<\/b>In an isosceles triangle PQR, \u2220P = 130$^\\circ$. If I is the in-centre of the triangle, then what is the value (in degrees) of \u2220QIR?<\/p>\n a)\u00a0130<\/p>\n b)\u00a0120<\/p>\n c)\u00a0155<\/p>\n d)\u00a0165<\/p>\n Question 32:\u00a0<\/b>In the given figure, EF = CE = CA, What is the value (in degrees) of $\\angle$EAC?<\/p>\n a)\u00a058<\/p>\n b)\u00a064<\/p>\n c)\u00a072<\/p>\n d)\u00a032<\/p>\n Question 33:\u00a0<\/b>In the given figure, O is the center of the circle, $\\ \\angle$PQR = 100$^\\circ\\ $and $\\angle$STR = 105$^\\circ$. What is the value (in degrees) of $\\ \\angle$OSP?<\/p>\n a)\u00a095<\/p>\n b)\u00a045<\/p>\n c)\u00a075<\/p>\n d)\u00a065<\/p>\n Question 34:\u00a0<\/b>What is the value of $\\ \\frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$?<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a0-1<\/p>\n d)\u00a02<\/p>\n 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n Question 35:\u00a0<\/b>If $\\ x^{2}\\ – 3x + 1 = 0$, then what is the value of $\\ x^{4}\\ $+ $\\ \\frac{1}{x^{4}}$?<\/p>\n a)\u00a011<\/p>\n b)\u00a018<\/p>\n c)\u00a047<\/p>\n d)\u00a051<\/p>\n Question 36:\u00a0<\/b>If $\\ \\frac{3x-1}{x}+\\frac{5y-1}{y}+\\frac{7z-1}{z}$= 0, then what is the value of $\\ \\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\ $?<\/p>\n a)\u00a0-3<\/p>\n b)\u00a00<\/p>\n c)\u00a015<\/p>\n d)\u00a021<\/p>\n Question 37:\u00a0<\/b>If $x^{2}- 2\\sqrt{10}x+ 1 = 0$, then what is the value of $x – \\frac{1}{x}$?<\/p>\n a)\u00a04<\/p>\n b)\u00a06<\/p>\n c)\u00a03<\/p>\n d)\u00a05<\/p>\n Question 38:\u00a0<\/b>If $x^{2} -7x + 1 = 0$, then what is the value of $x + \\frac{1}{x}$?<\/p>\n a)\u00a07<\/p>\n b)\u00a03<\/p>\n c)\u00a051<\/p>\n d)\u00a047<\/p>\n Question 39:\u00a0<\/b>If the angles of a triangle are (2x – 8)$^{o}$, (2x + 18)$^{o}$and 6x$^{o}$. What is the value of 3x (in degrees)?<\/p>\n a)\u00a017<\/p>\n b)\u00a034<\/p>\n c)\u00a051<\/p>\n d)\u00a060<\/p>\n Question 40:\u00a0<\/b>What is the value of $999\\frac{1}{3}+999\\frac{1}{6}+999\\frac{1}{12}+999\\frac{1}{20}+999\\frac{1}{30}$?<\/p>\n a)\u00a0$999\\frac{1}{6}$<\/p>\n b)\u00a0$999\\frac{5}{6}$<\/p>\n c)\u00a0$4995\\frac{1}{6}$<\/p>\n d)\u00a0$4995\\frac{4}{6}$<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n solving the equation for x, $\\frac{21}{2}$=$\\frac{13x}{6}$<\/p>\n $x$=$\\frac{63}{13}$<\/p>\n $13x$=63 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let X=$ \\sqrt{30+\\sqrt{30}+…}$<\/p>\n Above equation can be written as<\/p>\n