Mathematics · Quantitative Aptitude

Algebra and Arithmetic

406 Questions

Algebra and arithmetic questions cover fundamental mathematical operations, inequalities, and binomial products. They assess core quantitative reasoning skills required for various aptitude tests. Solving these problems strengthens the understanding of number systems and algebraic identities.

Binomial productsLinear inequalitiesLeast common multipleQuadratic equationsInteger properties

Algebra and Arithmetic Questions

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

If $\sqrt {x-1}- \sqrt {x+1}+1 =0$, then $4x$ is equal to ____. 

  1. $4 \sqrt {-1}$
  2. $0$
  3. $5$
  4. $1 \dfrac {1}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We can write it as $\sqrt{x-1} + 1 = \sqrt{x+1}$

Squaring both sides we get,
$x-1 + 1 +2\sqrt{x-1} = x+1$
$\Rightarrow 2\sqrt{x-1} = 1$

Squaring both sides, we get
$4(x-1) = 1$ 
$\therefore 4x = 5$

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

In the expression $\cfrac { x+1 }{ x-1 } $ each $x$ is replaced by $\cfrac { x+1 }{ x-1 } $. The resulting expression, evaluated for $x=\cfrac { 1 }{ 2 } $ equals:

  1. $3$
  2. $-3$
  3. $1$
  4. $\dfrac12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Expression is $\cfrac{x+1}{x-1}=y$

If $x$ is replaced by $\cfrac{x+1}{x-1}$
$\implies \cfrac{\cfrac{x+1}{x-1}+1}{\cfrac{x+1}{x-1}-1}$
$\implies \cfrac{x+1+x-1}{x+1-x+1}=\cfrac{2x}{2}$
The resultant expression is $x$.
When $x=\cfrac{1}{2}$
The value is $\cfrac{1}{2}$

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

If  $\sqrt{x+16} = x-4$, then the value of extraneous solution of the above equation is:

  1. $0$
  2. $4$
  3. $5$

  4. There are no extraneous solutions
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $\sqrt{x+16}=x-4$
Squaring both sides, we get
$x+16=(x-4)^{2}$
$\Rightarrow x+16=x^{2}-8x+16$
$\Rightarrow x^{2}-8x-x=16-16$
$\Rightarrow x^{2}-9x=0$
$\Rightarrow x(x-9)=0$
Then $x=0$ or $x=9$
Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

The values of a so that the equation $\Vert x - 2\vert - 1\vert = a \vert x \vert$ does not contain any solution lying in the interval {2, 3} are

  1. $a \ \epsilon(-\infty \dfrac{1}{2})$
  2. $a \ \epsilon (1, \infty)$
  3. $a \ \epsilon (-\infty, 0) \cup (\dfrac{1}{2}, \infty )$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The values of a so that the equation $\left| \left| x-2 \right| -1 \right| =a\left| x \right| $ does not contains any solution lying in the interval $\left{ 2,3 \right} $ are

$a\in (-\infty ,\cfrac { 1 }{ 2 } )\ \left| \left| x-2 \right| -1 \right| =a\left| x \right| \ =>\left| \left| 2-2 \right| -1 \right| =a\left| 2 \right| \ =>\left| -1 \right| =a\left| 2 \right| \ =>1=a\left| 2 \right| \ =>a=\cfrac { 1 }{ 2 } $
Obviously, it does not contains any solution lying in the interval $\left{ 2,3 \right} $ are $a\in (-\infty ,\cfrac { 1 }{ 2 } )$

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

If $(a+b):(b+c):(c+a)=6:7:8$ and $(a+b+c)=14$ , then the value of $c$ is

  1. $6$
  2. $7$
  3. $8$
  4. $14 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a+b:b+c:c+a=6:7:8$

