# Manhattan GMAT Challenge Problem of the Week – 29 June 2010

*by*Manhattan Prep, Jun 29, 2010

This is our latest Challenge Problem! As always, the problem and solution below were written by one of our fantastic instructors. Each challenge problem represents a 700+ level question. If you are up for the challenge, however, set your timer for 2 mins and go! (And don't forget to answer the Showdown for a chance to win our Strategy Guides.)

## Question

Ifv0, is |w| < |v|?(1)

w/v< 1(2) [pmath]w^2/v^2 < 1[/pmath]

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

## Answer

In words, we are asked whether the absolute value of *w* is less than the absolute value of *v*. We could rephrase this geometrically: is *w *closer to 0 on the number line than *v *is? However, we should also retain the original phrasing, which uses concise algebra. (By the way, the condition that *v *does not equal 0 is only there to prevent division by 0 in the statements.)

Statement (1): INSUFFICIENT. We should not cross-multiply this statement, since we dont know whether *v* is positive or negative. That is, we cannot claim that *w *< *v* results from *w*/*v* < 1, since we cant multiply both sides of an inequality by a variable and preserve the inequality sign as is, unless we know that the variable is positive.

Lets instead plug a couple of quick pairs of numbers, trying for a Yes case and a No case. (The Yes case would be one in which |*w*| is in fact less than |*v*|, and the No case would be one in which |*w*| is not less than |*v*|.) Remember to try negative numbers!

Yes case: *w* = 2, *v *= 3. 2/3 is indeed less than 1, satisfying the statement.

As for the question, we get |2| < |3|, giving us a Yes answer to the question.

No case: *w* = -5, *v* = 3. -5/3 is indeed less than 1, satisfying the statement.

But |-5| is NOT less than |3|, so we would answer the question No.

Since we have both a Yes case and a No case, this statement is insufficient.

Statement (2): SUFFICIENT. Here, we CAN cross-multiply, since [pmath]v^2[/pmath] is definitely positive. Thus, we can rephrase the statement as follows:

[pmath]w^2/v^2 < 1[/pmath]

[pmath]w^2 < v^2[/pmath]

Now, since both sides are positive (or at least not negative), we can take the positive square root of both sides. The positive square root of a square is in fact the absolute value, so we get

|*w*| < |*v*|

If you like, test numbers that satisfy [pmath]w^2 < v^2[/pmath]. In every case, you will find that |*w*| < |*v*|.

**The correct answer is (B).**

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