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# A contractor combined x tons of a gravel mixture that contai

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Director
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A contractor combined x tons of a gravel mixture that contai [#permalink]

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09 Apr 2006, 16:49
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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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09 Apr 2006, 16:57
C

g(x) = 10/100 * x
g(y) = 2/100 * y

g(x) + g(y) = g(z) = 5/100 * z

10x + 2y = 5z

so u need both z and y to find the value of x!

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09 Apr 2006, 21:42
Considering both the cases, we get the following equations:

Gravel: 10x/100 + 2y/100 = 5z/100
Non-Gravel: 90x/100 + 98y/100 = 95z/100

Case A: y = 10
=====

10x + 20 = 5z
90x + 98y = 95z

Substituting y = 10, we can solve for x = 6

Case B: z = 16
=====

10x + 2y = 80
90x + 98y = 1520

Substituting z = 16, we can solve for x = 6.

So, (A) and (B) are independently sufficient to answer the query.

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10 Apr 2006, 19:21
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.
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10 Apr 2006, 20:14

n1/n2 = (c2-cm)/(cm-c1)

c1, c2 and cm are concentrations , all given

n1 = x
n2 = y = statement A

statement b gives Z , which indirectly gives y in terms of x

So D

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13 Apr 2006, 01:00
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D is my choice

Question gives us x+y=z and also .10x+.02y=.05z = .05(x+y)

1) gives y=10
so we can solve .10x+.02y=.05(x+y)
so sufficient

2) x+y=16 so .10x+.02y=8 ... two unique equation can solve it
so sufficient

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14 Apr 2006, 10:33
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.
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Director
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16 Apr 2006, 12:35
I learnt a nice way of solving mixtures today and learnt that best to rewrite the problem..

So question is: x/10 + y/50 = 5/100(x+y)
z = x+y given.

1. y = 10, subs y, you get x. Suff.. So now down to A or D.
2. z = 16, so y = 16-x, subs in the question, you get X. so suff.

Hence D...

Hopefully this helps to a lot of us.. Under time pressure to get this right is whole another thing!

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28 Aug 2013, 04:42
Zooroopa wrote:
Considering both the cases, we get the following equations:

Gravel: 10x/100 + 2y/100 = 5z/100
Non-Gravel: 90x/100 + 98y/100 = 95z/100

Case A: y = 10
=====

10x + 20 = 5z
90x + 98y = 95z

Substituting y = 10, we can solve for x = 6

Case B: z = 16
=====

10x + 2y = 80
90x + 98y = 1520

Substituting z = 16, we can solve for x = 6.

So, (A) and (B) are independently sufficient to answer the query.

Ohhh this is crazy...
that means you need to guess, that there is also non gravel equation suppose to be.

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

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28 Aug 2013, 09:39
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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 ?

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.

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 contai   [#permalink] 28 Aug 2013, 09:39
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