Tag: simplest ratio

Questions Related to simplest ratio

The standard form of $\displaystyle\frac{192}{-168}$ is _________.

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $\displaystyle\frac{-2}{3}$

  2. $\displaystyle\frac{-8}{7}$

  3. $\displaystyle\frac{-1}{7}$

  4. $\displaystyle\frac{-6}{7}$

Show answer
Correct Option: B
Explanation:

the standard form of $ - \dfrac{192}{168}$

$=-\dfrac{2 \times 3\times 4\times 8}{2\times 3\times 4\times 7}$
$-\dfrac{8}{7}$

Which of the following sum is in the simplest form?

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $\dfrac{4}{9}+\dfrac{-5}{9}$

  2. $\dfrac{-2}{5}+\dfrac{13}{20}$

  3. $\dfrac{-5}{12}+\dfrac{11}{-12}$

  4. $\dfrac{-7}{8}+\dfrac{1}{12}+\dfrac{2}{3}$

Show answer
Correct Option: A
Explanation:
A. $\dfrac{4}{9} + \dfrac{-5}{9} = \dfrac{-1}{9}$ Simplest Form

B. $\dfrac{-2}{5} + \dfrac{13}{20} = \dfrac{-8+13}{20} = \dfrac{5}{20}$ Not the simplest form

C. $\dfrac{-5}{12} + \dfrac{11}{-12} = \dfrac{-5-11}{12} = \dfrac{-16}{12}$ Not the simplest form

D. $\dfrac{-7}{8} + \dfrac{1}{12} + \dfrac{2}{3} = \dfrac{-21+2+16}{24} = \dfrac{-3}{24}$ Not the simplest form

$\Rightarrow$ Only A gives answer in simplest form

Which of the following is not equivalent to $\dfrac{4}{8}$ ?

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $\cfrac { 1 }{ 2 } $

  2. $\cfrac { 16 }{ 32 } $

  3. $1$

  4. $\cfrac { 12 }{ 24 } $

Show answer
Correct Option: C
Explanation:

we have, $\dfrac{4}{8}=$ $\dfrac{1}{2}=$$\dfrac{16}{32}=$$\dfrac{12}{24}$

Among the given values, the one which is not equivalent to $\dfrac{4}{8}$ is 1.

Simplify: $\dfrac{5}{11} + 4\dfrac{3}{9} $

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $\dfrac{158}{33}$

  2. $\dfrac{168}{33}$

  3. $\dfrac{178}{33}$

  4. $\dfrac{148}{33}$

Show answer
Correct Option: A
Explanation:
$\displaystyle \frac{5}{11}+4\frac{3}{9}=\frac{5}{11}+\frac{36+3}{9}=\frac{5}{11}+\frac{39}{9}$

$\displaystyle =\frac{5\times 9+11\times 39}{9\times 11}=\frac{45+429}{99}=\frac{474}{99}=\frac{158}{33}$

Simplify: $\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} -\dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}$

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: B
Explanation:

$\dfrac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} -\dfrac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}\=\dfrac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{(\sqrt{10} + \sqrt{3})(\sqrt{10} - \sqrt{3})} - \dfrac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{(\sqrt{6} + \sqrt{5})(\sqrt{6} - \sqrt{5})} -\dfrac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{(\sqrt{15} - 3\sqrt{2})(\sqrt{15} + 3\sqrt{2})}\=\dfrac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{10-3} - \dfrac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{6-5} -\dfrac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{15-18}\=\dfrac{7\sqrt{3}(\sqrt{10} - \sqrt{3})}{7} - \dfrac{2\sqrt{5}(\sqrt{6} - \sqrt{5})}{1} +\dfrac{3\sqrt{2}(\sqrt{15} - 3\sqrt{2})}{3}$

$=\dfrac{21\sqrt{30}-63-425\sqrt{30}+210+21\sqrt{30}-18*7}{21}\=\dfrac{21}{21}=1$

Simplify:
$\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $4\sqrt{6}$

  2. $10$

  3. $2$

  4. $\dfrac{4\sqrt{6}}{5}$

Show answer
Correct Option: B
Explanation:

$\dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} + \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}=\dfrac{(\sqrt{3} + \sqrt{2})^2+(\sqrt{3} - \sqrt{2})^2}{3-2}=\dfrac{3+2+3+2}{1}=10$

