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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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