find the number.

Question:

If $\frac{3}{5}$ of a number exceeds its $\frac{2}{7}$ by 44, find the number.

Solution:

Let $x$ be the required number.

We know that $\frac{3}{5}$ of the number exceeds its $\frac{2}{7}$ by 44 .

That is,

$\frac{3}{5} \times x=\frac{2}{7} \times x+44$

$\frac{3}{5} \times x-\frac{2}{7} \times x=44$

$\left(\frac{3}{5}-\frac{2}{7}\right) \times x=44$

$\left(\frac{3}{5}+\right.$ Additive inverse of $\left.\frac{2}{7}\right) \times x=44$

$\left(\frac{21-10}{35}\right) \times x=44$

$\frac{11}{35} \times x=44$

$x=44 \div \frac{11}{35}$

$=44 \times \frac{35}{11}$

$=\frac{44}{1} \times \frac{35}{11}$

$=\frac{44 \times 35}{1 \times 11}$

$=\frac{1540}{11}$

$=140$

Hence, the number is 140.

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