Quantitative Aptitude

Number System and Simplification

585 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 power and exponent power of powers laws of exponents and powers law of indices

The value of $\cfrac { { 2 }^{ 2n-2 } }{ { 2 }^{ n(n-1) } }-\cfrac { { 8 }^{ n-1 } }{ { 2 }^{ (n-1)(n+1) } } $ will be

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

$\dfrac{2^{2n-2}}{2^{n(n-1)}} - \dfrac{8^{n-1}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^{2(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^{3(n-1)}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^2\times 2^{(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^3\times 2^{(n-1)}}{2^{(n-1)(n+1)}}$

$=\dfrac{2^2\times 2^{(n-1)}}{2^n \times 2^{(n-1)}} - \dfrac{2^3\times 2^{(n-1)}}{2^{(n-1)}2^{(n+1)}}$

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

$=\dfrac{2^2}{2^n}-\dfrac{2^3}{2^n \times 2^1}$

$=\dfrac{2^2}{2^n}-\dfrac{2^2}{2^n}$

$=0$
Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

If $y=mx+7\sqrt{3}$ is normal to $\dfrac{x^2}{18}-\dfrac{y^2}{24}=1$ then the value of m can be?

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

$7\sqrt{3}=\dfrac{42m}{\sqrt{24-18m^2}}\Rightarrow \sqrt{3}=\dfrac{\sqrt{6}m}{\sqrt{4-3m^2}}\Rightarrow 4-3m^2=2m^2$
$m=\dfrac{2}{\sqrt{5}}$.

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

To find out degree of freedom, the correct expression is :

  1. $f=\dfrac { 2 }{ \gamma -1 }$
  2. $f=\dfrac { \gamma +1 }{ 2 }$
  3. $f=\dfrac { 2 }{ \gamma +1 }$
  4. $f=\dfrac { 1 }{ \gamma +1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\because \gamma =1+\dfrac { 2 }{ f } $
$\Longrightarrow \dfrac { 2 }{ f } =\gamma -1\Longrightarrow f=\dfrac { 2 }{ \gamma -1 } $

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

In the final answer of the expression  $\dfrac { ( 29.2 - 20.2 ) \left( 1.79 \times 10 ^ { 5 } \right) } { 1.37 }.$  The number of significant figures is

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

29.2 - 20.2 = 9.0 (two sig figs). 9.0 * 1.79 = 16.11. 16.11 / 1.37 = 11.759. The result should be limited by the precision of the subtraction (two sig figs), but the options suggest three.

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $a = {\mathop{\rm cis}\nolimits} \alpha ,b = cis\beta ,c = cis\gamma $ then $\dfrac{{{a^3}{b^3}}}{{{c^2}}} = $

  1. $cis(3\alpha + 3\beta + 2\gamma )$
  2. $cis(3\alpha + 3\beta - 2\gamma )$
  3. $cis( - 3\alpha - 3\beta + 2\gamma )$
  4. $cis(3\alpha - 3\beta + 2\gamma )$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$a=cis\alpha=Cos\alpha+iSin\alpha= \ e^{i\alpha}$

$b=cis\beta=Cos\beta+iSin\beta= \ e^{i\beta}$      

$a=cis\gamma=Cos\gamma+iSin\gamma= \ e^{i\gamma}$

$\therefore \dfrac{a^3b^3}{c^2}=\dfrac{({e^{i\alpha}})^3({e^{i\beta}})^3}{({e^{i\gamma}})^2}$

$=\dfrac{e^{3i\alpha}e^{3i\beta}} {e^{2i\gamma}}=\ e^{i(3\alpha+3\beta-2\gamma)}$

$=Cos(3\alpha+3\beta-2\gamma)+iSin(3\alpha+3\beta-2\gamma)$

$=cis(3\alpha+3\beta-2\gamma)$

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

The value of $\displaystyle { \left( \frac { 1+i }{ \sqrt { 2 }  }  \right)  }^{ 8 }+{ \left( \frac { 1-i }{ \sqrt { 2 }  }  \right)  }^{ 8 }$ is equal to

  1. $4$
  2. $6$
  3. $8$
  4. $2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We have $\displaystyle { \left( \frac { 1+i }{ \sqrt { 2 }  }  \right)  }^{ 8 }+{ \left( \frac { 1-i }{ \sqrt { 2 }  }  \right)  }^{ 8 }$


$\displaystyle={ \left[ \cos { \frac { \pi  }{ 4 }  } +i\sin { \frac { \pi  }{ 4 }  }  \right]  }^{ 8 }+{ \left[ \cos { \frac { \pi  }{ 4 }  } -i\sin { \frac { \pi  }{ 4 }  }  \right]  }^{ 8 }$


$=\cos { 2\pi  } +i\sin { 2\pi  } +\cos { 2\pi  } -i\sin { 2\pi  } $      [by de-moivre's theorem]

$=2\cos { 2\pi  } =2\left( 1 \right) =2$  

Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

If $\displaystyle x=\frac{4\sqrt{2}}{\sqrt{2}+1}$ then find the value of $\displaystyle \frac{1}{\sqrt{2}}\left ( \frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x-2\sqrt{2}} \right )$

