A person at his first game loses 13 of his money- at the second 35 of the remainder- at the third 47 of the rest- what fraction of his original money has he left-

The correct option is C 435 Let x be the original money. Money lost = x3 Remainder = x x3 = 2x3 .Second lost = 35×2x3 = 2x5 Remainder ...

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Standard V

Mathematics

Multiplication of Fractions

A person at h...

Question

A person at his first game loses $\frac{1}{3}$ of his money, at the second $\frac{3}{5}$ of the remainder, at the third $\frac{4}{7}$ of the rest; what fraction of his original money has he left?

\frac{14}{35}

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\frac{24}{35}

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\frac{4}{35}

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\frac{4}{25}

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Solution

The correct option is C $\frac{4}{35}$
Let x be the original money. Money lost $= \frac{x}{3}$
$R e m a i n d e r = x - \frac{x}{3} = \frac{2 x}{3}$ .
Second lost $= \frac{3}{5} \times \frac{2 x}{3} = \frac{2 x}{5}$
$R e m a i n d e r = \frac{2 x}{3} - \frac{2 x}{5} = \frac{4 x}{15}$
Third lost $= \frac{4 x}{15} \times \frac{4}{7} = \frac{16 x}{105}$
Money left finally $= \frac{4 x}{15} - \frac{16 x}{105}$
$= \frac{12 x}{105}$ or $\frac{4 x}{35}$

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A person at his first game loses 13 of his money, at the second 35 of the remainder, at the third 47 of the rest; what fraction of his original money has he left?

The correct option is C 435Let x be the original money. Money lost =x3Remainder=x−x3=2x3.Second lost =35×2x3=2x5Remainder=2x3−2x5=4x15Third lost =4x15×47=16x105Money left finally =4x15−16x105=12x105 or 4x35

A person at his first game loses $\frac{1}{3}$ of his money, at the second $\frac{3}{5}$ of the remainder, at the third $\frac{4}{7}$ of the rest; what fraction of his original money has he left?