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 taylor's and maclaurin's series applications of differential calculus maths

If the sum of the series $\dfrac{3}{1!}+\dfrac{5}{2!}+\dfrac{7}{3!}+\dfrac{9}{4!}+...\infty=Ae+B$
Find the value of $A+B$

  1. $1$
  2. $7$
  3. $0$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Its general term is $\dfrac{2n+1}{n!}$

So we have to calculate $\sum^{n=\infty} _{n=0}\dfrac{2n+1}{n!}=2\sum^{k=\infty} _{k=0}\dfrac{1}{k!}+\sum^{n=\infty} _{n=0}\dfrac{1}{n!}-2=3e-2$ (using  taylor's expansion for $e^x$)

So A+B=1

Multiple choice maths indices negative indices law of indices laws of indices

$5^{-2}$ can also be expressed as

  1. $\dfrac{1}{25}$
  2. $\dfrac{1}{5^{-2}}$
  3. $\dfrac{1}{5}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$5^{-2}=5^{-1*2}$

$=(5^2)^{-1}$                       $\because  (a)^{mn}=(a^m)^n$

$=\dfrac{1}{5^2}$                              $\because a^{-n}=\dfrac{1}{a^n}$

$=\dfrac{1}{25}$
Multiple choice maths indices negative indices law of indices laws of indices

Evaluate : $(-4)^{-2}$---

  1. $\displaystyle \frac{1}{-16}$
  2. $\displaystyle \frac{1}{16}$
  3. $-16$
  4. $16$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$(-4)^{-2}=\dfrac{1}{(-4)^2}$            $\because a^{-m}=\dfrac{1}{a^m}$

$=\dfrac{1}{-4*-4}$

$=\dfrac1{16}$

$\therefore(-4)^{-2}=\dfrac1{16}$
Multiple choice maths indices negative indices law of indices laws of indices

If $\dfrac {p}{q} = \left (\dfrac {2}{3}\right )^{3} \div \left (\dfrac {3}{2}\right )^{-3}$, then the value of $\left (\dfrac {p}{q}\right )^{10}$ is _______.

  1. $1$
  2. $0$
  3. $-1$

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $\dfrac {p}{q} = \left (\dfrac {2}{3}\right )^{3} \div \left (\dfrac {3}{2}\right )^{-3}$
We see that, $\left (\dfrac{3}{2}\right )^{-3}$=$\left (\dfrac{2}{3}\right)^{3}$
So, $\left (\dfrac{2}{3}\right)^{3}$ / $\left (\dfrac{2}{3}\right)^{3}$ $= 1$ 
Thus, $\left(\dfrac {p}{q}\right)^{10}=1$
Hence, A is the right answer.
Multiple choice maths indices negative indices law of indices laws of indices

The value of $\left (\dfrac {32}{243}\right )^{-3/5}$ is _____.

  1. $\dfrac {27}{8}$
  2. $\dfrac {8}{27}$
  3. $\dfrac {16}{27}$
  4. $\dfrac {27}{16}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We need to find value of $\left (\dfrac {32}{243}\right )^{-3/5}$
It can be written as,
$\left(\dfrac {2^5}{3^5}\right)^{-3/5}$
$\Rightarrow \left (\dfrac {3}{2}\right)^{3} = \dfrac {27}{8}$
Multiple choice maths indices negative indices law of indices laws of indices

The value of ${1-[1-(1-n)^{-1}]^{-1}}^{-1}$ is

  1. $0$
  2. $1$
  3. $n$
  4. $\displaystyle\frac{1}{n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle{1-[1-(1-n)^{-1}]^{-1}}^{-1}=\left[1-\left(1-\frac{1}{1-n}\right)^{-1}\right]^{-1}$

$\displaystyle=\left[1-\left(\frac{1-n-1}{1-n}\right)^{-1}\right]^{-1}=\left[1+\left(\frac{n}{1-n}\right)^{-1}\right]^{-1}=\left(1+\frac{1-n}{n}\right)^{-1}$

$\displaystyle=\left(\frac{n+1-n}{n}\right)^{-1}=\left(\frac{1}{n}\right)^{-1}=n$

Multiple choice statistics measures of central tendency geometric and harmonic mean geometric mean mean

The harmonic mean of $\dfrac { a }{ 1-ab } and \dfrac { a }{ 1+ab }$ is:

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

Here, $n=2$

$x _1=\dfrac{a}{1-ab}$ and $x _2=\dfrac{a}{1+ab}$
Then, $\dfrac{1}{x _1}=\dfrac{1-ab}{a}$ and $\dfrac{1}{x _2}=\dfrac{1+ab}{a}$

$\Rightarrow$  Harmonic mean $=\dfrac{n}{\dfrac{1}{x _1}+\dfrac{1}{x _2}}$

                                  $=\dfrac{2}{\dfrac{1-ab}{a}+\dfrac{1+ab}{a}}$

                                  $=\dfrac{2a}{2}$

                                  $=a$

Multiple choice maths decimal numbers comparing and ordering of decimals more or less comparing decimals

Select the correct option which make the given expression true.
$12.5+5\displaystyle\frac{3}{8}+8\frac{1}{8}+9\frac{4}{5}+\square 2.9+15-6.88+11.08$.

  1. $<$
  2. $>$
  3. $=$

  4. Can't be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have, $\displaystyle 12.5+5\frac{3}{8}+8\frac{1}{8}+9\frac{4}{5}$

$=12.5+\dfrac{43}{8}+\dfrac{65}{8}+\dfrac{49}{5}$
$=12.5+5.375+8.125+9.8=35.8$ 
Now lets check for $2.9+15-6.88+11.08=22.1$
Since, $35.8 > 22.1$
Therefore, $ 12.5+\displaystyle 5\frac{3}{8}+8\frac{1}{8}+9\frac{4}{5} > 2.9+15-6.88+11.08$.

Multiple choice mathematical modelling proof by contradiction similar triangles

$\forall n\in N$, value of $\displaystyle \frac{n^{4}}{24}+\frac{n^{3}}{4}+\frac{11n^{2}}{24}+\frac{n}{4}$ is

  1. a rational number

  2. an integer

  3. a natural number

  4. a real number
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

 Given$ \dfrac{[n^{4}+6n^{3}+11n^{2}+6n]}{24}$


            $\dfrac{[n(n+1)(n+2)(n+3)]}{24}$

             $=^{(n+3)}{C _{4}}$
Thus the above number is always divisible by $24$ Thus all four options are correct.