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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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13 Mar 2018, 00:49
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47% (01:42) correct 53% (01:30) wrong based on 109 sessions

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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. $$-\frac{1}{16}$$

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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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13 Mar 2018, 03:07
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

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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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14 Mar 2018, 15:10
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.

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Posts: 13
Re: If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value  [#permalink]

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16 Mar 2018, 12:24
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.

why are we taking the max value of y^2 and not the lowest value ?
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Joined: 27 Apr 2015
Posts: 41
GMAT 1: 370 Q29 V13
Re: If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value  [#permalink]

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17 Mar 2018, 06:34
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.

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
Re: If -1/2 <= x <= -1/3 and -1/4 <= y <= -1/5, what is the minimum value &nbs [#permalink] 17 Mar 2018, 06:34
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