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 remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

The expression that should be subtracted from $\displaystyle 4x^{4}-2x^{3}-6x^{2}+x-5$ so that is may be exactly divisible by $\displaystyle 2x^{2}+x-2$ is

  1. $\displaystyle 3x+5$
  2. $\displaystyle -3x-5$
  3. $\displaystyle -3x+5$
  4. $\displaystyle 3x-5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
 $2x^2-2x$
$2x^2+x-2$  $4x^4-2x^3-6x^2+x-5$ $4x^4+2x^3-4x^2$
        $-4x^3-2x^2+x$      $-4x^3-2x^2+4x$
                             $-3x-5$

Thus, $-3x-5$ must be subtracted to make it exactly divisble

Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

If $\displaystyle \left ( x^{2}+4x-21 \right )$ is divided by  $x + 7$  then the quotient is

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

Consider the polynomial $f(x)=x^2+4x-21$ and factorise it as follows:


$f(x)=x^2+4x-21=(x^2+4x+4)-21-4=(x+2)^2-25=(x+2)^2-(5)^2=(x+2+5)(x+2-5)$
$=(x+7)(x-3)$

Therefore, $f(x)=(x+1)(x-1)(x-3)$

Let $g(x)=x+7$

Now divide $f(x)$ by $g(x)$ to get $q(x)$:

$q(x)=\frac { (x+7)(x-3) }{ (x+7) } =x-3$

Hence, the quotient is $x-3$.

Multiple choice maths multiplication and division of algebraic expressions linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

Evaluate: $\displaystyle \frac{a^3\, +\, b^3\, +\, c^3\, -\, 3abc}{a^2\, +\, b^2\, +\, c^2\, -\, ab\, -\, bc\, -\, ca}$

  1. $0$
  2. $a + b + c$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\cfrac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc }{ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca } \ =\cfrac { \left( a+b+c \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right)  }{ { (a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca) } \ =a+b+c$.

Multiple choice maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

Rationalising the denominator of $\dfrac {5}{\sqrt 3-\sqrt 5}$ is -

  1. $(\frac {5}{2}(\sqrt 3+\sqrt 5)$
  2. $(-\frac {5}{2}(\sqrt 3+\sqrt 5)$
  3. $(\frac {5}{2}(\sqrt 3-\sqrt 5)$
  4. $(-\frac {5}{2}(\sqrt 3-\sqrt 5)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

here, $\dfrac {5}{\sqrt 3-\sqrt 5}$

$=\dfrac {5}{\sqrt 3-\sqrt 5}\times \dfrac {\sqrt 3+\sqrt 5}{\sqrt 3+\sqrt 5}$

$=\dfrac {5(\sqrt 3+\sqrt 5)}{3-5}$


$=-\dfrac {5}{2}(\sqrt 3+\sqrt 5)$

Multiple choice maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

Which of the following numbers lies between $\dfrac {5}{24}$ and $\dfrac {3}{8}$?

  1. $\dfrac {7}{2}$
  2. $1$
  3. $\dfrac {7}{24}$
  4. $0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mean $= \dfrac {\dfrac {3}{8} + \dfrac {5}{24}}{2} = \dfrac {\dfrac {9 + 5}{24}}{2} = \dfrac{\left (\dfrac {14}{24}\right )}{2}$


$= \dfrac {7}{12}\times \dfrac {1}{2}$

$= \dfrac {7}{24}$

Mean of two numbers lies between the two numbers.  
So, $ \dfrac {7}{24}$ lies between $\dfrac {3}{8}$ and $\dfrac {5}{24}.$

Multiple choice maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

Which of the following numbers lies between $-1$ and $-2$?

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

Mean $= \dfrac {(-1) + (-2)}{2} = \dfrac {-1 -2}{2} = \dfrac {-3}{2}$.

Mean of two numbers always lies between the two numbers.
So, answer is option $B.$

Multiple choice maths fractions, decimals and rational numbers representation of rational numbers on number line rational numbers on the number line rational numbers between two rational numbers

What fraction lies exactly halfway between $\dfrac{2}{3}$ and $\dfrac{3}{4}$?

  1. $\dfrac{3}{5}$
  2. $\dfrac{5}{6}$
  3. $\dfrac{7}{12}$
  4. $\dfrac{9}{16}$
  5. $\dfrac{17}{24}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Consider $3 \times 4 = 12$, so 
$\dfrac 23 = \dfrac{8}{12}$


$\dfrac 34 = \dfrac{9}{12}$

Multiplying the numerator and denominator by $2$:
$\dfrac{16}{24}$ and $\dfrac{18}{24}$.

The mid point is $\dfrac{17}{24}$

Hence option $E$ is correct.

