Download important Algebra Questions PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Algebra Question and Answers for IBPS Clerk Exam.<\/p>\n
Download Algebra Questions For IBPS Clerk<\/a><\/p>\n Get 25 IBPS Clerk mocks for Rs. 149. Enroll here<\/a><\/p>\n Take Free IBPS Clerk Mock Test<\/a><\/p>\n Download IBPS Clerk Previous papers PDF<\/a><\/p>\n Go to Free Banking Study Material<\/a> (15,000 Solved Questions)<\/p>\n Question 1:\u00a0<\/b>If $\\frac{a}{b}=\\frac{2}{3}$, then the value of $(5a^3-2a^2b):(3ab^2-b^3)$ is:<\/p>\n a)\u00a016:27<\/p>\n b)\u00a032:29<\/p>\n c)\u00a034:19<\/p>\n d)\u00a027:16<\/p>\n Question 2:\u00a0<\/b>If $x + x^{-1} = 2$, then the value of $x^3 + x^{-3}$ is:<\/p>\n a)\u00a03<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a01<\/p>\n d)\u00a02<\/p>\n Question 3:\u00a0<\/b>If $(\\frac{x}{a}) + (\\frac{y}{b}) = 3$ and $(\\frac{x}{b}) – (\\frac{y}{a}) = 9$, then what is the value of $\\frac{x}{y}$?<\/p>\n a)\u00a0$\\frac{( b + 3 a)}{( a – 3 b)}$<\/p>\n b)\u00a0$\\frac{( a + 3 b)}{( b – 3 a)}$<\/p>\n c)\u00a0$\\frac{(1 + 3 a)}{( a + 3 b)}$<\/p>\n d)\u00a0$\\frac{( a + 3 b^2)}{( b – 3 a^2)}$<\/p>\n IBPS Clerk Online Mock Test<\/a><\/p>\n Question 4:\u00a0<\/b>If $x + y = 3$, then what is the value of $x^3 + y^3 + 9xy$?<\/p>\n a)\u00a015<\/p>\n b)\u00a081<\/p>\n c)\u00a027<\/p>\n d)\u00a09<\/p>\n Question 5:\u00a0<\/b>If $x \u00ad- 4y = 0$ and $x + 2y = 24$, then what is the value of $\\frac{(2x + 3y)}{(2x – 3y)}$?<\/p>\n a)\u00a0$\\frac{9}{5}$<\/p>\n b)\u00a0$\\frac{11}{5}$<\/p>\n c)\u00a0$\\frac{13}{7}$<\/p>\n d)\u00a0$\\frac{9}{7}$<\/p>\n Question 6:\u00a0<\/b>If $(3^{33} + 3^{33} + 3^{33})(2^{33} + 2^{33}) = 6^x$, then what is the value of x?<\/p>\n a)\u00a034<\/p>\n b)\u00a035<\/p>\n c)\u00a033<\/p>\n d)\u00a033.5<\/p>\n IBPS Clerk Previous Papers<\/a><\/p>\n Question 7:\u00a0<\/b>If $x_1x_2x_3 = 4(4 + x_1 + x_2 + x_3),$ then what is the value of $\\left[\\frac{1}{(2 + x_1)}\\right] + \\left[\\frac{1}{(2 + x_2)}\\right] + \\left[\\frac{1}{(2 + x_3)}\\right]$?<\/p>\n a)\u00a01<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a02<\/p>\n d)\u00a0$\\frac{1}{3}$<\/p>\n Question 8:\u00a0<\/b>If $a^3 + 3a^2 + 9a = 1$, then what is the value of $a^3 + (\\frac{3}{a})?$<\/p>\n a)\u00a031<\/p>\n b)\u00a026<\/p>\n c)\u00a028<\/p>\n d)\u00a024<\/p>\n Question 9:\u00a0<\/b>If $x + y + z = 0$, then what is the value of $\\frac{(3y^2 + x^2 + z^2)}{(2y^2 – xz)}?$<\/p>\n a)\u00a02<\/p>\n b)\u00a01<\/p>\n c)\u00a0$\\frac{3}{2}$<\/p>\n d)\u00a0$\\frac{5}{3}$<\/p>\n Question 10:\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 IBPS Clerk Important Questions PDF<\/a><\/p>\n Free Banking Study Material (15,000 Solved Questions)<\/a><\/p>\n Question 11:\u00a0<\/b>If $ x = \\sqrt[3]{7}+3$ then the value of $x^{3}-9x^{2}+27x-34$ is:<\/p>\n a)\u00a00<\/p>\n b)\u00a01<\/p>\n c)\u00a02<\/p>\n d)\u00a0-1<\/p>\n Question 12:\u00a0<\/b>If 3\u221a2 + \u221a18 + \u221a50 = 15.55, then what is the value of \u221a32 + \u221a72?