If $\frac{9}{5}$ of a number exceeds $\frac{5}{7}$ of that number by 70, find that number.

Given :

$\frac{9}{5}$ of a number exceeds $\frac{5}{7}$ of that number by 70.

To do :

We have to find the number.

Solution :

Let the number be x.

Therefore,

$\frac{9}{5}(x) = \frac{5}{7}(x)+70$

$\frac{9}{5}(x) -\frac{5}{7}(x)=70$

$x(\frac{9}{5} - \frac{5}{7}) = 70$

$x[\frac{(9\times 7-5\times 5)}{(5\times 7)}]=70$

$x[\frac{(63-25)}{35}]=70$

$x(\frac{38}{35})=70$

$x=70\times \frac{35}{38}$

$x=35\times \frac{35}{19}$

$x=\frac{1225}{19}$

The number is $\frac{1225}{19}$.

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Updated on: 10-Oct-2022

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