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A contractor combined x tons of a gravel mixture that contained 10 per

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A contractor combined x tons of a gravel mixture that contained 10 per [#permalink]

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New post 02 Sep 2009, 02:50
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A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x ?

(1) y = 10
(2) z = 16

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-contractor-combined-x-tons-of-a-gravel-mixture-that-102681.html
[Reveal] Spoiler: OA
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Re: A contractor combined x tons of a gravel mixture that contained 10 per [#permalink]

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New post 02 Sep 2009, 02:58
x.10%+y.2%=z.5% x+y=z so it makes
x.10%+y.2%=(x+y).5%
x.10%+y.2%=x.5%+y.5%=
x.5%=y.3%
5x=3y
x=3y/5
z=8y/5
so if we know either y or z we can find x.
i.e. 1 and 2 suff, D.
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Re: A contractor combined x tons of a gravel mixture that contained 10 per [#permalink]

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New post 03 Sep 2009, 22:17
One eq. will be 10x + 2y = 5z

Other eq, x+y=z

Stmt 1 : y =10 ..suff 3 equation...3 variable

Stmt 2 : z = 16 ..suff 3 equation...3 variable
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Re: A contractor combined x tons of a gravel mixture that contained 10 per [#permalink]

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New post 22 Apr 2011, 23:15
word translation problems requires simplification

and we should not miss to intrept that

z=x+y in this case
:-D :-D :-D
then simplified equation will be 5x=3y and z=x+y

:arrow: hence both staement are sufficent D
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Re: A contractor combined x tons of a gravel mixture that contained 10 per [#permalink]

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Re: A contractor combined x tons of a gravel mixture that contained 10 per [#permalink]

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New post 25 Aug 2015, 23:44
Expert's post
A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x ?

Set the equation: \(0.1x+0.02y=0.05(x+y)\), where \(x+y=z\) --> \(5x=3y\) --> Q: \(x=?\)

(1) \(y=10\) --> \(5x=3y=30\) --> \(x=6\). Sufficient.

(2) \(z=x+y=16\) --> \(y=16-x\) --> \(5x=3y=3(16-x)\) --> \(x=6\). Sufficient.

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-contractor-combined-x-tons-of-a-gravel-mixture-that-102681.html
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Re: A contractor combined x tons of a gravel mixture that contained 10 per   [#permalink] 25 Aug 2015, 23:44
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