X=$\\Rightarrow\\sqrt{30+X}$<\/p>\n Squaring on both sides<\/p>\n $X^{2}$=30+X<\/p>\n $X^{2}$-X-30=0<\/p>\n $X^{2}$-6X+5X-30=0<\/p>\n X(X-6)+5(X-6)=0<\/p>\n (X-6)(X+5)=0<\/p>\n X=-5,6<\/p>\n Taking positive value<\/p>\n X=6<\/p>\n 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $2\\frac{1}{5}x^{2}=2750$<\/p>\n ==> $\\frac{11}{5}x^{2}=2750$<\/p>\n ==> $x^{2}= \\frac{2750\\times5}{11}$<\/p>\n ==> $x^{2}= 1250$<\/p>\n ==> $x = \\sqrt{1250}=\\sqrt{625\\times2}$<\/p>\n ==> $x = 25\\sqrt{2}$<\/p>\n 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given sin A+$sin^{2}$ A=1<\/p>\n ==> sin A = 1-$sin^{2}$ A<\/p>\n ==> sin A = $cos^{2}$ A ($\\because cos^{2}A+sin^{2}A$=1)<\/p>\n $cos^{2}$ A=sin A ==> $cos^{4}A$=$sin^{2}A $<\/p>\n $\\therefore cos^{2}A+cos^{4}A=1 ( \\because sin A+sin^{2}A=1)$<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given x=\\sqrt[3]{5}+2<\/p>\n $\\ x^{3}-6x^{2}\\ $+ 12x – 13<\/p>\n = $(\\sqrt[3]{5}+2)^{3}-6(\\sqrt[3]{5}+2)^{2}+12(\\sqrt[3]{5}+2)-13$<\/p>\n = $(5+8+6\\times5^{\\frac{2}{3}}+12\\times5^{\\frac{1}{3}})-6[5^{\\frac{1}{3}}+4+4\\times5^{\\frac{1}{3}}]+12(5^{\\frac{1}{3}}+2)-13$<\/p>\n = $13+6\\times5^{\\frac{2}{3}}+12\\times5^{\\frac{1}{3}}-6\\times5^{\\frac{2}{3}}-24-24\\times5^{\\frac{1}{3}}+12\\times5^{\\frac{1}{3}}+24-13$<\/p>\n = 0<\/p>\n SSC GD FREE STUDY MATERIAL<\/a><\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given 2x+$\\frac{1}{x} =$ 3<\/p>\n 2(x+$\\frac{1}{x}) =$ 3<\/p>\n $\\Rightarrow$ x+$\\frac{1}{x} = \\frac{3}{2}$<\/p>\n Cubing on both sides<\/p>\n (x+$\\frac{1}{x})^{3} = \\frac{27}{8}$<\/p>\n x$^{3}$+$\\frac{1}{x^{3}}$+3$\\times$x$\\times$ $\\frac{1}{x}$(x+$\\frac{1}{x}$) $= \\frac{27}{8}$<\/p>\n $\\Rightarrow x^{3}+\\frac{1}{x^{3}}+3(\\frac{3}{2}) = \\frac{27}{8}$<\/p>\n $\\Rightarrow x^{3}+\\frac{1}{x^{3}} = \\frac{27}{8}-\\frac{9}{2} = \\frac{-9}{8}$<\/p>\n $x^{3}+\\frac{1}{x^{3}}+2 = \\frac{-9}{8}+2 = \\frac{7}{8}$<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\ a^{2}+b^{2}+c^{2}$= 2(a-b-c)-3 $ a^{2}-2a+1+b^{2}+2b+1+c^{2}+2c+1$=0 $(a-1)^{2}$=0$\\Rightarrow$a=1 4a-3b+5c= 4(1)-3(-1)+5(-1)=4+3-5=2<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\frac{3\\sqrt{2}}{(\\sqrt{3}+\\sqrt{6})}-\\frac{4\\sqrt{3}}{(\\sqrt{6}+\\sqrt{2})}+\\frac{\\sqrt{6}}{(\\sqrt{2}+\\sqrt{3})}$<\/p>\n = $\\frac{6\\sqrt{6}+18+6\\sqrt{2}+6\\sqrt{3}-(12\\sqrt{2}+24+12\\sqrt{3}+12\\sqrt{6})+6\\sqrt{3}+6+6\\sqrt{6}+6\\sqrt{2}}{(\\sqrt{3}+\\sqrt{6})(\\sqrt{6}+\\sqrt{2})(\\sqrt{2}+\\sqrt{3})}$<\/p>\n = $\\frac{6\\sqrt{6}+18+6\\sqrt{2}+6\\sqrt{3}-12\\sqrt{2}-24-12\\sqrt{3}-12\\sqrt{6}+6\\sqrt{3}+6+6\\sqrt{6}+6\\sqrt{2}}{(\\sqrt{3}+\\sqrt{6})(\\sqrt{6}+\\sqrt{2})(\\sqrt{2}+\\sqrt{3})}$<\/p>\n =0<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n To find : $y=\\sqrt{2\\sqrt[3]{4}\\sqrt{2\\sqrt[3]{4}}\\sqrt[4]{2\\sqrt[3]{4}}…..