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# Find the odd one out.$\dfrac{3}{5},{\text{ }}\dfrac{{12}}{{20}},{\text{ }}\dfrac{9}{{15}},{\text{ }}\dfrac{6}{{10}},{\text{ }}\dfrac{{15}}{{20}}{\text{ }}$

Last updated date: 27th Mar 2023
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Hint: - Convert the fraction into lowest form fraction.
As we know, the lowest form fraction is the fraction when fraction has no common factors except 1.

First fraction is $\dfrac{3}{5}$. This is written in lowest form as the fraction has no common factors except 1.

Second fraction is $\dfrac{{12}}{{20}}$, so convert it into the lowest form fraction by cancelling out its common factors.

$\therefore \dfrac{{12}}{{20}} = \dfrac{3}{5} \times \dfrac{4}{4} = \dfrac{3}{5}$, so this is in lowest form as the fraction has no common factors except 1.

Third fraction is $\dfrac{9}{{15}}$, so convert it into the lowest form fraction by cancelling out its common factors.

$\therefore \dfrac{9}{{15}} = \dfrac{3}{5} \times \dfrac{3}{3} = \dfrac{3}{5}$, so this is in lowest form as the fraction has no common factors except 1.

Fourth fraction is $\dfrac{6}{{10}}$, so convert it into the lowest form fraction by cancelling out its common factors.

$\therefore \dfrac{6}{{10}} = \dfrac{3}{5} \times \dfrac{2}{2} = \dfrac{3}{5}$, so this is in lowest form as the fraction has no common factors except 1.

Fifth fraction is $\dfrac{{15}}{{20}}$, so convert it into the lowest form fraction by cancelling out its common factors.

$\therefore \dfrac{{15}}{{20}} = \dfrac{3}{4} \times \dfrac{5}{5} = \dfrac{3}{4}$, so this is in lowest form as the fraction has no common factors except 1.

Now it is clear that from all of the above fractions are the same except the last one.
Therefore odd one from all of the above fraction is $\dfrac{{15}}{{20}} = \dfrac{3}{4}$
So, this is the required answer.

Note: -whenever we face such types of problems the key concept we have to remember is that convert all the fractions into its lowest form by cancelling out its common factors then check which fraction is different from the others then we will get the required answer.