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 negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Multiplication of one odd and one even integer is always :

  1. Even

  2. Odd

  3. Can't be determined

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

Even integer is an integer having unit digit as a multiple of $2$

So on multiplication with odd integer, unit digit will still remain a multiple of $2$, hence, multiplication of odd and even integer gives even integer.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

$a, b, c$ are even numbers and $x, y, z$ are odd numbers. Which of the following relationships can't be justified at any cost?
(a) $\dfrac{a \times b}{c} = x \times y$  (b)  $\dfrac{a \times b}{x} = yz$  (c) $\dfrac{xy}{z} = ab$

  1. Only a

  2. Only c

  3. All the three

  4. Only b and c
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Use the rules that 
(a) product of two odd or even numbers are odd and even respectively.
(b) the quotient of the odd number $ \div $ even number or vice-versa may or may not be strictly odd or even.

a. $ ab $ must be a even number and $ \dfrac{ab}{c} $ may be even number or a fraction or odd number and $ xy $ must be a odd number
therefore it can be justified in some case.

b. $ \dfrac { ab }{ x } $ is always a fraction or even number and $ yz $ is always a odd number
therefore it cannot be justified in any case

c.$ \dfrac { xy }{ z } $ maybe odd number or fraction but $ ab $ will always be even.
therefore it cannot be justified.

Relationships given in b and c can't be justified at any cost.
Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $f(x)=x^{2}+6x+c$, where $'c'$ is an integer, then $f(0)+f(-1)$ is

  1. an even integer
  2. an odd integer always disable by $3$
  3. an odd integer not divisible by $3$
  4. an odd integer may or not be divisible by $3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f(0) = c$

$f(-1) = c-5$

$f(0)+f(-1) = 2c-5 = 2(c-3) + 1$

As the above is of the form $2k+1$, it is always odd.

For $c=3$, the above is not divisible by 3 but for $c=4$, it is. Therefore, it may or may not be divisible by 3.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $|a|$ denotes the absolute value of an integer, then which of the following are correct?
1.$|ab| = |a| |b|$
2. $|a+b| \le |a|+|b|$
3. $|a-b| \ge| |a| -|b||$
Select the correct answer using the code given below.

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $\left| a \right| $ is the absolute value of an integer,

From the definition,
$\left| a \right| =a$ if $ a\ge 0$,
$=-a\quad $ is $a\le0$
$\therefore$ $\left| ab \right| =\left| a \right| \left| b \right| $ where $a,b$ are real numbers.
We know that from the triangle inequality sum of any two sides is always greater than the third side,
i.e.,$\left| a+b \right| \le \left| a \right| +\left| b \right| $,
We can also prove by considering 
Absolute part of the difference between any two sides is always less than the third side,
$\Longrightarrow \left| a-b \right| \ge \left| \left| a \right| -\left| b \right|  \right| $

Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

Evaluate:
$( b - c + d + a ) ( d + a - b + c ) + ( c - d + a + b ) ( b + c + d - a )$

  1. $4 ( a d + b c )$
  2. $2 ( a d + b c )$
  3. $3 ( a d + b c )$
  4. $ ( a d + b c )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(b-c+d+a)(d+a-b+c)+(c-d+a+b)(b+c+d-a)$

$=[(d+a)+(b-c)][(d+a)-(b-c)]+[(b+c)+(a-d)][(b+c)-(a-d)]$

$[\because (a-d)^2=(d-a)^2]$

$=(d+a)^2-(b-c)^2+(b+c)^2-(a-d)^2$

$=(d+a)^2-(d-a)^2+(b+c)^2-(b-c)^2$

$=4ad+4bc$

$=4(ad+bc)$.
Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

If ${a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca=0$, then

  1. $a+b=c$
  2. $b+c=a$
  3. $c+a=b$
  4. $a=b=c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$a^2+b^2+c^2-ab-bc-ca=0$

$\Rightarrow 2(a^2+b^2+c^2-ab-bc-ca)=2(0)$

$\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0$

$\Rightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0$

$(a-b)^2+(b-c)^2+(c-a)^2=0$

$\Rightarrow a-b=b-c=c-a=0$

$\therefore a=b=c$
Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

If $1\le a\le 2$, then $\sqrt { a-2\sqrt { a-1 }  } -\sqrt { a+2\sqrt { a-1 }  } =$.......

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

$\sqrt{a - 2 \sqrt{a - 1}} - \sqrt{a + 2 \sqrt{a - 1}}$

$\Rightarrow \sqrt{(\sqrt{a - 1})^2 - 2 (1) \sqrt{a - 1} + 1^2} - \sqrt{(\sqrt{(a - 1)})^2 + 2 \sqrt{a - 1} + 1^2}$
$\Rightarrow \sqrt{(\sqrt{a - 1} - 1)^2} - \sqrt{(\sqrt{a - 1} + 1)^2}$
$\Rightarrow (\sqrt{a - 1} - 1) - ( \sqrt{a - 1} + 1)$
$= -2$

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\cfrac{{x}^{3}+{x}^{2}+x+1}{{x}^{3}-{x}^{2}+x-1}=\cfrac{{x}^{2}+x+1}{{x}^{2}-x+1}$, then the number of real-value of $x$ satisfying are

