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# If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the

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If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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03 Jul 2016, 10:21
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If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the maximum value of (1/x - x/y)?

A. $$-3\frac{1}{8}$$
B. 0
C. 1/5
D. 5/18
E. 2
[Reveal] Spoiler: OA

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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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03 Jul 2016, 10:42
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max (1/x - x/y)

the value of the above expression will be maximum when value of x is least and y is maximum
x=3 and y = 9 (Since x and y are integers and their ranges are given )
1/3 - 3/9
= 0
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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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03 Jul 2016, 22:56
B

x should be min and y max

so 1/3-3/9=0

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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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09 Jul 2016, 18:54
I tackled this question by looking to make the denominators the same.
So Y*(1/X) - X*(X/Y)=? >> (Y-X^2)/XY=?
Now looking at the numerator, I see that as X gets larger, the smaller the overall value of the numerator will be.
You're going to want the Y value to be the largest it can be since X^2 will be subtracted from it. This means that the value of X will need to be the smallest it can be.
Therefore, Y=9 and X=3 >> (9-(3)^2)/(9*3)= 0/27 or 0.

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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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16 Oct 2016, 11:46
why can't x take value as 2 and y as 8?

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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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16 Oct 2016, 11:49
warriorguy wrote:
why can't x take value as 2 and y as 8?

x cannot be 2 because 2 < x ≤ 8.

But y can be 8 (2 < y ≤ 9).
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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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16 Oct 2016, 11:51
Bunuel wrote:
warriorguy wrote:
why can't x take value as 2 and y as 8?

x cannot be 2 because 2 < x ≤ 8.

But y can be 8 (2 < y ≤ 9).

Thanks Bunuel

Got it.

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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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16 Oct 2016, 12:50
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Bunuel wrote:
If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the maximum value of (1/x - x/y)?

A. $$-3\frac{1}{8}$$
B. 0
C. 1/5
D. 5/18
E. 2

1/x - x/y = y-x^2/xy , x^2 is key , x^2 value in the range 9<=x<= 64 , min value of x^2 is 9 that is the max value y can reach , thus to max y-x^2 , 0 is the max

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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the [#permalink]

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09 Dec 2017, 15:17
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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the   [#permalink] 09 Dec 2017, 15:17
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