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 calculating and mental strategies 3 written methods dividing decimals division of decimals

Simplify $\left[\displaystyle\frac{(0.333)^3}{(0.111)^2}-\frac{(0.222)^4}{(0.111)^3}\right]$.

  1. $1.331$
  2. $1.221$
  3. $1.484$
  4. $1.551$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, $[\dfrac{(0.333)^3}{(0.111)^2} - \dfrac{(0.222)^4}{(0.111)^3}]$
We can solve like 
= $[\dfrac{0.333 \times 0.333 \times 0.333}{0.111 \times 0.111} - \dfrac{0.222 \times 0.222 \times 0.222 \times 0.222}{0.111 \times 0.111 \times 0.111}]$
= ${3 \times 3 \times 0.333} - 2 \times  2 \times 2 \times 0.222$
= $2.997 - 1.776$
= $1.221$
Multiple choice maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

$\dfrac{p}{q}$ form of the number $0.\overline{3}$ is :

  1. $\dfrac{3}{10}$
  2. $\dfrac{3}{100}$
  3. $\dfrac{1}{3}$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $x$ = .33333......

Multiplying by 10 on both sides we get
$10x=3.3333....\ on\quad subtracting\quad both\quad equations\quad we\quad get,\ 9x=3\ x=\dfrac { 1 }{ 3 } $
So, correct answer is option C.

Multiple choice maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

The fraction, $\dfrac{1}{3}$

  1. equals $0.33333333$
  2. is less than $0.33333333\ by\ \dfrac{1}{3.10^{8}}$
  3. is less than $0.33333333\ by\ \dfrac{1}{3.10^{9}}$
  4. is greater than $0.33333333\ by\ \dfrac{1}{3.10^{8}}$
  5. is greater than $0.33333333\ by\ \dfrac{1}{3.10^{9}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\cfrac { 1 }{ 3 } -0.33333333=\cfrac { 1 }{ 3 } -\cfrac { 33333333 }{ { 10 }^{ 8 } } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { { 10 }^{ 8 }-99999999 }{ 3\cdot { 10 }^{ 8 } } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { 1 }{ 3\cdot { 10 }^{ 8 } } $

$\therefore \cfrac { 1 }{ 3 } $ is greater than 0.33333333 by $\cfrac { 1 }{ 3\cdot { 10 }^{ 8 } } $.

Multiple choice maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

Which one of the following is not a correct statement ?

  1. $\displaystyle 0.\overline{01}=\frac{1}{90}$
  2. $\displaystyle 0.\overline{1}=\frac{1}{9}$
  3. $\displaystyle 0.\overline{2}=\frac{2}{9}$
  4. $\displaystyle 0.\overline{3}=\frac{1}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{1}{90} = 0.0111111111 = 0.0\bar{1}$


$\dfrac{1}{9} = 0.11111111 = 0.\bar{1}$

$\dfrac{2}{9} = 0.222222222 = 0.\bar{2}$

$\dfrac{1}{3} = 0.3333333333= 0.\bar{3}$

Hence, option $A$ is not correct.

Multiple choice maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

If the denominator of a fraction has factors other then $2$ and $5$, the decimal expression ..............

  1. repeats

  2. is that of a whole number

  3. has equal numerator and denominator

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

If there are prime factors in the denominator other than $2$ or $5$, then the decimals repeat.
$\dfrac {1}{24} = \dfrac {1}{3\times 2\times 2\times 2}$ (there is a factor of $3$, the decimal will repeat.)
Therefore, $A$ is the correct answer.

Multiple choice maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

If the denominator of a fraction has only factors of $2$ and factors of $5$, the decimal expression ............. 

  1. has equal numerator and denominator

  2. becomes a whole number

  3. does not terminate

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

When the prime factorization of the denominator of a fraction has only factors of $2$ and factors of $5$, we can always express the decimal as terminating decimal. 
For examples $\dfrac {1}{25} = \dfrac {1}{5\times 5}$ repeats (just powers of $5$, the decimal terminates.)
Therefore, $D$ is the correct answer.

Multiple choice maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

$\dfrac {1}{2} = 0.5$
It is a terminating decimal because the denominator has a factor as ...........

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

If the prime factors in the denominator of a fraction has factors of $2$ and $5$, then the decimals terminate. The denominator has $2$ as a factor.
Therefore, $C$ is the correct answer.

Multiple choice maths real number first forms rational numbers as recurring/terminating decimals decimal expansions of real numbers

Given that $\dfrac {1}{7} = 0.\overline {142857}$, which is a repeating decimal having six different digits. If $x$ is the sum of such first three positive integers $n$ such that $\dfrac {1}{n} = 0.\overline {abcdef}$, where $a, b, c, d, e$ and $f$ are different digits, then the value of $x$ is

  1. $20$
  2. $21$
  3. $41$
  4. $42$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$1^{st}$ number
$x _{1} = 7$
$\Rightarrow \dfrac {1}{x _{1}} = 0.\overline {142857}$
such that,
$2^{nd}$ number
$x _{2} = 13$
$\Rightarrow \dfrac {1}{x _{2}} = 0.\overline {076923}$
$x _{3} = 21$
$\Rightarrow \dfrac {1}{x _{3}} = \dfrac {1}{21} = 0.\overline {047619}$
$x = x _{1} + x _{2} + x _{3}$
$\Rightarrow 7 + 13 + 21 = 41$.

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If $x - \dfrac {1}{x} = \sqrt {6}$, then $x^{2} + \dfrac {1}{x^{2}}$ is ________.

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $\dfrac {x - 1}{ x} = 6$

Multiplying and divide the above equation with $x - \dfrac {1}{x}$
Thus $ \dfrac{x-\dfrac{1}{x}\times x-\dfrac{1}{x}}{x-\dfrac{1}{x}} = \sqrt{6} $
Using $(a-b)^{2} = a^{2} + b^{2} - 2ab $
and substituting $x-\dfrac{1}{x} = \sqrt{6}$  in denominator, we get
$\dfrac{x^{2} + \dfrac{1}{x^{2}} - 2x\dfrac{1}{x}}{\sqrt{6}} = \sqrt{6}$
$\Rightarrow x^{2} + \dfrac{1}{x^{2}} - 2  =\sqrt{6}\times \sqrt{6}$
$\Rightarrow x^{2} + \dfrac{1}{x^{2}} = 6+2 = 8$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If $x^{2}+\dfrac{1}{^{x^2}}=18$, then the value of $\left(x+\dfrac{1}{x}\right)$ is ?

  1. $1$
  2. $3$
  3. $\sqrt {20}$
  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2+\dfrac{1}{x^2}=18$


$\Rightarrow$  $x^2+\dfrac{1}{x^2}=20-2$                           [ Since, $20-2=18$ ]

$\Rightarrow$  $x^2+\dfrac{1}{x^2}+2=20$

$\Rightarrow$  $x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}=20$

$\Rightarrow$  $\left(x+\dfrac{1}{x}\right)^2=20$                      [ Since, $a^2+b^2+2ab=(a+b)^2$ ]

$\Rightarrow$  $\left(x+\dfrac{1}{x}\right)=\sqrt{20}$