Download SSC CGL Algebra Previous Year 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 Previous Questions For SSC CGL PDF<\/a><\/p>\n Take a SSC CGL Free Mock Test<\/a><\/p>\n Question 1:\u00a0<\/b>What is the value of $\\frac{3}{4}+\\frac{8}{9}$ ?<\/p>\n a)\u00a0$\\frac{57}{27}$<\/p>\n b)\u00a0$\\frac{11}{13}$<\/p>\n c)\u00a0$\\frac{59}{36}$<\/p>\n d)\u00a0$\\frac{11}{9}$<\/p>\n Question 2:\u00a0<\/b>The sum of 2xy(3x + 4y – 5z) and 5yz(2x – 3y) is<\/p>\n a)\u00a0$6x^{2}y-8xy^{2}+15y^{2}z$<\/p>\n b)\u00a0$6x^{2}y+8xy^{2}-15y^{2}z$<\/p>\n c)\u00a0$6x^{2}y+8xy^{2}-15y^{2}z-20xyz$<\/p>\n d)\u00a0$6x^{2}y-8xy^{2}+15y^{2}z+20xyz$<\/p>\n Question 3:\u00a0<\/b>If 4x-7<x-2 and $5x+\\frac{2}{3}\\geq3x+1$; then x can take which of the following values?<\/p>\n a)\u00a02<\/p>\n b)\u00a0-1<\/p>\n c)\u00a0-2<\/p>\n d)\u00a01<\/p>\n Question 4:\u00a0<\/b>The solution set of 4x – 3y = 47 and 3x + y = 32 is<\/p>\n a)\u00a0{(15, 3)}<\/p>\n b)\u00a0{(4, 12)}<\/p>\n c)\u00a0{(11, -1)}<\/p>\n d)\u00a0{(12, 3)}<\/p>\n Question 5:\u00a0<\/b>What is the value of $\\frac{(4a^{2} + 8b + 14c + 2)}{2}$ ?<\/p>\n a)\u00a0$2a^{2} + 4b + 7c + 1$<\/p>\n b)\u00a0$a^{2} + 4b + 7c + 1$<\/p>\n c)\u00a0$2a^{2} + 4b + 7c + 2$<\/p>\n d)\u00a0$a^{2} + 4b + 7c + 2$<\/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>Coefficient of $x^{2}$ in (x + 9)(6 – 4x) is<\/p>\n a)\u00a054<\/p>\n b)\u00a0-4<\/p>\n c)\u00a0-30<\/p>\n d)\u00a04<\/p>\n Question 7:\u00a0<\/b>If 7x + 6y = 5xy and 10x – 4y = 4xy, then value of x and y is<\/p>\n a)\u00a03, 2<\/p>\n b)\u00a02, 3<\/p>\n c)\u00a04, 2<\/p>\n d)\u00a05, 6<\/p>\n Question 8:\u00a0<\/b>Coefficient of $x^{2}$ in $6x^{3}+4x^{2}+2x+3$ is<\/p>\n a)\u00a04<\/p>\n b)\u00a06<\/p>\n c)\u00a03<\/p>\n d)\u00a02<\/p>\n Question 9:\u00a0<\/b>On dividing $8a^{2}b^{2} c^{2}$ by $4a^{2}$, we get<\/p>\n a)\u00a0$2b^{2}$<\/p>\n b)\u00a0$2c^{2}$<\/p>\n c)\u00a0$2b^{2}c^{2}$<\/p>\n d)\u00a0$2$<\/p>\n Question 10:\u00a0<\/b>If 2x + 6y = 3xy and 10x -\u00ad 3y = 4xy, find x, y.<\/p>\n a)\u00a03, 2<\/p>\n b)\u00a02, 3<\/p>\n c)\u00a04, 6<\/p>\n d)\u00a06, 4<\/p>\n 18000+ Questions – Free SSC Study Material<\/a><\/p>\n FREE SSC EXAM YOUTUBE VIDEOS<\/a><\/p>\n Question 11:\u00a0<\/b>If $2x-3(4-2x)<4x-5<4x+\\frac{2x}{3}$, then x can take which of the following values?