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A contractor combined x tons of a gravel mixture that [#permalink]
28 Jan 2010, 14:07

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D

E

Difficulty:

65% (hard)

Question Stats:

49% (01:45) correct
51% (00:49) wrong based on 641 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 ?

Re: x tons of a gravel mixture [#permalink]
28 Jan 2010, 15:25

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ugimba wrote:

119.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

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

We have to find x. If you notice, we have two equations in 3 variables, so if we are given a value for either y OR z, this is sufficient to calculate x.

Re: Gravel Mixture [#permalink]
15 Aug 2010, 23:34

lalithajob wrote:

A contractor combined x tons of a gravel mixture that contained 10% gravel G, by weight, with y tons of a mixture that contained 2% gravel G, by weight, to produce z tons of a mixture that was 5% gravel G, by weight. What is the value of x?

1. y = 10 2. z = 16

x + y = z ( x tons of mixture1 + y tons of mixture2 = z tons of combined mixture) .1x + .02y = .05z (gravel in mixture1 + gravel in mixture2 = gravel in combined mixture)

x= ?

1) y=10. 3 equations(2 from the question + 1 from answer choice) with 3 unknowns. Can solve for x. Sufficient 2) z=16. 3 equations(2 from the question + 1 from answer choice) with 3 unknowns. Can solve for x. Sufficient

Answer D _________________

___________________________________ Please give me kudos if you like my post

Re: x tons of a gravel mixture [#permalink]
17 Aug 2010, 12:05

Question: For these types of questions can you always do Z=X+Y?

This is the first question that I've seen like this. I'm not sure if I would have gone with Z=X+Y, as Z could have been the combination X+Y and some other non-gravel material

Re: mixture problem [#permalink]
11 Oct 2010, 21:35

The answer in my opinion should be D)

By condition ,

10% X + 2% Y = 5 % Z ---(1) Also X + Y = Z ---(2)

Option (A) is sufficient to get the value of X by substituting the value of Y = 10 Option (B) is sufficient to get the value of X by substituting the value of Z = 16

Re: a contractor combined x tons ofa gravel mixture [#permalink]
26 Dec 2010, 22:06

5

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Expert's post

anilnandyala wrote:

a contractor combined x tons ofa gravel mixture that contained 10% gravel g by weight with y tons of a mixture that contained 2% gravel g by weight to produce z tons of a mixture that was 5% gravel by weight. what is the value of x?

y = 10 z = 16

Using scale method here, since 10% and 2% give weighted average of 5%, x:y = 3:5 We also know x + y = z.

1. y = 10. If y = 10, x = 6 since their ratios must be 3:5. Sufficient.

2. z = 16 If sum of x and y is 16, x must be 6 and y must be 10 to give a ratio of 3:5.

If this is not intuitive, think of it this way: x : y...... x + y 3 : 5...... 8 Since 8 in ratio terms is actually 16, 3 is actually 6 and 5 is actually 10.

Answer (D)

It will be worth your while if you understand the scale method. The time saving is huge and weighted average is a concept you will need to use time and again. For explanation of scale method, check this link: http://gmatclub.com/forum/tough-ds-105651.html#p828579 _________________

The answer is D. I was flummoxed on this one since at first I thought it was C, then I looked at it closer and thought it was B. Now I see how A and B are sufficient. Crazy DS - I HATE YOU!

Re: A contractor combined x tons of a gravel mixture that [#permalink]
15 Apr 2012, 15:40

Expert's post

ad20 wrote:

Can anyone please explain on X+Y=Z in que statement. I definitely missed it and ended up with answer C Where in question it says that x+y=z?

Sure. Question says: "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."

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

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

Merging similar topics. Please refer to the discussion above.

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