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A contractor combined x tons of a gravel mixture that [#permalink]
 28 Jan 2010, 15:07
Question Stats:
47% (01:46) correct
52% (00:56) wrong based on 6 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 ? (1) y = 10 (2) z = 16
Last edited by Bunuel on 31 Mar 2012, 12:02, edited 1 time in total.
Edited the question
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Re: x tons of a gravel mixture [#permalink]
28 Jan 2010, 16:25
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What does the prompt tell us?
x + y = z
0.1x + 0.02y = 0.05z
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.
1. Gives us y. Sufficient.
2. Gives us Z. Sufficient.
Pick D.
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Re: x tons of a gravel mixture [#permalink]
15 Jun 2010, 12:19
Goofed on this and answered B too quickly.
D it is...
we have .01x+.02y=.05z & x+y=z
1) y=10 means -> z-x=10, plug the numbers.
2) z=16 means -> x+y=16, plug the numbers
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Re: Gravel Mixture [#permalink]
16 Aug 2010, 00:34
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
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Re: Gravel Mixture [#permalink]
16 Aug 2010, 00:35
Hello,
From above we have:
0.1x + 0.02y = 0.05z
x+y = z
Thus,
0.1x + 0.02y = 0.05(x+y)
0.05x=0.03y
5x=3y
Statement 1
y=10 thus x =6 sufficient
Statement 2
z=16=x+y
and we have 5x=3y
Thus we have 2 equations with 2 unknowns; we can find x. Again sufficient.
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Re: x tons of a gravel mixture [#permalink]
17 Aug 2010, 13:05
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
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Re: mixture problem [#permalink]
11 Oct 2010, 22:35
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
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Re: a contractor combined x tons ofa gravel mixture [#permalink]
26 Dec 2010, 23:06
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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
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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!
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Re: GMAT Prep DS- Mixture problem [#permalink]
12 Mar 2011, 23:41
Easy way to solve this problem:
X lbs + Y lbs = Z lbs
Given 10% X + 2% Y = 5% Z
1) Y=10
2 eqns, 2 unknowns, solve!
2)Z= given.
Same as above, solve
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Re: mixture problem [#permalink]
13 Mar 2011, 11:38
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Re: mixture problem [#permalink]
23 Apr 2011, 00:09
remember X+Y=Z in que statement if you will not notice this statement you will land up to another answer
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Re: mixture problem [#permalink]
23 Apr 2011, 08:51
Using the method of alligations,
10:2
5
3:5
The quantities of x and y = x:y=3:5
z=8
So any one of the variables y and z will give the variable x
So answer is D
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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
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Re: A contractor combined x tons of a gravel mixture that [#permalink]
15 Apr 2012, 16:37
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?
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Re: A contractor combined x tons of a gravel mixture that [#permalink]
15 Apr 2012, 16:40
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Re: A contractor combined x tons of a gravel mixture that
[#permalink]
15 Apr 2012, 16:40
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