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 algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Divide : $\displaystyle \left( 51{ m }^{ 3 }{ p }^{ 2 }-34{ m }^{ 2 }{ p }^{ 3 } \right)$ by $17mp$

  1. $\displaystyle { m }^{ 2 }p$
  2. $\displaystyle m{ p }^{ 2 }$
  3. $\displaystyle 3{ m }^{ 2 }p-2m{ p }^{ 2 }$
  4. $\displaystyle mp$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \frac { 51{ m }^{ 3 }{ p }^{ 2 }-34{ m }^{ 2 }{ p }^{ 3 } }{ 17mp } $

$\displaystyle =\frac { 17mp\left( 3{ m }^{ 2 }p-2m{ p }^{ 2 } \right)  }{ 17mp } $
$\displaystyle = 3{ m }^{ 2 }p-2m{ p }^{ 2 }$

Multiple choice maths percentages finding one number as percentage of another money and metric measures as percentage expressing one quantity as a percentage of another

$\displaystyle 12\frac{1}{2}$% of .......... = 35% of 700

  1. $490$
  2. $500$
  3. $1960$
  4. $1800$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the blank space be $x$ and we solve the given equality $12\dfrac { 1 }{ 2 }$% of $x=35$% of $700$ as follows:


$\dfrac { 12\dfrac { 1 }{ 2 }  }{ 100 } \times x=\dfrac { 35 }{ 100 } \times 700$

$ \Rightarrow \dfrac { \dfrac { 25 }{ 2 }  }{ 100 } \times x=35\times 7$

$ \Rightarrow \dfrac { 25 }{ 200 } \times x=245$

$ \Rightarrow \dfrac { x }{ 8 } =245$

$ \Rightarrow x=245\times 8$

$ \Rightarrow x=1960$

Hence, $12\dfrac { 1 }{ 2 }$% of $1960=35$% of $700$.

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

The value of $\displaystyle\frac{2^{m+3}\times3^{2m-n}\times5^{m+n+3}6^{n+1}}{6^{m+1}\times10^{n+3}\times15^m}$ is equal to

  1. 0

  2. 1
  3. $2^m$

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

Numerator $=2^{m+3}\cdot3^{2m-n}\cdot5^{m+n+3}\cdot2^{n+1}\cdot3^{n+1}$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (i)
Denominator $=2^{m+1}\cdot3^{m+1}\cdot2^{n+3}\cdot5^{n+3}\cdot3^m\cdot5^m$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (ii)
Given expression $=1$

Multiple choice maths repeated multiplication exponentiation index notation and products of prime factors powers

Find the value of $\displaystyle\frac{5^0+5^{-1}}{5^0-5^{-1}}-\left(\frac{8}{27}\right)^{\displaystyle\frac{1}{3}}-\left(\frac{36}{25}\right)^{-\displaystyle\frac{1}{3}}$

  1. 0
  2. $\displaystyle\frac{1}{2}$

  3. 1

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

$\displaystyle\frac{\displaystyle1+\frac{1}{5}}{\displaystyle1-\frac{1}{5}}-\left[\left(\frac{2}{3}\right)^3\right]^{\displaystyle\frac{1}{3}}-\left[\left(\frac{6}{5}\right)^2\right]^{\displaystyle-\frac{1}{2}}=\frac{\displaystyle\frac{6}{5}}{\displaystyle\frac{4}{5}}-\left(\frac{2}{3}\right)^1-\left(\frac{6}{5}\right)^{-1}=\frac{6}{4}-\frac{2}{3}-\frac{5}{6}=\frac{18-8-10}{12}=0$

Multiple choice maths indices exponentiation index notation and products of prime factors powers

If $\displaystyle { 2 }^{ n }-{ 2 }^{ n-1 }=4$, then the value of $\displaystyle { n }^{ n }$ will be -

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

  3. 2

  4. 27
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle { 2 }^{ n }-{ 2 }^{ n-1 }=4$
$\displaystyle \therefore \quad { 2 }^{ n-1 }\left( 2-1 \right) =4$
$\displaystyle \therefore \quad { 2 }^{ n-1 }={ 2 }^{ 2 }$
$\displaystyle \therefore \quad n-1=2$
$\displaystyle \therefore \quad n=3$
$\displaystyle \therefore \quad { n }^{ n }={ 3 }^{ 3 }=27$

Multiple choice maths indices exponentiation index notation and products of prime factors powers

$\displaystyle \frac { { \left( 3.63 \right)  }^{ 2 }-{ \left( 2.37 \right)  }^{ 2 } }{ 3.63+2.37 } $ is simplified to -

