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 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

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

The sum of $1 + \dfrac {2}{5} + \dfrac {3}{5^{2}} + \dfrac {4}{5^{3}} + ....$ up to $n$ terms is

  1. $\dfrac {25}{16} - \dfrac {4n + 5}{16\times 5^{n - 1}}$
  2. $\dfrac {3}{4} - \dfrac {2n + 5}{16\times 5^{n + 1}}$
  3. $\dfrac {3}{7} - \dfrac {3n + 5}{16\times 5^{n - 1}}$
  4. $\dfrac {1}{2} - \dfrac {5n + 1}{3\times 5^{n + 2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $S=1+\cfrac { 2 }{ 5 } +\cfrac { 3 }{ { 5 }^{ 2 } } +\cfrac { 4 }{ { 5 }^{ 3 } } +...\quad \quad (1)$
Multiplying $S$ with $\cfrac{1}{5}$ we get
$\cfrac { 1 }{ 5 } S=\cfrac { 1 }{ 5 } +\cfrac { 2 }{ { 5 }^{ 2 } } +\cfrac { 3 }{ { 5 }^{ 3 } } +\cfrac { 4 }{ { 5 }^{ 4 } } +...\quad \quad (2)$
Subtracting $(2)$ from $(1)$
$\quad \quad \quad S\;\;=1+\cfrac { 2 }{ 5 } +\cfrac { 3 }{ { 5 }^{ 2 } } +\cfrac { 4 }{ { 5 }^{ 3 } } +....{ T } _{ n }\\ \underline { \quad \quad -\cfrac { 1 }{ 5 } S=-\left[ \cfrac { 1 }{ 5 } +\cfrac { 2 }{ { 5 }^{ 2 } } +\cfrac { 3 }{ { 5 }^{ 3 } } +...{ T } _{ n-1 } \right] -{ T } _{ n } } \\ \left( 1-\cfrac { 1 }{ 5 }  \right) S=1+\cfrac { 1 }{ 5 } +\cfrac { 1 }{ { 5 }^{ 2 } } +\cfrac { 1 }{ { 5 }^{ 3 } } +.....-{ T } _{ n-1 }\\ \left( 1-\cfrac { 1 }{ 5 }  \right) S=\cfrac { (1)\left( 1-\cfrac { 1 }{ { 5 }^{ n } }  \right)  }{ \left( 1-\cfrac { 1 }{ 5 }  \right)  } -\cfrac { n }{ { 5 }^{ n-1 } } \\ \cfrac { 4 }{ 5 } S=\cfrac { 5 }{ 4 } -\cfrac { (4n+5) }{ 4\times { 5 }^{ n } } \\ S=\cfrac { 25 }{ 16 } -\cfrac { (4n+5) }{ 16\times { 5 }^{ n-1 } } $
Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If $\displaystyle A=\left{ 2,4,5 \right} ,B=\left{ 7,8,9 \right} $ then $\displaystyle n\left( A \times B \right) $ is equal to

  1. $6$
  2. $9$
  3. $3$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle A=\left{ 2,4,5 \right} ,B=\left{ 7,8,9 \right} $

$\Rightarrow n(A)=3$  and  $n(B)=3$

$\therefore  n(A\times B)=n(A)n(B)=9$

Hence, option B.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

The GCD of $\displaystyle \frac{3}{16}$,$\displaystyle \frac{5}{12}$,$\displaystyle \frac{7}{18}$ is 

  1. $\displaystyle \frac{105}{48}$
  2. $\displaystyle \frac{1}{4}$
  3. $\displaystyle \frac{1}{48}$

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The greatest common divisor is same as the highest common factor that is GCD is same as HCF and,

HCF of two or more fractions is given by HCF of Numerators divided by LCM of Denominators 

HCF of the numerators $(3,5,7)=1$
LCM of the denominators $(16,12,18)=2\times 2\times 2\times 2\times 3\times 3=144$

Therefore, 

HCF$\left( \dfrac { 3 }{ 16 } ,\dfrac { 5 }{ 12 } ,\dfrac { 7 }{ 18 }  \right) =\dfrac { 1 }{ 144 }$

Hence, GCD of $\dfrac { 3 }{ 16 } ,\dfrac { 5 }{ 12 } ,\dfrac { 7 }{ 18 }$ is $\dfrac { 1 }{ 144 }$
Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

46.3=_______________

  1. $\displaystyle \frac{46}{10}$
  2. $\displaystyle \frac{460}{10}$
  3. $\displaystyle 46\frac{3}{10}$
  4. $\displaystyle \frac{463}{100}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

46.3 is the number we have to convert in a fraction. 

First we'll convert the given decimal in a fraction to eliminate the decimal point.
46.3=463/10
Now we divide 463 by 10.
 10×46=460 so the remainder is 3.
The resulting fraction will be, 46 whole 3/10.
So option C is the correct answer.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

If Rs 782 be divided into three parts proportional to $\dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4} $ then the first part is

  1. $Rs.\ 182$
  2. $Rs.\ 190$
  3. $Rs.\ 196$
  4. $Rs.\ 204$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given ratio = $\dfrac{1}{2} : \dfrac{2}{3} : \dfrac{3}{4} $     ...............   Multiplying by $12$ 


                    = $6:8:9$


$\therefore$ 1st part = Rs. $\left ( 782 \times \dfrac{6}{23} \right )$ = Rs. $204$

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

By what number should we multiply ${(-8)}^{-1}$ to obtain ${12}^{-1}$?

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

Let the number be $n$

then, 
        ${\left( { - 8} \right)^{ - 1}} \times x = {\left( {12} \right)^{ - 1}}$

        $= \dfrac{{ - 1}}{8} \times x = \dfrac{1}{{12}}$    $\because \left[ {{{\left( {\dfrac{a}{b}} \right)}^{ - n}} = {{\left( {\dfrac{b}{a}} \right)}^n}} \right]$

       $ = x = \dfrac{1}{{12}} \times  - 8$ 

       $ = x = \dfrac{{ - 2}}{3}$