If −1/3 ≤ x ≤−1/5 and -1/2 ≤ y ≤ −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
OA: D
My Sol: x will be positive, hence I choose x=-1/5 and y= -1/4 , hence got ans as A
HI Experts,
Can you pls explain this que where am I going wrong?
Thanks
Nandish
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If −1/3 ≤ x ≤−1/5 and -1/2 ≤ y ≤ −1/4
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Calculate using EVERY COMBINATION OF ENDPOINTS:NandishSS wrote:If −1/3 ≤ x ≤−1/5 and -1/2 ≤ y ≤ −1/4, what is the least possible value of x²y?
A. −1/100
B. −1/50
C. −1/36
D. −1/18
E. −1/6
(-1/3)²(-1/2) = -1/18.
(-1/3)²(-1/4) = -1/36.
(-1/5)²(-1/2) = -1/50.
(-1/5)²(-1/4) = -1/100.
The results above indicate the following:
The smallest possible product is -1/18.
The greatest possible product is -1/100.
Thus:
-1/18 ≤ x²y ≤ -1/100.
The correct answer is D.
On the number line:
-1......-1/2.....-1/18.....-1/100.....0
Since -1/18 is farther from 0 than is -1/100, -1/18 is LESS than -1/100.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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