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that


$\dfrac{x^3+x^2+x+1}{x^3-x^2+x+1}=\dfrac{x^2+x+1}{x^2-x+1}$

Performing Componendo Dividendo, we get

$\Rightarrow \dfrac{x^3+x^2+x+1+x^3-x^2+x+1}{x^3+x^2+x+1-x^3+x^2-x+1}=\dfrac{x^2+x+1+x^2-x+1}{x^2+x+1-x^2+x-1}$

$\Rightarrow \dfrac{2x^3+2x}{2x^2+2}=\dfrac{2x^2+2}{2x} $

$\Rightarrow \dfrac{x^3+x}{x^2+1}=\dfrac{x^2+1}{x} $

$\Rightarrow x(x^3+x)=(x^2+1)^2$

$\Rightarrow x^4+x^2=x^4+1+2x^2$

$\Rightarrow x^2=-1$

Hence no real values of $x$ satisfy this equation.

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\cfrac{a+3d}{a+9d}=\cfrac{a+d}{a+5d}=k$, then $k$ is equal to $(a,d> 0)$

  1. $\dfrac{1}{2}$
  2. $2$
  3. $6$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{a+3d}{a+9d}=\dfrac{a+d}{a+5d}$


Applying Componendo-Dividendo

$\Rightarrow \dfrac{a+3d+a+9d}{a+3d-a-9d}=\dfrac{a+d+a+5d}{a+d-a-5d}$ 

$\Rightarrow \dfrac{2a+12d}{-6d}=\dfrac{2a+6d}{-4d}$

$\Rightarrow \dfrac{a+6d}{-3}=\dfrac{a+3d}{-2}$

$\Rightarrow -2a-12d=-3a-9d$

$\Rightarrow a=3d$

Now, given that 

$\Rightarrow k=\dfrac{a+3d}{a+9d}$

$\Rightarrow k=\dfrac{3d+3d}{3d+9d}$

$\Rightarrow k=\dfrac{6d}{12d}$

$\Rightarrow k=\dfrac{1}{2}$

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $x=\cfrac { 4ab }{ a+b } $ then value of $\cfrac { x+2a }{ x-2a } +\cfrac { x+2b }{ x-2b } $

  1. $a$
  2. $b$
  3. $0$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $x = 4ab/(a+b)$

=> $x = 2a\times 2b / (a+b)$
=> $x/2a = 2b / (a+b)$
Using componendo and dividendo 
=> $x+2a / x-2a = 2b+a+b / 2b-a-b$
=> $x+2a / x-2a = 3b+a / b-a$

Similarly, $x+2b / x-2b = 3a+b / a-b$

Now, adding both,
$[x+2a / x-2a] + [x+2b / x-2b ] = [3b+a / b-a ] + [ 3a +b / a-b ]$
RHS => $[3b+a / b-a ] - [ 3a+b / b-a ]$
=> $3b+a-3a-b / b-a$
=> $2b-2a / b-a$
=> $2(b-a)/b-a$
=> $2$

Hence, LHS $= 2$ 

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\cfrac { { a }^{ 3 }+3a{ b }^{ 2 } }{ 3{ a }^{ 2 }b+{ b }^{ 3 } } =\cfrac { { x }^{ 3 }+3x{ y }^{ 2 } }{ 3{ x }^{ 2 }y+{ y }^{ 3 } } $ then

  1. $bx=ay$
  2. $by=ax$
  3. ${ b }^{ 2 }y={ a }^{ 2 }x$
  4. ${ b }^{ 2 }x={ a }^{ 2 }y$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{{{a^3} + 3a{b^2}}}{{3{a^2}b + {b^3}}} = \dfrac{{{x^3} + 3x{y^2}}}{{3{x^2}y + {y^3}}}$


Use of compodendo and Devidendo

$\dfrac{{{a^3} + 3a{b^2} + 3{a^2}b + {b^3}}}{{3{a^2}b + {b^3} - 3{a^2}b + {b^3}}} = \dfrac{{{x^3} + 3x{y^2} + 3{x^2}y + {y^3}}}{{3{x^2}y + {y^3} - 3{x^2}y + {y^3}}}$

$ \Rightarrow \dfrac{{{{\left( {a + b} \right)}^3}}}{{{{(a - b)}^3}}} = \dfrac{{{{\left( {x + y} \right)}^3}}}{{{{(x - y)}^3}}}$

$ \Rightarrow {\left( {\dfrac{{a + b}}{{a + b}}} \right)^3} = {\left( {\dfrac{{x + y}}{{x - y}}} \right)^3}$

$ \Rightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{x + y}}{{x - y}}$

$ \Rightarrow \left( {a + b} \right)\left( {x + y} \right) = \left( {x + y} \right)\left( {a - b} \right)$

$ \Rightarrow ax{\text{ }} - {\text{ }}ay{\text{ }} + {\text{ }}bx{\text{ }} - {\text{ }}by{\text{ }} = {\text{ }}xa{\text{ }} - {\text{ }}xb{\text{ }} + {\text{ }}ay{\text{ }} - {\text{ }}yb$

$ \Rightarrow bx{\text{ }} + {\text{ }}xb{\text{ }} = {\text{ }}ay{\text{ }} + {\text{ }}ay$

$ \Rightarrow 2bx{\text{ }} = {\text{ }}2ay$

$ \Rightarrow bx{\text{ }} = {\text{ }}ay$