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 fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

Observe the following pattern and fill in the missing number. 
$ \displaystyle 11^{2} =121$
$ \displaystyle 101^{2} =10201$
$ \displaystyle 10101^{2} =102030201$
$ \displaystyle 1010101^{2} =......................$

  1. $ \displaystyle 1010101^{2} $=10203030201
  2. $ \displaystyle 1010101^{2} $=10204040201
  3. $ \displaystyle 1010101^{2} $=1020304030201
  4. $ \displaystyle 1010101^{2} $=10204030201
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$11^{2}$$=$$121$

$101^{2}$$=$$10201$
$10101^{2}$$=$$10203020101$
$1010101^{2}$$=$$1020304030201$
$101010101^{2}$$=$$10203040504030201$
We will go up to number of  ones in the number numerically.
Hence, Option C is correct.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The value of $3 - 1 + \frac{1}{3} - \frac{1}{9} +  \ldots $ is equal to

  1. $\dfrac{{20}}{9}$
  2. $\dfrac{{9}}{20}$
  3. $\dfrac{{9}}{4}$
  4. $\dfrac{{4}}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The series is 3 - 1 + 1/3 - 1/9 + ... which can be split into 3 + (-1 + 1/3 - 1/9 + ...). The part in parentheses is a geometric series with first term a = -1 and common ratio r = -1/3. The sum is a / (1 - r) = -1 / (1 - (-1/3)) = -1 / (4/3) = -3/4. Adding the initial 3 gives 3 - 3/4 = 9/4.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The sum to infinity of the series $1 + \dfrac{2}{3} + \dfrac{6}{{{3^2}}} + \dfrac{{10}}{{{3^3}}} + \dfrac{{14}}{{{3^4}}} + ......,is$

  1. $3$
  2. $4$
  3. $6$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $S$=$1 + \cfrac{2}{3} + \cfrac{6}{{{3^2}}} + \cfrac{{10}}{{{3^3}}} + ......$

$\cfrac{S}{3} = \cfrac{1}{3} + \cfrac{2}{{{3^2}}} + \cfrac{6}{{{3^3}}} + ....$

$S - \cfrac{S}{3} = 1 + \cfrac{1}{3} + \cfrac{4}{{{3^2}}} + ....$

$\cfrac{{2S}}{3} = \cfrac{{\cfrac{4}{3} }}{{1 - \cfrac{1}{3}}}$

$ = \dfrac{\cfrac{4}{3}} { \cfrac{2}{3}} = \cfrac{4}{2} = 2$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If the sum of the series $2+\frac {\displaystyle 5}{\displaystyle x}+\frac {\displaystyle 25}{\displaystyle x^2}+\frac {\displaystyle 125}{\displaystyle x^3}+....$ is finite, then-

  1. $\mid x\mid > 5$

  2. -5 < x < 5
  3. $\mid x\mid < 5/2$
  4. $\mid x\mid > 5/2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We can rewrite the series as
$1+1+\dfrac {5}{x}+(\dfrac {5}{x})^2+(\dfrac {5}{x})^3+.....$
We can sum up this series if $\mid 5/x\mid < 1$
$\Leftrightarrow \mid x\mid > 5$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The series $\dfrac{2x}{x+3}+(\dfrac{2x}{x+3})^{2}+(\dfrac{2x}{x+3})^{3}+........\infty$ will have a definite sum when  

  1. $x<3$
  2. $x>3$
  3. $x=0$
  4. $x=-3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\dfrac{2x}{x+3}+\left(\dfrac{2x}{x+3}\right)^{2}+\left(\dfrac{2x}{x+3}\right)^{3}.......\infty $
$\therefore a=\dfrac{2x}{x+3}$   $r=\dfrac{2x}{x+3}$
$\therefore s=\dfrac{a}{1-r}=\dfrac{\dfrac{2x}{x+3}}{1-\dfrac{2x}{x+3}}$
$=\dfrac{2x}{x+3-2x}=\dfrac{2x}{3-x}$
Now, to have definite sum
$r < 1$
$\therefore \dfrac{2x}{x+3} < 1$
$\therefore 2x < x+3$
$\therefore x < 3$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If sum of an infinite geometric series is $\dfrac{4}{3}$ and its Ist term is $\dfrac{3}{4}$, then its common ratio is

  1. $\dfrac{7}{16}$
  2. $\dfrac{9}{16}$
  3. $\dfrac{1}{9}$
  4. $\dfrac{7}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that sum of infinite geometric series $=\dfrac{a}{1-r}$
Where $a=\text{first term}$ and $r=\text{common ratio}$.
$\dfrac{a}{1-r}=\dfrac{4}{3}$
Then, $\dfrac{\dfrac{3}{4}}{1-r}=\dfrac{4}{3}\Rightarrow\,r=1-\dfrac{9}{16}=\dfrac{7}{16}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

What is the sum of the series $ 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + ....$ equal to ?

