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A contractor combined x tons of gravel mixture that cotained

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A contractor combined x tons of gravel mixture that cotained [#permalink] New post 17 Oct 2007, 18:39
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50% (01:55) correct 50% (00:23) wrong based on 15 sessions
A contractor combined x tons of gravel mixture that cotained 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 misture that was 5 percent gravel G,
by weight. What is the value of X?

1) y=10
2) z=16
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 [#permalink] New post 17 Oct 2007, 18:41
Here is what i wrote down, but I don't think its the right way to do this. I really bad at mixture probs.

x+y=z

.10x+.02y=.05z

Knowng the value of X,Y, or Z, we can solve for any of the variables this problem given the equations above.


I get D this way, but again i dunno if its just coincidence.
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 [#permalink] New post 18 Oct 2007, 08:35
Your approach looks right to me.

This note is for Killersquirrel....

Killersquirrel, can you solve this problem using your approach? I could not solve it that way.
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 [#permalink] New post 18 Oct 2007, 11:01
hey guys, i got D also, but my approach was a bit different, call it old fashion if you'd like.

I figured from the problem that x=10% gravel (G) y=2%G, and Z=5%G
we know that z=x+y

if we know x=10, therefore 10=10%G, so we find out how much gravel is in the mixture. After we find G, z=5%G, and it proves suff. Same goes for y, also suff, becasue we can find the gravel from it.

How do you guys feel about this approach? Is it dangerous to think it that way?
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 [#permalink] New post 18 Oct 2007, 12:35
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see attachment:

The new ratio is 3:5

statement 1

when you know that Y=10 then

10/X = 5/3

30 = 5X

6 = X

sufficient

statement 2

since Z = 16 and you know that 16/(5+3) = 2 then X=3*2 Y=5*2

sufficient

the answer is (D)
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 [#permalink] New post 18 Oct 2007, 14:41
KillerSquirrel wrote:
see attachment:

The new ratio is 3:5

statement 1

when you know that Y=10 then

10/X = 5/3

30 = 5X

6 = X

sufficient

statement 2

since Z = 16 and you know that 16/(5+3) = 2 then X=3*2 Y=5*2

sufficient

the answer is (D)


Hey, KS...would you mind explaining this a little further?

Thanks!
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 [#permalink] New post 18 Oct 2007, 17:37
z/20 = x/10 + y/50
5z = 10x + 2y
Since z = x+y,
5x + 5y = 10x+2y
x = 3y/5

St1:
y = 10. Sufficient. Can solve for x.

St2:
z = 16.

5z = 10x+2y
80 = 10x + 2y

Possible sets:
x = 7, y = 5 (x+y = 12. Out)
x = 6, y = 10 (x+y = 16. In)
x = 5, y = 15 (x+y = 20. Out)
x = 4, y = 20 (x+y = 24. Out)
x = 3, y = 25 (x+y = 28. Out)
....

So only one set works, x = 6, y = 10. Sufficient.

Ans D
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Re: GMAT Prep DS- Mixture problem [#permalink] New post 15 Aug 2009, 19:04
I was lost without you ywilfred !!
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Re: GMAT Prep DS- Mixture problem [#permalink] New post 17 Feb 2010, 16:57
I have to give kudos to Killer Squirrel for posting his approach. I am not a math person so the algebraic method makes no sense to me. KS's approach is helping me destroy these problems. Just one questing is there any limitations to this approach?
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Re: GMAT Prep DS- Mixture problem [#permalink] New post 22 Apr 2011, 22:51
ANSWER IS d

there is a similar question on forum in mixture prob
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lets start again

Re: GMAT Prep DS- Mixture problem   [#permalink] 22 Apr 2011, 22:51
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