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Difficulty:
65%
(hard)
Question Stats:
46% (01:35) correct
54%
(01:46)
wrong
based on 48
sessions
History
Date
Time
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If \({-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}\) and \({-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6
OPEN DISCUSSION OF THIS QUESTION IS HERE: 12-easy-pieces-or-not-126366-20.html
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6
OPEN DISCUSSION OF THIS QUESTION IS HERE: 12-easy-pieces-or-not-126366-20.html
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09 Apr 2014, 01:40
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If \({-\frac{1}{3}}\leq{x}\leq{-\frac{1}{5}}\) and \({-\frac{1}{2}}\leq{y}\leq{-\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6
To get the least value of \(x^2*y\), which obviously will be negative, try to maximize absolute value of \(x^2*y\), as more is the absolute value of a negative number "more" negative it is (the smallest it is).
To maximize \(|x^2*y|\) pick largest absolute values possible for \(x\) and \(y\): \((-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}\). Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).
Answer: D.
OPEN DISCUSSION OF THIS QUESTION IS HERE: 12-easy-pieces-or-not-126366-20.html
A. -1/100
B. -1/50
C. -1/36
D. -1/18
E. -1/6
To get the least value of \(x^2*y\), which obviously will be negative, try to maximize absolute value of \(x^2*y\), as more is the absolute value of a negative number "more" negative it is (the smallest it is).
To maximize \(|x^2*y|\) pick largest absolute values possible for \(x\) and \(y\): \((-\frac{1}{3})^2*(-\frac{1}{2})=-\frac{1}{18}\). Notice that: -1/18<-1/36<-1/50<-1/100, so -1/100 is the largest number and -1/18 is the smallest number (we cannot obtain -1/6 from x^2*y or else it would be the correct answer).
Answer: D.
OPEN DISCUSSION OF THIS QUESTION IS HERE: 12-easy-pieces-or-not-126366-20.html
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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