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

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Joined: 02 Dec 2012
Posts: 177
In a weight-lifting competition, the total weight of Joe's t  [#permalink]

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17 Dec 2012, 07:47
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Difficulty:

5% (low)

Question Stats:

89% (01:39) correct 11% (01:52) wrong based on 608 sessions

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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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Posts: 50613
Re: In a weight-lifting competition, the total weight of Joe's t  [#permalink]

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17 Dec 2012, 07:50
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1
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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08 Sep 2014, 00:25
1
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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21 Jun 2016, 09:34
1
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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06 Feb 2018, 11:07
Top Contributor
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's solve the question using ONE VARIABLE.

The total weight of Joe's two lifts was 750 pounds.
Let x = weight of first lift
So, 750 - x = weight of second lift

...twice the weight of his first lift was 300 pounds more than the weight of his second lift
In other words, (2)(first lift) = second lift + 300
Or...(2)(x) = (750 - x) + 300
Simplify: 2x = 1050 - x
Rearrange: 3x = 1050
x = 350

Answer: D

Cheers,
Brent
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Re: In a weight-lifting competition, the total weight of Joe's t &nbs [#permalink] 06 Feb 2018, 11:07
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# In a weight-lifting competition, the total weight of Joe's t

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