Download SSC CGL Algebra questions with answers PDF based on previous papers very useful for SSC CGL exams. 20 Very important Algebra objective questions (MCQ’s) for SSC exams.<\/p>\n
Download Algebra Questions For SSC CGL Set-2 PDF<\/a><\/p>\n Take a SSC CGL Free Mock Test<\/a><\/p>\n Question 1:\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 2:\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 3:\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 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>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 SSC CGL Previous Papers Download PDF<\/a><\/p>\n Get 780+ mocks for just Rs. 299<\/a><\/p>\n Question 6:\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 7:\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 8:\u00a0<\/b>Find the number of even factors of 16200.<\/p>\n a)\u00a060<\/p>\n b)\u00a042<\/p>\n c)\u00a036<\/p>\n d)\u00a045<\/p>\n Question 9:\u00a0<\/b>Find the number of even factors of 3920.<\/p>\n a)\u00a024<\/p>\n b)\u00a042<\/p>\n c)\u00a036<\/p>\n d)\u00a030<\/p>\n Question 10:\u00a0<\/b>Find the number of odd factors of 13500.<\/p>\n a)\u00a024<\/p>\n b)\u00a032<\/p>\n c)\u00a016<\/p>\n d)\u00a048<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Question 11:\u00a0<\/b>Find the number of prime factors of 5544.<\/p>\n a)\u00a07<\/p>\n b)\u00a04<\/p>\n c)\u00a06<\/p>\n d)\u00a03<\/p>\n Question 12:\u00a0<\/b>What is the square root of $214 – 78\\sqrt{5}$<\/p>\n a)\u00a0$13-3\\sqrt{5}$<\/p>\n b)\u00a0$17-4\\sqrt{5}$<\/p>\n c)\u00a0$13-6\\sqrt{5}$<\/p>\n d)\u00a0$17-3\\sqrt{5}$<\/p>\n Question 13:\u00a0<\/b>If $(9^{x})(3^y) = 81$ and $(2^{x})(4^y) = 32$, then find $(x, y)$<\/p>\n a)\u00a0(1,4)<\/p>\n b)\u00a0(2,1)<\/p>\n c)\u00a0(1,2)<\/p>\n d)\u00a0(4,1)<\/p>\n Question 14:\u00a0<\/b>If p = 21, q = -16, r = -5, then find the value of $\\large\\frac{8p^{3}+8q^{3}+8r^{3}}{9pqr}$<\/p>\n a)\u00a0$\\Large\\frac{8}{3}$<\/p>\n b)\u00a0$\\Large\\frac{16}{3}$<\/p>\n c)\u00a0$\\Large\\frac{8}{9}$<\/p>\n d)\u00a0$\\Large\\frac{8}{16}$<\/p>\n Question 15:\u00a0<\/b>Find the value of $\\sqrt{12+\\sqrt{12+\\sqrt{12+……}}}$<\/p>\n a)\u00a04<\/p>\n b)\u00a03<\/p>\n c)\u00a06<\/p>\n d)\u00a07<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n SSC CHSL Free Mock Test<\/a><\/p>\n Question 16:\u00a0<\/b>What is the units digit of $437^{79}$ ?<\/p>\n a)\u00a07<\/p>\n b)\u00a09<\/p>\n c)\u00a01<\/p>\n d)\u00a03<\/p>\n Question 17:\u00a0<\/b>If $(p-q)^{2} = 9$ and $(p+q)^{2} = 121$ $p,q>0$, then find the value of $\\frac{pq}{p+q}$<\/p>\n a)\u00a0$\\frac{22}{13}$<\/p>\n b)\u00a0$\\frac{24}{13}$<\/p>\n c)\u00a0$\\frac{24}{11}$<\/p>\n d)\u00a0$\\frac{28}{11}$<\/p>\n Question 18:\u00a0<\/b>What is the units digit of $639^{147}$ ?<\/p>\n a)\u00a07<\/p>\n b)\u00a09<\/p>\n c)\u00a01<\/p>\n d)\u00a03<\/p>\n Question 19:\u00a0<\/b>If $3p – (1\/2)(p -3) = 14$, find p ?<\/p>\n a)\u00a02<\/p>\n b)\u00a03<\/p>\n c)\u00a04<\/p>\n d)\u00a05<\/p>\n Question 20:\u00a0<\/b>A fraction is greater than its reciprocal by $\\Large\\frac{9}{20}$. What is the fraction?<\/p>\n a)\u00a0$\\Large\\frac{5}{4}$<\/p>\n b)\u00a0$\\Large\\frac{-4}{5}$<\/p>\n c)\u00a0$\\Large\\frac{1}{5}$<\/p>\n d)\u00a0Either A or B.<\/p>\n 1500+ Free SSC Questions & Answers<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\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 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given 2x+y = 3 and x+2y = 3 3)\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 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(C)<\/strong><\/p>\n we have $(a-b)^{2}$=$a^{2}+b^{2}-2ab$ 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We have to factorise the number into prime factors i.e 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We have to factorise the number into prime factors i.e 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We have to factorise the number into prime factors i.e Hence, Option D is correct