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If −1/3 ≤ x ≤−1/5 and -1/2 ≤ y ≤ −1/4

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If −1/3 ≤ x ≤−1/5 and -1/2 ≤ y ≤ −1/4

by NandishSS » Sat Feb 03, 2018 5:55 am
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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least value

by GMATGuruNY » Sat Feb 03, 2018 6:26 am
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
Calculate using EVERY COMBINATION OF ENDPOINTS:
(-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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