Bunuel wrote:
Official Solution:
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 don't know whether v is positive or negative. That is, we cannot claim that \(w \lt v\) results from \(\frac{w}{v} \lt 1\), since we can't multiply both sides of an inequality by a variable and preserve the inequality sign as is, unless we know that the variable is positive.
Let's 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: \(w = 2\), \(v = 3\). \(\frac{2}{3}\) is indeed less than 1, satisfying the statement. As for the question, we get \(|2| \lt |3|\), giving us a "Yes" answer to the question.
No case: \(w = -5\), \(v = 3\). \(-\frac{5}{3}\) 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 \(v^2\) is definitely positive. Thus, we can rephrase the statement as follows:
\(\frac{w^2}{v^2} \lt 1\)
\(w^2 \lt v^2\)
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
\(|w| \lt |v|\)
If you like, test numbers that satisfy \(w^2 \lt v^2\). In every case, you will find that \(|w| \lt |v|\).
Answer: B
Hi Bunuel,
When |w| < |v| given, can't we square both the inequalities. ( as we do |w| = |v| , we can square both to get rid of the modulus sign).
Is there any way to get rid of modulus for < or > sign in between inequalities ?
Please let us know.