answer<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We have to factorise the number into prime factors i.e 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We have to factorise the number into prime factors i.e 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We have to factorise the number into prime factors i.e 15000 Questions – Free SSC Study Material<\/a><\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n we have $(a-b)^{2}$=$a^{2}+b^{2}-2ab$<\/p>\n Comparing this with $214-78\\sqrt{5}$ = $a^{2}+b^{2}-2ab$<\/p>\n We have 214=$a^{2}+b^{2}$<\/p>\n & $ab = 39\\sqrt{5}$<\/p>\n For a=13 and b=3$\\sqrt{5}$; 2ab=2*13*3$\\sqrt{5}$<\/p>\n And so required answer is 13-3$\\sqrt{5}$<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, $(9^{x})(3^y) = 81$ and $(2^{x})(4^y) = 32$ 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given p = 21, q = -16, r = -5, We know that if p+q+r = 0, then $p^{3}+q^{3}+r^{3} = 3pqr$<\/p>\n Then, $\\large\\frac{8p^{3}+8q^{3}+8r^{3}}{9pqr}$<\/p>\n =$\\large\\frac{8p^{3}+8q^{3}+8r^{3}}{9pqr}$<\/p>\n = $\\Large\\frac{8(3pqr)}{9pqr}$<\/p>\n = $\\Large\\frac{8}{3}$<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let\u00a0$\\sqrt{12+\\sqrt{12+\\sqrt{12+……}}}$\u00a0= X<\/p>\n Then, $\\sqrt{12+X} = X$<\/p>\n Squaring on both sides,<\/p>\n $12+X = X^{2}$<\/p>\n \u21d2 $X^{2}-X-12 = 0$<\/p>\n \u21d2 $X^{2}-4X+3X-12 = 0$<\/p>\n \u21d2 $X(X-4)+3(X-4) = 0$<\/p>\n \u21d2 $(X-4)(X+3) = 0$<\/p>\n \u21d2 $X = 4$ or $X = -3$<\/p>\n X cannot be negative when all the terms are positive. <\/p>\n 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n In the cycle of 7 power we get 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given $(p-q)^{2} = 9$ and $(p+q)^{2} = 121$<\/p>\n $(p+q)^{2} = (p-q)^{2}+4pq$<\/p>\n $121 = 9+4pq$<\/p>\n => $4pq = 112$<\/p>\n $pq = 28$<\/p>\n $(p+q)^{2} = 121$<\/p>\n => $p+q = 11$ As,\u00a0$p,q>0$<\/p>\n Hence, $\\frac{pq}{p+q} = \\frac{28}{11}$<\/p>\n <\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $639^{147}$<\/p>\n In the cycle of 9 power we get $639^{2(73)+1}$ ends with 9<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given, $3p – (1\/2)(p -3) = 14$<\/p>\n $(6p – p +3)\/2 = 14$<\/p>\n $(5p +3)\/2 = 14$<\/p>\n $(5p +3) = 28$<\/p>\n $5p = 25$<\/p>\n $ p = 5$<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let that fraction be $\\dfrac{1}{f}$<\/p>\n $ (\\dfrac{1}{f}) = f + \\dfrac{9}{20}$<\/p>\n $\\Rightarrow 20f^{2}+9f-20 = 0$<\/p>\n $\\Rightarrow(4f+5)(5f-4)=0$<\/p>\n $\\Rightarrow(f+\\frac{5}{4})(f-\\frac{4}{5})=0$<\/p>\n $\\Rightarrow f = \\dfrac{-5}{4}$ or $\\dfrac{4}{5}$<\/p>\n $\\Large\\Rightarrow$ fraction $=$ $\\Large\\frac{1}{f}$
\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
\nSolving above equations,
\nWe get x = 1 and y = 1.
\nHence, the lines intersect at (1,1)<\/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
\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
\n14560=$2^{5}*5*13*7$
\nThere are 4 different prime factors namely 2,5,7 and 13.<\/p>\n
\n15680=$2^{6}*5*7^{2}$
\nNo of even factors =6*(1+1)*(2+1)
\n=36<\/p>\n
\n16200=$2^{3}*5^{2}*3^{4}$
\nNo of even factors =3*(2+1)*(4+1)
\n=45<\/p>\n
\n3920=$2^{4}*5*7^{2}$
\nNo of even factors =4*(1+1)*(2+1)
\n= 24<\/p>\n
\n13500=$2^{2}*3^{3}*5^{3}$
\nNo of odd factors =1*(3+1)*(3+1)
\n= 16<\/p>\n
\n5544 = $2^{3}*3^{2}*7*11$
\nNo of prime factors =4 (2, 3, 7, 11)<\/p>\n
\n$(9^{x})(3^y) = 81$
\n=> $3^{2x+y} = 3^{4}$
\n=> $2x+y = 4$ — (1)
\n$(2^{x})(4^y) = 32$
\n=> $(2^{x})((2^2)^y) = 2^5$
\n=> $2^x 2^{2y} = 2^5$
\n=> $x+2y = 5$ — (2)
\nSolving (1) and (2)
\n=> $y = 2$
\nSubstituting y = 2 in (1) –> x = 1
\nTherefore, $(x,y) = (1,2)$<\/p>\n
\nThen p+q+r = 0.<\/p>\n
\nHence, $X = 4$<\/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$437^{4(19)+3}$ has 3 as its unit digit.<\/p>\n
\n$9^{1}=9$
\n$9^{2}=81$
\n$9^{3}=729$
\n$9^{4}=6561$
\nAnd this cycle repeats. Cyclicity =2<\/p>\n
\n$=$ $\\Large\\frac{-4}{5}$ (or) $\\Large\\frac{5}{4}$<\/p>\n