Mathematics · Quantitative Aptitude

Algebra and Arithmetic

333 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 parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

Which of the following statements is INCORRECT?

  1. Zero has a reciprocal

  2. The product of two negative rational numbers is always positive

  3. The reciprocal of a positive rational number is always positive

  4. The product of two positive rational numbers is always positive
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Zero can't have a reciprocal because
$1/0$ is not defined.
The rest statements are
$(-a)\times (-b)=ab$
Reciprocal of $a$ is $1/a$
$a\times b=ab$
where $a$ and $b$ are positive.

Rest all statements are true except for option A.

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

Which of the following statements is true?

  1. 1 and -1 are reciprocal of themselves.

  2. Zero has no reciprocal.

  3. The product of the two middle rational numbers is a rational number.

  4. All of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Option A

Reciprocal of a number is the number obtained by dividing it by $1$. 
Here, reciprocal of $1 = \dfrac{1}{1} = 1$ and reciprocal of $-1 = \dfrac{1}{-1} = -1$
Hence, they are both the reciprocals of themselves.

Option B
Reciprocal of $0$ can be denoted as $\dfrac{1}{0}$ which isn't defined. Anything divided by $0$ is not defined.

Option C
Addition, subtraction, multiplication or division of a rational number with another rational number always gives a rational number.

$\therefore$ All the statements are correct.

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

The product of a fractional number and its multiplicative inverse is

  1. $0$
  2. $1$

  3. number itself

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If the product of two numbers is $1$, then each number is known as multiplicative inverse or the reciprocal of one another. To find the multiplicative inverse of a proper or improper fraction, interchange the numerator and denominator. For example, the multiplicative inverse of the fraction $\dfrac {5}{18}$ is $\dfrac {18}{5}$.


Now, the product of the fractional number $\dfrac {5}{18}$ and its multiplicative inverse $\dfrac {18}{5}$ is as follows:

$\dfrac { 5 }{ 18 } \times \dfrac { 18 }{ 5 } =1$

Hence, the product of a fractional number and its multiplicative inverse is $1$.

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

The value of $x$ for which $\cfrac{x-3}{4}--x< \cfrac{x-1}{2}-\cfrac{x-2}{3}$ and $2-x> 2x-8$

  1. $[-1,10/3]$
  2. $(1,10/3)$
  3. $R$

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

$\dfrac{x-3}{4}-x<\dfrac{x-1}{2}-\dfrac{x-2}{3}$


$\dfrac{-3{x}-3}{4}<\dfrac{x+1}{6}$


$\implies x>-1$

$2-x>2{x}-8\implies x<\dfrac{10}{3}$

$\implies x\in (-1,10/3)$

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

(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).