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A contractor combined x tons of a gravel mixture that

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

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New post Updated on: 31 Mar 2012, 12:02
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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

Originally posted by ugimba on 28 Jan 2010, 15:07.
Last edited by Bunuel on 31 Mar 2012, 12:02, edited 1 time in total.
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Re: x tons of a gravel mixture  [#permalink]

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New post 28 Jan 2010, 16:25
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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: \(x=?\)

(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.

Answer: D.
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Re: gmat prep DS  [#permalink]

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New post 15 Jun 2010, 05:53
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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]

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New post 15 Jun 2010, 12:19
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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]

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New post 16 Aug 2010, 00:34
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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: mixture problem  [#permalink]

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New post 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]

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New post 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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New post 12 Mar 2011, 23:38
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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]

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New post 12 Mar 2011, 23:41
1
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]

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New post 23 Apr 2011, 00:09
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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: A contractor combined x tons of a gravel mixture that  [#permalink]

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New post 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]

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New post 15 Apr 2012, 16:40
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."

Hope it's clear.
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Re: A contractor combined x tons of a gravel mixture that  [#permalink]

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New post 13 Jun 2013, 02:15
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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New post 09 May 2014, 06:02
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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.

Answer: D.
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A contractor combined x tons of a gravel mixture that  [#permalink]

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New post 14 Jan 2016, 11:16
ugimba wrote:
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


It's a weighted average problem.
X(10)....5....5.....3.....Y(2)
\(\frac{x}{y}=\frac{3}{5}\)

(1) \(\frac{x}{10}=\frac{3}{5}\), \(x=6\)
(2) \(\frac{x}{y}=\frac{3}{5}\) --> \(\frac{x}{16}=\frac{3}{8}\), \(x=6\)
Answer D
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Re: A contractor combined x tons of a gravel mixture that  [#permalink]

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New post 14 Jan 2016, 18:34
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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


When you modify the original condition and the question, they becomes x+y=z and (10/100)x+(2/100)y=(5/100)z, 10x+2y=5z. Then there are 3 variables(x,y,z) and 2 equations, which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), when y=10, x+y=z, 10x+2y=5z → x+10=z, 10x+20=z. Since the value of x is unique, it is sufficient.
For 2), when z=16, the value of x is also unique in x+y=16, 10x+2y=80, which is unique and sufficient. Therefore, the answer is D.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: A contractor combined x tons of a gravel mixture that  [#permalink]

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New post 09 May 2016, 07:32
A super quick way of solving mixture questions is using the rule of alligation
(In this case, it took me +/- 1 minute)

Difference between the diagonals is the numbers in red

X (Current proportion of gravel G) _______________ 10/100____________ 3 ______ (5-2)

Z (Desired proportion of gravel G) ________________________ 5/100

Y (Proportion of gravel G in the mix to be added)_____2/100____________ 5 ______ (10-5)


We already know the proportion between X, Y and Z from the question stem.

If we have the value of either Y or Z we can solve for X.

Each statement independently is therefore sufficient. D


Not required, but for completeness:


Statement 1
X:Y
3:5
Y=10
X = 10*3/5 = 6

Statement 2
X:Z
3:8
Z=16
X= 16*3/8 = 6
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Re: A contractor combined x tons of a gravel mixture that  [#permalink]

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New post 28 Jan 2017, 15:44
x = 10% gravel
y = 2% gravel.
z = the MIXTURE of x and y = 5% gravel.

To determine the required ratio of x to y, use ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for x and y on the ends and the percentage for mixture z in the middle.
x 10%-----------5%-----------2% y

Step 2: Calculate the distances between the percentages.
x 10%-----5-----5%----3-----2% y

Step 3: Determine the ratio in the mixture.
The required ratio of x to y is equal to the RECIPROCAL of the distances in red.
x:y = 3:5.

Since x:y = 3:5, and 3+5 = 8, every 8 tons of mixture z is composed of 3 tons of x and 5 tons of y.

Statement 1: y=10
Since x:y = 3:5 = 6:10, x=6.
SUFFICIENT.

Statement 2: z=16
Since x:y = 3:5 = 6:10, and 6+10 = 16, the 16 tons of mixture z must be composed of 6 tons of x and 10 tons of y.
SUFFICIENT.

The correct answer is D.
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Re: A contractor combined x tons of a gravel mixture that  [#permalink]

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New post 04 Aug 2018, 08:38
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ugimba wrote:
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


Let's use some weighted averages to solve this question
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Target question: What is the value of x ?

Given: 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.
First, we can write: x + y = z

Also, the total weight of the mixture = z (aka x + y)
So, when we apply the above formula, we get: 5% = (x/z)(10%) + (y/z)(2%)
Ignore the % symbols: 5 = (x/z)(10) + (y/z)(2)
Multiply both sides by z to get: 5z = 10x + 2y
Since x + y = z, we can rewrite the above equation as: 5(x +y) = 10x + 2y
Expand: 5x + 5y = 10x + 2y
Simplify to get: 5x - 3y = 0

Now onto the statements!!!!!

Statement 1: y = 10
Replace y with 10 to get: 5x - 3(10) = 0
Solve to get, x = 6
Since we can answer the target question with certainty, statement 1 is SUFFICIENT


Statement 2: z = 16
In other words, x + y = 16

So, we have:
5x - 3y = 0 and x + y = 16
Since we have 2 linear equations with 2 variables, we COULD solve the system for x, which means we COULD answer the target question
So, statement 2 is SUFFICIENT

Answer: D

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Re: A contractor combined x tons of a gravel mixture that &nbs [#permalink] 04 Aug 2018, 08:38
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