Pls explain this Algebra PS

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by Uva@90 » Sat Oct 26, 2013 8:45 pm
[email protected] wrote:If x, y, and k are positive numbers such that x/(x+y)*10 + y/(x+y)*20 = 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

Ans-18
Hi Shibriz,
10*x/(x+y) + 20*y/x+y = k
The above equation can be re-written as

(10x+20Y)/x+y => 10(x+y+y)/x+y => 10(1 + y/(x+y))

Question says y>x,so it mean y/(x+y) will always have values greater than .5 and less than 1

so 10(1 + y/(x+y)) = > 10*(1+ {>.5 and < 1} )
Hence k should be >15 and <20
from options we have only 18
Answer is D

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Uva.
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by GMATGuruNY » Sat Oct 26, 2013 11:26 pm
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.
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by Brent@GMATPrepNow » Sun Oct 27, 2013 6:01 am
As you can see, Mitch's first solution uses weighted averages to quickly find the correct answer. If you're interested, we have a free video on eighted averages: 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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