A contractor combined x tons of a gravel mixture

[Guest]

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

Answer is D. Can someone please explain how to get to the answer?

Thanks

[Sumit]

you can write the problem statement as 10%x +2%y=5%z

if you are given values of y and z, it can be solved for x.

so both statements together are sufficient. hence D.

[RonPurewal]

you can write the problem statement as 10%x +2%y=5%z

if you are given values of y and z, it can be solved for x.

so both statements together are sufficient. hence D.

yeah, but that's not what (d) means. (d) means that each of the statements, individually and alone, is sufficient to solve the problem.

here's the easiest way to do this:
FACT ABOUT WEIGHTED AVERAGES: if you have the weighted average and both endpoints, then you also have the RATIO of the weights in the problem. that ratio is the reciprocal of the ratio of the distances between the endpoints and the weighted average.
in this problem, we have this:
(endpoint Y 2%)-------distance=3-------(weighted average 5%)----------distance=5----------(endpoint X 10%)
so
since distance Y : distance X = 3 : 5, the ratio of the weights (literal "weights" in tons, in this problem) of Y : X must be 5 : 3.

because you have this ratio, specifying even one of the quantities is sufficient to determine everything - just use the ratio to figure out the rest.
therefore, either of the choices will be sufficient.

[accguy]

you can write the problem statement as 10%x +2%y=5%z

if you are given values of y and z, it can be solved for x.

so both statements together are sufficient. hence D.

yeah, but that's not what (d) means. (d) means that each of the statements, individually and alone, is sufficient to solve the problem.

here's the easiest way to do this:
FACT ABOUT WEIGHTED AVERAGES: if you have the weighted average and both endpoints, then you also have the RATIO of the weights in the problem. that ratio is the reciprocal of the ratio of the distances between the endpoints and the weighted average.
in this problem, we have this:
(endpoint Y 2%)-------distance=3-------(weighted average 5%)----------distance=5----------(endpoint X 10%)

Hi Ron,

Is there an algebraic way to solve this question?

I started with 0.1x + .02y = .05z and wrongly got C as the answer as it allows us to the solve the equation.

[StaceyKoprince]

Weighted averages are very tough.

You can write the equation 0.1x + .02y = .05z, but the problem gives more info than that (not just the statements - the problem itself). First, that equation tells us how much gravel G we have, right? 10% of x is a certain amount of gravel G and 2% of y is a certain amount of gravel G and so on.

Let's call x and y "inputs" and z the "mixture."

If we were to input the exact same amount of x (which is 10% gravel G) and y (which is 2% gravel G), what percentage of the resulting mixture z would be gravel G? Average 10 and 2: 6%.

If you want to prove with real numbers, take 100 grams of x (which gives us 10 grams of gravel G) and 100 grams of y (which gives us 2 grams of gravel G). The resulting mixture has 12 grams of gravel G and 200 grams of material total, or 12/200 = 6/100 = 6%.

But x and y are two different variables, so we don't input the same amount of each. The problem says the mixture percentage is 5%, not 6%, so do we input more of x or more of y in that mixture? We input more y, because the mixture percentage is closer to y's percentage than it is to x's percentage. With me so far?

So this is what Ron was getting at up above: the problem also gives us something about the ratio of x to y. Figure out how far away the mixture percentage is from each of the two starting percentages. 5 is three away from y's percentage of 2, and 5 is five away from x's percentage of 10. The "three away" and "five away" numbers tell us how to write the ratio. The larger number goes with the input that's contributing more; in this case, y is contributing more, so the 5 goes with the y. The smaller number goes with the input that's contributing less; in this case, x is contributing less, so the 3 goes with x. The ratio of x to y, then, is 3 to 5.

And that's where we get to write a second equation: x/y = 3/5. Plug that into your first equation and now you see that you have only two variables, so one variable given is sufficient to answer the question.

[Khalid]

[
since distance Y : distance X = 3 : 5, the ratio of the weights (literal "weights" in tons, in this problem) of Y : X must be 5 : 3.

Ron, I am confused on the above a bit

Y:distance x = 3:5

How did we get to

Y:X = 5:3

thx

[RonPurewal]

[
since distance Y : distance X = 3 : 5, the ratio of the weights (literal "weights" in tons, in this problem) of Y : X must be 5 : 3.

