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 complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

The value of ${ i }^{ \frac { 1 }{ 3 }  }$ is:

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

The cube roots of i can be found using De Moivre's Theorem. i = cos(pi/2) + i sin(pi/2). The roots are cos((pi/2 + 2k*pi)/3) + i sin((pi/2 + 2k*pi)/3) for k=0, 1, 2. For k=0, we get cos(pi/6) + i sin(pi/6) = sqrt(3)/2 + i/2.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

The value of $\displaystyle\sum _{ n=0 }^{ 100 }{ { i }^{ n! } } $ equals ( where $i=\sqrt { -1 } $  ):

  1. $-1$
  2. $i$
  3. $2i + 95$
  4. $96 + i$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\sum _{0}^{n} i^{n!}=i^0+i^1+i^2+i^{1\times2\times3}+.....+ i^{1\times 2.....\times 100}$
We know that , $i^{4n}=1$ , $i^{4n+1}=i$ , $i^{4n+2}=-1$ , $i^{4n+3}=-i$
So starting from n=4 every number will be 1
So our sum shortens to $1+i+(-1)+(-1)+97=96+i$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

 Find the value of $\dfrac{i^6 + i^7 + i^8 + i^9}{i^2 + i^3}$

  1. $ 0
    $
  2. $ 1
    $
  3. $ -1
    $
  4. $ None.
    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $Z=\dfrac{i^6+i^7+i^8+i^9}{i^2+i^3}$


$=\dfrac{(i^2)^3-(i^2)^3i+(i^4)^2+(i^4)^2i}{-1-i}$


We know that, $i^2=-1$    and     $i^4=1$

$=\dfrac{-1-i+1+i}{-1-i}$

$Z=0$


$\therefore \dfrac{i^6+i^7+i^8+i^9}{i^2+i^3}=0$

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

The value of the sum $\displaystyle \sum _{n=1}^{13}(i^n+i^{n+1})$, where $i=\sqrt {-1}$, equals

  1. i

  2. i-1

  3. -i

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

We have $i^2 = -1$
Thus, $\displaystyle \sum _{n=1}^{4}(i^n+i^{n+1}) = (i^1 + i^2) + (i^2 + i^3) + (i^3 + i^4) + (i^4 + i^5) = (i - 1) + (-1 - i) + (-i + 1) + (1 + i) = 0$
\Rightarrow $\displaystyle \sum _{n=1}^{12}(i^n+i^{n+1}) = 0$
Now, only remains is $i^{13} + i^{14} = i - 1$

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

What is the value of the sum
$\displaystyle \sum _{ n=2 }^{ 11 }{ \left( { i }^{ n }+{ i }^{ n+1 } \right)  } $ where $i=\sqrt { -1 } $?

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

$\displaystyle \sum _{n = 1}^{11} (i^{n} + i^{n + 1}) i = \sqrt {-1}$

Now $i^{2} = -1, i^{3} = -i, i^{4} = 1, i^{5} = i$ after this values repeats
$i^{2} + i^{3} + i^{4} + i^{5} = 0$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} = i^{11} + i^{12} = -i + 1$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} + i^{n + 1} = -2i$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

Given that $\displaystyle {\frac{-6p\, -\, 9}{3}\, =\, \frac{2p\, +\, 9}{5}},$ find the value of p

  1. -4

  2. -2

  3. 3

  4. 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We shall Apply cross multiplication method.

$\dfrac { -6p-9 }{ 3 } =\dfrac { 2p+9 }{ 5 } \ \Longrightarrow 5\times \left( -6p-9 \right) =3\times \left( 2p+9 \right) \ \Longrightarrow -30p-6p=27+45\ \Longrightarrow -36p=72\ \Longrightarrow p=-2$
Ans- Option B.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

$\displaystyle\left(\displaystyle\frac{2}{3}\right)^{rd}$ of a number when multiplied by $\displaystyle\frac{3}{4}$ of the same number make $338$. The number is ___________.

  1. $18$
  2. $24$
  3. $36$
  4. $26$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the number is $x$

Thus according to given problem, we have
$\dfrac{2}{3} x\times  \dfrac{3}{4} x= 338$
$x^2=338\times 2$
$x^2= 676=26^2$
$x=26$
Therefore, the number is $26$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

If $\left(\displaystyle\frac{2}{3}\right)^{rd}$ of a number is $20$ less than the original number, then the number is ___________.

