# Help to solve a problem please. In three small groups everything together Was 48 counters. At first from the first small group shifted in the second so many counters How many in it already was, then from the second pack shifted in the third so many counters How many in it already was, and, at last, IS shifted 3 small groups in the first so many counters How many in it already was after that in all small groups it appeared counters equally. How many counters were in each barrel originally?

Artem

200

Here an analog it is also simpler to solve This problem with matches since the end. As after all shifting the number of matches in small groups became identical, in their each small group there were 48:3 = 16 pieces. Before it in the first a small group added so many matches how many in it was, i.e. 8 pieces. These took 8 matches from the third small group, i.e. there before the last shifting there were 16 + 8 = 24 matches. But these we receive 24 matches shifting from the second small group in a third of such quantity of matches what in the third small group already was. I.e. doubling of matches. Means before the second shifting in the third small group there were 12 matches, and in the second 16 + 12 = 28 matches. Arguing it is similarly received that in the second small group of 14 matches, and in the first 8 + 14 = 22 matches. X+x+x=48 3x=48 x=48:3 x=16

18

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Bogdana

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