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16 Sep 2014, 01:25
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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54% (01:39) correct
46%
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If \(-\frac{1}{3} \le x \le -\frac{1}{5}\) and \(-\frac{1}{2} \le y \le -\frac{1}{4}\), what is the smallest possible value of \(x^2y\)?
A. \(-\frac{1}{100}\)
B. \(-\frac{1}{50}\)
C. \(-\frac{1}{36}\)
D. \(-\frac{1}{18}\)
E. \(-\frac{1}{6}\)
A. \(-\frac{1}{100}\)
B. \(-\frac{1}{50}\)
C. \(-\frac{1}{36}\)
D. \(-\frac{1}{18}\)
E. \(-\frac{1}{6}\)
21 Mar 2020, 06:43
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NandishSS
Hello, NandishSS, and thank you for tagging me. A little number sense goes a long way on this one. In fact, I did nothing more than a little mental math and arrived at the answer in 38 seconds. (I am not a numbers genius. I just apply common sense and core knowledge of mathematical principles, the type of reasoning that is often tested on the GMAT™.) My approach:
Bunuel
Since we are looking for the least possible value for \(x^2*y\), we have to deal with a few matters, conceptually:
1) Both x and y must be negative
2) Both x and y must be fractions
3) The answer must be negative (since the square will produce a positive value, which we then need to multiply by a negative value)
4) The larger the number after a negative sign, the smaller the value
5) Fractions with a larger denominator than numerator tend to get smaller when multiplied. For example,
\(\frac{1}{6} * \frac{1}{6} = \frac{1}{36}\)
With these statements in mind, we can approach the extrema of the given variables one by one to produce the least value. Start with
\(x^2\)
It would not make sense to test any values other than the extrema, since they would necessarily fall in between the uppermost and least values. Test the two extrema:
\((-\frac{1}{3})^2 = \frac{1}{9}\)
\((-\frac{1}{5})^2 = \frac{1}{25}\)
Clearly, one-ninth is greater than one-twenty-fifth, so hang onto the larger value. Why do we want the large one, you might wonder? Because we are going to multiply it by a negative number, and again, the larger the value of a negative number, the further from 0 it is, and the smaller its value becomes. Thus, you want the largest positive times the negative that is about to follow to yield the smallest product. Now repeat the process by multiplying our earlier square by an extreme value of y:
\(\frac{1}{9} * -\frac{1}{2} = -\frac{1}{18}\)
\(\frac{1}{9} * -\frac{1}{4} = -\frac{1}{36}\)
Since negative one-thirty-sixth is closer to 0 than negative one-eighteenth (or if it helps you to picture as negative two-thirty-sixths), the latter is the lesser value and is thus our answer. Choice (D) it is. I hope that helps. If you have further questions, feel free to ask.
- Andrew
16 Sep 2014, 01:25
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Official Solution:
If \(-\frac{1}{3} \le x \le -\frac{1}{5}\) and \(-\frac{1}{2} \le y \le -\frac{1}{4}\), what is the smallest possible value of \(x^2y\)?
A. \(-\frac{1}{100}\)
B. \(-\frac{1}{50}\)
C. \(-\frac{1}{36}\)
D. \(-\frac{1}{18}\)
E. \(-\frac{1}{6}\)
To find the smallest value of \(x^2y\), which will be negative, try to maximize the absolute value of \(x^2y\). The larger the absolute value of a negative number, the smaller it is.
To maximize \(|x^2y|\), choose the largest absolute values possible for \(x\) and \(y\): \((-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}\). Notice that: \(-\frac{1}{18} < -\frac{1}{36} < -\frac{1}{50} < -\frac{1}{100}\), so \(-\frac{1}{100}\) is the largest number and \(-\frac{1}{18}\) is the smallest number (we cannot obtain \(-\frac{1}{6}\) from \(x^2y\) or else it would be the correct answer).
Answer: D
If \(-\frac{1}{3} \le x \le -\frac{1}{5}\) and \(-\frac{1}{2} \le y \le -\frac{1}{4}\), what is the smallest possible value of \(x^2y\)?
A. \(-\frac{1}{100}\)
B. \(-\frac{1}{50}\)
C. \(-\frac{1}{36}\)
D. \(-\frac{1}{18}\)
E. \(-\frac{1}{6}\)
To find the smallest value of \(x^2y\), which will be negative, try to maximize the absolute value of \(x^2y\). The larger the absolute value of a negative number, the smaller it is.
To maximize \(|x^2y|\), choose the largest absolute values possible for \(x\) and \(y\): \((-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}\). Notice that: \(-\frac{1}{18} < -\frac{1}{36} < -\frac{1}{50} < -\frac{1}{100}\), so \(-\frac{1}{100}\) is the largest number and \(-\frac{1}{18}\) is the smallest number (we cannot obtain \(-\frac{1}{6}\) from \(x^2y\) or else it would be the correct answer).
Answer: D
