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In a weight-lifting competition, the total weight of Joe's t

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In a weight-lifting competition, the total weight of Joe's two lifts was 750 pounds. If twice the weight of his first lift was 300 pounds more than the weight of his second lift, what was the weight, in pounds, of his first lift?

(A) 225
(B) 275
(C) 325
(D) 350
(E) 400
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Re: In a weight-lifting competition, the total weight of Joe's t [#permalink]

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New post 17 Dec 2012, 07:50
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Walkabout wrote:
In a weight-lifting competition, the total weight of Joe's two lifts was 750 pounds. If twice the weight of his first lift was 300 pounds more than the weight of his second lift, what was the weight, in pounds, of his first lift?

(A) 225
(B) 275
(C) 325
(D) 350
(E) 400


Let x and y be the weights of first and second lifts, respectively. Then:

x + y = 750 (the total weight of Joe's two lifts was 750 pounds);
2x = y + 300 (twice the weight of his first lift was 300 pounds more than the weight of his second lift);

Solving gives x = 350.

Answer: D.
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Re: In a weight-lifting competition, the total weight of Joe's t [#permalink]

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New post 08 Sep 2014, 00:25
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Let the first lift = x

Second lift = 750 - x

Setting up the equation

2x = 750 - x + 300

x = 250+100 = 350

Answer = D
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Re: In a weight-lifting competition, the total weight of Joe's t [#permalink]

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Walkabout wrote:
In a weight-lifting competition, the total weight of Joe's two lifts was 750 pounds. If twice the weight of his first lift was 300 pounds more than the weight of his second lift, what was the weight, in pounds, of his first lift?

(A) 225
(B) 275
(C) 325
(D) 350
(E) 400


This problem is a general word translation. We first define variables and then set up equations.

We can define the following variables:

F = the weight of the first lift

S = the weight of the second lift

We are given that the total weight of Joe's two lifts was 750 pounds. We sum the two variables to obtain:

F + S = 750

We are also given that twice the weight of his first lift was 300 pounds more than the weight of his second lift. We express this as:

2F = 300 + S

2F – 300 = S

We can now plug in (2F – 300) for S into the first equation, so we have:

F + 2F – 300 = 750

3F = 1,050

F = 350

Answer is D.
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Re: In a weight-lifting competition, the total weight of Joe's t [#permalink]

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New post 16 Jul 2017, 02:21
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Re: In a weight-lifting competition, the total weight of Joe's t   [#permalink] 16 Jul 2017, 02:21
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