  1. 6

  2. 1.36

  3. 2.26

  4. 1.26
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \frac { { \left( 3.63 \right)  }^{ 2 }-{ \left( 2.37 \right)  }^{ 2 } }{ 3.63+2.37 } $
$\displaystyle =\frac { \left( 3.63+2.37 \right) \left( 3.63-2.37 \right)  }{ 3.63+2.37 } $
$\displaystyle =3.63-2.37$
$\displaystyle =1.26$

Multiple choice maths indices exponentiation index notation and products of prime factors powers

Simplest form of the Expression $\displaystyle { \left( { x }^{ 6 }.{ y }^{ { -5 }/{ 4 } } \right)  }^{ { -4 }/{ 3 } }$ will be-

  1. $\displaystyle { x }^{ -24 }y$
  2. $\displaystyle { x }^{ -8 }{ y }^{ { 5 }/{ 3 } }$
  3. $\displaystyle { x }^{ 8 }{ y }^{ { -5 }/{ 3 } }$
  4. $\displaystyle { x }^{ -8 }{ y }^{ { -5 }/{ 3 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle { \left( { x }^{ 6 }.{ y }^{ { -5 }/{ 4 } } \right)  }^{ { -4 }/{ 3 } }$
$\displaystyle ={ x }^{ -6\times { 4 }/{ 3 } }{ y }^{ \dfrac { -5 }{ 4 } \times -\dfrac { 4 }{ 3 }  }$
$\displaystyle ={ x }^{ -8 }{ y }^{ { 5 }/{ 3 } }$

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

The value of $\displaystyle \frac{1}{1+\frac{1}{1+\frac{1}{1+1/2}}}$ on simplification is

  1. 5/8

  2. 6/7

  3. 7/8

  4. 8/6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}=\frac{1}{1+\frac{1}{1+\frac{1}{\frac{2+1}{2}}}}=$
=$\frac{1}{1+\frac{1}{1+\frac{2}{3}}}= \frac{1}{1+\frac{1}{\frac{3+2}{3}}}$
=$\frac{1}{1+\frac{3}{5}}=\frac{1}{\frac{8}{5}}$
=$\frac{5}{8}$
Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

Subtract  $ - \frac{2}{3}{y^3}-\frac{2}{7}{y^2} - 5$ from $\frac{1}{3}{y^3} + \frac{5}{7}{y^2} - 2$, then the resultant value is . 

  1. ${y^3} + {y^2} - 7$
  2. ${y^3} + {y^2} + 3$
  3. ${y^3} + {y^2} - 3$
  4. ${y^3} + {y^2} + 7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To subtract the first expression from the second, compute ((1/3)y^3 + (5/7)y^2 - 2) - (-(2/3)y^3 - (2/7)y^2 - 5). Distributing the negative sign gives ((1/3) + (2/3))y^3 + ((5/7) + (2/7))y^2 + (-2 + 5), which simplifies to y^3 + y^2 + 3.

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

If $\displaystyle b=6-\left [ \frac{4b+3}{2a-5} \right ]$, express a in terms of b.

  1. $\displaystyle a=\frac{33-b}{2\left ( 6-b \right )}$
  2. $\displaystyle a=\frac{b-33}{2\left ( 6-b \right )}$
  3. $\displaystyle a=\frac{33-b}{2\left ( b-6 \right )}$

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

Given,
$ b = 6 - \left[ \frac { 4b+3 }{ 2a-5 }  \right]  $
$ => \left[ \frac { 4b+3 }{ 2a-5 }  \right] = 6 - b $
$ => 4b + 3 = (6-b)(2a-5) $
$ => 4b + 3 = 12a - 30 - 2ab +5b $
$ => 12a - 33  -2ab + b = 0 $
$ =>  a(12 -2b) = 33 - b $
$ => 2a(6-b) = 33-b $
$ => a = \frac {33-b}{2(6-b)} $

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

Given $\displaystyle b=\frac{2a}{a-2}$ and $\displaystyle c=\frac{3b-4}{4b+3}$, express c in terms of a.

  1. $\displaystyle c=\frac{2a+8}{11a+6}$
  2. $\displaystyle c=\frac{2a-8}{11a-6}$
  3. $\displaystyle c=\frac{2a+8}{11a-6}$

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

Given, $ c = \frac {3b-4}{4b + 3} $

But, $ b = \frac {2a}{a-2} $
Hence, $ c = \frac {3(\frac {2a}{a-2} )-4}{4(\frac {2a}{a-2} ) + 3} $
$ => c = \frac {6a-4a + 8}{8a +3a - 6} $
$ => c = \frac {2a+8}{11a -6} $