Download important Algebra PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Algebra for IBPS Clerk Exam.<\/p>\n
Download Algebra Questions For IBPS Clerk Set-3 PDF<\/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 $x = 2 +\\surd3, y = 2 – \\surd3$ and $z = 1$, then what is the value of $\\left(\\frac{x}{yz}\\right) + \\left(\\frac{y}{xz}\\right) + \\left(\\frac{z}{xy}\\right) + 2 \\left[\\left(\\frac{1}{x}\\right) + \\left(\\frac{1}{y}\\right) + \\left(\\frac{1}{z}\\right)\\right]$?<\/p>\n a)\u00a025<\/p>\n b)\u00a022<\/p>\n c)\u00a017<\/p>\n d)\u00a043<\/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 $\\frac{(a + b)}{c} = \\frac{6}{5}$ and $\\frac{(b + c)}{a} = \\frac{9}{2}$, then what is the value of $\\frac{(a + c)}{b}$?<\/p>\n a)\u00a0$\\frac{9}{5}$<\/p>\n b)\u00a0$\\frac{11}{7}$<\/p>\n c)\u00a0$\\frac{7}{11}$<\/p>\n d)\u00a0$\\frac{7}{4}$<\/p>\n Question 9:\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 IBPS Clerk Important Questions PDF<\/a><\/p>\n Free Banking Study Material (15,000 Solved Questions)<\/a><\/p>\n Question 10:\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 11:\u00a0<\/b>What is the value of\u00a0 $\\frac{(1.2)^3 + (0.8)^3 + (0.7)^3 – 2.016}{1.35[(1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56]}$ ?<\/p>\n a)\u00a0$\\frac{1}{4}$<\/p>\n b)\u00a0$\\frac{1}{2}$<\/p>\n c)\u00a01<\/p>\n d)\u00a02<\/p>\n Question 12:\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 13:\u00a0<\/b>Out of the given responses, one of the factors of $(a^{2}-b^{2})^3+(b^{2}-c^{2})^3+(c^{2}-a^{2})^{3}$is<\/p>\n a)\u00a0(a + b) (a – b)<\/p>\n b)\u00a0(a + b) (a + b)<\/p>\n c)\u00a0(a – b) (a – b)<\/p>\n d)\u00a0(b – c) (b – c)<\/p>\n Question 14:\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 15:\u00a0<\/b>The value of $\\frac{a}{a-b}+\\frac{b}{b-a}$ is<\/p>\n a)\u00a0(a+b)\/(a-b)<\/p>\n b)\u00a0-1<\/p>\n c)\u00a02ab<\/p>\n d)\u00a01<\/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(A)<\/strong><\/p>\n $x = 2 +\\surd3, y = 2 – \\surd3$ $(\\frac{x}{yz}$=$(2 +\\surd3)\/(2 -\\surd3)$<\/p>\n =$(2 +\\surd3)^{2}$<\/p>\n $(\\frac{y}{xz}$=$(2 – \\surd3)\/((2 +\\surd3))$<\/p>\n =$(2 -\\surd3)^{2}$<\/p>\n $(\\frac{z}{xy}$=1<\/p>\n $(\\frac{x}{yz} + \\left(\\frac{y}{xz}\\right) + \\left(\\frac{z}{xy}\\right) + 2 \\left[\\left(\\frac{1}{x}\\right) + \\left(\\frac{1}{y}\\right) + \\left(\\frac{1}{z}\\right)\\right]$<\/p>\n =$(2 +\\surd3)^{2} +(2 -\\surd3)^{2}+1+2(2 – \\surd3+2 + \\surd3+1)$ 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(D)<\/strong><\/p>\n $\\frac{(a + b)}{c} = \\frac{6}{5}$ 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $a^3 + 3a^2 + 9a = 1$ 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution 1: Get 25 IBPS Clerk mocks for Rs. 149. Enroll here<\/a><\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $x^3 + y^3 + z^3 -3xyz$=$(x + y + z )(x^{2}+y^{2}+z^{2}-xy-yz-zx)$ =$\\frac{((2.7)((1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56)}{1.35[(1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56]}$<\/p>\n =2.7\/1.35<\/p>\n =2<\/p>\n 12)\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 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let, X = $a^{2} – b^{2}$, Y = $b^{2} – c^{2}$, Z = $c^{2} – a^{2}$<\/p>\n Then, X + Y + Z = 0 (i.e $a^{2} – b^{2}$ +\u00a0$b^{2} – c^{2}$ +\u00a0$c^{2} – a^{2}$ = 0)<\/p>\n We know that,<\/p>\n X$^{3}$ + Y$^{3}$ + Z$^{3}$ = 3XYZ i.e,<\/p>\n $(a^{2}-b^{2})^3+(b^{2}-c^{2})^3+(c^{2}-a^{2})^{3}$ = 3 ($a^{2} – b^{2}) (b^{2} – c^{2}) (c^{2} – a^{2}$)<\/p>\n One of the factors is,<\/p>\n $a^{2} – b^{2} (or) (a + b)(a – b)$<\/p>\n Hence, option A is the correct answer.<\/p>\n 14)\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 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Expression\u00a0:\u00a0$\\frac{a}{a-b}+\\frac{b}{b-a}$<\/p>\n Taking (-) common from second term<\/p>\n = $\\frac{a}{a-b}-\\frac{b}{a-b}$<\/p>\n = $\\frac{a-b}{a-b}=1$<\/p>\n => Ans – (D)<\/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
\n$(1\/x)=(2 – \\surd3)$
\n$(1\/y)=(2 +\\surd3)$<\/p>\n
\n=14+1+2(5)
\n=14+1+10
\n=245<\/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
\n5a+5b=6c
\n$\\frac{(b + c)}{a} = \\frac{9}{2}$
\n2b+2c=9a
\n9a-2b=2c
\n27a-6b=6c
\n5a+5b=6c
\n27a-6b=5a+5b
\n22a=11b
\nb=2a
\n4a+2c=9a
\n2c=5a
\nc=(5\/2)a
\n$\\frac{(a + c)}{b}$
\n=((a+(5\/2)a))\/2a
\n=7a\/4a
\n=7\/4<\/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
\nx=1.2 y=0.8 z=0.7
\n$\\frac{(1.2)^3 + (0.8)^3 + (0.7)^3 – 2.016}{1.35[(1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56]}$<\/p>\n