Multiple choice maths congruence introduction to shapes similarity of triangles introduction to similar triangles

It is given that $\Delta ABC \sim \Delta PQR$ with $\dfrac{BC}{QR} = \dfrac{1}{3}$. Then $\dfrac{ar (\Delta PQR)}{ar (\Delta ABC)}$ is equal to

  1. $9$
  2. $3$
  3. $\dfrac{1}{3}$
  4. $\dfrac{1}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
 If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Since, $\Delta ABC \sim \Delta PQR$

$\therefore \dfrac{ar (\Delta PQR)}{ar (\Delta ABC)} = \dfrac{PR^2}{AC^2} = \dfrac{QR^2}{BC^2} = \dfrac{9}{1} =9 \ \ \ ..........  \left [ \therefore \dfrac{QR}{BC} = \dfrac{3}{1} \right ]$
Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

Order of $\frac { 1 } { 8 \times 10 ^ { 9 } }$ is:

  1. $1.77\times {{10}^{-10}}$
  2. $1.25\times {{10}^{-10}}$
  3. $1.55\times {{10}^{-10}}$
  4. $1.95\times {{10}^{-10}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac{1}{8\times 10^{9}}=\left(\dfrac{1}{8}\right)\times 10^{-9}=(0.125)\times 10^{-9}$
$=1.25\times 10^{-10}$
Order of $\dfrac{1}{8\times 10^{9}}$ is $10^{-10}$
Option $B$ is correct
Multiple choice multiplication and division vedic methods of multiplication vedic mathematics history of mathematics maths

Which of the following statements is CORRECT?

  1. The product of $\dfrac{231}{119}$ and $\dfrac{117}{118}$ is greater than $\dfrac{231}{119}$
  2. The product of $\dfrac{17}{25}$ and $\dfrac{117}{225}$ is greater than $\dfrac{17}{25}$
  3. The product of $\dfrac{1735}{2001}$ and $\dfrac{2734}{2724}$ is greater than $\dfrac{1735}{2001}$
  4. $\dfrac{1}{3}$ of $\dfrac{4}{5}$ is greater than $\dfrac{3}{4}$ of $\dfrac{8}{7}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Option $[A]$  :  $\dfrac{117}{118}<1$.  
The product of any number with a number less than $1$ is less than that number.
So,  $\dfrac{231}{119}\times\dfrac{117}{118}<\dfrac{231}{119}$.

$\Rightarrow[A]$ is not correct.

Option $[B]$  :  $\dfrac{117}{225}<1$
The product of any number with a number less than $1$ is less than that number.
So,  $\dfrac{17}{25}\times\dfrac{117}{225}<\dfrac{17}{25}$.
$\Rightarrow[B]$ is not correct.

Option $[C]$  :  $\dfrac{2734}{2724}>1$
The product of any number with a number greater than $1$ is greater than that number.
So,  $\dfrac{1735}{2001}\times\dfrac{2734}{2724}>\dfrac{1735}{2001}$.
$\Rightarrow[C]$ is correct.

Option $[D]$  :
$\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{15}\approx0.267$   and   $\dfrac{3}{4}\times\dfrac{8}{7}=\dfrac{6}{7}\approx0.857$.
But  $0.857>>0.267$.
So, $[D]$ is not correct.


$\therefore$  The correct answer is  $[C]$.

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

The value of $\dfrac { { 2 }^{ m+3 }\times { 3 }^{ 2m-n }\times { 5 }^{ m+n+3 }\times { 6 }^{ n+1 } }{ { 6 }^{ m+1 }\times { 10 }^{ n+3 }\times { 15 }^{ m } } $ is equal to 

  1. $0$
  2. $1$
  3. $2^ {m}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Now,

$\dfrac { { 2 }^{ m+3 }\times { 3 }^{ 2m-n }\times { 5 }^{ m+n+3 }\times { 6 }^{ n+1 } }{ { 6 }^{ m+1 }\times { 10 }^{ n+3 }\times { 15 }^{ m } } $ 
$=\dfrac { { 2 }^{ m+3 }\times { 3 }^{ 2m-n }\times { 5 }^{ m+n+3 }\times(2^{n+1}\times { 3 }^{ n+1 }) }{ (2^{m+1}\times { 3 }^{ m+1 })\times (2^{n+3}\times { 5 }^{ n+3 })\times (3^{m}\times { 5 }^{ m }) } $ 
$=\dfrac { { 2 }^{ m+n+4 }\times { 3 }^{ 2m+1 }\times { 5 }^{ m+n+3 } }{ { 2 }^{ m+n+4 }\times { 3 }^{ 2m+1 }\times { 5 }^{ mm+n+3 } } $ 
$=1$.

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Evaluate : $\displaystyle \left( \frac{3}{4} \right)^0 \times 2 \frac{1}{4} - \left( 2 \frac{1}{4} \right)^0 \times \frac{3}{4}$--

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

$1 \times \dfrac{9}{4} - 1 \times \dfrac{3}{4}$

$=\dfrac{9}{4} - \dfrac{3}{4}$

$=\dfrac{6}{4}$

$=\dfrac{3}{2}$