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 500,000 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:

If x, y, and k are positive numbers such that [#permalink]

Show Tags

15 Oct 2009, 16:08

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

64% (00:57) correct
36% (02:07) wrong based on 144 sessions

HideShow timer Statistics

If x, y, and k are positive numbers such that \((\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k\) and if x < y, which of the following could be the value of k?

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 \((\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k\) and if x < y, which of the following could be the value of k?

A. 10 B. 12 C. 15 D. 18 E. 30

Couldn't find a way to solve it.

Ans D. the above expression can be written as [10(x+y)/(x+y)] + 10y/(x+y) = 10 + 10y/(x+y) now we know y>x then to simplify we can take values for x and y such that they sum up to 10 and we are left with y. If x=2 and y=8 we get 18. Other options will not satisfy.

We can pick nos. for this and solve it. Let x=1 & y=4, solve the equation answer is 18, Option D Let x=2 & y=3,solve the equation, answer is 16, no option available. Let x=1 & y=9, solve the equation, answer is 19, no option available.

If x, y, and k are positive numbers such that \((\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k\) and if x < y, which of the following could be the value of k?

A. 10 B. 12 C. 15 D. 18 E. 30

Given that: \(x < y\) \((\frac{x}{x+y})< (\frac{y}{x+y})< 1\)

Also given that: \((\frac{x}{x+y})(10)+(\frac{y}{x+y})(20) = k\) \((10)(\frac{x+2y}{x+y}) = k\) \((10)(\frac{x+y+y}{x+y}) = k\) \((10)(1+\frac{y}{x+y}) = k\)

The value of the above expression range from 10 to 20 but with the given fact, it cannot be 10, 12 and 30 because all are out of scope. Now its between 15 and 18 however 15 is also out for being the average of 10 and 20. That would be true only if x and y were equal. That leaves out 18 as the only choice as answer..
_________________

PussInBoots's Solution is also good enough. If you this x boxes that weight 10lbs y boxes that weight 20lbs, and x<y, then the number is obviously greater than the average of 10 and 20 (if there were same number of boxes). Since 20lb boxes are more in number, the average weight must be between 15-20 lbs. Less than 1 min solution