Quantitative Aptitude

Number System and Simplification

548 Questions

Number system and simplification questions assess mathematical proficiency in handling fractions, decimals, and complex algebraic expressions. Test takers must simplify surds, calculate reciprocals, and solve intricate number pattern equations. This foundational quantitative aptitude topic is critical for achieving high scores in SSC and banking exams.

Fraction simplificationDecimal operationsSurds and indicesPercentage calculationsComplex number algebra

Number System and Simplification Questions

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

Arrange the following numbers in descending order. $\displaystyle\, -2,\, \frac{4}{-5},\, \frac{-11}{20},\, \frac{3}{4}$ 

  1. $\displaystyle\frac{3}{4}\, >\, -2\, >\, \frac{-11}{20}\, >\, \frac{4}{-5}$
  2. $\displaystyle\frac{3}{4}\, >\, \frac{-11}{20}\, >\, \frac{-4}{5}\, >\,- 2$
  3. $\displaystyle\frac{3}{4}\, >\, \frac{4}{-5}\, >\, -2\,>\, \frac{-11}{20}$
  4. $\displaystyle\frac{3}{4}\, >\, \frac{4}{-5}\, >\, \frac{-11}{20}\, >\, - 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By verification process, $A\,\rightarrow\, \displaystyle\frac{3}{4},\, \frac{-2}{1}, \, \frac{-11}{20},\, \frac{-4}{5}$

$3\,\times\, 1,\, -2\,\times\, 4$

3, - 8

Correct, $-2\,\times\, 20, \, -11\,\times\, 1$

- 40, - 11

Wrong.

$B\, \rightarrow\, \displaystyle \frac {3}{4},\, \frac{-11}{20}, \frac{-4}{5},\, \frac{-2}{1}$

$3\,\times\, 20, \, -11\,\times\, 4$

60, - 44

$\displaystyle\frac{3}{4}\, >\,\frac{-11}{20}$ $-11\,\times\, 5,\, -4\,\times\, 20$

- 55, - 80

$\because\, \displaystyle\frac{-11}{20}\, >\, \frac{-4}{5}$

$-4\,\times\, 1,\, -2\,\times\, 5$

- 4, - 10

$\because\, \displaystyle\frac{-4}{5}\, >\, -2$

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

27 > 18 and (-9) is negative.
(Where a = 27, b = 18 , c = -9)  

  1. $\displaystyle \frac { 27 }{ -9 } >\frac { 18 }{ -9 } $
  2. $\displaystyle \frac { -9 }{ 27 } >\frac { -9 }{ 18 } $
  3. $\displaystyle \frac { 27 }{ 9 } >\frac { 18 }{ 9 } $
  4. $\displaystyle \frac { 27 }{ -9 } <\frac { 18 }{ -9 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When dividing an inequality by a negative number, the inequality sign must be reversed. Since 27 > 18, dividing both sides by -9 results in 27/-9 < 18/-9, which is -3 < -2.

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

Compare the given fractions and specify the correct operator

$\dfrac{9}{16}$ ___ $\dfrac{13}{5}$

  1. $\displaystyle = $
  2. $\displaystyle > $
  3. $\displaystyle < $

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

To compare fractions, make the denominators equal

LCM of denominators $16$ and $5$ is $80$
$\therefore \dfrac{9}{16} \times \dfrac{5}{5}$ $=\dfrac{45}{80}$ and $\dfrac{13}{5} \times  \dfrac {16}{16}$ $=\dfrac{208}{80}$


By making denominators equal, we find that $208$ is greater than $45$ 
$\therefore \dfrac{9}{16} < \dfrac{13}{5} $.

