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

The value of ( a + b)2+(a -b)2 is

  1. 2a2b2

  2. 2a2 + b2

  3. 2( a2+b2)

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

Expand both squares: (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2). The 2ab and -2ab cancel out, leaving 2a^2 + 2b^2, which is 2(a^2 + b^2).

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

$if\,A\, = \sqrt 7  - \sqrt 6 \,and\,B = \,\sqrt 6  - \sqrt {5,} \,then\,$

  1. $A > B$
  2. $A = B$
  3. $A < B\,$
  4. $A \geqslant B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given numbers are $A=\sqrt{7}-\sqrt{6}$ and $B=\sqrt{6}- \sqrt{5}$
Let $x =\sqrt{5}$ and $y= \sqrt{7}$ then by
$A.M$ and $G.M$
$\dfrac{x+y}{2} \le \sqrt{\dfrac{x^{2}+y^{2}}{2}}$
$\Rightarrow \dfrac{\sqrt{5}+ \sqrt{7}}{2} \le \sqrt{6}$
$\Rightarrow \sqrt{5}+ \sqrt{7} \le 2 \sqrt{6}$
$\Rightarrow \sqrt{7}- \sqrt{6} \le \sqrt{6} - \sqrt{5}$
Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $a+b+c=0$ then $a^3+b^3+c^3$ is equal to

  1. 3abc
  2. $\displaystyle\frac{3}{abc}$
  3. $3a^3b^3c^3$

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

$Using\quad { a }^{ 3 }+\quad { b }^{ 3 }+\quad { c }^{ 3 }-3abc=\left( a+b+c \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right) $

$Using\quad { a }^{ 3 }+\quad { b }^{ 3 }+\quad { c }^{ 3 }-3abc=0\times \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right) $
$Using\quad { a }^{ 3 }+\quad { b }^{ 3 }+\quad { c }^{ 3 }=3abc$
Here if a+b+c is 0 then answer will be 3abc


Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $\cfrac{a}{2}=\cfrac{b}{3}=\cfrac{c}{4}$, then $a:b:c=$

  1. $2:3:4$
  2. $4:3:2$
  3. $3:2:4$

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

Given, $\displaystyle \frac{a}{2} = \frac{b}{3} = \frac{c}{4}$


Lets take $\displaystyle \frac{a}{2} = \frac{b}{3} = \frac{c}{4} = k$


So, $\dfrac{a}{2}  = k$

$a = 2k$

$\dfrac{b}{3}  = k$

$b = 3k$

$\dfrac{c}{4} = k$

$c = 4k$

i.e., $a : b: c = 2k : 3k : 4k$

$a : b; c = 2 : 3 : 4$  

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $a:b=5:7$ and $b:c=6:11$, then $a:b:c=$

  1. $35:49:66$
  2. $30:42:77$
  3. $30:42:55$

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

$\dfrac{a}{b} = \dfrac{5}{7} $       $\dfrac{b}{c} = \dfrac{6}{11}$


$\left.\begin{matrix} a = \dfrac{5}{7} b \end{matrix}\right|\begin{matrix} c = \dfrac{11 b}{6} \end{matrix}$

$a : b : c$

$\dfrac{5}{7} b : b : \dfrac{11b}{6}$

$\dfrac{5}{7} : 1 : \dfrac{11}{6}$

L.C.M of $7, 6$ is $42$

$\dfrac{5}{7} \times 42 : 42 \times 1 : \dfrac{11}{6} \times 42$

$5 \times 6 : 42 : 11 \times 7$

$30 : 42 : 77$

Multiple choice composition of ratios types of ratios ratio and proportions ratio and proportion maths

If $\cfrac{1}{a}:\cfrac{1}{b}:\cfrac{1}{c}=3:4:5$ then $a:b:c$

  1. $5:4:3$
  2. $20:15:12$
  3. $9:12:15$
  4. $12:15:20$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\displaystyle \frac{1}{a} : \frac{1}{b} : \frac{1}{c} = 3 : 4 : 5$


As, $\dfrac{1}{a} = 3$      So, $\dfrac{1}{3} = a$,


$\dfrac{1}{b} = 4$            So,  $\dfrac{1}{4} = b$,


$\dfrac{1}{c} = 5$             So, $\dfrac{1}{5} = c$


i.e., a : b : c = $\displaystyle \frac{1}{3} : \frac{1}{4} : \frac{1}{5}$

LCM of $3, 4$ and $5$ is $60$

So, multiply with $60$

We get, $\displaystyle \frac{60}{3} : \frac{60}{4} : \frac{60}{5}$

$= 20 : 15 : 12$

Multiple choice

a2+b3×(1/a)-2 for a=√3 and b=∛4

  1. 12

  2. 10

  3. 71/3

  4. 15
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given a = sqrt(3) and b = cbrt(4). The expression is a^2 + b^3 * (1/a)^-2. This becomes (sqrt(3))^2 + (cbrt(4))^3 * (a^2). Substituting values: 3 + 4 * (sqrt(3))^2 = 3 + 4 * 3 = 3 + 12 = 15.

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

If x is a positive integer less than 100, then the number of x which make $\displaystyle \sqrt{1+2+3+4+x}$ an integer is

  1. 6

  2. 7

  3. 8

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let, $\sqrt{1+2+3+4+5+x}=l$, where $ l $ is an positive integer.
$\Rightarrow \sqrt{10+x}=l$
$\Rightarrow 10+x=l^2$
$\Rightarrow x=l^2-10$
Given, $1<x<100$
$\Rightarrow 1<l^2-10<100$
$\Rightarrow 11<l^2<110$
$\Rightarrow \sqrt{11}<l<\sqrt{110}$
$\Rightarrow 3.32<l<10.49$
$\Rightarrow l=4,5,6,7,8,9,10$              ($\because l\text{ is an integer.}$)
Therefore, there are 7 values of $x$.

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

If ${2^a} = 3$ and ${9^b} = 4$ then the value of $a.b$ is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 ${ 2 }^{ a }=3$$ a\log _{ 10 }{ 2 } =\log _{ 10 }{ 3 } $$ a= \dfrac { \log _{ 10 }{ 3 }  }{ \log _{ 10 }{ 2 }  }   $  ${ 9 }^{ b }=4$$ b\log _{ 10 }{ 9 } =\log _{ 10 }{ 4 } $$b=\dfrac { \log _{ 10 }{ 4 }  }{ \log _{ 10 }{ 9 }  } $$ b=\dfrac { \log _{ 10 }{ { 2 }^{ 2 } }  }{ \log _{ 10 }{ { 3 }^{ 3 } }  } $$ b=\dfrac { \log _{ 10 }{ { 2 } }  }{ \log _{ 10 }{ { 3 } }  } $

$\therefore a.b= \dfrac { \log _{ 10 }{ 3 }  }{ \log _{ 10 }{ 2 }  }   \times \dfrac { \log _{ 10 }{ { 2 } }  }{ \log _{ 10 }{ { 3 } }  }$


$\therefore a.b=1$