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03 Jan 2018, 21:22
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B
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Difficulty:
35%
(medium)
Question Stats:
77% (02:30) correct
23%
(02:35)
wrong
based on 61
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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
A. n/2
B. 4n/3 - 100
C. 2n/3 + 350
D. 2n/6 + 350
E. (4n + 50)/3
03 Jan 2018, 22:10
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Average of two weights
\(L_1\) + \(L_2\) =2n -----(1)
\(L_1\) = 2\(L_2\) - 300
\(L_1\)- 2\(L_2\) = - 300 -----(2)
Multiply eq 1 with 2
2\(L_1\) + 2\(L_2\) =4n -----(3)
Adding eq no 2 and 3
3\(L_1\) = 4n-300
\(L_1\) = 4n/3-100
Ans. B
\(L_1\) + \(L_2\) =2n -----(1)
\(L_1\) = 2\(L_2\) - 300
\(L_1\)- 2\(L_2\) = - 300 -----(2)
Multiply eq 1 with 2
2\(L_1\) + 2\(L_2\) =4n -----(3)
Adding eq no 2 and 3
3\(L_1\) = 4n-300
\(L_1\) = 4n/3-100
Ans. B
03 Jan 2018, 22:40
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Bunuel
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
