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If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value

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If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value  [#permalink]

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New post 13 Mar 2018, 01:49
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D
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Question Stats:

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If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value  [#permalink]

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New post 13 Mar 2018, 04:07
Bunuel wrote:
If \(-\frac{1}{2} \leq x \leq -\frac{1}{3}\) and \(-\frac{1}{4} \leq y \leq -\frac{1}{5}\), what is the minimum value of \(x*y^2\)?


A. \(-\frac{1}{75}\)

B. \(-\frac{1}{50}\)

C. \(-\frac{1}{48}\)

D. \(-\frac{1}{32}\)

E. \(-\frac{1}{16}\)


For Minimum Value of x*y^2

1) The absolute value of x*y^2 must be Maximum and
2) Sign of x*y^2 must be negative


@x = -1/2 (Max absolute value), y = -1/4 (Max absolute value)

x*y^2 = -1/32

Answer: option D
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Re: If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value  [#permalink]

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New post 14 Mar 2018, 16:10
Bunuel wrote:
If \(-\frac{1}{2} \leq x \leq -\frac{1}{3}\) and \(-\frac{1}{4} \leq y \leq -\frac{1}{5}\), what is the minimum value of \(x*y^2\)?


A. \(-\frac{1}{75}\)

B. \(-\frac{1}{50}\)

C. \(-\frac{1}{48}\)

D. \(-\frac{1}{32}\)

E. \(-\frac{1}{16}\)



We want the value of x to be as small as possible and the value of y^2 (which is positive) to be as large as possible.

The smallest value of x is -1/2, and the largest value of y^2 is (-1/4)^2 = 1/16.

Thus, the minimum value of x * y^2 is -1/2 x (-1/4)^2 = -1/2 x 1/16 = -1/32.

Answer: D
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Re: If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value  [#permalink]

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New post 16 Mar 2018, 13:24
ScottTargetTestPrep wrote:
Bunuel wrote:
If \(-\frac{1}{2} \leq x \leq -\frac{1}{3}\) and \(-\frac{1}{4} \leq y \leq -\frac{1}{5}\), what is the minimum value of \(x*y^2\)?


A. \(-\frac{1}{75}\)

B. \(-\frac{1}{50}\)

C. \(-\frac{1}{48}\)

D. \(-\frac{1}{32}\)

E. \(-\frac{1}{16}\)



We want the value of x to be as small as possible and the value of y^2 (which is positive) to be as large as possible.

The smallest value of x is -1/2, and the largest value of y^2 is (-1/4)^2 = 1/16.

Thus, the minimum value of x * y^2 is -1/2 x (-1/4)^2 = -1/2 x 1/16 = -1/32.

Answer: D



why are we taking the max value of y^2 and not the lowest value ?
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Re: If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value  [#permalink]

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New post 17 Mar 2018, 07:34
1
ishpreetanand wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
If \(-\frac{1}{2} \leq x \leq -\frac{1}{3}\) and \(-\frac{1}{4} \leq y \leq -\frac{1}{5}\), what is the minimum value of \(x*y^2\)?


A. \(-\frac{1}{75}\)

B. \(-\frac{1}{50}\)

C. \(-\frac{1}{48}\)

D. \(-\frac{1}{32}\)

E. \(-\frac{1}{16}\)



We want the value of x to be as small as possible and the value of y^2 (which is positive) to be as large as possible.

The smallest value of x is -1/2, and the largest value of y^2 is (-1/4)^2 = 1/16.

Thus, the minimum value of x * y^2 is -1/2 x (-1/4)^2 = -1/2 x 1/16 = -1/32.

Answer: D



why are we taking the max value of y^2 and not the lowest value ?


Hi Ispreet

Sharing my thoughts ....

On the number line larger the magnitude or larger the absolute value of a -ve number SMALLER is the number
=>Eg -10 is smaller than -5 i.e -10<-5 BUT Mod of -10 is larger than -5 i.e |-10|>|-5|
=>Eg \(\frac{-1}{2}=-0.5\) < \(\frac{-1}{3}=-0.3\) BUT |\(\frac{-1}{2}|=0.5\) > |\(\frac{-1}{3}|=0.3\)

In the above Ques in discussion we have to find the MINIMUM Value of (\(x*y^2\))
=> Now (\(x*y^2\)) will be -ve (since 'x' is -ve & \(y^2\) is +ve) MINIMUM Value of (\(x*y^2\)) can be obtained WHEN |\(x*y^2\)| is MAXIMUM
=> Since 'x' is -ve we take the MINIMUM value of 'x' i.e \(x=\frac{-1}{2}\) AND
=> Since \(y^2\) is +ve we take the MAXIMUM value of \(y^2\) i.e \(y^2=(\frac{-1}{4})*(\frac{-1}{4})=\frac{1}{16}\)
=> The above combination gives MAXIMUM |\(x*y^2\)| i.e MINIMUM (\(x*y^2\))

Regards
Dinesh
GMAT Club Bot
Re: If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value   [#permalink] 17 Mar 2018, 07:34
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