Monthly GMAT Challenge Problem Showdown: January 13, 2013


challenge problem
We invite you to test your GMAT knowledge for a chance to win! The second week of every month, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that month’s drawing for free Manhattan GMAT prep materials. Tell your friends to get out their scrap paper and start solving!

Here is this month’s problem:

If pq, and r are different positive integers such that p + q + r = 6, what is the value of x ?

(1) The average of xp and xq is xr.

(2) The average of xp and xr is not xq.

GMAT Challenge ProblemTo see the answer choices, and to submit your answer, visit our Challenge Problem Showdown page on our site.

Discuss this month’s problem with like-minded GMAT takers on our Facebook page.

The monthly winner, drawn from among all the correct submissions, will receive One Year of Access to our Challenge Problem Archive, AND the GMAT Navigator, AND Our Six Computer Adaptive Tests ($92 value).

  1. Bharadwaj March 27, 2016 at 3:14 am

    The answer is E
    There can be 3 possible equations based on the values of p,q and r. They fall in the set (1, 2, 3).
    Accordingly using statement 1,
    value of x can be -1/2 or 1 based on 1 equation.
    1 based on second equation
    -2 or 1 based on another equation
    Even if wetake statement2, we can eliminate any 1 of the above 3 equations which leaves us with at least 2 different values of ‘x’ in each case.
    Hence even by combining both statements, we cannot solve the value of ‘x’.
    So answer is E

  2. pradeep June 26, 2015 at 4:59 am


  3. akshay January 12, 2015 at 1:38 am


  4. rupali41 March 14, 2014 at 1:34 am

    I do not think option B helps in anyways because it is said that x, y and z are different integers. But it is true that X is 1. But since the X=1 does not suffice the statement, the answer should be c.
    Please correct me if wrong

  5. saurabh jha February 23, 2014 at 10:12 am