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 1/2 of the diamonds he had then and 2 more besides. He escaped with one diamond. How many did he steal originally?

Answer: Option A

At last thief is left with one diamond.
Hence, the number of diamonds before he gave some diamonds to the third watchman,
$$\begin{array}{rl}& \Rightarrow x-\left(\frac{x}{2}+2\right)=1\\ & \text{or, }\frac{x-4}{2}=1\\ & \text{or, }x=6\end{array}$$
Hence, he had 6 diamonds before he gave 5 to the third watchman.
Similarly number of diamonds before giving to second watchman,
$$\begin{array}{rl}& \frac{x-4}{2}=6\\ & \text{or,}x=16\end{array}$$
And number of diamonds before giving to the first watchman,
$$\begin{array}{rl}& \frac{x-4}{2}=16\\ & \text{or, }x=36\end{array}$$
∴ The thief has stolen 36 diamonds originally