}$<\/p>\n Let $2\\sqrt[3]4=x$<\/p>\n => $y=\\sqrt{(x)\\times(\\sqrt{x})\\times(\\sqrt[4]{x})\\times…….}$<\/p>\n => $y^2=(x)^{[1+\\frac{1}{2}+\\frac{1}{4}+……+\\infty]}$<\/p>\n Now, sum of infinite G.P. = $\\frac{a}{(1-r)}$, where first term = $a=1$ and common ratio = $r=\\frac{1}{2}$<\/p>\n => $y^2=(x)^{\\frac{1}{1-\\frac{1}{2}}}$<\/p>\n => $y^2=(x)^2$<\/p>\n => $y=x$<\/p>\n $\\therefore$ $\\sqrt{2\\sqrt[3]{4}\\sqrt{2\\sqrt[3]{4}}\\sqrt[4]{2\\sqrt[3]{4}}…..}=2\\sqrt[3]4$<\/p>\n => Ans – (A)<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n cosec$^{2}\\ 18^\\circ\\ – \\frac{1}{cot^{2}72^\\circ}\\ $<\/p>\n = cosec$^{2} 18^\\circ – tan^{2} 72^\\circ$ ($\\because \\frac{1}{cot^{2} \\ominus}$=$tan^{2}\\ominus$)<\/p>\n = cosec$^{2} 18^\\circ$ – $tan^{2} (90-72)^\\circ$ 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $x^{2}-3x+1$=0<\/p>\n Taking ‘x’ common<\/p>\n x(x-3+$\\frac{1}{x})$=0<\/p>\n $\\Rightarrow x+\\frac{1}{x}$=$3\\rightarrow(1)$<\/p>\n Squaring on both sides<\/p>\n $x^{2}+\\frac{1}{x^{2}}+2\\times x\\times\\frac{1}{x}$=9<\/p>\n $\\Rightarrow x^{2}+\\frac{1}{x^{2}}$=$7\\rightarrow(2)$<\/p>\n Cubing equation(1) on both sides<\/p>\n $x^{3}+\\frac{1}{x^{3}}+3\\times x\\times\\frac{1}{x}(x+\\frac{1}{x})$=27<\/p>\n $x^{3}+\\frac{1}{x^{3}}$+$3\\times 1\\times3$=27($\\because x+\\frac{1}{x}$=3)<\/p>\n $x^{3}+\\frac{1}{x^{3}}$=27-9=$18\\rightarrow(3)$<\/p>\n Squaring equation(2) on both sides<\/p>\n $x^{4}+\\frac{1}{x^{4}}+2\\times x^{2}\\times\\frac{1}{x^{2}}$=49<\/p>\n $x^{4}+\\frac{1}{x^{4}}$=$47\\rightarrow(4)$<\/p>\n Multiplying equation(1) and equation(4)<\/p>\n $(x^{4}+\\frac{1}{x^{4}})(x+\\frac{1}{x}$)=$47\\times3$<\/p>\n $x^{5}+\\frac{1}{x^{5}}+x^{3}+\\frac{1}{x^{3}}$=$47\\times3$=141<\/p>\n $x^{5}+\\frac{1}{x^{5}}+18$=141($\\because x^{3}+\\frac{1}{x^{3}}$)<\/p>\n $\\therefore x^{5}+\\frac{1}{x^{5}}$=123<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : $(0.65\\times0.65)+(0.35\\times0.35)+(0.70\\times0.65)$<\/p>\n = $(0.65)^2+(0.35)^2+2(0.35)(0.65)$<\/p>\n Comparing with : $(x)^2+(y)^2+2(x)(y)=(x+y)^2$<\/p>\n = $(0.65+0.35)^2=(1)^2=1$<\/p>\n => Ans – (B)<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\frac{75.8^{2}-35.8^{2}}{40}$=$\\frac{(75.8+35.8)(75.8-35.8)}{40}$<\/p>\n =$\\frac{111.6\\times40}{40}$=111.6<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : $tan(\\theta)tan(5\\theta)=1$<\/p>\n Using, $tan(A+B)=\\frac{tanA+tanB}{1-tanAtanB}$<\/p>\n $tan(\\theta+5\\theta)=\\frac{tan(\\theta)+tan(5\\theta)}{1-tan(\\theta)tan(5\\theta)}$<\/p>\n => $tan(6\\theta)=\\frac{tan(\\theta)+tan(5\\theta)}{1-1}$<\/p>\n => $tan(6\\theta)=\\infty$<\/p>\n => $tan(6\\theta)=tan(90^\\circ)$<\/p>\n => $6\\theta=90^\\circ$<\/p>\n => $\\theta=\\frac{90^\\circ}{6}=15^\\circ$<\/p>\n $\\therefore$ $sin(2\\theta)=sin(2\\times15^\\circ)$<\/p>\n = $sin(30^\\circ)=\\frac{1}{2}$<\/p>\n => Ans – (B)<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression : $\\ \\frac{2sin^{3}\\ \\theta\\ – sin\\theta}{cos\\theta\\ -\\ 2\\ cos^{3}\\ \\theta}$<\/p>\n = $\\ \\frac{sin\\ \\theta(2sin^{2}\\ \\theta\\ – 1)}{cos\\ \\theta(2 -\\ 2\\ cos^{2}\\ \\theta)}$<\/p>\n = $\\frac{sin\\ \\theta(cos2\\ \\theta)}{cos\\ \\theta(cos2\\ \\theta)}$<\/p>\n = $\\frac{sin\\ \\theta}{cos\\ \\theta}=tan\\ \\theta$<\/p>\n => Ans – (A)<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $tan15^\\circ=\\frac{\\sqrt3-1}{\\sqrt3+1}$<\/p>\n Expression = $\\ \\frac{x-xtan^{2}15^\\circ}{1+tan^{2}15^\\circ}$= sin $60^\\circ\\ $+ cos 30$^\\circ\\ $<\/p>\n => $\\ \\frac{x(1-tan^{2}15^\\circ)}{1+tan^{2}15^\\circ}= \\frac{\\sqrt3}{2}+\\frac{\\sqrt3}{2}$<\/p>\n => $x\\times\\frac{1-(\\frac{\\sqrt3-1}{\\sqrt3+1})^2}{1+(\\frac{\\sqrt3-1}{\\sqrt3+1})^2}=\\sqrt3$<\/p>\n => $x\\times\\frac{(\\sqrt3+1)^2-(\\sqrt3-1)^2}{(\\sqrt3+1)^2+(\\sqrt3-1)^2}=\\sqrt3$<\/p>\n => $x\\times\\frac{(3+1+2\\sqrt3)-(3+1-2\\sqrt3)}{(3+1+2\\sqrt3)+(3+1-2\\sqrt3)}=\\sqrt3$<\/p>\n => $x\\times\\frac{4\\sqrt3}{8}=\\sqrt3$<\/p>\n => $x=\\frac{8}{4}=2$<\/p>\n => Ans – (A)<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression = $\\ \\sqrt{\\frac{sec^{2} \\theta+cosec^{2} \\theta}{4}}$<\/p>\n = $\\ \\sqrt{\\frac{(\\frac{1}{cos^2\\ \\theta})+(\\frac{1}{sin^2\\ \\theta})}{4}}$<\/p>\n = $\\ \\sqrt{\\frac{(\\frac{sin^2\\ \\theta+cos^2\\ \\theta)}{sin^2\\ \\theta.cos^2\\ \\theta}}{4}}$<\/p>\n = $\\sqrt{\\frac{1}{4sin^2\\ \\theta cos^2\\ \\theta}}$<\/p>\n = $\\sqrt{(\\frac{1}{2sin\\ \\theta cos\\ \\theta})^2}$<\/p>\n = $\\frac{1}{sin2\\theta}=cosec2\\theta$<\/p>\n => Ans – (A)<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let $\\angle Q=x$, => $\\angle P=2x$<\/p>\n Using exterior angle property in $\\triangle$ PQR,<\/p>\n => $\\angle$ P + $\\angle$ Q = $\\angle$ PRS<\/p>\n => $2x+x=105^\\circ$<\/p>\n => $x=\\frac{105^\\circ}{3}=35^\\circ$<\/p>\n $\\therefore$ $\\angle P=2\\times35^\\circ=70^\\circ$<\/p>\n => Ans – (C)<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : \u2220A : \u2220B : \u2220C = 3 : 3 : 4 and PQ is parallel to BC<\/p>\n To find : \u2220AQP – \u2220APQ = ?