Manhattan GMAT Challenge Problem of the Week – 8 October 2012

by on October 8th, 2012

Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people that enter our challenge, the better the prizes!


If x, y, and z lie between 0 and 1 on the number line, with no two variables equal, then the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables is between

A. 0 and 1/3
B. 1/3 and 2/3
C. 2/3 and 1
D. 1 and 5/3
E. 5/3 and 7/3


First, decode the language describing the expression you care about. “The product of all three variables” is xyz. Meanwhile, “the sum of all the distinct products of exactly two of the three variables” is xy + yz + xz. So the expression is xyz/(xy + yz + xz).

At this point, there are two paths forward: doing algebra and picking smart numbers. The latter may turn out to be faster in this case, because the expression is tough to manipulate on its own (we’ll show how in a minute). Say that x = ¼, y = ½, and z = ¾ (all three are supposed to be different). Then the product of all three is 3/32, while xy + yz + xz = (¼)(½) + (½)(¾) + (¼)(¾) = 1/8 + 3/8 + 3/16 = 2/16 + 6/16 + 3/16 = 11/16. So the value you ultimately want equals 3/32 divided by 11/16, or 3/32 times 16/11, which equals 3/22. 3/22 is definitely between 0 and 1/3. Now you know that the question must be well-formed—in other words, there can’t be two right answers. For any legal values of x, y, and z chosen according to the criteria (between 0 and 1, and no two values alike), you must get the same answer to the question. So you only need to check one set of values. The answer is (A).

If you’re interested, here’s one way to manipulate the expression to prove that it must lie between 0 and 1/3. Say that xyz/(xy + yz + xz) = S, to give the expression a variable name. You can’t split denominators that are sums, but you can split numerators. So take the reciprocal of both sides:

(xy + yz + xz)/xyz = 1/S

Now you can split the numerator, making three different fractions. Canceling common terms in each, you get this:

1/z + 1/x + 1/y = 1/S

Since each variable x, y, and z is less than 1 (but still positive), the reciprocal of each of those variables is greater than 1. So the sum of those three reciprocals is greater than 1+1+1=3. Since 1/S is greater than 3, S must be less than 1/3. Meanwhile, S is evidently positive (all the variables are positive and nothing’s being subtracted), so the proper boundaries are 0 < S < 1/3. The two algebra tricks to learn here are (1) naming a whole expression as a new variable, so that you can transform both sides of the equation, and (2) taking the reciprocal of both sides of an equation.

The correct answer is A.

Special Announcement: If you want to win prizes for answering our Challenge Problems, try entering our Challenge Problem Showdown. Each week, we draw a winner from all the correct answers. The winner receives a number of our our Strategy Guides. The more people enter, the better the prize. Provided the winner gives consent, we will post his or her name on our Facebook page.

Ask a Question or Leave a Reply

The author Manhattan_Prep gets email notifications for all questions or replies to this post.

Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a Gravatar to have your pictures show up by your comment.