Express each of the following as a fraction in simplest form.
(i)$25\% $
(ii)$625\% $
(iii)$33\dfrac{1}{2}\% $
(iv) $66\dfrac{2}{3}\% $
Answer
606k+ views
Hint: First write the percentage into fraction by writing it in the form-$x\% = \dfrac{x}{{100}}$ .
To find the fraction in simplest form divide both numerator and denominator by the same number that can divide them both numbers completely. Divide till the numbers cannot be divided anymore and still stay whole numbers.
Complete step-by-step answer:
(i)We can write $25\% $as-
$ \Rightarrow 25\% = \dfrac{{25}}{{100}}$
On dividing the numerator and denominator by $5$ we get,
$ \Rightarrow 25\% = \dfrac{5}{{20}}$
Again dividing by $5$ we get,
$ \Rightarrow 25\% = \dfrac{1}{4}$
This is the fraction in its simplest form.
Answer-$25\% = \dfrac{1}{4}$
(ii)We can divide $625\% $as-
$ \Rightarrow 625\% = \dfrac{{625}}{{100}}$
On dividing the numbers by $5$ again and again till they cannot be divided we get,
$ \Rightarrow 625\% = \dfrac{{125}}{{20}} = \dfrac{{25}}{4}$
This is the fraction in its simplest form.
Answer-$625\% = \dfrac{{25}}{4}$
(iii) First we have to convert the mixed fraction$33\dfrac{1}{2}\% $into whole fraction
We can write$33\dfrac{1}{2}$ =$33 + \dfrac{1}{2}$
On taking LCM we get,
$ \Rightarrow 33\dfrac{1}{2} = \dfrac{{66 + 1}}{2} = \dfrac{{67}}{2}$
Then we can write-
$ \Rightarrow 33\dfrac{1}{2}\% = \dfrac{{67}}{2}\% $
We can write$\dfrac{{67}}{2}\% $ as-
$ \Rightarrow \dfrac{{67}}{2}\% = \dfrac{{\dfrac{{67}}{2}}}{{100}}$
On solving we get,
$ \Rightarrow \dfrac{{67}}{2}\% = \dfrac{{67}}{{2 \times 100}}$
On solving we get,
$ \Rightarrow \dfrac{{67}}{2}\% = \dfrac{{67}}{{200}}$
This is the simplest form of fraction.
Answer-$33\dfrac{1}{2}\% = \dfrac{{67}}{{200}}$
(iv) First we have to convert the mixed fraction$66\dfrac{2}{3}\% $ into the whole fraction. So we can write-
$ \Rightarrow 66\dfrac{2}{3} = 66 + \dfrac{2}{3}$
On taking LCM,
$ \Rightarrow 66\dfrac{2}{3} = \dfrac{{198 + 2}}{3} = \dfrac{{200}}{3}$
Now we can express$66\dfrac{2}{3}\% $as-
\[ \Rightarrow 66\dfrac{2}{3}\% = \dfrac{{\dfrac{{200}}{3}}}{{100}}\]
On solving we get,
$ \Rightarrow 66\dfrac{2}{3}\% = \dfrac{{200}}{{3 \times 100}} = \dfrac{{200}}{{300}}$
On dividing both numbers by$100$, we get,
$ \Rightarrow 66\dfrac{2}{3}\% = \dfrac{2}{3}$
This is the simplest form of fraction.
$66\dfrac{2}{3}\% = \dfrac{2}{3}$ .
Note: A fraction is in its simplest form when the numerator and denominator cannot be any smaller while still being whole numbers. We can also convert the numbers by dividing them by the greatest common factor.
In (i), we know the greatest common factor of $25$ and $100$ is $25$ so divide the numbers of numerator and denominator by $25$, we get,
$ \Rightarrow 25\% = \dfrac{{25}}{{100}} = \dfrac{1}{4}$
In (ii), we know that the greatest common factor of $625$ and $100$is$25$ so divide the numbers of numerator and denominator by $25$, we get,
$ \Rightarrow 625\% = \dfrac{{625}}{{100}} = \dfrac{{25}}{4}$
To find the fraction in simplest form divide both numerator and denominator by the same number that can divide them both numbers completely. Divide till the numbers cannot be divided anymore and still stay whole numbers.