<\/p>\n Solution : Let $\\angle A=3x$, $\\angle B=3x$ and $\\angle C=4x$<\/p>\n Thus, in $\\triangle$ ABC,<\/p>\n => $\\angle A+\\angle B+\\angle C=180^\\circ$<\/p>\n => $3x+3x+4x=180^\\circ$<\/p>\n => $x=\\frac{180^\\circ}{10}=18^\\circ$<\/p>\n $\\because$ PQ $\\parallel$ BC, => $\\angle$ APQ = $\\angle$ B and $\\angle$ AQP = $\\angle$ C (Corresponding angles)<\/p>\n $\\therefore$ $\\angle$ AQP – $\\angle$ APQ = $4x-3x=x=18^\\circ$<\/p>\n => Ans – (B)<\/p>\n 21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x=5-\\frac{1}{x}$<\/p>\n => $x+\\frac{1}{x}=5=k$<\/p>\n Now, $x^5+\\frac{1}{x^5}=[(x^3+\\frac{1}{x^3})\\times(x^2+\\frac{1}{x^2})]-(x+\\frac{1}{x})$<\/p>\n = $[(x+\\frac{1}{x})^3-3(x+\\frac{1}{x})\\times(x+\\frac{1}{x})^2-2(x)(\\frac{1}{x})]-(x+\\frac{1}{x})$<\/p>\n = $[(k^3-3k)\\times(k^2-2)]-(k)$<\/p>\n = $[(125-15)\\times(25-2)]-(5)$<\/p>\n = $(110\\times23)-5$<\/p>\n = $2530-5=2525$<\/p>\n => Ans – (C)<\/p>\n SSC GD Important Questions & Answers PDF<\/a><\/p>\n 22)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x^3-y^3=112$ ————–(i)<\/p>\n Also, $x-y=4$ ————-(ii)<\/p>\n Cubing both sides, we get :<\/p>\n => $(x-y)3=(4)^3$<\/p>\n => $(x^3-y^3)-3(x)(y)(x-y)=64$<\/p>\n Substituting values from equations (i) and (ii),<\/p>\n => $112-3xy(4)=64$<\/p>\n => $12xy=112-64=48$<\/p>\n => $xy=\\frac{48}{12}=4$ ———–(iii)<\/p>\n Now, squaring equation (ii), we get :<\/p>\n => $x^2+y^2-2xy=16$<\/p>\n => $x^2+y^2=16+8=24$<\/p>\n => Ans – (C)<\/p>\n 23)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : $3x-\\frac{1}{3x}=9$<\/p>\n Dividing both sides by 3, => $x-\\frac{1}{9x}=3$<\/p>\n Squaring both sides, we get :<\/p>\n => $(x-\\frac{1}{9x})^2=(3)^2$<\/p>\n => $x^2+\\frac{1}{81x^2}-2(x)(\\frac{1}{9x})=9$<\/p>\n => $(x^2+\\frac{1}{81x^2})-\\frac{2}{9}=9$<\/p>\n => $(x^2+\\frac{1}{81x^2})=9+\\frac{2}{9}$<\/p>\n => $(x^2+\\frac{1}{81x^2})=\\frac{83}{9}$<\/p>\n => Ans – (B)<\/p>\n 24)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given : $\\ x^{4}+\\frac{1}{x^{4}}\\ =198$<\/p>\n => $(x^2-\\frac{1}{x^2})^2+2(x^2)(\\frac{1}{x^2})=198$<\/p>\n => $(x^2-\\frac{1}{x^2})^2=198-2=196$<\/p>\n => $x^2-\\frac{1}{x^2}=\\sqrt{196}=14$<\/p>\n => Ans – (A)<\/p>\n 25)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x+y=4$ ———–(i)<\/p>\n Cubing both sides, we get :<\/p>\n => $(x+y)^3=(4)^3$<\/p>\n => $x^3+y^3+3xy(x+y)=64$<\/p>\n => $x^3+y^3+3xy(4)=64$<\/p>\n => $x^3+y^3+12xy=64$<\/p>\n => Ans – (C)<\/p>\n 26)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : $N=\\frac{\\sqrt7-\\sqrt5}{\\sqrt7+\\sqrt5}$<\/p>\n => $\\frac{1}{N}=\\frac{\\sqrt7+\\sqrt5}{\\sqrt7-\\sqrt5}$<\/p>\n Rationalizing the denominator, we get :<\/p>\n = $\\frac{\\sqrt7+\\sqrt5}{\\sqrt7-\\sqrt5}\\times\\frac{\\sqrt7+\\sqrt5}{\\sqrt7+\\sqrt5}$<\/p>\n = $\\frac{(\\sqrt7+\\sqrt5)^2}{(\\sqrt7-\\sqrt5)(\\sqrt7+\\sqrt5)}$<\/p>\n = $\\frac{7+5+2(\\sqrt7)(\\sqrt5)}{7-5}$<\/p>\n = $\\frac{12+2\\sqrt{35}}{2}=6+\\sqrt{35}$<\/p>\n => Ans – (B)<\/p>\n 27)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : $cosec\\theta+3sec\\theta=5cosec\\theta$<\/p>\n => $3sec\\theta=5cosec\\theta-cosec\\theta$<\/p>\n => $3sec\\theta=4cosec\\theta$<\/p>\n => $\\frac{3}{cos\\theta}=\\frac{4}{sin\\theta}$<\/p>\n => $\\frac{cos\\theta}{sin\\theta}=\\frac{3}{4}$<\/p>\n => $cot\\theta=\\frac{3}{4}$<\/p>\n => Ans – (B)<\/p>\n 28)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression : $\\ \\frac{7}{sec^{2} \\theta}+ \\frac{3}{1+cot^{2} \\theta}+ 4\\ sin^{2} \\theta$<\/p>\n = $7cos^2\\ \\theta+\\frac{3}{cosec^2\\ \\theta}+4sin^2\\ \\theta$<\/p>\n = $7cos^2\\ \\theta+3sin^2\\ \\theta+4sin^2\\ \\theta$<\/p>\n = $7cos^2\\ \\theta+7sin^2\\ \\theta$<\/p>\n = $7(cos^2\\ \\theta+sin^2\\ \\theta)$<\/p>\n = $7\\times1=7$<\/p>\n => Ans – (D)<\/p>\n 29)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given : $\\sqrt5tan\\ \\theta=5sin\\ \\theta$<\/p>\n => $\\frac{sin\\ \\theta}{cos\\ \\theta}=\\sqrt5sin\\ \\theta$<\/p>\n => $cos\\ \\theta=\\frac{1}{\\sqrt5}$<\/p>\n => $cos^2\\ \\theta=\\frac{1}{5}$ —————(i)<\/p>\n Now, $sin^2\\ \\theta=1-cos^2\\ \\theta$<\/p>\n => $sin^2\\ \\theta=1-\\frac{1}{5}=\\frac{4}{5}$ ————–(ii)<\/p>\n Subtracting equation (i) from (ii), we get :<\/p>\n $\\therefore$ $(sin^2\\ \\theta-cos^2\\ \\theta)=\\frac{4}{5}-\\frac{1}{5}=\\frac{3}{5}$<\/p>\n => Ans – (A)<\/p>\n 30)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $tan\\ \\theta=\\frac{2}{3}$<\/p>\n => $\\frac{sin\\ \\theta}{cos\\ \\theta}=\\frac{2}{3}$<\/p>\n Let $sin\\ \\theta=2$ and $cos\\ \\theta=3$<\/p>\n To find : $\\ \\frac{15sin^{2} \\theta-3cos^{2} \\theta}{5sin^{2} \\theta+3cos^{2} \\theta}$<\/p>\n = $\\frac{15(2)^2-3(3)^2}{5(2)^2+3(3)^2}$<\/p>\n = $\\frac{60-27}{20+27}$<\/p>\n = $\\frac{33}{47}$<\/p>\n => Ans – (C)<\/p>\n 31)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : I is the incentre of $\\triangle$ PQR and $\\angle$ BAC = 130\u00b0<\/p>\n To find : $\\angle$ QIR = $\\theta$ = ?