# Try this DS Problem from Veritas Prep’s New GMAT Question Bank

by on October 18th, 2012

Earlier this month we launched our new GMAT Question Bank. This new resource contains hundreds of realistic, completely free GMAT practice questions.

The reaction so far has been extremely positive, but the one question that keeps coming up is, “What’s the catch? When will you start charging for this?” (Okay, that’s two questions.) The answers are:

• There is no catch.
• We currently have no plans to charge for the GMAT Question Bank.

We created this tool and opened it up to everyone so that we can collect data on our questions. We will use the data we collect to measure and refine our questions, which will then go into new generations of our GMAT practice tests. In effect, by answering these questions, you are helping our system learn about the questions — which ones are easy, which ones are hard, which ones are confusing and need to be refined, etc. The system is also learning about each user.  It’s an iterative process that helps it measure users by seeing how they did on certain questions, and it assesses those questions by seeing how well certain users performed on those questions.

The Veritas Prep GMAT Question Bank not only contains hundreds of GMAT problems, but it also contains a complete solution for each problem. You can come back and log in at any time to review individual questions and track your progress against that of other users. And best of all, you can try out hundreds of new practice questions, all designed to keep up with the always-evolving GMAT and to provide students with practice on the topics that give them the most trouble.  For example, try this problem, which tests factors, multiples, and “must be true logic” in the GMAT’s tricky way, from our bank:

For integers x and y, if , which of the following must be true?

I.  y > x
II. is an integer
III. The cube root of x is an integer

A) I only
B) II only
C) III only
D) I and II
E) II and III

Solution: B.  Statement 1 is not necessarily true. x could equal -8 and y could equal -91, for example, in which case the equation holds but to equal , then the prime factorization: . y must then be able to account for the prime factor of 7 on the other side of the equation. And statement 3 is not necessarily true. While x must account for the factors , it could also include a non-cubed factor as well. For example, x could be and y could be . The equation would hold because the extra 5 is accounted for on both sides. x MUST account for , but need not be limited to just that, as x and y could have duplicate factors on either side. Beware the statements that look very likely to be true when you face these “must be true” problems – the GMAT is a master of misdirection.

The number of questions in the system will vary over time as the system validates questions. Once the system deems a question good enough to be in one or our 15 GMAT practice tests, it may disappear from the Question Bank and only be available to Veritas Prep GMAT students. But, rest assured that you can come back and view your past results at any time, even if some of the questions you previously answered have “graduated” from the system and have been added to our students-only practice tests.

Finally, right now you will see the five question types that are in the computer-adaptive parts of the GMAT: Critical Reasoning, Sentence Correction, Reading Comprehension, Data Sufficiency, and Problem Solving. We will also add Integrated Reasoning questions shortly… so stay tuned!

What are you waiting for? Visit the Veritas Prep GMAT Question Bank, register, and get started!

• I didn't understood the solution provided by you for statement II and III.

• Let's tackle these one at a time. First off, the fact that 91x = 8y means that the left hand side of the equation has to be EXACTLY the same as the right hand side. and since we know that the left hand side includes all the factors of 91, in order to be equal the right hand side must include all of those, too.

This might be a little easier to conceptualize if you use smaller numbers. If 3a = 2b (and a and b are integers), then we know that since the right hand side is even (it's 2 times an integer), then the left hand side must be, too. So we'd know that a must be even.

Well, the same is true of the problem as given - since the left hand side is known to be divisible by 91 (which is 7 * 13), then the right hand side must also be divisible by 7 and 13. Which means that for statement 2, it must be true. 8 on the right hand side is not, itself, divisible by 7, so that factor of 7 must come from y. And this means that y/7 must be an integer.

• For statement 3, similar logic to statement 2 applies, but there's a little bit of a trap embedded there. x must be divisible by 8, since 91 itself is not, but we know that since the right hand side is divisible by 8, the left hand side must be too. However - we don't know that it's EXACTLY 8. It could be 16, for example, in which case we'd be able to "balance" the equation by making y divisible by 2. This would still hold true:

91 * (16) = 8 * (91*2)

So III "could be" true but it doesn't have to be.

• exp for II: condition given in the question is that x and y are integers. so "y" can be an integer only if "x" is a multiple of 8. therefore, y/7 = 13x/8. since we have shown that x is a multiple of 8, y/7 will be an integer.
exp for III: cub root x = 2xcub root.(y/91). x can be an integer only if y is multiple of 91. if y is 91, then cube root of x is integer. if y is 182, then cube root of x is not an integer. so statement III cannot be true.

i hope the explanation helps.