<\/p>\n a)\u00a02<\/p>\n b)\u00a08<\/p>\n c)\u00a00<\/p>\n d)\u00a0-8<\/p>\n Question 12:\u00a0<\/b>If a – b = 11 and ab = 24, then value of $a^{2}+b^{2}$ \u00a0is<\/p>\n a)\u00a0169<\/p>\n b)\u00a037<\/p>\n c)\u00a073<\/p>\n d)\u00a048<\/p>\n Question 13:\u00a0<\/b>The simpli\ufb01ed form of \u00a0$(x+3)^{2}+(x-1)^{2}$ \u00a0is<\/p>\n a)\u00a0$(x^{2}+2x+5)$<\/p>\n b)\u00a0$2(x^{2}+2x+5)$<\/p>\n c)\u00a0$(x^{2}-2x+5)$<\/p>\n d)\u00a0$2(x^{2}-2x+5)$<\/p>\n Question 14:\u00a0<\/b>What should be added to 5(2x-y) to obtain 4(2x – 3y) + 5(x + 4y)?<\/p>\n a)\u00a03x – 13y<\/p>\n b)\u00a03x + 13y<\/p>\n c)\u00a013x – 3y<\/p>\n d)\u00a013x + 3y<\/p>\n Question 15:\u00a0<\/b>If 3(2 – 3x) < 2 – 3x \u2265 4x -6; then x can take which of the following values?<\/p>\n a)\u00a02<\/p>\n b)\u00a0-1<\/p>\n c)\u00a0-2<\/p>\n d)\u00a01<\/p>\n SSC CGL Free Mock Test<\/a><\/p>\n SSC CHSL Free Mock Test<\/a><\/p>\n Question 16:\u00a0<\/b>If (4x – 3) – (2x + 1) = 4, then the value of x is<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a04<\/p>\n d)\u00a03<\/p>\n Question 17:\u00a0<\/b>Which of the following equations has real and distinct roots?<\/p>\n a)\u00a0$3x^{2} – 6x + 2 = 0$<\/p>\n b)\u00a0$3x^{2} – 6x + 3 = 0$<\/p>\n c)\u00a0$x^{2} – 8x + 16 = 0$<\/p>\n d)\u00a0$4x^{2} – 8x + 4 = 0$<\/p>\n Question 18:\u00a0<\/b>Value of $(4a^{2}+12ab+9b^{2}\/(2a+3b)$ is<\/p>\n a)\u00a02a – 3b<\/p>\n b)\u00a02a + 3b<\/p>\n c)\u00a02a<\/p>\n d)\u00a03b<\/p>\n Question 19:\u00a0<\/b>Coef\ufb01cient of $x^{2}$\u00a0 in (x + 9)(6 – 4x)(4x – 7) is<\/p>\n a)\u00a0216<\/p>\n b)\u00a0-4<\/p>\n c)\u00a0-92<\/p>\n d)\u00a0108<\/p>\n Question 20:\u00a0<\/b>Given: 5x -3(2x-7) > 3x – 1 < 7 + 4x; then x can take which of the following values?<\/p>\n a)\u00a06<\/p>\n b)\u00a09<\/p>\n c)\u00a0-6<\/p>\n d)\u00a0-9<\/p>\n 1500+ Free SSC Questions & Answers<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0:\u00a0$\\frac{3}{4}+\\frac{8}{9}$<\/p>\n = $\\frac{3(9)+8(4)}{36}$<\/p>\n = $\\frac{27+32}{36} = \\frac{59}{36}$<\/p>\n => Ans – (C)<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Sum of\u00a02xy(3x + 4y – 5z) and 5yz(2x – 3y)<\/p>\n = $(6x^2y+8xy^2-10xyz)+(10xyz-15y^2z)$<\/p>\n = $6x^2y+8xy^2-15y^2z$<\/p>\n => Ans – (B)<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression 1\u00a0:\u00a04x – 7 < x – 2<\/p>\n => $4x-x$ < $7-2$<\/p>\n => $3x$ < $5$<\/p>\n => $x$ < $\\frac{5}{3}$ ————(i)<\/p>\n