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If -2 < x < 1/2 and 7 < y 10, what

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If -2 < x < 1/2 and 7 < y 10, what  [#permalink]

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New post 30 Dec 2017, 21:21
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A
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C
D
E

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Question Stats:

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If \(-2 < {x} \le {\frac{1}{2}}\) and \(7 < {y} \le 10\), what is the least value of \(x^2*y\) possible?

A. \(2\)
B. \(\frac{7}{4}\)
C. \(0\)
D. \(-8\)
E. \(-14\)

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If -2 < x < 1/2 and 7 < y 10, what  [#permalink]

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New post 30 Dec 2017, 21:28
chetan2u wrote:
If \(-2 < {x} \le {\frac{1}{2}}\) and \(7 < {y} \le 10\), what is the least value of \(x^2*y\) possible?

A. \(2\)
B. \(\frac{7}{4}\)
C. \(0\)
D. \(-8\)
E. \(-14\)


\(x^2\) can be \(0\) or positive only & \(y\) is positive, so negative options are not possible

as we have to minimize \(x^2*y\) (which is either \(0\) or a positive function) so any value closer or equal to \(0\) will be minimum.

as \(-2 < {x} \le {\frac{1}{2}}\), so \(x=0\) will yield the minimum value of \(x^2*y\).

Answer: \(0\)

option C
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Re: If -2 < x < 1/2 and 7 < y 10, what  [#permalink]

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New post 31 Dec 2017, 00:32
chetan2u wrote:
If \(-2 < {x} \le {\frac{1}{2}}\) and \(7 < {y} \le 10\), what is the least value of \(x^2*y\) possible?

A. \(2\)
B. \(\frac{7}{4}\)
C. \(0\)
D. \(-8\)
E. \(-14\)


The complementary approach to the one provided above is to just use the answers (they are right there in front of us!)
This is an Alternative approach.

Let's start with the smallest possible value in the answers: -14.
Trying out different numbers for x and y we'll just SEE that all of our results are positive.
Therefore we can never get to -14 so (E) is elimianted.
Similarly, we can never get to -8 so (D) is elimianted.
The next smallest answer is 0 and we'll get to it if we choose x=0.
Therefore (C) is possible so it must be our answer.
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Re: If -2 < x < 1/2 and 7 < y 10, what  [#permalink]

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New post 31 Dec 2017, 08:05
chetan2u wrote:
If \(-2 < {x} \le {\frac{1}{2}}\) and \(7 < {y} \le 10\), what is the least value of \(x^2*y\) possible?

A. \(2\)
B. \(\frac{7}{4}\)
C. \(0\)
D. \(-8\)
E. \(-14\)



Irrespective of the range of x, the variable x^2 will always non-negative. If we take into consideration the value of x, which is between -2 and 1/2, we can very easily conclude that the least value of x^2 can be 0, when x = 0.

Since we need to find the least value of x^2 * y, we can easily say it has to be zero, since 0 * anything is always 0.

The correct answer is Option C.
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If -2 < x < 1/2 and 7 < y 10, what  [#permalink]

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New post 31 Dec 2017, 08:49
chetan2u wrote:
If \(-2 < {x} \le {\frac{1}{2}}\) and \(7 < {y} \le 10\), what is the least value of \(x^2*y\) possible?

A. \(2\)
B. \(\frac{7}{4}\)
C. \(0\)
D. \(-8\)
E. \(-14\)



From the asked question, we can conclude that whatever be the value of "x", it can't be negative as x has even exponent.

"y" has a positive range. Therefore eliminate (D) & (E) as answer will always be positive or zero.

Least value of x will be "0" as all the other values (negative or positive) will give positive result for x^2 .

Take any value for "y" from the specified range,

Take x = 0 , y = 8

Least value of x^2 * y will always be "0"

(C)
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If -2 < x < 1/2 and 7 < y 10, what &nbs [#permalink] 31 Dec 2017, 08:49
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