Algebra Questions For IBPS Clerk Set-3 PDF
Download important Algebra PDF based on previously asked questions in IBPS Clerk and other Banking Exams. Practice Algebra for IBPS Clerk Exam.
Download Algebra Questions For IBPS Clerk Set-3 PDF
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Question 1: If $\frac{a}{b}=\frac{2}{3}$, then the value of $(5a^3-2a^2b):(3ab^2-b^3)$ is:
a) 16:27
b) 32:29
c) 34:19
d) 27:16
Question 2: If $x + x^{-1} = 2$, then the value of $x^3 + x^{-3}$ is:
a) 3
b) $\frac{1}{2}$
c) 1
d) 2
Question 3: If $(\frac{x}{a}) + (\frac{y}{b}) = 3$ and $(\frac{x}{b}) – (\frac{y}{a}) = 9$, then what is the value of $\frac{x}{y}$?
a) $\frac{( b + 3 a)}{( a – 3 b)}$
b) $\frac{( a + 3 b)}{( b – 3 a)}$
c) $\frac{(1 + 3 a)}{( a + 3 b)}$
d) $\frac{( a + 3 b^2)}{( b – 3 a^2)}$
Question 4: If $x + y = 3$, then what is the value of $x^3 + y^3 + 9xy$?
a) 15
b) 81
c) 27
d) 9
Question 5: If $x = 2 +\surd3, y = 2 – \surd3$ and $z = 1$, then what is the value of $\left(\frac{x}{yz}\right) + \left(\frac{y}{xz}\right) + \left(\frac{z}{xy}\right) + 2 \left[\left(\frac{1}{x}\right) + \left(\frac{1}{y}\right) + \left(\frac{1}{z}\right)\right]$?
a) 25
b) 22
c) 17
d) 43
Question 6: If $(3^{33} + 3^{33} + 3^{33})(2^{33} + 2^{33}) = 6^x$, then what is the value of x?
a) 34
b) 35
c) 33
d) 33.5
Question 7: If $x_1x_2x_3 = 4(4 + x_1 + x_2 + x_3),$ then what is the value of $\left[\frac{1}{(2 + x_1)}\right] + \left[\frac{1}{(2 + x_2)}\right] + \left[\frac{1}{(2 + x_3)}\right]$?
a) 1
b) $\frac{1}{2}$
c) 2
d) $\frac{1}{3}$
Question 8: If $\frac{(a + b)}{c} = \frac{6}{5}$ and $\frac{(b + c)}{a} = \frac{9}{2}$, then what is the value of $\frac{(a + c)}{b}$?
a) $\frac{9}{5}$
b) $\frac{11}{7}$
c) $\frac{7}{11}$
d) $\frac{7}{4}$
Question 9: If $a^3 + 3a^2 + 9a = 1$, then what is the value of $a^3 + (\frac{3}{a})?$
a) 31
b) 26
c) 28
d) 24
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Question 10: If $x + y + z = 0$, then what is the value of $\frac{(3y^2 + x^2 + z^2)}{(2y^2 – xz)}?$
a) 2
b) 1
c) $\frac{3}{2}$
d) $\frac{5}{3}$
Question 11: What is the value of $\frac{(1.2)^3 + (0.8)^3 + (0.7)^3 – 2.016}{1.35[(1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56]}$ ?
a) $\frac{1}{4}$
b) $\frac{1}{2}$
c) 1
d) 2
Question 12: If $ x = \sqrt[3]{7}+3$ then the value of $x^{3}-9x^{2}+27x-34$ is:
a) 0
b) 1
c) 2
d) -1
Question 13: 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
a) (a + b) (a – b)
b) (a + b) (a + b)
c) (a – b) (a – b)
d) (b – c) (b – c)
Question 14: If 3√2 + √18 + √50 = 15.55, then what is the value of √32 + √72?