Expression 2 :\u00a0$5x+\\frac{2}{3}\\geq3x+1$<\/p>\n => $5x-3x \\geq 1-\\frac{2}{3}$<\/p>\n => $2x \\geq \\frac{1}{3}$<\/p>\n => $x \\geq \\frac{1}{6}$ ———–(ii)<\/p>\n Combining inequalities (i) and (ii), we get\u00a0: $\\frac{1}{6} \\leq x$ < $\\frac{5}{3}$<\/p>\n The only value that $x$ can take among the options = 1<\/p>\n => Ans – (D)<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Equation 1\u00a0:\u00a04x – 3y = 47<\/p>\n Equation 2\u00a0: 3x + y = 32<\/p>\n Multiplying equation (ii) by 3 and adding it to equation (i)<\/p>\n => $(4x+9x)+(-3y+3y)=(47+96)$<\/p>\n => $13x=143$<\/p>\n => $x = \\frac{143}{13}=11$<\/p>\n Substituting it in equation (ii), => $y=32-3(11) = 32 – 33 = -1$<\/p>\n $\\therefore (x,y)=(11,-1)$<\/p>\n => Ans – (C)<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Expression\u00a0:\u00a0$\\frac{(4a^{2} + 8b + 14c + 2)}{2}$<\/p>\n = $\\frac{2(2a^2+4b+7c+1)}{2}$<\/p>\n = $2a^2+4b+7c+1$<\/p>\n => Ans – (A)<\/p>\n 100+ Free GK Tests for SSC Exams<\/a><\/p>\n Download Free GK PDF<\/a><\/p>\n 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg :\u00a0In $ax^2$, coefficient is $a$<\/p>\n Expression\u00a0: $(x + 9)(6 – 4x)$<\/p>\n = $6x – 4x^2 + 54 – 36x$<\/p>\n = $-4x^2 – 30x + 54$<\/p>\n $\\therefore$ Coefficient of $x^2$ = -4<\/p>\n => Ans – (B)<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Equation 1\u00a0: 7x + 6y = 5xy<\/p>\n Equation 2\u00a0: 10x -\u00ad 4y = 4xy<\/p>\n Dividing both equations by $(xy)$<\/p>\n => $\\frac{7}{y} + \\frac{6}{x} = 5$<\/p>\n and $\\frac{10}{y}-\\frac{4}{x} = 4$<\/p>\n Let $\\frac{1}{y} = u$ and $\\frac{1}{x} = v$<\/p>\n => $7u+6v=5$ ————(iii)<\/p>\n and $10u-4v=4$ ————(iv)<\/p>\n Multiplying equation (iv) by 3 and equation (iii) by 2 and adding them, we get\u00a0:<\/p>\n => $(14u+30u) = (10+12)$<\/p>\n => $u = \\frac{22}{44} = \\frac{1}{2}$<\/p>\n Substituting it in equation (iv), => $4v = 5 – 4 = 1$<\/p>\n => $v = \\frac{1}{4}$<\/p>\n $\\therefore (x,y) = (4,2)$<\/p>\n => Ans – (C)<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg :\u00a0In $ax^2$, coefficient is $a$<\/p>\n Expression\u00a0:\u00a0$6x^{3}+4x^{2}+2x+3$<\/p>\n => Coefficient of $x^2 = 4$<\/p>\n => Ans – (A)<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n On dividing $8a^{2}b^{2} c^{2}$ by $4a^{2}$<\/p>\n = $\\frac{8a^2b^2c^2}{4a^2}$<\/p>\n = $\\frac{8}{4} \\times \\frac{a^2}{a^2} \\times (b^2c^2)$<\/p>\n = $2b^2c^2$<\/p>\n => Ans – (C)<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Equation 1\u00a0:\u00a02x + 6y = 3xy<\/p>\n Equation 2\u00a0: 10x -\u00ad 3y = 4xy<\/p>\n Dividing both equations by $(xy)$<\/p>\n => $\\frac{2}{y} + \\frac{6}{x} = 3$<\/p>\n and $\\frac{10}{y}-\\frac{3}{x} = 4$<\/p>\n Let $\\frac{1}{y} = u$ and $\\frac{1}{x} = v$<\/p>\n => $2u+6v=3$ ————(iii)<\/p>\n and $10u-3v=4$ ————(iv)<\/p>\n Multiplying equation (iv) by 2 and adding it to equation (iii), we get\u00a0:<\/p>\n => $22u = 11$<\/p>\n => $u = \\frac{11}{22} = \\frac{1}{2}$<\/p>\n Substituting it in equation (iii), => $6v = 3 – 1 = 2$<\/p>\n => $v = \\frac{2}{6} = \\frac{1}{3}$<\/p>\n $\\therefore (x,y) = (3,2)$<\/p>\n => Ans – (A)<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression 1\u00a0:\u00a02x – 3(4 – 2x) < 4x – 5<\/p>\n => $2x-12+6x$ < $4x-5$<\/p>\n => $8x-4x$ < $-5+12$<\/p>\n => $4x$ < $7$<\/p>\n => $x$ < $\\frac{7}{4}$ ———–(i)<\/p>\n Expression 2\u00a0:\u00a04x – 5 < 4x + 2x\/3<\/p>\n => $\\frac{2x}{3}$ > $-5$<\/p>\n => $x$ > $\\frac{-15}{2}$ ———-(ii)<\/p>\n Combining inequalities (i) and (ii), we get\u00a0: $\\frac{-15}{2}$ < $x$ < $\\frac{7}{4}$<\/p>\n The only value that $x$ can take among the options = 0<\/p>\n => Ans – (C)<\/p>\n 12)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given\u00a0: $(a – b) = 11$ and $ab = 24$<\/p>\n Using $(a – b)^2 = a^2 + b^2 – 2ab$<\/p>\n => $(11)^2 = (a^2 + b^2) – (2 \\times 24)$<\/p>\n => $(a^2 + b^2) = 121 + 48 = 169$<\/p>\n => Ans – (A)<\/p>\n 15000 Questions – Free SSC Study Material<\/a><\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression\u00a0:\u00a0$(x+3)^{2}+(x-1)^{2}$<\/p>\n = $(x^2+9+6x)+(x^2+1-2x)$<\/p>\n = $2x^2+4x+10$<\/p>\n = $2(x^2+2x+5)$<\/p>\n => Ans – (B)<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Let $m$ should be added to\u00a05(2x-y) to obtain 4(2x – 3y) + 5(x + 4y)<\/p>\n => $(m) + [5(2x-y)] = 4(2x-3y)+5(x+4y)$<\/p>\n => $m + 10x-5y=8x-12y+5x+20y$<\/p>\n => $m + 10x-5y=13x+8y$<\/p>\n => $m = (13x-10x)+(8y+5y)$<\/p>\n => $m=3x+13y$<\/p>\n => Ans – (B)<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression 1\u00a0:\u00a03(2 – 3x) < 2 – 3x<\/p>\n => $6-9x$ < $2-3x$<\/p>\n => $9x-3x$ > $6-2$<\/p>\n => $6x$ > $4$<\/p>\n => $x$ > $\\frac{2}{3}$ ————(i)<\/p>\n Expression 2 :\u00a02 – 3x \u2265 4x -6<\/p>\n => $4x+3x \\leq 2+6$<\/p>\n => $7x \\leq 8$<\/p>\n => $x \\leq \\frac{8}{7}$ ———–(ii)<\/p>\n Combining inequalities (i) and (ii), we get\u00a0: $\\frac{2}{3}$\u00a0<\u00a0$x \\leq \\frac{8}{7}$<\/p>\n The only value that $x$ can take among the options = 1<\/p>\n => Ans – (D)<\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0:\u00a0(4x – 3) – (2x + 1) = 4<\/p>\n => $4x-3-2x-1=4$<\/p>\n => $2x-4=4$<\/p>\n => $2x=4+4=8$<\/p>\n => $x=\\frac{8}{2}=4$<\/p>\n => Ans – (C)<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n A quadratic equation : $ax^2 + bx + c = 0$ has real and distinct roots iff Discriminant, $D = b^2 – 4ac$ > $0$<\/p>\n (A) : $3x^{2} – 6x + 2 = 0$<\/p>\n => D = $(-6)^2 – 4(3)(2) = 36 – 24 = 12$<\/p>\n (B) :\u00a0$3x^{2} – 6x + 3 = 0$<\/p>\n =>\u00a0D = $(-6)^2 – 4(3)(3) = 36 – 36 = 0$<\/p>\n (C) :\u00a0$x^{2} – 8x + 16 = 0$<\/p>\n => D = $(-8)^2 – 4(1)(16) = 64 – 64 = 0$<\/p>\n (D) :\u00a0$4x^{2} – 8x + 4 = 0$<\/p>\n =>\u00a0D = $(-8)^2 – 4(4)(4) = 64 – 64 = 0$<\/p>\n Thus, the equation :\u00a0$3x^{2} – 6x + 2 = 0$ has real and distinct roots.<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression\u00a0:\u00a0$(4a^{2}+12ab+9b^{2}\/(2a+3b)$<\/p>\n = $\\frac{(2a)^2+(3b)^2+(2.2a.3b)}{(2a+3b)}$<\/p>\n =\u00a0$\\frac{(2a+3b)^2}{(2a+3b)}$<\/p>\n = $2a+3b$<\/p>\n => Ans – (B)<\/p>\n 19)\u00a0Answer\u00a0(C)<\/strong><\/p>\n A coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression. Eg :\u00a0In $ax^2$, coefficient is $a$<\/p>\n Expression\u00a0: $(x + 9)(6 – 4x)(4x – 7)$<\/p>\n = $(6x – 4x^2 + 54 – 36x)(4x – 7)$<\/p>\n = $(-4x^2 – 30x + 54)(4x – 7)$<\/p>\n =\u00a0$4x(-4x^2 – 30x + 54) – 7(-4x^2 – 30x + 54)$<\/p>\n = $-16x^3 – 120x^2 + 216x + 28x^2 + 210x – 378$<\/p>\n = $-20x^3 – 92x^2 + 426x – 378$<\/p>\n $\\therefore$ Coefficient of $x^2$ = -92<\/p>\n => Ans – (C)<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression 1\u00a0:\u00a05x -3(2x-7) > 3x – 1<\/p>\n => $5x-6x+21$ > $3x-1$<\/p>\n => $3x+x$ < $21+1$<\/p>\n => $4x$ < $22$<\/p>\n => $x$ < $\\frac{11}{2}$ ———-(i)<\/p>\n Expression 2\u00a0:\u00a03x – 1 < 7 + 4x<\/p>\n => $4x-3x$ > $-1-7$<\/p>\n => $x$ > $-8$ ———-(ii)<\/p>\n Combining inequalities (i) and (ii), we get\u00a0: $-8$ < $x$ < $\\frac{11}{2}$<\/p>\n The only value that $x$ can take among the options = -6<\/p>\n => Ans – (C)<\/p>\n