  1. $\dfrac{1}{2}$
  2. $\dfrac{3}{2}$
  3. $2$
  4. $\dfrac{2}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$1,\dfrac { -1 }{ 2 } ,\dfrac { 1 }{ 4 } ,\dfrac { -1 }{ 8 } ,..$ is in G.P series. 
So, sum of infinite terms of a G.P is $\dfrac { a }{ 1-r } $, where $a$ is first term $=1$
$r$ is common difference $=-1/2=-0.5$
Thus, $1+\dfrac { -1 }{ 2 } +\dfrac { 1 }{ 4 } +\dfrac { -1 }{ 8 } ,..=$ $\dfrac { 1 }{ 1-\left( \frac { -1 }{ 2 }  \right)  } =\dfrac { 1 }{ \left( \frac { 3 }{ 2 }  \right)  } =\dfrac { 2 }{ 3 } $
Hence, D is correct.
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

$1 + x + x^2 + x^3 +......$ = ?

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

Sum of a GP $a,ar,ar^2......$

$=\dfrac{a(r^n -1)}{r-1}$    (for $n$ terms)
For the given series,
$a=1, r=x, n \to \infty$
Sum of the series is:
$\text{sum} = \lim _{n \to \infty} \dfrac{x^n-1}{x-1}$
For $x>1$ 
$\text{sum} \to \infty$
For $x<1$
$x^n \to 0$
$\Rightarrow \text{sum} = \dfrac{-1}{x-1}$
$\Rightarrow \text{sum} = \dfrac{1}{1-x}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Sum to infinity of the series $\displaystyle \frac { 2 }{ 3 } -\frac { 5 }{ 6 } +\frac { 2 }{ 3 } -\frac { 11 }{ 24 } +...$ is

  1. $\displaystyle \frac { 4 }{ 9 } $
  2. $\displaystyle \frac { 1 }{ 3 } $
  3. $\displaystyle \frac { 2 }{ 9 } $

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

Let $\displaystyle S=\frac { 2 }{ 3 } -\frac { 5 }{ 6 } +\frac { 2 }{ 3 } -\frac { 11 }{ 24 } +...$ to $\infty$   ...(1)


Multiplying both sides by $\displaystyle -\frac { 1 }{ 2 } $, the common ratio $G.P.$


$\displaystyle -\frac { 1 }{ 2 } S=-\frac { 2 }{ 6 } +\frac { 5 }{ 12 } -\frac { 8 }{ 24 } +...$ to $\infty$    ....(2)

Subtracting (2) from (1), we have

$\displaystyle \frac { 3 }{ 2 } S=\frac { 2 }{ 3 } -\frac { 3 }{ 6 } +\frac { 3 }{ 12 } -\frac { 3 }{ 24 } +...$ to $\infty$

$\displaystyle =\frac { 2 }{ 3 } -\left( \frac { 1 }{ 2 } -\frac { 1 }{ 4 } +\frac { 1 }{ 8 } +... \right) $

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

$\displaystyle \therefore S=\frac { 1 }{ 3 } \times \frac { 2 }{ 3 } =\frac { 2 }{ 9 } $

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

$\displaystyle2+1+\frac{1}{2}+\frac{1}{4}+\cdots\cdots\infty$ is

  1. 1

  2. 2

  3. 3

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

The given series is a Geometric Progression, with first terms $ a =2$ and common ratio $ r = \dfrac {T _2}{T _1} = \dfrac {1}{2} $

For a GP, sum to infinity is given by the formula $ \dfrac {a}{1-r} $

So, for the given series, $ S _\infty  = \dfrac {2}{1-\dfrac {1}{2}} = 4 $

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The value of the infinite product $6^{\frac{1}{2}}\times 6^{\frac{1}{2}}\times 6^{\frac{3}{8}}\times 6^{\frac{1}{4}}\times .........$ is