<\/p>\n a)\u00a013.22<\/p>\n b)\u00a010.83<\/p>\n c)\u00a014.13<\/p>\n d)\u00a016.54<\/p>\n Question 13:\u00a0<\/b>If $7^{m+1}=2401$, then find the value of $2^{2m+2}$<\/p>\n a)\u00a0224<\/p>\n b)\u00a0256<\/p>\n c)\u00a0264<\/p>\n d)\u00a0286<\/p>\n Question 14:\u00a0<\/b>If a(x + y) = b(x – y) = 2ab, then the value of 2($x^{2} + y^{2}$) is<\/p>\n a)\u00a02($a^{2} – b^{2}$)<\/p>\n b)\u00a02($a^{2}+b^{2}$)<\/p>\n c)\u00a04($a^{2} – b^{2}$)<\/p>\n d)\u00a04($a^{2} + b^{2}$)<\/p>\n Question 15:\u00a0<\/b>If $( x – 5)^{2}$ + $(y – 2)^{2}$ + $(z – 9)^{2}$ = 0 , then value of (x + y – z) is<\/p>\n a)\u00a016<\/p>\n b)\u00a0-1<\/p>\n c)\u00a0-2<\/p>\n d)\u00a012<\/p>\n Daily Free Banking Online Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let a = 2 and b = 3 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given, $x+\\dfrac{1}{x} = 2$<\/p>\n Cubing on both sides<\/p>\n $(x+\\dfrac{1}{x})^3 = 2^3$<\/p>\n => $x^3+\\dfrac{1}{x^3}+3\\times x\\times \\dfrac{1}{x}(x+\\dfrac{1}{x}) = 8$<\/p>\n => $x^3+\\dfrac{1}{x^3}+3(2) = 8$<\/p>\n Therefore,\u00a0$x^3+\\dfrac{1}{x^3} = 8-6 = 2$<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $(\\frac{x}{a}) + (\\frac{y}{b}) = 3$ 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n x+y=3 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given x-4y=0 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $(3^{33} + 3^{33} + 3^{33})(2^{33} + 2^{33}) = 6^x$ 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $x_1x_2x_3 = 4(4 + x_1 + x_2 + x_3),$ 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $a^3 + 3a^2 + 9a = 1$ 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution 1: 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression\u00a0:\u00a0$1^{2}+3^{2}+5^{2}+……..+17^{2}$<\/p>\n =\u00a0$[1^{2}+2^{2}+3^{2}+4^{2}……..+16^{2}+17^{2}]$ $-[2^2+4^2+………+16^2]$<\/p>\n =\u00a0$[1^{2}+2^{2}+3^{2}+4^{2}……..+16^{2}+17^{2}]$ $-(2^2)[1^2+2^2+3^2………+8^2]$<\/p>\n = $[\\frac{17(17+1)+(34+1)}{6}]-[4\\times\\frac{8(8+1)(16+1)}{6}]$<\/p>\n =\u00a0$[\\frac{17(17+1)+(34+1)}{6}]-[4\\times\\frac{8(8+1)(16+1)}{6}]$<\/p>\n = $[51\\times35]-[48\\times17]$<\/p>\n = $17\\times(105-48)=969$<\/p>\n => Ans – (D)<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Given\u00a0:\u00a0$ x = \\sqrt[3]{7}+3$<\/p>\n => $x-3=\\sqrt[3]7$<\/p>\n Cubing both sides, we get\u00a0:<\/p>\n => $(x-3)^3=(\\sqrt[3]7)^3$<\/p>\n => $x^3-27-3(3x)(x-3)=7$<\/p>\n => $x^3-27-9x^2+27x-7=0$<\/p>\n =>\u00a0$x^{3}-9x^{2}+27x-34=0$<\/p>\n => Ans – (A)<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given\u00a0: $3\\sqrt2+\\sqrt{18}+\\sqrt{50}=15.55$<\/p>\n => $3\\sqrt2+3\\sqrt2+5\\sqrt2=15.55$<\/p>\n => $\\sqrt2=\\frac{15.55}{11}=1.413$ ———–(i)<\/p>\n To find\u00a0: $\\sqrt{32}+\\sqrt{72}$<\/p>\n = $4\\sqrt2+6\\sqrt2=10\\sqrt2$<\/p>\n = $10\\times1.413=14.13$<\/p>\n => Ans – (C)<\/p>\n 13)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given\u00a0:\u00a0$7^{m+1}=2401$<\/p>\n =>\u00a0$7^{m+1}=7^4$<\/p>\n => $m+1=4$<\/p>\n => $m=4-1=3$<\/p>\n $\\therefore$\u00a0$2^{2m+2}=2^{2\\times3+2}$<\/p>\n = $2^8=256$<\/p>\n => Ans – (B)<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Given\u00a0:\u00a0a(x + y) = b(x – y) = 2ab<\/p>\n => $a(x+y)=2ab$<\/p>\n => $(x+y)=2b$<\/p>\n Squaring both sides,<\/p>\n => $(x+y)^=(2b)^2$<\/p>\n => $x^2+y^2+2xy=4b^2$ ———–(i)<\/p>\n Similarly, $(x-y)=2a$<\/p>\n Squaring both sides,<\/p>\n => $(x-y)^=(2a)^2$<\/p>\n => $x^2+y^2-2xy=4a^2$ ———–(i)<\/p>\n Adding equations (i) and (ii), we get\u00a0:<\/p>\n => $2x^2+2y^2=4a^2+4b^2$<\/p>\n => $2(x^2+y^2)=4(a^2+b^2)$<\/p>\n => Ans – (D)<\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Expression\u00a0:\u00a0$( x – 5)^{2}$ + $(y – 2)^{2}$ + $(z – 9)^{2}$ = 0<\/p>\n $\\because$ If sum of three positive numbers is zero, then the numbers are equal to zero<\/p>\n => $(x-5)^2=0$<\/p>\n => $(x-5)=0$<\/p>\n => $x=5$<\/p>\n Similarly, $y=2$ and $z=9$<\/p>\n To find\u00a0: $(x+y-z)$<\/p>\n = $5+2-9=-2$<\/p>\n => Ans – (C)<\/p>\n Highly Rated Free Preparation App for Banking Exams<\/a><\/p>\n
\nThen,\u00a0$(5a^3-2a^2b):(3ab^2-b^3) = (5\\times2^3 – 2\\times2^2\\times3) : (3\\times2\\times3^2 – 3^3)$
\n$= 5\\times8 – 2\\times4\\times3 : 3\\times2\\times9 – 27$
\n$= 40-24 : 54-27 = 16 : 27$<\/p>\n
\nbx+ay=3ab
\n3bx+3ay=9ab
\n$(\\frac{x}{b}) – (\\frac{y}{a}) = 9$
\nax-by=9ab
\n3bx+3ay=ax-by
\n3bx-ax=-by-3ay
\nx(3b-a)=y(-b-3a)
\ny\/x =(a-3b)\/(3a+b)
\nx\/y=(3a+b)(a-3b)<\/p>\n
\nCubing on both sides
\n$x^{3}+3xy(x+y)+y^{3}$=27
\n$x^{3}+3xy(3)+y^{3}$=27
\n$x^{3}+9xy+y^{3}$=27<\/p>\n
\nx=4y
\nx+2y=24
\n6y=24
\ny=4
\nx=16
\n$\\frac{(2x + 3y)}{(2x – 3y)}$=$\\frac{(2*(16) + 3*(4))}{(2*(16) – 3*(4))}$
\n=44\/20
\n=11\/5<\/p>\n
\n$(3*3^{33})(2*2^{33}) = 6^x$
\n$(3^{34})(2^{34})=6^x$
\n$6^{34}=6^x$
\nx=34<\/p>\n
\nFrom clear observation we can say that\u00a0$x_1=4,x_2=4,x_3=4 $ will satisfy the equation
\ni.e 4*4*4=4(4+12)
\n64=64
\nTherefore\u00a0$\\left[\\frac{1}{(2 + x_1)}\\right] + \\left[\\frac{1}{(2 + x_2)}\\right] + \\left[\\frac{1}{(2 + x_3)}\\right]$=3(1\/6)
\n=1\/2<\/p>\n
\n$a(a^2 + 3a + 9)=1$
\n$a^2 + 3a + 9=1\/a$
\n$(a^3-b^3)$=$(a-b)(a^2+ab+b^2)$
\nfor b=3
\nwe have $(a^3-3^3)$=$(a-3)(a^2+3a+9)$
\n$(a^3-27)$=$(a-3)(1\/a)$
\n$a^3+(3\/a)=1+27$
\n$a^3+(3\/a)=28$<\/p>\n
\nAs the answer is independent of variables and so we can assume values for x,y and z an solve
\nlet x=1,y=-1,z=0 therefore x+y+z=1-1+0=0
\n$\\frac{(3y^2 + x^2 + z^2)}{(2y^2 – xz)}$
\n=$\\frac{(3(-1)^2 + 1^2 + 0^2)}{(2(-1)^2 – 1*(0))}$
\n=$\\frac{4}{2}$
\n=2
\nSolution 2:$\\frac{(3y^2 + x^2 + z^2)}{(2y^2 – xz)}$=k
\n$(3y^2 + x^2 + z^2)$=$k(2y^2 – xz)$
\n$x^2 + z^2+kxz$=$2ky^2-3y^2$
\nWe know x+y+z=0
\nwe can see that for k=2
\nwe get $(x+z)^{2}=y^{2}$
\nx+z+y=0
\nTherefore value of k=2<\/p>\n