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 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 fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

Let $x\;\in\;Q,\;y\;\in\;Q^c$, which of the following statement is always WRONG ?

  1. $xy\;\in\;Q^c$
  2. $y/x\;\in\;Q$, whenever defined
  3. $\sqrt{2}x+y\;\in\;Q$
  4. $x/y\;\in\;Q^c$, whenever defined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $x=1,\;y=\sqrt{2}$
Then $xy=\sqrt{2}\;\in\;Q^c$
Obvious
$x=-1,\;y=\sqrt{2}$ then $\sqrt{2}x+y=0\;\in\;Q$
$x=1,\;y=\sqrt{2}$ then $x/y=\displaystyle\frac{1}{\sqrt{2}}\;\in\;Q^c$

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 } }  } $

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


$\therefore a.b=1$

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

whether the following relation is${{ \frac{1}{{{x^{a - b}}}}} ^{\frac{1}{{a - c}}}}{{ \frac{1}{{{x^{b - c}}}}} ^{\frac{1}{{b - a}}}}{{ \frac{1}{{{x^{c - a}}}}} ^{^{\frac{1}{{c - b}}}}} = 1$

  1. True

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

Using exponent rules, each term simplifies to x^(b-a)/(a-c) * x^(c-b)/(b-a) * x^(a-c)/(c-b). Adding the exponents (b-a)/(a-c) + (c-b)/(b-a) + (a-c)/(c-b) results in 0, and x^0 = 1.