# Answer to Question #4552 in Calculus 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.

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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