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 ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $\displaystyle\frac{a}{b}=\displaystyle\frac{7}{9},\displaystyle\frac{b}{c}=\displaystyle\frac{3}{5}$, then what is the value of $a\,\colon\,b\,\colon\,c$?

  1. $7:9:15$
  2. $9:7:15$
  3. $7:9:14$
  4. $1:9:15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $\dfrac{a}{b}=\dfrac{7}{9}$ and$\dfrac{b}{c}=\dfrac{3}{5}$
Then $\displaystyle \frac{a}{b}\times \frac{b}{c}=\frac{7}{9}\times \dfrac{3}{5}\Rightarrow \frac{a}{c}=\frac{7}{15}$
$\Rightarrow \dfrac{b}{c}=\dfrac{3}{5}=\dfrac{9}{15}$
$\Rightarrow a:b=7:9$ and $b:c=9:15$

$\Rightarrow \dfrac{a}{b}:\dfrac{b}{c}=\dfrac{7}{9}:\dfrac{9}{15}$
In this $9$ is common. 
Then $ a:b:c=7:9:15$

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $a:b = \displaystyle \frac {2}{9} : \frac {1}{3}, b:c = \frac {2}{7}: \frac {5}{14} , d:c = \frac {7}{10} : \frac {3}{5},$ then find $a :b:c:d$.

  1. $2 : 12 : 28 : 30$
  2. $10 : 12 : 18 : 39$
  3. $16 : 24 : 30 : 35$
  4. $9 : 18 : 20 : 31$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \frac {a}{b} = \frac {2}{9} \div \frac {1}{3}=  \frac {2}{9} \times \frac {3}{1}, \frac {b}{c}= \frac {2}{7} \div \frac {5}{14} = \frac {2}{7} \times \frac {14}{5}= \frac {4}{5}$

$\displaystyle \frac {d}{c}=\frac {7}{10} \div \frac {3}{5} = \frac {7}{10} \times \frac {5}{3} = \frac {7}{6} \Rightarrow \frac {c}{d} = \frac {6}{7} \Rightarrow a = \frac {2b}{3}, c= \frac {5b}{4}, d= \frac {7c}{6}=\frac {7}{6} \times \frac {5b}{4} = \frac {35b}{24}$
$\therefore a:b:c:d = \displaystyle \frac {2b}{3}: b : \frac {5b}{4}\times 24 : \frac {35b}{24} = 16 : 24: 30 : 35.$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Arrange the following in ascending order of magnitude: 

$\displaystyle \sqrt[3]{4}, \sqrt[4]{5}, \sqrt{3}$ 

  1. $\displaystyle \sqrt[4]{5} < \sqrt[3]{4} < \sqrt{3}$
  2. $\displaystyle \sqrt[4]{5} > \sqrt[3]{4} > \sqrt{3}$
  3. $\displaystyle \sqrt[4]{5} > \sqrt[3]{4} < \sqrt{3}$
  4. $\displaystyle \sqrt[4]{5} < \sqrt[3]{4} > \sqrt{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Convert to 12th roots: sqrt[3]{4} = 4^(4/12) = 256^(1/12). sqrt[4]{5} = 5^(3/12) = 125^(1/12). sqrt{3} = 3^(6/12) = 729^(1/12). Ordering 125 < 256 < 729 gives the correct sequence.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which is the greatest out of the following ?

  1. $\displaystyle \sqrt[3]{1.728}$
  2. $\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}$
  3. $\displaystyle \left ( \frac{1}{2} \right )^{-2}$
  4. $\displaystyle \frac{17}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\Rightarrow  \sqrt[3]{1.728}=1.2$

$\Rightarrow \cfrac{\sqrt{3}-1}{\sqrt{3}+1}=\cfrac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\cfrac{3+1-2\sqrt{3}}{3-1}$
$ =\cfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}$
$ =2-1.732=0.268$

$\Rightarrow \left ( \cfrac{1}{2} \right )^{-2}=2^{2}=4$

$\Rightarrow \cfrac{17}{8}=2.2125$

$ \therefore \left ( \cfrac{1}{2} \right )^{-2}$ is the greatest. 
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which is greater $\displaystyle (\sqrt{7}+\sqrt{10})$ or $\displaystyle (\sqrt{3}+\sqrt{19})$?

  1. $\displaystyle \sqrt{7}+\sqrt{10}$
  2. $\displaystyle \sqrt{3}+\sqrt{19}$

  3. Both are equal

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

$(\sqrt{7}+\sqrt{10})$
$2.6457+3.1622=5.8079$
$(\sqrt{3}+\sqrt{19})$
$1.732+4.358=6.090$
Hence $(\sqrt{3}+\sqrt{19})$is greater.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

$\displaystyle \sqrt[4]{3},\sqrt[6]{10},\sqrt[12]{25}$, when arranged in descending order will be 

  1. $\displaystyle \sqrt[4]{3},\sqrt[6]{10},\sqrt[12]{25}$
  2. $\displaystyle \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$
  3. $\displaystyle \sqrt[6]{10},\sqrt[12]{25},\sqrt[4]{3}$
  4. $\displaystyle \sqrt[4]{3},\sqrt[12]{25},\sqrt[6]{10}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

LCM of $4, 6$ and $12 = 12$.
$ \therefore $ Raising each of the given number to power $12$, we have 
$ (3^{1/4})^{12},(10^{1/6})^{12},(25^{1/12})^{12}$
$= 3^{3},10^{2},25$
$= 27, 100, 25$
Arranging in descending order, the numbers are $ 100, 27, 25$
$\Rightarrow \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Write $\displaystyle \sqrt[4]{6},\sqrt{2},\sqrt[3]{4}$ in ascending order