  1. $60$
  2. $40$
  3. $80$
  4. $120$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the number is $x$

Therefore, $x-\dfrac{2}{3}x =20$
$\Rightarrow \dfrac{1}{3}x=20$
$\Rightarrow x=60$
Therefore, the  number is $60$.

Multiple choice maths equations and simple functions further equations solving linear equations with variable on both sides solution of linear equations in one variable

Solve: $\displaystyle \frac{2x\, +\,1}{10}\, -\, \frac{3\, -\, 2x}{15}\, =\, \frac{x\, -\, 2}{6}$.


Hence, find y, if $\displaystyle \frac{1}{x}\, +\, \frac{1}{y}\, +\, 1\, = 0$.

  1. $\displaystyle x\, =\, -\frac{7}{5}; \, y\, =\, -\frac{7}{2}$
  2. $\displaystyle x\, =\, -\frac{2}{5}; \, y\, =\, \frac{7}{2}$
  3. $\displaystyle x\, =\, -\frac{6}{5}; \, y\, =\, -\frac{7}{2}$
  4. $\displaystyle x\, =\, -\frac{12}{5}; \, y\, =\, \frac{7}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \dfrac {2x + 1}{10} - \dfrac {3-2x}{15} = \dfrac {x-2}{6} $

On taking LCM and simplifying, we get

$ \dfrac {6x + 3 - 6 + 4x}{30} = \dfrac {x - 2}{6} $

$ => \dfrac {10x - 3}{30} = x - 2 $


$ => 10x - 3 = 5x - 10 $

$ 5x = -7 $

$ x = -\dfrac {7}{5} $

Now, substituting x in $ \dfrac {1}{x} + \dfrac {1}{y} + 1 = 0 $, we get

$ - \dfrac {5}{7} + \dfrac {1}{y} + 1 = 0 $

$ => \dfrac {1}{y} = - 1 + \dfrac {5}{7} =  - \dfrac {2}{7} $

$ => y = -\dfrac {7}{2} $

Multiple choice physics learning how to measure measurement of small and large distances measurement of distance unit of fundamental quantities

Convert $1$ $ \displaystyle g/cm^{3} $ to $\displaystyle kg/m^{3} $

  1. 1000

  2. 100

  3. 0.1

  4. 10
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that  $1 \ g = \dfrac{1}{1000} \ kg$
Also,  $1\ cm = \dfrac{1}{100} \ m$
Thus   $1g/cm^3 = 1 \ g\times\dfrac{1  \ kg}{1000 \ g}\times \dfrac{1}{1 \ cm^3}\times \dfrac{(100)^3 \ cm^3}{1 \ m^3} = 1000 \ kg/m^3$

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

If $\frac {4+3\sqrt 5}{\sqrt 5}=a+b\sqrt 5$ then, the value of b is

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

$\frac {4+3\sqrt 5}{\sqrt 5}=a+b\sqrt 5$
$\frac {(4+3\sqrt 5)\sqrt 5}{\sqrt 5\times \sqrt 5}=\frac {4\sqrt 5+15}{5}=\frac {4\sqrt 5}{5}+\frac {15}{5}$
$=3+\frac {4\sqrt 5}{5}$
Clearly, $a=3$
$b=\frac {4}{5}$.

Multiple choice maths operations on rational numbers rational numbers on number line rational numbers and their decimal expansions rational numbers between two rational numbers

Which of the following numbers lies between $2\dfrac {1}{7}$ and $3\dfrac {1}{7}$?

  1. $\dfrac {37}{7}$
  2. $\dfrac {14}{7}$
  3. $\dfrac {37}{14}$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Mean = \dfrac{\left (\dfrac {15}{7} + \dfrac {22}{7}\right )}{2} = \left (\dfrac {15 + 22}{7}\right ) \times \dfrac {1}{2} = \dfrac {37}{7}\times \dfrac {1}{2}$
$= \dfrac {37}{14}$

Mean of two numbers always lies between the two numbers.
So, answer is option $C.$