  1. $\displaystyle \sqrt{2}$
  2. $12+8\displaystyle \sqrt{2}/5$
  3. $12-8\displaystyle \sqrt{2}$
  4. $\displaystyle \frac{16\sqrt{2}+24}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x=\frac{4\sqrt{2}}{\sqrt{2+1}}$
$\frac{1}{\sqrt{2}}\left ( \frac{x+2}{x-2}+\frac{x+2\sqrt{2}}{x-2\sqrt{2}} \right )$
Put the value of x
$\frac{1}{\sqrt{2}}\left ( \frac{\frac{4\sqrt{2}}{\sqrt{2+1}}+2}{\frac{4\sqrt{2}}{\sqrt{2+1}}-2}+\frac{\frac{4\sqrt{2}}{\sqrt{2+1}}+2\sqrt{2}}{\frac{4\sqrt{2}}{\sqrt{2+1}}-2\sqrt{2}} \right )$
=$\frac{1}{\sqrt{2}}\left ( \frac{4\sqrt{2}+2\sqrt{2}+2}{4\sqrt{2}-2\sqrt{2}-2} \right )+\left ( \frac{4\sqrt{2}+4+2\sqrt{2}}{4\sqrt{2}-4-2\sqrt{2}} \right )$
=$\frac{6\sqrt{2}+2}{2\sqrt{2-2}}+\frac{6\sqrt{2}+4}{2\sqrt{2}-4}$
=$\frac{1}{\sqrt{2}}\left ( \frac{32-24\sqrt{2}}{16-12\sqrt{2}} \right )$
=$\sqrt{2}$

Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

If P : Q : R = 6 : 5 : 4 and $\displaystyle P^{2}+Q^{2}+R^{2}=192500$ then find $\displaystyle \frac{(P+Q-R)}{2}$

  1. 175

  2. 165

  3. 185

  4. 200
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$
:Q:R\quad =\quad 6:5:4\ Let\quad P\quad \quad =\quad 6x\ Q=\quad 5x\ R\quad =\quad 4x\ { P }^{ 2 }+{ Q }^{ 2 }{ +\quad R }^{ 2 }\quad =\quad 192500\ { (6x) }^{ 2 }+{ (5x) }^{ 2 }+(4x)^{ 2 }\quad =\quad 192500\ 77{ x }^{ 2 }\quad =\quad 192500\ { x }^{ 2 }\quad =\quad 2500\ x\quad =\quad 50\ \ \frac { P+Q-R }{ 2 } \quad =\quad \frac { 6x+5x-4x }{ 2 } \quad =\quad \frac { 7x }{ 2 } \quad =\quad \frac { 7\times 50 }{ 2 } \quad =\quad 175
$

Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

If a=2, b=3, c=4, then the difference between $\displaystyle 2\frac{3}{4}$ and $\displaystyle b\frac{a}{c}$ is

  1. $-1/4$
  2. $1/4$
  3. $-3/4$
  4. $5/4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a=2 ,b=3 and c=4

Then $b\tfrac{a}{c}=3\tfrac{2}{4}=3\tfrac{1}{2}$
Then difference between=$2\tfrac{3}{4}-3\tfrac{1}{2}=\frac{11}{4}-\frac{7}{2}=\frac{11-14}{4}=\frac{-3}{4}$

Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

$\left (1 - \dfrac {1}{2}\right ) + \left (\dfrac {3}{4} - \dfrac {1}{4}\right )=$

  1. $0$
  2. $1$
  3. $\dfrac {1}{2}$
  4. $\dfrac {3}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\left ( 1-\dfrac{1}{2} \right )+\left ( \dfrac{3}{4}-\dfrac{1}{4} \right )$
$=\left ( \dfrac{2-1}{2} \right )+\left ( \dfrac{3-1}{4} \right )$
$=\dfrac{1}{2}+\dfrac{2}{4}= \dfrac{1}{2}+\dfrac{1}{2}= 1$
Multiple choice maths average arithmetic mean of ap introduction to averages means

If $A _1,A _2$ be two arithmetic means between $\dfrac{1}{3}$ and $\dfrac{1}{24}$, then their value are 

  1. $\dfrac{7}{72},\dfrac{5}{36}$
  2. $\dfrac{17}{72},\dfrac{5}{36}$
  3. $\dfrac{7}{36},\dfrac{5}{72}$
  4. $\dfrac{5}{72},\dfrac{17}{72}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For two means A1, A2 between 1/3 and 1/24, the common difference d = (1/24 - 1/3) / (2 + 1) = (-7/24) / 3 = -7/72. A1 = 1/3 - 7/72 = 17/72. A2 = 17/72 - 7/72 = 10/72 = 5/36.

Multiple choice maths average arithmetic mean of ap introduction to averages means

The arithmetic mean of 1, 2, 3, ..., n, is

  1. $\displaystyle \frac{n-1}{2}$
  2. $\displaystyle \frac{n+1}{2}$
  3. $\displaystyle \frac{n}{2}$
  4. $\displaystyle \frac{n}{2}+1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow$   We have the sequence, $1,2,3.......n$

$\Rightarrow$   This is an AP, with the initial term $a=1$ and the common difference $d=1$.
$\therefore$   The sum of $n$ terms of an AP is given by,
$\Rightarrow$  $S _n=\dfrac{n}{2}[2a+(n-1)d]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[2\times 1+(n-1)\times 1]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[2+(n-1)]$

$\Rightarrow$  $S _n=\dfrac{n}{2}[n+1]$
$\rightarrow$   Arithmetic mean of $n$ numbers $a _1,a _2,a _3,a _4,... a _n$ is given by the formula
$\Rightarrow$  $Arithmetic\,mean=\dfrac{a _1+a _2+a _3+a _4+...+a _n}{n}$

$\Rightarrow$  $Arithmetic\,mean=\dfrac{S _n}{n}$

$\Rightarrow$  $Arithmetic\, mean=\dfrac{\dfrac{n}{2}[n+1]}{n}$

$\therefore$     $Arithmetic\, mean=\dfrac{n+1}{2}$