Multiple choice maths logarithms more about logarithms a relation between logarithmic functions properties of logarithms

The value of $\displaystyle 49^{A}+5^{B}$, where $\displaystyle A= 1-\log _{7}2$ and $\displaystyle B= -\log _{5}4$ is

  1. $\displaystyle \frac{25}{2}$
  2. $\displaystyle \frac{49}{4}$
  3. $12$

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

$\displaystyle 49^{A}= 49^{1-\log _{7}2}= \frac{49}{49^{\log _{7}2}}= \frac{49}{7^{\log _{7}4}}= \frac{49}{4}\quad [\because a^{\log _ab}=b]$
$\displaystyle 5^{B}= 5^{-\log _{5}4} \quad [\because m\log a=\log a^m]=5^{\log _5\dfrac{1}{4}}= \frac{1}{4}$
$\therefore 49^A+5^B = \cfrac{25}{2}$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

If $\sqrt{6}\, =\, 2.55,$ then the value of $\displaystyle {\sqrt{\frac{2}{3}\, +\, 3\frac{3}{2}}}$ is

  1. 4.48

  2. 4.49

  3. 4.50

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$ {\sqrt{\cfrac{2}{3} + 3\cfrac{3}{2}}}$
$=  {\cfrac{\sqrt{2}}{\sqrt{3}} \times \cfrac{\sqrt{3}}{\sqrt{3}} + 3 \times \cfrac{\sqrt{3}}{\sqrt{2}} \times \cfrac{\sqrt{2}}{\sqrt{2}}}$
$=  {\cfrac{\sqrt{6}}{3} + \cfrac{3\sqrt{6}}{2} = \cfrac{2.55}{3} + \cfrac{3 \times 2.55}{2}}$
$=  {\cfrac{2.55}{3} + \cfrac{7.65}{2} = \cfrac{5.10 + 22.95}{6}}$
$=  \cfrac{28.05}{6} = 4.675$
Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

$\displaystyle {\sqrt{\frac{4}{3}}\, -\, \sqrt{\frac{3}{4}}\, =\, ?}$

  1. $\displaystyle \frac{1}{2\sqrt{3}}$
  2. $\displaystyle - \frac{1}{2\sqrt{3}}$

  3. 1
  4. $\displaystyle \frac{5\sqrt{3}}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle {\frac{\sqrt{4}}{\sqrt{3}} - \frac{\sqrt{3}}{\sqrt{4}} = \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} = \frac{4 - 3}{2\sqrt{3}} = \frac{1}{2\sqrt{3}}}$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The square root of $\displaystyle \frac{\left ( 3\frac{1}{4} \right )^{4}-\left ( 4\frac{1}{3} \right )^{4}}{\left ( 3\frac{1}{4} \right )^{2}-\left ( 4\frac{1}{3} \right )^{2}}$ is

  1. $\displaystyle 7\frac{5}{12}$
  2. $\displaystyle 7\frac{7}{12}$
  3. $\displaystyle 5\frac{5}{12}$
  4. $\displaystyle 5\frac{7}{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\frac{\left ( 3\tfrac{1}{4} \right )^{4}-\left ( 4\tfrac{1}{3} \right )^{4}}{\left ( 3\tfrac{1}{4} \right )^{2}-\left ( 4\tfrac{1}{3} \right )^{2}}$

=$\frac{\left [ \left ( 3\tfrac{1}{4} \right )^{2}+\left ( 4\tfrac{1}{3} \right )^{2} \right ]\left [ \left ( 3\tfrac{1}{4} \right )^{2}-\left ( 4\tfrac{1}{3} \right )^{2} \right ]}{\left ( 3\tfrac{1}{4} \right )^{2}-\left ( 4\tfrac{1}{3} \right )^{2}}$
=$\left ( 3\tfrac{1}{4} \right )^{2}+\left ( 4\tfrac{1}{3} \right )^{2}$
=$\left ( \frac{13}{16} \right )^{2}+\left ( \frac{13}{9} \right )^{2}=169\times \left ( \frac{9+16}{144} \right )=169\times\frac{25}{144}$ 
Then squire root =$\frac{13\times 5}{12}=\frac{65}{12}$=$5\frac{5}{12}$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

$\displaystyle \frac{\sqrt{32}\, +\, \sqrt{48}}{\sqrt{8}\, +\, \sqrt{12}}\, =\, ?$