Complete step-by-step answer:
(i)We can write $25\% $as-
$ \Rightarrow 25\% = \dfrac{{25}}{{100}}$
On dividing the numerator and denominator by $5$ we get,
$ \Rightarrow 25\% = \dfrac{5}{{20}}$
Again dividing by $5$ we get,
$ \Rightarrow 25\% = \dfrac{1}{4}$
This is the fraction in its simplest form.
Answer-$25\% = \dfrac{1}{4}$
(ii)We can divide $625\% $as-
$ \Rightarrow 625\% = \dfrac{{625}}{{100}}$
On dividing the numbers by $5$ again and again till they cannot be divided we get,
$ \Rightarrow 625\% = \dfrac{{125}}{{20}} = \dfrac{{25}}{4}$
This is the fraction in its simplest form.
Answer-$625\% = \dfrac{{25}}{4}$
(iii) First we have to convert the mixed fraction$33\dfrac{1}{2}\% $into whole fraction
We can write$33\dfrac{1}{2}$ =$33 + \dfrac{1}{2}$
On taking LCM we get,
$ \Rightarrow 33\dfrac{1}{2} = \dfrac{{66 + 1}}{2} = \dfrac{{67}}{2}$
Then we can write-
$ \Rightarrow 33\dfrac{1}{2}\% = \dfrac{{67}}{2}\% $
We can write$\dfrac{{67}}{2}\% $ as-
$ \Rightarrow \dfrac{{67}}{2}\% = \dfrac{{\dfrac{{67}}{2}}}{{100}}$
On solving we get,
$ \Rightarrow \dfrac{{67}}{2}\% = \dfrac{{67}}{{2 \times 100}}$
On solving we get,
$ \Rightarrow \dfrac{{67}}{2}\% = \dfrac{{67}}{{200}}$
This is the simplest form of fraction.
Answer-$33\dfrac{1}{2}\% = \dfrac{{67}}{{200}}$
(iv) First we have to convert the mixed fraction$66\dfrac{2}{3}\% $ into the whole fraction. So we can write-
$ \Rightarrow 66\dfrac{2}{3} = 66 + \dfrac{2}{3}$
On taking LCM,
$ \Rightarrow 66\dfrac{2}{3} = \dfrac{{198 + 2}}{3} = \dfrac{{200}}{3}$
Now we can express$66\dfrac{2}{3}\% $as-
\[ \Rightarrow 66\dfrac{2}{3}\% = \dfrac{{\dfrac{{200}}{3}}}{{100}}\]
On solving we get,
$ \Rightarrow 66\dfrac{2}{3}\% = \dfrac{{200}}{{3 \times 100}} = \dfrac{{200}}{{300}}$
On dividing both numbers by$100$, we get,
$ \Rightarrow 66\dfrac{2}{3}\% = \dfrac{2}{3}$
This is the simplest form of fraction.
$66\dfrac{2}{3}\% = \dfrac{2}{3}$ .
Note: A fraction is in its simplest form when the numerator and denominator cannot be any smaller while still being whole numbers. We can also convert the numbers by dividing them by the greatest common factor.
In (i), we know the greatest common factor of $25$ and $100$ is $25$ so divide the numbers of numerator and denominator by $25$, we get,
$ \Rightarrow 25\% = \dfrac{{25}}{{100}} = \dfrac{1}{4}$
In (ii), we know that the greatest common factor of $625$ and $100$is$25$ so divide the numbers of numerator and denominator by $25$, we get,
$ \Rightarrow 625\% = \dfrac{{625}}{{100}} = \dfrac{{25}}{4}$
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 7 Social Science: Engaging Questions & Answers for Success

Master Class 7 Science: Engaging Questions & Answers for Success

Master Class 7 Maths: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
Full Form of IASDMIPSIFSIRSPOLICE class 7 social science CBSE

What are the factors of 100 class 7 maths CBSE

List of coprime numbers from 1 to 100 class 7 maths CBSE

What is a subcontinent class 7 social science CBSE

How many thousands make a crore class 7 maths CBSE

When phenolphthalein is added toNaOH the colour of class 7 chemistry CBSE