a) 13.22
b) 10.83
c) 14.13
d) 16.54
Question 15: The value of $\frac{a}{a-b}+\frac{b}{b-a}$ is
a) (a+b)/(a-b)
b) -1
c) 2ab
d) 1
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Answers & Solutions:
1) Answer (A)
Let a = 2 and b = 3
Then, $(5a^3-2a^2b):(3ab^2-b^3) = (5\times2^3 – 2\times2^2\times3) : (3\times2\times3^2 – 3^3)$
$= 5\times8 – 2\times4\times3 : 3\times2\times9 – 27$
$= 40-24 : 54-27 = 16 : 27$
2) Answer (D)
Given, $x+\dfrac{1}{x} = 2$
Cubing on both sides
$(x+\dfrac{1}{x})^3 = 2^3$
=> $x^3+\dfrac{1}{x^3}+3\times x\times \dfrac{1}{x}(x+\dfrac{1}{x}) = 8$
=> $x^3+\dfrac{1}{x^3}+3(2) = 8$
Therefore, $x^3+\dfrac{1}{x^3} = 8-6 = 2$
3) Answer (A)
$(\frac{x}{a}) + (\frac{y}{b}) = 3$
bx+ay=3ab
3bx+3ay=9ab
$(\frac{x}{b}) – (\frac{y}{a}) = 9$
ax-by=9ab
3bx+3ay=ax-by
3bx-ax=-by-3ay
x(3b-a)=y(-b-3a)
y/x =(a-3b)/(3a+b)
x/y=(3a+b)(a-3b)
4) Answer (C)
x+y=3
Cubing on both sides
$x^{3}+3xy(x+y)+y^{3}$=27
$x^{3}+3xy(3)+y^{3}$=27
$x^{3}+9xy+y^{3}$=27
5) Answer (A)
$x = 2 +\surd3, y = 2 – \surd3$
$(1/x)=(2 – \surd3)$
$(1/y)=(2 +\surd3)$
$(\frac{x}{yz}$=$(2 +\surd3)/(2 -\surd3)$
=$(2 +\surd3)^{2}$
$(\frac{y}{xz}$=$(2 – \surd3)/((2 +\surd3))$
=$(2 -\surd3)^{2}$
$(\frac{z}{xy}$=1
$(\frac{x}{yz} + \left(\frac{y}{xz}\right) + \left(\frac{z}{xy}\right) + 2 \left[\left(\frac{1}{x}\right) + \left(\frac{1}{y}\right) + \left(\frac{1}{z}\right)\right]$
=$(2 +\surd3)^{2} +(2 -\surd3)^{2}+1+2(2 – \surd3+2 + \surd3+1)$
=14+1+2(5)
=14+1+10
=245
6) Answer (A)
$(3^{33} + 3^{33} + 3^{33})(2^{33} + 2^{33}) = 6^x$
$(3*3^{33})(2*2^{33}) = 6^x$
$(3^{34})(2^{34})=6^x$
$6^{34}=6^x$
x=34
7) Answer (B)
$x_1x_2x_3 = 4(4 + x_1 + x_2 + x_3),$
From clear observation we can say that $x_1=4,x_2=4,x_3=4 $ will satisfy the equation
i.e 4*4*4=4(4+12)
64=64
Therefore $\left[\frac{1}{(2 + x_1)}\right] + \left[\frac{1}{(2 + x_2)}\right] + \left[\frac{1}{(2 + x_3)}\right]$=3(1/6)
=1/2
8) Answer (D)
$\frac{(a + b)}{c} = \frac{6}{5}$
5a+5b=6c
$\frac{(b + c)}{a} = \frac{9}{2}$
2b+2c=9a
9a-2b=2c
27a-6b=6c
5a+5b=6c
27a-6b=5a+5b
22a=11b
b=2a
4a+2c=9a
2c=5a
c=(5/2)a
$\frac{(a + c)}{b}$
=((a+(5/2)a))/2a
=7a/4a
=7/4
9) Answer (C)
$a^3 + 3a^2 + 9a = 1$
