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1. The owner of a local jewellery store hired 3 watchmen to guard his diamonds, but a thief still got in and stole some diamonds. On the way out, the thief met each watchman, one at a time. To each he gave $\frac{1}{2}$ of the diamonds he had then and 2 more besides. He escaped with one diamond. How many did he steal originally?

A. 36

B. 25

C. 640

D. None of these

Solution:

At last thief is left with one diamond.
Hence, the number of diamonds before he gave some diamonds to the third watchman,

\begin{aligned} \Rightarrow x - (\frac{x}{2} + 2) = 1 \\ or, \frac{x - 4}{2} = 1 \\ or, x = 6 \end{aligned}

Hence, he had 6 diamonds before he gave 5 to the third watchman.
Similarly number of diamonds before giving to second watchman,

\begin{aligned} \frac{x - 4}{2} = 6 \\ or, x = 16 \end{aligned}

And number of diamonds before giving to the first watchman,

\begin{aligned} \frac{x - 4}{2} = 16 \\ or, x = 36 \end{aligned}

∴ The thief has stolen 36 diamonds originally

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