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If −2 < x ≤ 1/2 and 7< y ≤ 10......

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by M7MBA » Sat Apr 28, 2018 1:48 am

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Answers

A

B

C

D

E

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Difficulty—

If −2 < x ≤ 1/2 and 7< y ≤ 10, what is the least value of y*x^2 possible?

A. 2
B. 7/4
C. 0
D. −8
E. −14

The OA is C.

Shouldn't be -14 the correct answer? Can anyone explain it to me? Thanks in advance.
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by GMATGuruNY » Sat Apr 28, 2018 2:51 am
M7MBA wrote:If −2 < x ≤ 1/2 and 7< y ≤ 10, what is the least value of y*x^2 possible?

A. 2
B. 7/4
C. 0
D. −8
E. −14
Since y is POSITIVE and x² must be NONNEGATIVE, only two cases are possible:
Case 1: yx² = (positive)(0) = 0.
Case 2: yx² = (positive)(positive) = positive.

Case 1 is possible if y=8 and x=0.
Thus, the least possible value for yx² is 0.

The correct answer is C.
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by Vincen » Sat Apr 28, 2018 5:03 am
Hello gmat_mission.

This is how I'd solve it:

Since -2 < x ≤ 1/2 then $$0\ \le\ x^2\ <4.$$ On the other hand 7< y ≤ 10, therefore when $$y\cdot x^2\ \ge0\ always\ \ \Rightarrow\ \ The\ least\ value\ is\ 0.$$ Hence, the correct answer is the option C.

I hope it helps.
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by Jeff@TargetTestPrep » Wed May 02, 2018 9:33 am
M7MBA wrote:If −2 < x ≤ 1/2 and 7< y ≤ 10, what is the least value of y*x^2 possible?

A. 2
B. 7/4
C. 0
D. −8
E. −14
Since x^2 is nonnegative and we are given that y is positive, the smallest value of y*x^2 is when x^2 is can be made as small as possible. We see that the range of values of x includes 0; thus, we can make x = 0 and hence x^2 = 0. Thus, the smallest possible value of y*x^2 is 0.

Answer: C

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