harsh.champ wrote:A thief was stealing diamonds from a jewellery store. On his way out, he encountered three guards, each was given half of the existing diamonds and two over it by the thief. In the end, he was left with one diamond. How many did the thief steal?


(1)40
(2)36
(3)32
(4)28
(5)26


___________
Hey Stuart and Ian,
I solved this equation by taking no. of diamonds initially= x and solving the corresponding equations.
Is there any shortcut method of solving these type of questions because the above method took some time??

ajith wrote:

harsh.champ wrote:A thief was stealing diamonds from a jewellery store. On his way out, he encountered three guards, each was given half of the existing diamonds and two over it by the thief. In the end, he was left with one diamond. How many did the thief steal?


(1)40
(2)36
(3)32
(4)28
(5)26


___________
Hey Stuart and Ian,
I solved this equation by taking no. of diamonds initially= x and solving the corresponding equations.
Is there any shortcut method of solving these type of questions because the above method took some time??

Well you can start by 1 diamond left in the end add 2 to get 3; double it 6; add two 8; double it 16; add 2 18; double it to get 36; which may be faster

harsh.champ wrote: ____________________
Thanks ajith,
Got the shortcut technique.Solving by the 1st method ,it took me around 3-4 minutes but when I applied your approach,I got the answer within 45 seconds.Way to go!!That means I save time for 1 additional question.Also,is it not advisable to approach the questions with the conventional "considering x" technique??
Thanks again.