Aptitude Questions on Algebra for Placements

Algebra is a vast topic and you will benefit from reviewing the Algebra Concepts first:

Let’s go through Algebra questions for placement aptitude tests now!

Q. 1 If $( x + \frac{1}{x} = 3 )$, what is the value of $( x^3 + \frac{1}{x^3} )$?

A) 15

B) 18

C) 19

D) 21

Check Solution

Ans: B) 18

We know that $( x + \frac{1}{x} = 3 )$.
Squaring both sides:
$[x^2 + 2 + \frac{1}{x^2} = 9]$
$[x^2 + \frac{1}{x^2} = 7]$
Now, cube both sides of the original equation:
$[x^3 + \frac{1}{x^3} = (x + \frac{1}{x})(x^2 + \frac{1}{x^2}) – (x + \frac{1}{x})]$
Substitute values:
$[x^3 + \frac{1}{x^3} = 3 \cdot 7 – 3 = 18]$

Q. 2 If $( x^2 + 6x + 9 = 0 )$, find the value of $( x )$.

A) -3

B) 3

C) 0

D) None

Check Solution

Ans: A) -3

The given equation can be rewritten as $( (x + 3)^2 = 0 )$.
Therefore, $( x = -3 )$.

Q. 3 If $( y = 2x + 3 )$ and $( x = 4 )$, what is the value of $( y^2 – x^2 )$?

A) 125

B) 100

C) 95

D) 85

Check Solution

Ans: C) 105

Substitute $( x = 4 )$ into the equation for $( y )$:
$[y = 2 \cdot 4 + 3 = 8 + 3 = 11]$
Now, $( y^2 – x^2 = 11^2 – 4^2 = 121 – 16 = 105 )$.

Q. 4 If $( x + y = 5 )$ and $( x^2 + y^2 = 13 )$, find the value of $( xy )$.

A) 6

B) 8

C) 10

D) 12

Check Solution

Ans: A) 6

We know that $( (x + y)^2 = x^2 + y^2 + 2xy )$.
Substitute values:
$[5^2 = 13 + 2xy]$
$[25 = 13 + 2xy]$
$[2xy = 12 \Rightarrow xy = 6]$

Q. 5 If $( a + b = 7 )$ and $( ab = 10 )$, what is the value of $( a^3 + b^3 )$?

A) 133

B) 157

C) 175

D) 191

Check Solution

Ans: A) 133

We know that $( a^3 + b^3 = (a + b)(a^2 – ab + b^2) )$.
$[a^2 + b^2 = (a + b)^2 – 2ab = 49 – 20 = 29]$
So, $( a^3 + b^3 = 7(29 – 10) = 7 \cdot 19 = 133 )$.

Q. 6 Solve for $( x ): ( 2^{x+2} = 64 )$.

A) 3

B) 4

C) 5

D) 6

Check Solution

Ans: B) 4

Rewrite ( 64 ) as a power of ( 2 ):
$[64 = 2^6]$
So, $( 2^{x+2} = 2^6 )$.
Therefore, $( x + 2 = 6 )$ and $( x = 4 )$.

Q. 7 If $( (x – 1)(x + 1) = 12 )$, find the value of $( x^2 )$.

A) 11

B) 12

C) 13

D) 14

Check Solution

Ans: C) 13.

Expand the left side:
$[x^2 – 1 = 12]$
So, $( x^2 = 13 )$.

Q. 8 If $( x^2 + 4x + k = 0 )$ has real roots, what is the range of values for $( k )$?

A) $( k \leq 4 )$

B) $( k \geq 4 )$

C) $( k \leq -4 )$

D) $( k \geq -4 )$

Check Solution

Ans: A) $( k \leq 4 )$

For real roots, the discriminant $( b^2 – 4ac \geq 0 )$:
Here, $( a = 1, b = 4, c = k )$.
$[4^2 – 4 \cdot 1 \cdot k \geq 0]$
$[16 – 4k \geq 0]$
$[k \leq 4]$

Q. 9 If $( 3x + y = 7 )$ and $( x – y = 5 )$, find the value of $( x + y )$.

A) 6

B) 7

C) 8

D) 9

Check Solution

Ans: C) 8

Solve the second equation for ( y ):
$[y = x – 5]$
Substitute into the first equation:
$[3x + (x – 5) = 7]$
$[4x – 5 = 7 \Rightarrow 4x = 12 \Rightarrow x = 3]$
Substitute $( x = 3 )$ back into $( y = x – 5 )$:
$[y = -2]$

Q. 10 If $( x = 2 )$ is a root of $( x^2 + px + 4 = 0 )$, what is the value of $( p )$?

A) -6

B) 4

C) 2

D) -4

Check Solution

Ans: D) -4

Substitute $( x = 2 )$ into the equation:
$[2^2 + 2p + 4 = 0]$
$[4 + 2p + 4 = 0 \Rightarrow 2p = -8 \Rightarrow p = -4]$

Refer Topic wise Aptitude Questions with Solutions

Refer Questions for next topic: https://www.learntheta.com/aptitude-questions-permutation-combination/

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