$a(a^2 + 3a + 9)=1$
$a^2 + 3a + 9=1/a$
$(a^3-b^3)$=$(a-b)(a^2+ab+b^2)$
for b=3
we have $(a^3-3^3)$=$(a-3)(a^2+3a+9)$
$(a^3-27)$=$(a-3)(1/a)$
$a^3+(3/a)=1+27$
$a^3+(3/a)=28$
10) Answer (A)
Solution 1:
As the answer is independent of variables and so we can assume values for x,y and z an solve
let x=1,y=-1,z=0 therefore x+y+z=1-1+0=0
$\frac{(3y^2 + x^2 + z^2)}{(2y^2 – xz)}$
=$\frac{(3(-1)^2 + 1^2 + 0^2)}{(2(-1)^2 – 1*(0))}$
=$\frac{4}{2}$
=2
Solution 2:$\frac{(3y^2 + x^2 + z^2)}{(2y^2 – xz)}$=k
$(3y^2 + x^2 + z^2)$=$k(2y^2 – xz)$
$x^2 + z^2+kxz$=$2ky^2-3y^2$
We know x+y+z=0
we can see that for k=2
we get $(x+z)^{2}=y^{2}$
x+z+y=0
Therefore value of k=2
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11) Answer (D)
$x^3 + y^3 + z^3 -3xyz$=$(x + y + z )(x^{2}+y^{2}+z^{2}-xy-yz-zx)$
x=1.2 y=0.8 z=0.7
$\frac{(1.2)^3 + (0.8)^3 + (0.7)^3 – 2.016}{1.35[(1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56]}$
=$\frac{((2.7)((1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56)}{1.35[(1.2)^2 + (0.8)^2 + (0.7)^2 – 0.96 – 0.84 – 0.56]}$
=2.7/1.35
=2
12) Answer (A)
Given : $ x = \sqrt[3]{7}+3$
=> $x-3=\sqrt[3]7$
Cubing both sides, we get :
=> $(x-3)^3=(\sqrt[3]7)^3$
=> $x^3-27-3(3x)(x-3)=7$
=> $x^3-27-9x^2+27x-7=0$
=> $x^{3}-9x^{2}+27x-34=0$
=> Ans – (A)
13) Answer (A)
Let, X = $a^{2} – b^{2}$, Y = $b^{2} – c^{2}$, Z = $c^{2} – a^{2}$
Then, X + Y + Z = 0 (i.e $a^{2} – b^{2}$ + $b^{2} – c^{2}$ + $c^{2} – a^{2}$ = 0)
We know that,
X$^{3}$ + Y$^{3}$ + Z$^{3}$ = 3XYZ i.e,
$(a^{2}-b^{2})^3+(b^{2}-c^{2})^3+(c^{2}-a^{2})^{3}$ = 3 ($a^{2} – b^{2}) (b^{2} – c^{2}) (c^{2} – a^{2}$)
One of the factors is,
$a^{2} – b^{2} (or) (a + b)(a – b)$
Hence, option A is the correct answer.
14) Answer (C)
Given : $3\sqrt2+\sqrt{18}+\sqrt{50}=15.55$
=> $3\sqrt2+3\sqrt2+5\sqrt2=15.55$
=> $\sqrt2=\frac{15.55}{11}=1.413$ ———–(i)
To find : $\sqrt{32}+\sqrt{72}$
= $4\sqrt2+6\sqrt2=10\sqrt2$
= $10\times1.413=14.13$
=> Ans – (C)
15) Answer (D)
Expression : $\frac{a}{a-b}+\frac{b}{b-a}$
Taking (-) common from second term
= $\frac{a}{a-b}-\frac{b}{a-b}$
= $\frac{a-b}{a-b}=1$
=> Ans – (D)
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