Question

In gambling a man first lost 75 percent of his money and then 75 percent of the remaining sum in second venture and again 75 percent of the remaining amount in the third venture.He returned home with only r.S. 2 left.What sum had he at first

Answer 1

Let the original sum of money be y.

He lost 75% of y,

$\sf \: \frac{75}{100} \: \times \: y\: = \: 0.75y$

The second time he lost 75% of the remaining sum.

$\sf \: \frac{75}{100} \: (y - 0.75y) \: = \: 0.75(0.25y)$

After second gambling, the amount remaining is equal to

$\sf \: y - 0.75y - 0.75 \times 0.25y \: = \: 0.0625y$

The third time again, 76% of the remaining after second gambling.

$\sf \: \frac{75}{100} \: \times \: (0.0625y) \: = \: 0.0469y$

The remaining amount is equal to 2.

$\sf \: 0.0625y \: - \: 0.0469 \: = \: 2 \\ \\ \sf \: 0.0156y \: = \: 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf \: y \: = \: 128.21 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

The total amount that he had, in the beginning, is equal to Rs. 128.21.