Bunuel wrote:

In a weight-lifting competition, the average (arithmetic mean) of Pierre's two lifts was n pounds. If the weight of his first lift was 300 less pounds than twice the weight of his second lift, what is the weight, in pounds of his first lift?

A. n/2

B. 4n/3 - 100

C. 2n/3 + 350

D. 2n/6 + 350

E. (4n + 50)/3

First lift = F

Second lift = S

The average of Pierre's two lifts was \(n\)

\(\frac{F + N}{2}= n\)

\(F + S = 2n\)

His first lift was 300 less pounds than twice the weight of his second lift

--We want to solve for F

--Define S in terms of F

\(F = 2S - 300\)

\(F + 300 = 2S\)

\(2S = F + 300\)

\(S = \frac{F+300}{2}\)

Back to the equation above, substitute for S

\(F + S = 2n\)

\(F +

\frac{F+300}{2} = 2n\)

\(2F + F + 300 = 4n\)

\(3F = 4n - 300\)

\(F = \frac{4}{3}n - \frac{300}{3}\)

\(F = \frac{4}{3}n - 100\)

Answer

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