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Answer on Calculus Question for ong

Question #4552
Box B contained 1/2 as many buttons as Box A.Box C contained 2/3 as many buttons as Box B.The average number of buttons per box was 99.How many butttons were there in Box C?
Expert's answer
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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