<\/p>\n Incentre of a triangle = $90^\\circ+\\frac{\\angle P}{2}$<\/p>\n => $\\theta=90^\\circ+\\frac{130^\\circ}{2}$<\/p>\n => $\\theta=90^\\circ+65^\\circ$<\/p>\n => $\\theta=155^\\circ$<\/p>\n => Ans – (C)<\/p>\n 32)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : EF = CE = CA<\/p>\n => $\\angle$ CAE = $\\angle$ CEA = $x$ and $\\angle$ ECF = $\\angle$ EFC = $y$<\/p>\n To find : $\\angle$ EAC = $x=?$<\/p>\n Solution : Using exterior angle property, => $\\angle$ CAE + $\\angle$ CFE = $\\angle$ ACD<\/p>\n => $x+y=96^\\circ$ —————–(i)<\/p>\n Also, $\\angle$ CEF = $(180^\\circ-2y)=180^\\circ-x$<\/p>\n => $x=2y$ ————(ii)<\/p>\n Substituting it in equation (i), => $2y+y=3y=96^\\circ$<\/p>\n => $y=\\frac{96}{3}=32^\\circ$<\/p>\n $\\therefore$ $x=2\\times32=64^\\circ$<\/p>\n => Ans – (B)<\/p>\n 33)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given : $\\ \\angle$PQR = 100$^\\circ\\ $and $\\angle$STR = 105$^\\circ$<\/p>\n To find : $\\ \\angle$OSP = ?<\/p>\n Solution : Quadrilateral PQRS is cyclic quadrilateral, hence opposite angles are supplementary.<\/p>\n => $\\angle$ PQR + $\\angle$ PSR = $180^\\circ$<\/p>\n => $\\angle$ PSR = $180^\\circ-100^\\circ=80^\\circ$ ————–(i)<\/p>\n Also, angle at the centre is double the angle at any point on the circumference of the circle in the same segment.<\/p>\n => reflex ($\\angle$ SOR) = $2$ $\\times$ $\\angle$ STR<\/p>\n => reflex ($\\angle$ SOR) = $2\\times105^\\circ=210^\\circ$<\/p>\n Thus, $\\angle$ SOR = $360^\\circ-210^\\circ=150^\\circ$<\/p>\n Now, in $\\triangle$ SOR, OS = OR = radius<\/p>\n => $\\angle$ OSR = $\\angle$ ORS = $15^\\circ$ ———-(ii)<\/p>\n Subtracting equation (ii) from (i), we get :<\/p>\n $\\therefore$ $\\angle$ OSP = $80^\\circ-15^\\circ=65^\\circ$<\/p>\n => Ans – (D)<\/p>\n 34)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression : $\\ \\frac{(a^{2}+b^{2})(a-b)-(a-b)^{2}}{a^{2}b-ab^{2}}$<\/p>\n = $\\ \\frac{(a-b)[(a^{2}+b^{2})-(a-b)]}{(ab)(a-b)}$<\/p>\n = $\\frac{(a^2+b^2)-(a-b)}{ab}$<\/p>\n = $\\frac{(a-b)^2+2ab-(a-b)}{ab}$<\/p>\n 35)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $x^2-3x+1=0$<\/p>\n Dividing both sides by $’x’$<\/p>\n => $x+\\frac{1}{x}=3$<\/p>\n Squaring both sides, we get :<\/p>\n => $x^2+\\frac{1}{x^2}+2(x)(\\frac{1}{x})=9$<\/p>\n => $x^2+\\frac{1}{x^2}=9-2=7$<\/p>\n Again squaring both sides,<\/p>\n => $x^4+\\frac{1}{x^4}+2(x^2)(\\frac{1}{x^2})=49$<\/p>\n => $x^4+\\frac{1}{x^4}=49-2=47$<\/p>\n => Ans – (C)<\/p>\n 36)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given : $\\ \\frac{3x-1}{x}+\\frac{5y-1}{y}+\\frac{7z-1}{z}=0$<\/p>\n => $(3-\\frac{1}{x})+(5-\\frac{1}{y})+(7-\\frac{1}{z})=0$<\/p>\n => $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=3+5+7$<\/p>\n => $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=15$<\/p>\n => Ans – (C)<\/p>\n 37)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given : $x^2-2\\sqrt{10}x+1=0$<\/p>\n Dividing both sides by $’x’$<\/p>\n => $x+\\frac{1}{x}=2\\sqrt{10}$<\/p>\n Squaring both sides, we get :<\/p>\n => $x^2+\\frac{1}{x^2}+2(x)(\\frac{1}{x})=40$<\/p>\n => $x^2+\\frac{1}{x^2}=40-2=38$<\/p>\n => $(x-\\frac{1}{x})^2+2(x)(\\frac{1}{x})=38$<\/p>\n => $(x-\\frac{1}{x})^2=38-2=36$<\/p>\n => $x-\\frac{1}{x}=\\sqrt{36}=6$<\/p>\n => Ans – (B)<\/p>\n 38)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given : $x^2-7x+1=0$<\/p>\n Dividing both sides by $’x’$<\/p>\n => $x-7+\\frac{1}{x}=0$<\/p>\n => $x+\\frac{1}{x}=7$<\/p>\n => Ans – (A)<\/p>\n 39)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Sum of angles of a triangle = $180^\\circ$<\/p>\n => $(2x-8)^\\circ+(2x+18)^\\circ+(6x)^\\circ=180^\\circ$<\/p>\n => $10x+10=180$<\/p>\n => $10x=180-10=170$<\/p>\n => $x=\\frac{170}{10}=17$<\/p>\n $\\therefore$ $3x=3\\times17=51$<\/p>\n => Ans – (C)<\/p>\n 40)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression : $999\\frac{1}{3}+999\\frac{1}{6}+999\\frac{1}{12}+999\\frac{1}{20}+999\\frac{1}{30}$<\/p>\n = $(999+999+999+999+999)+(\\frac{1}{3}+\\frac{1}{6}+\\frac{1}{12}+\\frac{1}{20}+\\frac{1}{30})$<\/p>\n = $(4995)+(\\frac{20+10+5+3+2}{60})$<\/p>\n = $4995+\\frac{40}{60}$<\/p>\n = $4995\\frac{4}{6}$<\/p>\n => Ans – (D)<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/37_3xIRG00.png\")
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\nWe get $\\frac{21}{2}$=$2x+\\frac{x}{6}$<\/p>\n
\nFactorising 63 we get 63=$3^{2}\\times 7$
\nNumber of factors=(2+1)(1+1)
\n=6<\/p>\n
\nWe can write the above equation as<\/p>\n
\n$\\Rightarrow (a-1)^{2}+(b+1)^{2}+(c+1)^{2}$=0<\/p>\n
\n$(b+1)^{2}$=0$\\Rightarrow$ b=-1
\n$(c+1)^{2}$=0$\\Rightarrow$ c=-1<\/p>\n
\n= cosec$^{2} 18^\\circ$ – $sec^{2} 18^\\circ$ ($\\because sec^{2}\\ominus= tan^{2}(90^\\circ-\\ominus)$)
\ncosec$^{2} 18^\\circ$ – $sec^{2} 18^\\circ$=1($\\because cosec^{2} \\ominus$ – $sec^{2} \\ominus$=1)<\/p>\n<\/figure>\n
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