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 multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

What is the number to be multiplied by $(-7)^{-1}$ so as to get $10^{-1}$ as the product?

  1. $\displaystyle\frac{-7}{10}$
  2. $\displaystyle\frac{7}{10}$
  3. $\displaystyle\frac{9}{10}$
  4. $\displaystyle\frac{-3}{10}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
let the number be $ x$
According to the question:
$\Rightarrow(-7)^{-1} x = 10 ^{-1}$
$\Rightarrow\dfrac{1}{-7^{1}}x = \dfrac{1}{10}$
Applying cross multiplication
$\Rightarrow x  = \dfrac{-7}{10}$


Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The sum $1+\dfrac { 2 }{ x } +\dfrac { 4 }{ { x }^{ 2 } } +\dfrac { 8 }{ { x }^{ 3 } } +....\left( up\ to\ \infty  \right) ,x\neq 0,$ is finite if

  1. $\left| x \right| < 2$
  2. $\left| x \right| > 2$
  3. $\left| x \right| < 1$
  4. $2\left| x \right| < 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$1 + \dfrac{2}{x} + \dfrac{4}{{{x^3}}} + \dfrac{8}{{{x^3}}}.......\,upto\,\,\infty $
It is infinite $G.P$ with Common ratio $r=\dfrac{2}{x}$
It's sum is infinite if $\left| r \right| < 1$
i-e    
 $\left| {\dfrac{2}{x}} \right| < 1$
 $ \Rightarrow \dfrac{2}{{\left| x \right|}} < 1$
 $ \Rightarrow 2 < \left| x \right|$
$ \Rightarrow \left| x \right| > 2$       
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The sum of the infinite series $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+......$

  1. Cannot be determined.
  2. Equals $\dfrac{15}{8}$
  3. Equals $2$
  4. Will be higher than $2$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Given\>series\>is\>an\>infinite\>GP\>with\>first\>term\>=1\>and\>common\>ratio=1/2\\\therefore\>sum=(\frac{a}{1-r})\\=(\frac{a}{1-1/2})=2$

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

$4, \dfrac{8}{3}, \dfrac{16}{9}, \dfrac{32}{27}..$ is a

  1. arithmetic sequence

  2. geometric sequence

  3. geometric series

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

lets check the ratio between the consecutive terms.
$\dfrac {\frac {8}{3}}{4}=\dfrac {8}{12}=\dfrac {2}{3}$
Again take the ratio between next consecutive terms.
$\dfrac {\frac {16}{9}}{\frac {8}{3}}=\dfrac {16\times 3}{9\times 8}=\dfrac {2}{3}$
Here the common ratio is same $\dfrac{2}{3}$ throughout.
Hence, $4, \dfrac{8}{3}, \dfrac{16}{9}, \dfrac{32}{27}..$ is a geometric sequence.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The sum of infinity of $\frac{1}{7} + \frac{2}{7^2} + \frac{1}{7^3} + \frac{2}{7^4} + ......$ is:

  1. $\frac{1}{5}$
  2. $\frac{1}{24}$
  3. $\frac{5}{48}$
  4. $\frac{3}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series can be split into two geometric series: S1 = 1/7 + 1/7^3 + 1/7^5... and S2 = 2/7^2 + 2/7^4 + 2/7^6... For S1, a=1/7, r=1/49, sum = (1/7)/(1-1/49) = 7/48. For S2, a=2/49, r=1/49, sum = (2/49)/(1-1/49) = 2/48. Total sum = 9/48 = 3/16.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

If $S=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+....\infty$.
then, the sum of the given series is $2$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,

$S=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+....\infty$

Then,
$a=1$, $r=\dfrac{1}{2}$

We know that
$S=\dfrac{a}{1-r}$

$S=\dfrac{1}{1-\dfrac{1}{2}}$

$S=\dfrac{1}{\dfrac{1}{2}}$

$S=2$

Hence, this is the answer.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

Sum the series: $\displaystyle {1\, -\, \frac{1}{3}\, +\, \frac{1}{3^2}\, -\, \frac{1}{3^3}\, +\, \frac{1}{3^4}.......\infty}$

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

We know, $S _{\infty }=\dfrac{a}{1-r}$
From the given series, $a=1 ,r=-\dfrac{1}{3}$
$\therefore S _{\infty }=\dfrac{1}{1-\left ( -\dfrac{1}{3} \right )}$
$= \dfrac{1}{1+\dfrac{1}{3}}$


$ =\dfrac{1}{\dfrac{4}{3}}$

$= \dfrac{3}{4}$

Multiple choice

What is the value of the expression (\frac{1}{2} + \frac{1}{3} + \frac{1}{6})?

  1. 1

  2. 1.1

  3. 1.2

  4. 1.3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To evaluate the expression (\frac{1}{2} + \frac{1}{3} + \frac{1}{6}), we can find a common denominator: (\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1). Therefore, the value of the expression is 1.1.

Multiple choice

What is the Brahmagupta formula?

  1. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  2. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
  3. $x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
  4. $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{4a}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The Brahmagupta formula is given by the equation $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

Multiple choice

What is the general formula for the Madhava series for the arctangent function?

  1. $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$
  2. $$\arctan x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
  3. $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$
  4. $$\arctan x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

The general formula for the Madhava series for the arctangent function is $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$.