Reduce the following fractions to their lowest forms.
a. $\dfrac{36}{144}$


b. $\dfrac{65}{117}$

c. $\dfrac{180}{120}$

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $a=\dfrac{1}{4}, b=\dfrac{65}{117}, c=\dfrac{2}{3}$

  2. $a=\dfrac{1}{4}, b=\dfrac{5}{9}, c=\dfrac{3}{2}$

  3. $a=\dfrac{1}{4}, b=\dfrac{65}{117}, c=\dfrac{3}{2}$

  4. $a=\dfrac{3}{4}, b=\dfrac{65}{117}, c=\dfrac{2}{3}$

Show answer
Correct Option: B
Explanation:

$\\(a.)(\frac{36}{144})=(\frac{3}{12})=(\frac{1}{4})\\(b.)(\frac{65}{117})=(\frac{13\cdot 5}{13\cdot 9})=(\frac{5}{9})\\(c.)(\frac{180}{120})=(\frac{60\cdot 3}{60\cdot 2})=(\frac{3}{2})$

The Simplified form of $0.35$ is

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $\dfrac {7}{20}$

  2. $\dfrac {4}{20}$

  3. $\dfrac {35}{100 }$

  4. $None$

Show answer
Correct Option: A
Explanation:

$0.35$

$=\dfrac{35}{100}=\dfrac{7}{20}$

Given that  $n$  $AM's$  are inserted between two sets of numbers  $a , 2 b$  and  $2 a , b$  where  $a , b \in R .$  Suppose further that  $mth$  mean between these sets of numbers is same, then the ratio  $a : b$  is equal to

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $( n - m + 1 ) : m$

  2. $( n - m + 1 ) : n$

  3. $n : ( n - m + 1 )$

  4. $m : ( n - m + 1 )$

Show answer
Correct Option: D
Explanation:

Let the common difference be d. As there are $n\;AM's$ between $a$ and $b$ and total number of terms in the sequence is $=n+2$ 

$\Rightarrow nth \;term\;b=a+\left( n-1\right)d$
      $d=\dfrac{\left( b-a\right)}{n+1}$
so, 2nd term that is first term $A\left( 1\right) =a+\left[ \dfrac{\left( b-a\right)}{\left( n+1\right)}\right]$
3rd term that is second mean $A\left(2\right)=a+\left[ 2\times \dfrac{ \left( b-a\right)}{\left( n+1\right)}\right]$
In the way $r^{th}$ mean $=a+\left[ r\times \dfrac{ \left( b-a\right) }{\left(n+1\right)}\right]$
In the first sequence first term is a $n^{th}$ term $=2b$ and $n\;AM's$ between them. 
As such from above concept  -
$\Rightarrow r^{th}$ term is $a+\left[ r\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]$
       $m^{th}$ term is $a+\left[ m\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]$
$ii)$ Similarly for second sequence -
    $m^{th}$ mean $=2a+\left[ m\times \dfrac{\left( b-2a\right)}{\left( n+1\right)}\right]$
$iii).$ Since the $m^{th}$ mean, are equal equation like the above. 
$=a+\left[ m\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]$
$=2a+m\times \dfrac{\left( b-a\right) }{\left( n+1\right)}$
$=\left[ m\times \dfrac{\left( 2b-a\right)}{\left( n+1\right)}\right]-\left[ m\times \dfrac{\left( b-2a\right)}{\left( n+1\right)}\right]$
$=2a-a$
$\Rightarrow \dfrac{m}{\left( n+1\right)} \times \left( 2b-a-b+2a\right)=a$
$\dfrac{a+b}{a}=\dfrac{n+1}{m}$
subtracting on both sides 
$\Rightarrow \dfrac{a+b}{a}-1=\dfrac{n+1}{m}-1$
$\Rightarrow \dfrac{a+b-a}{a}=\dfrac{n+1-m}{m}$

$\Rightarrow \dfrac{b}{a}=\dfrac{n+1-m}{m}$

$\Rightarrow \dfrac {a}{b}=\dfrac{m}{n-m+1}$
Hence, the answer is $\dfrac{m}{n-m+1}.$

The lowest form of $3.5$ is 

maths fraction lowest form of a fraction simplest ratio lowest form of fractions

  1. $\frac{7}{20}$

  2. $\frac{4}{20}$

  3. $\frac{35}{100}$

  4. $None$

Show answer
Correct Option: A