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.

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:

From the first equation of the setup we learn that either y = 0 or u = c. If y = 0, then yj = 0. If u = c then from the second equation j = 0 as c < k. Taking two equations combined, we will arrive to the conclusion that yj = 0.

From the first equation of the setup we learn that either y = 0 or u = c. If y = 0, then yj = 0. If u = c then from the second equation j = 0 as c < k. Taking two equations combined, we will arrive to the conclusion that yj = 0.

How is colored portion true ?? pls explain.

In this from first equation either j=0 or u=k and from second equation either y=0 or u=c now it is given c < k so u=k & u=c cannot be true at same time. Thts why if j=0 then u=c. else if y=0 then u=k.

How do we arrive at saying: if y(u-c)=0, then u = c? it's like saying if y(x) = 0, then x = 0; I don't think so. x (expression inside the braces) can be any value, but the equation (independent variables) reduce to zero.

Am i missing something? _________________

KUDOS me if you feel my contribution has helped you.

If y(u - c) = 0 and j(u - k) = 0, which of the following must be true, assuming c < k?

A. yj < 0 B. yj > 0 C. yj = 0 D. j = 0 E. y = 0

y(u - c) = 0 --> u=c or y=0; j(u - k) = 0 --> u=k or j=0;

Now, the first option (u=c and u=k) cannot be simultaneously correct for both equations because if it is, then it would mean that u=c=k, but we are given that c<k. So, only one can be correct so either y=0 or j=0, which makes yj = 0 is always true.