M26-18

I think this is a high-quality question and I don't agree with the explanation. Isn't -1/100 lesser in value than -1/18? And -1/100 should be the answer as it could be obtained by using x=-1/5 and y=-1/4

No. -1/100 > -1/18.

I think this is a high-quality question and I agree with explanation. High quality question.

Using the approach of number ranges to solve the question. Two points to be noted when using range approach.

- both multiplying ranges should either be +ve or -ve
- when multiplying one range with a negative number, inequality signs are reversed

Given,

-1/3 < x < -1/5
--> 1/9 > x2 > 1/25

-1/2 < y < -1/4
--> 1/2 > -y > 1/4

Now since the range of both x2 and -y are positive we can multiply the ranges 1/9 > x2 > 1/25 and 1/2 > -y > 1/4
--> 1/18 > -x2y > 1/100

Multiply by -1 and change signs
--> -1/18 < x2y < -1/100

Hence the smallest value of x2y is -1/18


------ Crave Your Rave -------

I rephrased this question:
-3>=x>=-5 and -2>=y>=-4
What is the greatest value possible?

Obviously, it is -18

HI GMATGuruNY, MentorTutoring

How to deal with these kinds of problems? I was very confused.

Calculate \(x^2*y\) using every combination endpoints.

\(x=-\frac{1}{3}\) and \(y=-\frac{1}{2}\) --> \((-\frac{1}{3})^2 * -\frac{1}{2} = -\frac{1}{18}\)

\(x=-\frac{1}{3}\) and \(y=-\frac{1}{4}\) --> \((-\frac{1}{3})^2 * -\frac{1}{4} = -\frac{1}{36}\)

\(x=-\frac{1}{5}\) and \(y=-\frac{1}{2}\) --> \((-\frac{1}{5})^2 * -\frac{1}{2} = -\frac{1}{50}\)

\(x=-\frac{1}{5}\) and \(y=-\frac{1}{4}\) --> \((-\frac{1}{5})^2 * -\frac{1}{4} = -\frac{1}{100}\)

Of the four results, the least \(= -\frac{1}{18}\)

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:


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

I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I think this is a high-quality question and I agree with explanation.