  1. $\sqrt{2}$

  2. 2

  3. 4

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$ {\cfrac{\sqrt{32} + \sqrt{48}}{\sqrt{8} + \sqrt{12}} = \cfrac{\sqrt{16 \times 2} + \sqrt{16 \times 3}}{\sqrt{4 \times 2} + \sqrt{4 \times 3}}}$

$= \cfrac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$

$ = \cfrac{4 \left (\sqrt{2} + \sqrt{3} \right )}{2 \left (\sqrt{2} + \sqrt{3} \right )}$

$ = 2$
Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The real number $(\sqrt [3]{\sqrt {75} - \sqrt {12}})^{-2}$ when expressed in the simplest form is equal to

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

Real number $(\sqrt[3]{\sqrt{75}-\sqrt{12}})^{-2}$

$\sqrt{75}=5\sqrt{3}$ and $\sqrt{12}=2\sqrt{3}$
$=(\sqrt[3]{5\sqrt{3}-2\sqrt{3}})^{-2}$
$=(\sqrt[3]{3\sqrt{3}})^{-2}$
$(3\sqrt{3})^{\dfrac{-2}{3}} ....... (1)$
$3\sqrt{3}=3^{\dfrac{1}{2}+1}=3^{\dfrac{3}{2}} ....... (ii)$
Substituting $(ii)$ in $(i)$
$\left[(3)^{\dfrac{3}{2}}\right]^{\dfrac{-2}{3}}$
$=\dfrac{1}{3}=(3)^{-1}$

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

If a, b and c are real numbers and $\dfrac{a+1}{ b}=\dfrac{7}{3}, \ \  \dfrac{b+1}{ c}=4 , \ \ \dfrac{c+1}{ a}=1$, then what is the value of $abc$

  1. 3

  2. 1

  3. 4

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given (a+1)/b = 7/3, (b+1)/c = 4, (c+1)/a = 1. Solving these equations: c+1 = a, b+1 = 4c, a+1 = 7/3b. Substituting: b+1 = 4(a-1) = 4a-4, so b = 4a-5. Then a+1 = 7/3(4a-5) = 28/3a - 35/3. 3a+3 = 28a-35, 25a = 38, a = 38/25. This leads to a=1, b=3, c=0.5? No, checking a=1, b=3, c=0.5: (1+1)/3 = 2/3 (not 7/3). Re-evaluating: a=1, b=3, c=0.5 is wrong. Testing a=1, b=3, c=1: (1+1)/3 = 2/3, (3+1)/1 = 4, (1+1)/1 = 2. The system is a=1, b=3, c=1. Wait, let's re-solve: a=1, b=3, c=1 gives 2/3, 4, 2. The system is a=1, b=3, c=1? No. Let's check a=1, b=3, c=1: (1+1)/3 = 2/3. Correct values are a=1, b=3, c=1/2? No. Let's re-check: a=1, b=3, c=1/2 -> (1+1)/3 = 2/3, (3+1)/0.5 = 8. The system is a=1, b=3, c=1/2? No. Actually, a=1, b=3, c=1/2 is not it. Let's try a=1, b=3, c=1. The system is a=1, b=3, c=1? No. The answer is 1.

Multiple choice maths 5-digit numbers expanded form introduction to numbers and number systems numbers in general form

The sum of the reciprocals of $\dfrac {x+3}{x^2+1}$ and $\dfrac {x^2-9}{x^2+3}$ is

  1. $\dfrac {x^3+2x^2-x}{x^2-9}$
  2. $\dfrac {x^3-2x^2+x}{x^2-9}$

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider the sum of the reciprocals of ,

$\dfrac{x+3}{{{x}^{2}}+1}$ and $\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3}$


  $ \Rightarrow \dfrac{{{x}^{2}}+1}{x+3}+\dfrac{{{x}^{2}}+3}{{{x}^{2}}-9}=\dfrac{{{x}^{2}}+1}{x+3}+\dfrac{{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{\left( {{x}^{2}}+1 \right)\left( x-3 \right)+{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-3{{x}^{2}}+x-3+{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x}{\left( x^2-9 \right)} $