2013 AMC 12A Problem 16

A,

B,

and

C

are three piles of rocks. The mean weight of the rocks in

A

40

pounds, the mean weight of the rocks in

B

50

pounds, the mean weight of the rocks in the combined piles

A

and

B

43

pounds, and the mean weight of the rocks in the combined piles

A

and

C

44

pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles

B

and

C?

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Solution

Let

a, b, c

be the numbers of rocks in the piles. From

\dfrac{40a + 50b}{a + b} = 43,

we get

7b = 3a,

a = 7k

and

b = 3k.

Let

\mu_{BC}

be the mean of

B

and

C.

Using the

A, C

mean

44

to express

\mu_C = \dfrac{28k + 44c}{c},

we find

\mu_{BC} = \dfrac{178k + 44c}{3k + c},

(\mu_{BC} - 44)c = (178 - 3\mu_{BC})k.

Since

B

is heavier than

A,

the mean of

B

and

C

exceeds

44,

forcing

178 - 3\mu_{BC} \gt 0,

i.e.

\mu_{BC} \lt \tfrac{178}{3} = 59\tfrac13.

The value

59

is attainable by taking

k=15c;

then pile

C

has mean

464

and the displayed formula gives

\mu_{BC}=59.

Thus the greatest integer mean is

59.

Thus, the correct answer is E.

Tagged: mean · optimization