Ron, I am confused on the above a bit

Y:distance x = 3:5

How did we get to

Y:X = 5:3

thx

do you follow the theorem that's stated in boldface? if not, then please explain the difficulty you have in understanding it, and we'll help you out.
if you understand the statement of that result (the one labeled "FACT ABOUT WEIGHTED AVERAGES"), then this is a direct application of that fact.

thanks.

[exit8284bk]

Hi Ron, Stacey,

Can we calculate like this:

10x+2y=5Z or 5(x+y)

1. y=10 sufficient to know x
2. z=(x+y)=16 sufficient to know x

Therefore, D.

[RonPurewal]

Hi Ron, Stacey,

Can we calculate like this:

10x+2y=5Z or 5(x+y)

1. y=10 sufficient to know x
2. z=(x+y)=16 sufficient to know x

Therefore, D.

yes, that's an excellent way to do it.

BUT
DO NOT, EVER, LEAVE THE SAME VARIABLE ON BOTH SIDES OF A LINEAR EQUATION.

in this case, this means that you should simplify 10x + 2y = 5(x + y) to 10x + 2y = 5x + 5y, which in turn simplifies to 5x = 3y.
in that case, it's a lot easier to see why the statements are sufficient; they're both easy substitutions at this point.

if you leave the equation as 10x + 2y = 5(x + y) and then are given x + y = 16, there's a dangerous tendency to just plug 16 in for (x + y), giving 10x + 2y = 80, and then think that's all you can do (and therefore that that statement is insufficient). that's an incorrect approach.

[supshalu]

Yet another easy method..
use alligation..

X Y
10 2

5

3 5

Therefore STMT1 and STMT 2 individually help us find the answer

Isn't it RON ?!

[sunny.jain]

I think DS questions mean to check whether data is sufficient or not...!
most of time we are not suppose to solve the Question.

this Question for example, involves 3 variable: X, Y and Z.
After reading the question, we have:
X + Y = Z .......(1)

0.1X + 0.02Y = 0.05Z ........(2)

Clearly, high school fundamental, in order to solve equation of n variables we need n different equations. Here two equations are given, we need 3rd one.

Both the answer choices give one equation. So answer has to be D.

[Nishant.Chandra]

Ron and Stacey,

Great explanations on Weigted averages. Stacey's explanation provided conceptual framework, wheras Ron's technique is ideal for test as it saves a lot of time.

My question is: will this work even if there are 3 or more variables/ weights in a weighted average problem. If so how.....

[RonPurewal]

My question is: will this work even if there are 3 or more variables/ weights in a weighted average problem. If so how.....

the number line shortcut won't work anymore, no.

if you're especially hardcore, you can look at articles on a concept called "mass points" (usually used in physics, rather than pure math). that will tell you how to generalize the concept, although the generalized versions will be way way WAY beyond the scope of the gmat.
in fact, it's highly unlikely that you'll see a weighted average of more than two things, so you shouldn't worry.

--

of course, the advice about SIMPLIFYING still holds.

[david.khoy]

Hi Ron, Stacey,

Can we calculate like this:

10x+2y=5Z or 5(x+y)

1. y=10 sufficient to know x
2. z=(x+y)=16 sufficient to know x

Therefore, D.

yes, that's an excellent way to do it.

BUT
DO NOT, EVER, LEAVE THE SAME VARIABLE ON BOTH SIDES OF A LINEAR EQUATION.

in this case, this means that you should simplify 10x + 2y = 5(x + y) to 10x + 2y = 5x + 5y, which in turn simplifies to 5x = 3y.
in that case, it's a lot easier to see why the statements are sufficient; they're both easy substitutions at this point.

if you leave the equation as 10x + 2y = 5(x + y) and then are given x + y = 16, there's a dangerous tendency to just plug 16 in for (x + y), giving 10x + 2y = 80, and then think that's all you can do (and therefore that that statement is insufficient). that's an incorrect approach.

Well, no, because once you get:

10x + 2y = 80, you know you can do more since x + y = 16.

Just plug in 16 for (x+y) again:

8x + 2x + 2y = 80
8x + 2(x+y) = 80
8x + 2*16 = 80
8x = 48
x = 6

[akhp77]

By applying allegation as explained above we can have something like this
x : y : z = 3 : 5 : 8

if we know any one of them we can find other two