  1. $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$
  2. $\displaystyle \sqrt[4]{6}$, $\sqrt{2}$ and $\displaystyle \sqrt[3]{4}$
  3. $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{4}$ and $\sqrt[4]{6}$

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

$\displaystyle 6^\cfrac 14,2^\cfrac 12,4^\cfrac 13$
The surds are of the order $4, 2$ and $3$ respectively The L.C.M. is $12$ So, we change each surd of the order $12$
The terms now are $\displaystyle \left ( 6^{3} \right )^\cfrac {1}{12},\left ( 2^{6} \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 4^{4} \right )^\cfrac{1}{12}$
$\displaystyle \Rightarrow \left ( 216 \right )^\cfrac{1}{12},\left ( 64 \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 256 \right )^\cfrac{1}{12}$
$\displaystyle \therefore $ Ascending order is $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

The sum of the reciprocals of $\displaystyle\frac{x+3}{x^2+1}$ and $\displaystyle\frac{x^2-9}{x^2+3}$ is

  1. $\displaystyle\frac{x^3+2x^2-x}{x^2-9}$
  2. $\displaystyle\frac{x^3-2x^2+x}{x^2-9}$

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reciprocals will be $\frac { { { x }^{ 2 } }+1 }{ x+3 } $,$\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
Their sum will be
$\frac { { { x }^{ 2 } }+1 }{ x+3 } +\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { \left( x-3 \right) \left( { x }^{ 2 }+1 \right) +{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
$=\frac { { x }^{ 3 }+x-3{ x }^{ 2 }-3+{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { { x }^{ 3 }-2{ x }^{ 2 }+x }{ { x }^{ 2 }-9 } $

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle x^{2}-3x+1=0$ then the value of $\displaystyle x-\frac{1}{x}$ is

  1. $\displaystyle \sqrt{5}$
  2. $\displaystyle \sqrt{3}$
  3. $\displaystyle \sqrt{2}$
  4. $\displaystyle \sqrt{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^{2}-3x+1$

$\therefore x^{2}+1=3x\Rightarrow \frac{x^{2}+1}{x}=\frac{3x}{x}\Rightarrow x+\frac{1}{x}=3$
$x^{2}+\frac{1}{x}^{2}=\left ( x+\frac{1}{x} \right )^{2}-2=(3)^{2}-=9-2=7$
We know 
$\left ( x-\frac{1}{x} \right )^{2}=x^{2}+\frac{1}{x}^{2}-2\Rightarrow 7-2=5$
$\therefore x-\frac{1}{x}=\sqrt{5}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle x-\frac{1}{x}=3$; then the value of $\displaystyle \frac{3x^{2}-3}{x^{2}+2x-1}$ is

  1. 9/5

  2. 8/5

  3. 7/5

  4. 6/5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $x+\frac{1}{x}=3$ multiply by x both sides

Then $x^{2}-1=3x$
$\Rightarrow x^{2}-3x-1=0$
So $\frac{3x^{2}-3}{x^{2}+2x-1}=\frac{3(x^{2}-1)}{x^{2}-3x-1+5x}= \frac{3\times 3x}{0+5x}= \frac{9x}{5x}=\frac{9}{5}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle x+\frac{a}{x}=b$ then the value of $\displaystyle \frac{x^{2}+bx+a}{bx^{2}-x^{3}}$ is

  1. $\displaystyle \frac{6b}{a}$
  2. $\displaystyle \frac{5b}{a}$
  3. $\displaystyle \frac{2b}{a}$
  4. $\displaystyle \frac{4b}{a}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $x+\frac{a}{x}=b$ Multiply by x both sides

$x^{2}+a=bx$
$\Rightarrow x^{2}-bx+a=0$
So $\frac{x^{2}+bx+a}{bx^{2}-x^{3}}=\frac{x^{2}-bx+a+2bx}{-x(x^{2}-bx)}=\frac{0+2bx}{-x(-a)}=\frac{2bx}{ax}=\frac{2b}{a}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle a-\frac{1}{3}=\frac{1}{a}$ then the value of $\displaystyle a^{3}-\frac{1}{a^{3}}$ is

  1. $\displaystyle 1\frac{1}{27}$
  2. $\displaystyle 1\frac{2}{27}$
  3. $\displaystyle 1\frac{3}{27}$
  4. $\displaystyle 1\frac{4}{27}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $a-\frac{1}{3}=\frac{1}{a}$

=$a-\frac{1}{a}=\frac{1}{3}$
We know that
$x^{3}-y^{3}=(x-y)^{3}+3xy(x+y)$
Then $a^{3}-\left ( \frac{1}{a} \right )^{3}=(a-\left ( \frac{1}{a} \right ))^{3}+3a\left ( \frac{1}{a} \right )(a+\left ( \frac{1}{a} \right ))$ 
=$\left ( \frac{1}{3} \right )^{3}+3\times\left (  \frac{1}{3} \right )$
=$\frac{1}{27}+1=\frac{28}{27}=1\frac{1}{27}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\dfrac{2+3}{x}=\dfrac{2+x}{3}$
What one value for $x$ can be correctly entered into the answer grid?

  1. -5

  2. 3

  3. -3

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\dfrac{2+3}{x}=\dfrac{2+x}{3}$

$\Rightarrow 2x+x^2=15$
$\Rightarrow x^2+2x-15=0$
$\Rightarrow x^2+5x-3x-15=0$
$\Rightarrow x(x+5)-3(x+5)=0$
$\Rightarrow (x+5)(x-3)=0$
$\Rightarrow x=-5,3$
Value of $x$ is not negative, so $x=3$.