  1. 6

  2. 36

  3. 216
  4. $\infty$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$6^{\frac12}\times6^{\frac12}\times6^{\frac38}\times6^{\frac14}\times...............$
$=6^{\frac12+\frac24+\frac38+\frac4{16}+...................}$
$=6^{\frac12\left(1+\frac22+\frac3{2^2}+\frac4{2^3}+......................\right)}$

Let $S=1+\cfrac22+\cfrac3{2^2}+\cfrac4{2^3}+............$, then
$\cfrac S2=S-\cfrac S2$
or, $\cfrac S2=1+\cfrac12+\cfrac1{2^2}+\cfrac1{2^3}+...........$
or, $\cfrac S2=\cfrac1{1-\cfrac12}$ ..... [Using sum of infinite terms of an G.P]
or, $S=4$
Then we have,
$6^{\frac12}\times6^{\frac12}\times6^{\frac38}\times6^{\frac14}\times...............$
$=6^{\frac12\times4}=6^2=36$
Hence, B is the correct option.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Calculate the sum of the infinite series: $1 - \dfrac {1}{3} + \dfrac {1}{9} - \dfrac {1}{27} + .....$.

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

Given series is $1,-\dfrac{1}{3},+\dfrac{1}{9},-\dfrac{1}{27}......................$

Then common ratio $=$ $ (-\dfrac{1}{3})/1=-\dfrac{1}{3}$
Then sum of infinite series $S=$ $\dfrac{a _{1}}{1-r}=\dfrac{1}{1-(-\frac{1}{3})}=\dfrac{1}{\frac{1+3}{3}}=\dfrac{3}{4}$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

If the sum of infinite G.P. $p, 1, \dfrac{1}{p}, \dfrac{1}{p^2}, ......., $ is $\dfrac{9}{2}$. Then find the value of $p$.

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

Sum of infinite series of a GP $=\dfrac {a} {1-r}$, where $a$ is the first term and $r$ is the common ratio

 
Here $a=p$ and $r=\dfrac {1}{p}$

$\Rightarrow \dfrac {p} {1-\dfrac {1}{p}}=\dfrac {9}{2}$

$\Rightarrow \dfrac {p^{2}}{p-1}=\dfrac {9}{2}$

$\Rightarrow 2p^{2}-9p+9=0$

$\Rightarrow 2p^{2}-6p-3p+9=0$

$\Rightarrow (2p-3)(p-3)=0$

$\Rightarrow p=3,\dfrac{3}{2}$    (because common ratio $r=1/p$ must be less than 1) 

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Find the value of the sum $\displaystyle \sum _{r=1}^{n}\,$ $\displaystyle \sum _{s=1}^{n}\, \delta _{rs}\, 2^r\, 3^s$ where $ \delta _{rs}$ is zero if $r \neq s$ & $\delta _{rs}$ is one if $r=s$

  1. $ \dfrac {6(6^n-1)}{5}$
  2. $ \dfrac {6(6^n+1)}{5}$
  3. $ \dfrac {5(6^n+1)}{6}$
  4. $ \dfrac {n(6^n-1)}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given  $\delta _{rs}\, =\, 0   if \quad r\neq s\quad \delta  _{rs}=1\quad if \quad r=s$

$\therefore \displaystyle \sum _{r=1}^{n}\, \sum _{s=1}^{n}\, \delta _{rs}\, 2^r\, 3^s$

$=\displaystyle

\sum _{r=1}^{n} 2^r3^r\, =\, \displaystyle \sum _{r=1}^{n} 6^r\, \,

=6+6^2\, +\, 6^3+..6^n\,=\, \displaystyle \frac {6(6^n-1)}{5}$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The A.M. of the series $1, 2, 4, 8, 16, ......, 2$$^n$ is

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

$\displaystyle A.M.=\frac { 1+2+4+8+...+{ 2 }^{ n } }{ n+1 } =\frac { { 2 }^{ n+1 }-1 }{ \left( n+1 \right) \left( 2-1 \right)  } =\frac { { 2 }^{ n+1 }-1 }{ n+1 } $

Ans: B