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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.

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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

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

Let a = 2 and b = 3
Then, $(5a^3-2a^2b):(3ab^2-b^3) = (5\times2^3 – 2\times2^2\times3) : (3\times2\times3^2 – 3^3)$
$= 5\times8 – 2\times4\times3 : 3\times2\times9 – 27$
$= 40-24 : 54-27 = 16 : 27$

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, $x^3+\dfrac{1}{x^3} = 8-6 = 2$

$(\frac{x}{a}) + (\frac{y}{b}) = 3$
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)

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

$x = 2 +\surd3, y = 2 – \surd3$
$(1/x)=(2 – \surd3)$
$(1/y)=(2 +\surd3)$

$(\frac{x}{yz}$=$(2 +\surd3)/(2 -\surd3)$

=$(2 +\surd3)^{2}$

$(\frac{y}{xz}$=$(2 – \surd3)/((2 +\surd3))$

=$(2 -\surd3)^{2}$

$(\frac{z}{xy}$=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

$(3^{33} + 3^{33} + 3^{33})(2^{33} + 2^{33}) = 6^x$
$(3*3^{33})(2*2^{33}) = 6^x$
$(3^{34})(2^{34})=6^x$
$6^{34}=6^x$
x=34

$x_1x_2x_3 = 4(4 + x_1 + x_2 + x_3),$
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

$\frac{(a + b)}{c} = \frac{6}{5}$
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

$a^3 + 3a^2 + 9a = 1$
$a(a^2 + 3a + 9)=1$
$a^2 + 3a + 9=1/a$
$(a^3-b^3)$=$(a-b)(a^2+ab+b^2)$
for b=3
we have $(a^3-3^3)$=$(a-3)(a^2+3a+9)$
$(a^3-27)$=$(a-3)(1/a)$
$a^3+(3/a)=1+27$
$a^3+(3/a)=28$

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

$x^3 + y^3 + z^3 -3xyz$=$(x + y + z )(x^{2}+y^{2}+z^{2}-xy-yz-zx)$
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

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)

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.

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)

Expression : $\frac{a}{a-b}+\frac{b}{b-a}$
= $\frac{a}{a-b}-\frac{b}{a-b}$
= $\frac{a-b}{a-b}=1$