Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If x, y, and k are positive [#permalink]
27 Feb 2012, 07:50

26

This post received KUDOS

Expert's post

6

This post was BOOKMARKED

BANON 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

\(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)

\(10*\frac{x+2y}{x+y}=k\)

\(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\)

Finally we get: \(10*(1+\frac{y}{x+y})=k\)

We know that \(x<y\)

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]
14 Apr 2012, 00:19

Bunuel, I didn't understand how you got the 15 and 20 at the extreme ends in the equation. Could you please explain? thanks! 15<10*(1+(y/x+y)<20 _________________

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
14 Apr 2012, 01:09

2

This post received KUDOS

Expert's post

mymbadreamz wrote:

Bunuel, I didn't understand how you got the 15 and 20 at the extreme ends in the equation. Could you please explain? thanks! 15<10*(1+(y/x+y)<20

Since \(x<y\) then \(\frac{y}{x+y}\) is more than \(0.5\) and less than \(1\): \(0.5<\frac{y}{x+y}<1\).

In this case for \(1+\frac{y}{x+y}\) is more than 1.5 and less than 2, so \(10*(1+\frac{y}{x+y})\) is more than 15 and less than 20: \(15<10*(1+\frac{y}{x+y})<20\).

Re: If x, y, and k are positive [#permalink]
14 May 2012, 09:35

Bunuel wrote:

BANON 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

\(10*\frac{x}{x+y}+20*\frac{y}{x+y}=k\)

\(10*\frac{x+2y}{x+y}=k\)

\(10*(\frac{x+y}{x+y}+\frac{y}{x+y})=k\)

Finally we get: \(10*(1+\frac{y}{x+y})=k\)

We know that \(x<y\)

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)

Very well solved by Bunuel, this is question no. 148 in OG 12 P.S, which has been solved in a very complicated manner in O.G.

(A) \(\frac{-10}{0}\) Out! x and y are non-zeroes. (B) \(\frac{x}{y}=\frac{-8}{-2}=\frac{4}{1}\) Out! x should be less than y. (C) \(\frac{-5}{-5}\) Out! x should not be equal to y. (D) \(\frac{-2}{-8}=\frac{1}{4}\) Bingo!

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
11 Dec 2012, 03:10

I use this approach: X=1 Y=2 1/3*10 + 2/3*20 10/3+40/3= 50/3--> 16...... Te only outcome greater than 15 is 18 (D). 30 is too much. am I correct? thanks

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
14 Dec 2012, 23:05

fabrizio1983 wrote:

I use this approach: X=1 Y=2 1/3*10 + 2/3*20 10/3+40/3= 50/3--> 16...... Te only outcome greater than 15 is 18 (D). 30 is too much. am I correct? thanks

The expression is all about recognizing that it represents "weighted averages" and that k could have a range of values. In this case, the range of values is 15 < k < 20. Perhaps this link would be helpful: Weighted Averages Pattern on the GMAT _________________

Re: Question to OG 12th Edition PS 148. [#permalink]
29 Jan 2013, 03:05

0haxx wrote:

Hey guys, this is my first post here. Please pardon me if I am doing anything wrong I am currently preparing for the test and I stumbled upon the task in the title. It is like \(\frac{x}{x+y} * 10 + \frac{y}{x+y} * 20 = k ; y>x\) What is a possible value of k? The OG has a very complicated and time consuming explanation for this and I hope that it can be solved more quickly. I also don't know if it is really that difficult.. Can anyone help?

Thank you in advance!

The above equation can be written as \(\frac{10x}{x+y} + \frac{20y}{x+y} = k\)

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
06 Sep 2013, 11:37

1

This post received KUDOS

Another very easy way to do this question is to understand that (x+y) is the total sum in a sample space, and treat this as a question of "Averages".. Now, applying the weighted average concept, 'k', i.e. the average should ideally be closer to 20 (between 10 and 20) - since we have just once choice that is between 15 and 20, the answer can easily be deduced to option (D), i.e. 18. _________________

Re: If x, y, and k are positive numbers such that (x/(x+y))(10) [#permalink]
12 Jan 2014, 08:24

1

This post received KUDOS

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? Solved by trail and error The denominator, x+y which is common to both the terms must be a factor of 10 and 20 to get a integer value. First minimum value of x+y which can be used is 2, and using 2 with the possible values of x and y ==> (0,2) or (1,1) is not possible as 0 is not a positive integer in the first possible set and x is equal to y in the second set. The next minimum value of x+y possible is 5. The possible values of x and y with sum of 5 are (2,3) and (1,4) By substituting (2,3) as (x,y) in the equation, we get k=16. This is not given in the answer options By substituting (1,4) as (x,y) in the equation, we get k=18. This is given in the answer option and hence the possible value of k with all conditions satisfied. So, option D

gmatclubot

Re: If x, y, and k are positive numbers such that (x/(x+y))(10)
[#permalink]
12 Jan 2014, 08:24

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...

I started running back in 2005. I finally conquered what seemed impossible. Not sure when I would be able to do full marathon, but this will do for now...