Algebra Questions For IBPS Clerk
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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 $x - 4y = 0$ and $x + 2y = 24$, then what is the value of $\frac{(2x + 3y)}{(2x – 3y)}$?
a) $\frac{9}{5}$
b) $\frac{11}{5}$
c) $\frac{13}{7}$
d) $\frac{9}{7}$
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 $a^3 + 3a^2 + 9a = 1$, then what is the value of $a^3 + (\frac{3}{a})?$
a) 31
b) 26
c) 28
d) 24
Question 9: 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 10: 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:
a) 1785
b) 1700
c) 980
d) 969
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Question 11: 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 12: 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 13: If $7^{m+1}=2401$, then find the value of $2^{2m+2}$
a) 224
b) 256
c) 264
d) 286
Question 14: If a(x + y) = b(x – y) = 2ab, then the value of 2($x^{2} + y^{2}$) is
a) 2($a^{2} – b^{2}$)
b) 2($a^{2}+b^{2}$)
c) 4($a^{2} – b^{2}$)
d) 4($a^{2} + b^{2}$)
Question 15: If $( x – 5)^{2}$ + $(y – 2)^{2}$ + $(z – 9)^{2}$ = 0 , then value of (x + y – z) is
a) 16
b) -1
c) -2
d) 12
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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 (B)
Given x-4y=0
x=4y
x+2y=24
6y=24
y=4
x=16
$\frac{(2x + 3y)}{(2x – 3y)}$=$\frac{(2*(16) + 3*(4))}{(2*(16) – 3*(4))}$
=44/20
=11/5
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 (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 $(a^3-3^3)$=$(a-3)(a^2+3a+9)$
$(a^3-27)$=$(a-3)(1/a)$
$a^3+(3/a)=1+27$
$a^3+(3/a)=28$
9) 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
10) Answer (D)
Expression : $1^{2}+3^{2}+5^{2}+……..+17^{2}$
= $[1^{2}+2^{2}+3^{2}+4^{2}……..+16^{2}+17^{2}]$ $-[2^2+4^2+………+16^2]$
= $[1^{2}+2^{2}+3^{2}+4^{2}……..+16^{2}+17^{2}]$ $-(2^2)[1^2+2^2+3^2………+8^2]$
= $[\frac{17(17+1)+(34+1)}{6}]-[4\times\frac{8(8+1)(16+1)}{6}]$
= $[\frac{17(17+1)+(34+1)}{6}]-[4\times\frac{8(8+1)(16+1)}{6}]$
= $[51\times35]-[48\times17]$
= $17\times(105-48)=969$
=> Ans – (D)
11) 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)
12) 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)
13) Answer (B)
Given : $7^{m+1}=2401$
=> $7^{m+1}=7^4$
=> $m+1=4$
=> $m=4-1=3$
$\therefore$ $2^{2m+2}=2^{2\times3+2}$
= $2^8=256$
=> Ans – (B)
14) Answer (D)
Given : a(x + y) = b(x – y) = 2ab
=> $a(x+y)=2ab$
=> $(x+y)=2b$
Squaring both sides,
=> $(x+y)^=(2b)^2$
=> $x^2+y^2+2xy=4b^2$ ———–(i)
Similarly, $(x-y)=2a$
Squaring both sides,
=> $(x-y)^=(2a)^2$
=> $x^2+y^2-2xy=4a^2$ ———–(i)
Adding equations (i) and (ii), we get :
=> $2x^2+2y^2=4a^2+4b^2$
=> $2(x^2+y^2)=4(a^2+b^2)$
=> Ans – (D)
15) Answer (C)
Expression : $( x – 5)^{2}$ + $(y – 2)^{2}$ + $(z – 9)^{2}$ = 0
$\because$ If sum of three positive numbers is zero, then the numbers are equal to zero
=> $(x-5)^2=0$
=> $(x-5)=0$
=> $x=5$
Similarly, $y=2$ and $z=9$
To find : $(x+y-z)$
= $5+2-9=-2$
=> Ans – (C)
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