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?
76
68
60
52
Assume number of coins with A, B, C and D before giving as A, B, C & D respectively. The number of coins with D = 10 (smallest 2 digit number). Let’s work backwards now. D = (C/2) + 2 Or C/2 = 10 – 2 = 8 Or C = 16 Similarly, C = (B/2) + 2 Or B/2 = 16 – 2 = 14 Or B = 28 And finally, B = (A/2) + 2 Or A/2 = 28 – 2 = 26 Or A = 52 So, A initially had 52 coins.
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