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If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]
17 Jul 2010, 11:57
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Question Stats:
55% (02:24) correct
45% (02:31) wrong based on 162 sessions
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
I posted this question earlier but somehow it disappeared from the forum... I dont know why... Anyhow, I am posting again. Can someone come up with some solution???
Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)
\(0.5<\frac{y}{x+y}<1\)
So, \(15<10*(1+\frac{y}{x+y})<20\)
Only answer between \(15\) and \(20\) is \(18\).
Answer: D (18)
There can be another approach:
We have: \(\frac{10x+20y}{x+y}=k\), if you look at this equation you'll notice that it's a weighted average.
There are \(x\) red boxes and \(y\) blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?
k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.
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
The explantion seems to be time consuming. Help for easy process. _________________
Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)
\(0.5<\frac{y}{x+y}<1\)
So, \(15<10*(1+\frac{y}{x+y})<20\)
Only answer between \(15\) and \(20\) is \(18\).
Answer: D (18)
There can be another approach:
We have: \(\frac{10x+20y}{x+y}=k\), if you look at this equation you'll notice that it's a weighted average.
There are \(x\) red boxes and \(y\) blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?
k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.
Answer: D (18)
actually, you don't need to write all them down to find the answer is D. especially in your test you don't have much time, so just pick any number you want. In this case, just pick x as 1 and y as 2, so k would be around 18.333, which is near to 18. If you little bit hesitate then try different numbers, such x as 1 and y as 3. _________________
Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]
15 Jan 2015, 23:16
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]
16 Jan 2015, 17:02
Expert's post
Hi All,
You can take advantage of the answer choices and some math "logic" to get to the correct answer. Let's TEST THE ANSWERS....
We're told that X, Y and K are POSITIVE and that X < Y.
We're also told that 10X/(X+Y) + 20Y/(X+Y) = K. We're asked what COULD be the value of K, which means that there's more than one possible answer. Since the answers are NUMBERS, one of the them MUST be a possible answer.
We can manipulate the given equation into:
10X + 20Y = K(X+Y)
Since all of the variables are POSITIVE, K MUST be between 10 and 20. Here's the proof:
IF.....K=10, then the equation becomes... 10X + 20Y = 10X + 10Y 20Y = 10Y Since Y is positive, 20Y = 10Y is NOT possible. Eliminate Answer A
In that same way, K can't be 20 (or greater) because the end equation would be an impossibility.
With the remaining 3 answers, we can TEST the possibilities...
IF...K = 12, then the equation becomes... 10X + 20Y = 12X + 12Y 8Y = 2X 4Y = X In this scenario, X > Y which is the OPPOSITE of what we were told. This is NOT the answer. Eliminate B.
IF....K=15, then the equation becomes... 10X + 20Y = 15X + 15Y 5Y = 5X Y = X In this scenario, X = Y, which is NOT a match for what we were told. This is NOT the answer. Eliminate C.
IF....K=18, then the equation becomes.... 10X + 20Y = 18X + 18Y 2Y = 8X Y = 4X Here, X < Y, which IS a match for what we were told. This IS a POSSIBLE answer.
Hence \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\)
\(0.5<\frac{y}{x+y}<1\)
So, \(15<10*(1+\frac{y}{x+y})<20\)
Only answer between \(15\) and \(20\) is \(18\).
Answer: D (18)
There can be another approach:
We have: \(\frac{10x+20y}{x+y}=k\), if you look at this equation you'll notice that it's a weighted average.
There are \(x\) red boxes and \(y\) blue boxes. Red box weight is 10kg and blue box weight 20kg, what is the average weight of x red boxes and y blue boxes?
k represents the weighted average. As y>x, then the weighted average k, must be closer to 20 than to 10. 18 is the only choice satisfying this condition.
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