Let's mark Na, Nb and Nc number of buttons from boxes A, B and C respectfully. So,
Nb = Na/2,
Nc = 2Nb/3,
(Na + Nb + Nc)/3 = 99.
Therefore,
Nc = 2Nb/3 = 2(Na/2)/3 = Na/3.
So,
Na + Nb + Nc = Na + Na/2 + Na/3 = 11Na/6.
From the other side,
Na + Nb + Nc = 3*99 = 297.
Hence, 11Na/6 = 297 and Na = 297*6/11 = 162.
At last, Nc = Na/3 = 162/3 = 54.
So, there were 54 buttons in box C.
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