Product of two rationals is $\dfrac{4}{9}$ and one is $\dfrac{2}{3}$. The other is
Product of two rationals is $\dfrac{4}{9}$ and one is $\dfrac{2}{3}$. The other is
- $\dfrac{2}{3}$
- $\dfrac{4}{6}$
- $\dfrac{8}{27}$
- $\dfrac{3}{2}$
Solution
$\dfrac{4}{9} \div \dfrac{2}{3} = \dfrac{2}{3}$.
Asked in: IMO
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