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 part number dividing fractions division of a fractions division of a fraction

Evaluate the following :

$\displaystyle  8 \times \dfrac {\dfrac{5}{24}}{\dfrac {7}{12}} $.

  1. $\displaystyle \frac {5 }{24}$
  2. $\displaystyle \frac {20}{7}$
  3. $\displaystyle \frac {2}{7}$
  4. $\displaystyle \frac {21}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $ 8 \times \displaystyle \frac {\frac {5}{24}}{\frac {7}{12}} $

$ \therefore 8 \times \displaystyle \frac {\frac {5 }{24}}{\frac {7}{12}} = 8 \times \frac {5\times 12}{7 \times 24}   $

$=\displaystyle \frac {5 \times 4}{7}$

$= \displaystyle \frac {20}{7} $

Multiple choice maths part number dividing fractions division of a fractions division of a fraction

$I = \displaystyle \frac{3}{4}\div \frac{5}{6}$ 

$II = 3\displaystyle \div $[($4\displaystyle \div 5$)$\displaystyle \div 6$]
$III = [3\displaystyle \div (4\displaystyle \div 5)]\displaystyle \div 6$
$IV = 3\displaystyle \div 4(5\displaystyle \div 6)$
Select the correct option from the following.

  1. I and II are equal

  2. I and IV are equal

  3. I and III are equal

  4. All are equal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ I=\cfrac{3}{4}\times \cfrac{6}{5}=\cfrac{9}{10}$


$II=3\div \left [ \cfrac{4}{5}\times \cfrac{1}{6} \right ]=3\times \cfrac{15}{2}=\cfrac{45}{2}$
$III=\left [ 3\div \cfrac{4}{5} \right ]\div 6=3\times \cfrac{5}{4}\times \cfrac{1}{6}=\cfrac{5}{8}$
$ IV=3\div 4\times\cfrac{5}{6}=3\times \cfrac{3}{10}=\cfrac{9}{10}$

Hence, $I$ and $IV$ are equal.

Multiple choice maths part number dividing fractions division of a fractions division of a fraction

The least fraction that must be added to $\displaystyle1\frac{1}{3}\div 1\frac{1}{2}\div 1\frac{1}{9}$  to make the result an integer is:

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

$\displaystyle 1\frac{1}{3}\div 1\frac{1}{2}=\frac{4}{3}\div \frac{3}{2}\div \frac{10}{9}$
$\displaystyle =\frac{4}{3}\times \frac{2}{3}\div \frac{10}{9}=\frac{8}{9}\times \frac{9}{10}=\frac{4}{5}$
$\displaystyle \therefore \frac{1}{5}$ should be added to $\displaystyle \frac{4}{5}$ to make it an integer

Multiple choice mathematics and statistics introduction to set union and intersections union and intersection of sets basic operations on sets

If X=(multiples of $2$ ), Y = ( multiples of $5$) , Z= (multiples of $10$), then $ \displaystyle X \cap(Y\cap Z)    $ is equal to 

  1. Multiples of $10$
  2. Multiples of $5$
  3. Multiples of $2$
  4. Multiples of $7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$x= (multiples\ of\ 2\ is\ 2, 4, 6, 8, 10.............)$
$y=(multiples\ of\ 5\ is\ 5, 10,15,20,25.........)$
$z=(multiples\ of\ 10\ is\ 10,20,30,40...........)$
Then, $ X\cap(Y\cap Z)$

Apply the value
$=(2, 4, 6, 8, 10,.......)\cap[ (5, 10, 15, 20, 25,........)\cap (10, 20, 30, 40,.......)]$
$=(2, 4, 6, 8, 10,.......)\cap (10, 20, 30,........)$
$=(10, 20, 30,........)$

Hence, this is multiple of $10$.


Hence, this is the answer.

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

$\displaystyle i+\frac{1}{i}=$

  1. $1$
  2. $-1$
  3. $0$
  4. $2i$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $Z= i+\dfrac{1}{i}$


Mutiplying numerator and denominator by i. We get,

          $=i+\dfrac{i}{i^{2}}$

          $=i+\dfrac{i}{-1} \quad \dots (i^2=-1)$

          $=i-i$

      $Z=0$

Hence, 

$i+\dfrac{1}{i}=0$.

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 $\displaystyle \left( 4+2i \right) \left( 4-2i \right) $ given that $\displaystyle { i }^{ 2 }=-1$. 

  1. $12$
  2. $20$
  3. $\displaystyle 16-4i$
  4. $\displaystyle 4+16i$
  5. $\displaystyle 12-16i$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

After expanding, we get $(4+2i)(4-2i)=16+8i-8i-4i^4$

According to the question $i^2=-1$
$\Rightarrow 16-4i^4$
$\Rightarrow 16-4 ( -1)$
$\Rightarrow 16+4=20$

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 $\displaystyle \sum _{ n=1 }^{ 13 }{ \left( { i }^{ n }+{ i }^{ n+1 } \right)  }$. 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

Given:

$\displaystyle \sum _{n=1}^{13}(i^n+i^{n+1})$

So,
$\Rightarrow(i^1+i^{2})+(i^2+i^3)+(i^3+i^4)+........+(i^{13}+i^{14})$

We know that
$i^2=-1$
$i^3=-i$
$i^4=1$
$i^5=i$
$i^6=-1$
$i^7=-i$
$i^8=1$

Therefore,
$\Rightarrow(i-1)+(-1-i)+(-i+1)+........+(i-1)$

Same cycle upto $4^{th}$ term.

Therefore,

$\Rightarrow(i-1)+(-1-i)+(-i+1)+(1+i)+........+(i-1)$

So, all terms will cancel out with each other up to $12th$ term.

Therefore,
$\Rightarrow i-1$

Hence, this is the answer.

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
$i^{2} + i^{3} + i^{4} + i^{5} = 0$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} = i^{11} + i^{12} = -i + 1$
$\therefore \displaystyle \sum _{n = 2}^{11} i^{n} + i^{n + 1} = -2i$.