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If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]

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17 Jul 2010, 12:57

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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]

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16 Jan 2015, 00:16

Hello from the GMAT Club BumpBot!

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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.

Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]

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11 Apr 2016, 03:04

Hello from the GMAT Club BumpBot!

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]

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01 Apr 2017, 09:58

amitjash 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

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???

10 and 20 multiplied by fractions so sum has to be >10 and < 30 - A, E out. 15 possible ONLY when both the ratios are 1/2 -> not possible as x != y - C Out Greater number y is being multiplied by 20 so > 15 makes sense - B Out

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???

Let \(f(x) = \frac{10x}{x + y} + \frac{20y}{x+y}\) be a continuous function on \([0,y]\). Then \(f(0) = 20\) and \(f(y) = \frac{10y}{2y} + \frac{20y}{2y} = 15\). For any number between \(15\) and \(20\), there must be a number \(k\) in \((0,20)\) by Intermediate Value Theorem.

Only \(18\) from the choices is between \(15\) and \(20\). Therefore \(18\) is the correct answer.

The following is Intermediate Value Theorem. It simply says that for an intermediate value between \(y\) values at end points, there is another point between \(x\) values at end points that has the intermediate value as a function value.

Let \(f\) be a continuous function \([a,b]\) and \(f(a) < f(b)\), where \(a < b\). For any \(k\) in \((f(a),f(b))\), there must be \(c\) in \((a,b)\) such that \(f(c) = k\). In the case that \(f(a) > f(b)\), it is also valid.
_________________

Re: If x, y, and k are positive numbers such that [X/(X+Y)][10] [#permalink]

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02 Apr 2017, 04:56

amitjash 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

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???

hi i am happy to share my thoughts on this problem as kaplan says when we see which of the following in PS problems , we should go with D and E options means we should always check them first., well here is my shot see if we solve the above equation we come with this solution E.10x+20y=30x+30y ( taken k =30) now we simplify this we get -2x=y , as it is mentioned in stem that x,y or k are +ve numbers. so here x is -ve so not an answer. D. taken k=18 and after simplifying as above we get 4x=y here we can say x would be +ve as well as would be less than y ,hold on C. take k=15 we get x=y so wrong (as x must be less than Y) B. take k=12 now we get x=4y (here also x is not less than y) wrong A. take k=10 in this we get y=0 so wrong as like B and C reason .

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