$a+b+c=14$

$\therefore \dfrac{a+b}{b+c}=\dfrac{6}{7}$ ……..$(1)$

$\therefore \dfrac{b+c}{c+a}=\dfrac{7}{8}$ ……..$(2)$

$\therefore \dfrac{a+b}{c+a}=\dfrac{6}{8}$ ...........$(3)$

From $(1)$ & $(2)$

$\left.\begin{matrix}If & a+b=6x\\ then & b+c=7x\end{matrix}\right\}\rightarrow x\in R$

then $c+a=8x$

$\therefore 2(a+b+c)=6x+7x+8x$

$\therefore a+b+c=\dfrac{21x}{2}$

$\therefore a+b+c=10.5x$

$\therefore c=10.5x-6x$

$\therefore c=4.5x$

Also, $10.5x=14$

$\therefore x=\dfrac{14}{10.5}$

$\therefore c=\dfrac{4.5\times 14}{10.5}$

$\therefore c=6$.
Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

Find the value of A and B in the following sum:
$3 B$
$\underline {\times A}$
$\underline {2 5 2}$

  1. $B=6,A=7$
  2. $B=4,A=3$
  3. $B=3,A=4$
  4. $B=7,A=6$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Now as $A \times B$ gives 2 in the product, the combinations could be

$2\times 1$ or $1 \times 2, 6 \times 2$ or $2\times 6, 3 \times 4$

or $4 \times 3, 6 \times 7$ or $7 \times 6$ or $8 \times 4$ or

$4\times 8, 9 \times 8$ or $8\times 9$. But to get 2 in hundred's

place and 5 in ten's place in the product, B has to be 6 and A has to be

7. Then $6\times 7=42$.
Write 2 and carry over 4. then $7\times 3= 21$ and add 4 to get 2 and 5 in the product.
The value of B is 6 and A is 7.

Multiple choice

Given x is an integer, which of the following is a valid conditional?

  1. if (x > 2 or < 5)

  2. if (x = 4)

  3. if (x < 2 && x > 5)

  4. if (x)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In Java, conditional statements inside an if-statement must evaluate to a boolean value. Option C uses the valid logical AND operator (&&) to combine two valid boolean comparisons, even though the condition itself can never be true. Other options either use invalid syntax or try to use an integer as a boolean.

Multiple choice

A signed integer allows...

  1. Negative numbers

  2. No negative numbers

  3. Neither
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In computer science, a signed integer is a data type that can represent both positive and negative whole numbers, using one bit (usually the most significant bit) to indicate the sign. Unsigned integers, by contrast, can only represent non-negative values.

Multiple choice

(½)1 ÷ (½)0 x (½)2 is equivalent to:

  1. (½)3

  2. 1

  3. (½)1

  4. ¼
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using exponent rules: (1/2)^1 / (1/2)^0 * (1/2)^2. Since (1/2)^0 = 1, the expression becomes (1/2)^1 * (1/2)^2. Adding the exponents (1+2) gives (1/2)^3.

Multiple choice

Anything to the power of 0 is:

  1. 0

  2. undefined

  3. 1

  4. itself
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By definition, any non-zero number raised to the power of 0 is 1. This is a fundamental rule of exponents.

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If ${x+y}{ax+by}=\dfrac{y+z}{ay+bz}=\dfrac{z+x}{az+bx}$, then "each of these ratio is equal to $\dfrac{2}{a+b}$, unless $x+y+z=0$." this statement is ____

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the property of ratios, if a/b = c/d = e/f, then each ratio is equal to (a+c+e)/(b+d+f). Adding the numerators gives 2(x+y+z) and adding the denominators gives (a+b)(x+y+z). Canceling (x+y+z) yields 2/(a+b).

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $A : B = 2 : 3$ and $B : C = 4 : 5$, then $C : A$ is equal to ________.

  1. $15 : 8$
  2. $12 : 10$
  3. $8 : 5$
  4. $8 : 15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $\dfrac{A}{B}=\dfrac{2}{3}$


$\Rightarrow B=\dfrac{3A}{2}$


Now, $\dfrac{B}{C}=\dfrac{4}{5}$
putting the value of B in terms of A we get

$\Rightarrow \dfrac{3A}{2C}=\dfrac{4}{5}$

$\Rightarrow \dfrac{C}{A}=\dfrac{15}{8}$