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A contractor combined x tons of a gravel mixture that contai [#permalink]
09 Apr 2006, 15:49

1

This post was BOOKMARKED

00:00

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B

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Difficulty:

95% (hard)

Question Stats:

45% (02:16) correct
55% (01:26) wrong based on 110 sessions

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 ?

It is quite simple. Match weights of gravel and non-gravel.

Gravel --> 10% of x + 2% of y = 5% of z
Therefore everything that remains must be non-gravel or whatever constituent is used in the mixture.

So,

Non-Gravel -->

90% of x (this is what remains from the first mixture) + 98% of y (this is what remains from the second mixture) = 95% of z (this is what remains from the third mixture)

So you get the equation that you have in bold. _________________

Answer = B
Here is what i think
if you take total X + Y = Z (over all mass balance)
and 0.1X + 0.02 Y = 0.05 Z (Gravel balance)
We have 3 variable and two equations. But if you know the value of Z as in the second case, you can find out X and Y.
So I think Z = 16 should be sufficient to solve the equations. _________________

Re: A contractor combined x tons of a gravel mixture that contai [#permalink]
28 Aug 2013, 08:39

1

This post received KUDOS

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: \(x=?\)

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