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A certain factory produces buttons and buckles at a uniform weight. If

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Bunuel
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A certain factory produces buttons and buckles at a uniform weight. If [#permalink] New post   05 Jul 2016, 12:18
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A certain factory produces buttons and buckles at a uniform weight. If the total weight of 2 buttons and 2 buckles is one third of 11 buckles and 3 buttons, then the weight of 3 buttons and 2 buckles is how many times that of 5 buckles and 6 buttons?

A. 7/15.
B. 4/9.
C. 6/11.
D. 5/9.
E. 8/15.
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Re: A certain factory produces buttons and buckles at a uniform weight. If [#permalink] New post   05 Jul 2016, 20:43
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2buttons + 2buckles = 1/3(11buckles + 3buttons)
6buttons + 6buckles = 11buckles + 3buttons
3buttons = 5buckles

Weight of 3buttons and 2buckles = 5buckles + 2buckles = 7buckles
Weight of 6buttons and 5buckles = 10buckles + 5buckles = 15buckles

(3buttons and 2buckles) = x * (6buttons and 5buckles)
7 = x*15
x = 7/15

Answer: A
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Re: A certain factory produces buttons and buckles at a uniform weight. If [#permalink] New post   24 Mar 2018, 23:15
Vyshak wrote:
2buttons + 2buckles = 1/3(11buckles + 3buttons)
6buttons + 6buckles = 11buckles + 3buttons
3buttons = 5buckles

Weight of 3buttons and 2buckles = 5buckles + 2buckles = 7buckles
Weight of 6buttons and 5buckles = 10buckles + 5buckles = 15buckles

(3buttons and 2buckles) = x * (6buttons and 5buckles)
7 = x*15
x = 7/15

Answer: A


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Re: A certain factory produces buttons and buckles at a uniform weight. If [#permalink] New post   23 Apr 2019, 19:56
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Bunuel wrote:
A certain factory produces buttons and buckles at a uniform weight. If the total weight of 2 buttons and 2 buckles is one third of 11 buckles and 3 buttons, then the weight of 3 buttons and 2 buckles is how many times that of 5 buckles and 6 buttons?

A. 7/15.
B. 4/9.
C. 6/11.
D. 5/9.
E. 8/15.


Let b = the weight of one button and k = the weight of one buckle. We can create the equation:

2b + 2k = (1/3)(3b + 11k)

6b + 6k = 3b + 11k

3b = 5k

b = 5k/3

In terms of k, 3 buttons and 2 buckles would weigh

3b + 2k = 3(5k/3) + 2k = 5k + 2k = 7k

and 5 buckles and 6 buttons would weigh

6b + 5k = 6(5k/3) + 5k = 10k + 5k = 15k.

Therefore, the weight of 3 buttons and 2 buckles is 7k/(15k) = 7/15 of the weight of 5 buckles and 6 buttons.

Answer: A
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Re: A certain factory produces buttons and buckles at a uniform weight. If [#permalink] New post   16 Mar 2023, 19:10
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A certain factory produces buttons and buckles at a uniform weight. If [#permalink]
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