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A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning?

A has some coins. He gives half of the coins and 2 more to B. B gives half of the coins and 2 more to C. C gives half of the coins and 2 more to D. The number of coins D has now, is the smallest two-digit number. How many coins does A have in the beginning? Correct Answer 52

Calculation:

Let A has x number of coins.

B = A/2 + 2 = x/2 + 2

C = B/2 + 2 = x/4 + 2/2 + 2 = x/4 + 3

D = C/2 + 2 = x/8 + 3/2  + 2 = x/8 + 7/2

It is given that D has the number of coins which is equal to lowest two-digit number i.e. 10.

D = x/8 + 7/2 = 10

x/8 = 13/2

x = 52

Hence, option 4 is correct.

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