simple math problem question

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simple math problem question

by kuzzden » Fri May 17, 2013 3:59 am
Hi everyone, I can't understand how to get to the solution, though it seems to be pretty easy... Can someone please help:

If x, y and k are positive numbers such that 10*x/(x+y)+20*y/(x+y)=k and if x<y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30

I see that answer should be either B or D, but can't find the right solution
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by GMATGuruNY » Fri May 17, 2013 4:32 am
If x, y and k are positive numbers such that 10*x/(x+y)+20*y/(x+y)=k and if x<y, which of the following could be the value of k?
A. 10
B. 12
C. 15
D. 18
E. 30
Putting the sum over a common denominator, we get:
(10x + 20y) / (x+y) = k.

Let x = the number of $10 shirts purchased at a certain store.
Let y = the number of $20 shirts purchased at a certain store.
Total cost of the $10 shirts = 10x.
Total cost of the $20 shirts = 20y.
Total number of shirts purchased = x+y.
Thus, the AVERAGE cost per shirt is equal to the following:

(10x + 20y) / (x+y).

In the problem above, the value of k is equal to the AVERAGE cost per shirt.
Since each shirt costs either $10 or $20, the average cost per shirt must be BETWEEN 10 and 20.
Since y>x, the number of $20 shirts purchased is GREATER than the number of $10 shirts purchased, with the result that the average cost per shirt must be CLOSER TO 20 than to 10.
Of the answer choices, the only viable option is k=18.

The correct answer is D.

An alternate approach is to plug in the answers, which represent the value of k.
Let x=1.
When we plug in the correct answer choice for k, x<y.

Answer choice C: k=15
(10*1 + 20y)/(1+y) = 15
10 + 20y = 15 + 15y
5y = 5
y = 1.
Since x=y, eliminate C.

Answer choice D: k=18
(10*1 + 20y)/(1+y) = 18
10 + 20y = 18 + 18y
2y = 8
y = 4.
Since x<y, success!

The correct answer is D.
Last edited by GMATGuruNY on Sat May 18, 2013 2:04 am, edited 1 time in total.
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by Atekihcan » Fri May 17, 2013 5:19 am
k = 10*x/(x+y)+20*y/(x+y) = (10x + 20y)/(x + y)

This means k is nothing but the weighted average of 10 and 20, x and y being the weights, respectively. As x < y, the weight or contribution of 10 in the average is less than that of 20. So, k must be closer to 20 than 10.

Only possible correct answer is D.


Now, if you can't identify/understand this logic, you can proceed as follows also...
k = (10x + 20y)/(x + y) = [10(x + y) + 10y]/(x + y) = 10 + (10y)/(x + y)

Now, y > x
So, 2y > (x + y)
So, y/(x + y) > 1/2
So, 10y/(x + y) > 10/2
So, 10y/(x + y) > 5

And, y < (x + y)

So, 2y > (x + y)
So, y/(x + y) < 1
So, 10y/(x + y) < 10

So, 5 < 10y/(x + y) < 10
So, 10 + 5 < 10 + 10y/(x + y) < 10 + 10
So, 15 < k < 20

Answer : D

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by Brent@GMATPrepNow » Fri May 17, 2013 5:31 am
As you can see, Mitch's and Atekihcan's awesome (i.e., fast) solutions both use weighted averages.
If you're interested, we have a free video on that subject: https://www.gmatprepnow.com/module/gmat- ... ics?id=805

Here are some additional practice questions related to weighted averages:
- https://www.beatthegmat.com/weighted-ave ... 17237.html
- https://www.beatthegmat.com/weighted-ave ... 14506.html
- https://www.beatthegmat.com/average-weig ... 57853.html
- https://www.beatthegmat.com/averages-que ... 87118.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by kuzzden » Sat May 18, 2013 9:46 pm
Now I see the logic, thank you very much!
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