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Question:

A thief somehow managed to steal some chocolates from a Mall but while coming out of it at the first door he was caught by the guard and he successfully dealt him by giving 1 chocolate plus half of the rest chocolates. Further he had to pay 2 chocolates and half of the rest to the second guard. Once again at the third gate he gave 3 chocolates and then half of the rest. After that he was left with only one chocolate. How many chocolates had he stolen?

Explanation:

Let the number of chocolates initially with him = x.

To the first guard he gives 1 chocolate and half of the remaining. Hence, no. of chocolates left with the man is (x – 1)/2.

To the second guard he gives 2 chocolate and half of the remaining. Hence, no. of chocolates left with the man is $\frac{\frac{x - 1}{2} - 2}{2}$ .

To the third guard he gives 3 chocolate and half of the remaining. Hence, no. of chocolates left with the man is $\frac{[\frac{\frac{x - 1}{2} - 2}{2} - 3]}{2}$ .

Finally, he is left with only 1 chocolate, hence $\frac{[\frac{\frac{x - 1}{2} - 2}{2} - 3]}{2} = 1$

⇒ $\frac{\frac{x - 1}{2} - 2}{2} - 3 = 2$

⇒ $\frac{\frac{x - 1}{2} - 2}{2} = 5$

⇒ $\frac{x - 1}{2} - 2 = 10$

⇒ $\frac{x - 1}{2} = 12$

⇒ x = 25

Hence, 25.

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