Answer to Question #4552 in Calculus for ong

Question

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?

Answer 1

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.