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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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New post 03 Jul 2016, 10:21
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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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New post 03 Jul 2016, 10:42
4
3
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
Answer B
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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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New post 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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New post 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.
Answer is B
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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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New post 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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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the  [#permalink]

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

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New post 24 Jun 2018, 05:12
Abhishek009 niks18 pushpitkc gmatbusters Skywalker18

I did not understand below quote:
Quote:
max (1/x - x/y)

the value of the above expression will be maximum when value of x is least and y is maximum

How do I break the question stem knowing that x and y are positive integers?

1.What can I say about 1/x
2. What can I say about x/y
What can I say about difference between (1) and (2)
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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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New post 24 Jun 2018, 05:27
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adkikani wrote:
Abhishek009 niks18 pushpitkc gmatbusters Skywalker18

I did not understand below quote:
Quote:
max (1/x - x/y)

the value of the above expression will be maximum when value of x is least and y is maximum

How do I break the question stem knowing that x and y are positive integers?

1.What can I say about 1/x
2. What can I say about x/y
What can I say about difference between (1) and (2)


Hi adkikani

From the given ranges of x & y we know that both are positive. Now we need Maximum value of 1/x-x/y.

There are two parts to the equation. we have 1/x and then you are subtracting x/y from 1/x. so essentially the value of the resulting equation will be less than 1/x. But the question asks us to maximize 1/x-x/y, so 1/x has to attain the maximum value possible.

1/x will be maximum when x has least value because in that case its reciprocal i.e 1/x will be max. For eg. if you are given x=3 & x=4, then out of these two values max 1/x will be when x=3 because 1/3=0.333 & 1/4=0.25. So now you know that we need the least value of x for the first part.

Coming to y, note that in x/y, y is in denominator and we are subtracting this value from 1/x. now how can we make the impact of subtraction as low as possible. again as explained above if y is maximum then 1/y will be minimum. so to make x/y minimum we need maximum y.

Hence x=3 & 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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New post 24 Jun 2018, 07:53
Thanks niks18 for your two cents. I had an additional query.


Quote:
There are two parts to the equation. we have 1/x and then you are subtracting x/y from 1/x. so essentially the value of the resulting equation will be less than 1/x. But the question asks us to maximize 1/x-x/y, so 1/x has to attain the maximum value possible.
1/x will be maximum when x has least value because in that case its reciprocal i.e 1/x will be max. For eg. if you are given x=3 & x=4, then out of these two values max 1/x will be when x=3 because 1/3=0.333 & 1/4=0.25. So now you know that we need the least value of x for the first part.


The highlighted text is an universal fact which holds true even when 1/x reduces to an integer.
E.g. x = 1 (although I do realize that with additional constraints in Q stem, x can not take value of 1,
but can you validate my reasoning.)
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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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New post 24 Jun 2018, 20:59
adkikani wrote:
Thanks niks18 for your two cents. I had an additional query.


Quote:
There are two parts to the equation. we have 1/x and then you are subtracting x/y from 1/x. so essentially the value of the resulting equation will be less than 1/x. But the question asks us to maximize 1/x-x/y, so 1/x has to attain the maximum value possible.
1/x will be maximum when x has least value because in that case its reciprocal i.e 1/x will be max. For eg. if you are given x=3 & x=4, then out of these two values max 1/x will be when x=3 because 1/3=0.333 & 1/4=0.25. So now you know that we need the least value of x for the first part.


The highlighted text is an universal fact which holds true even when 1/x reduces to an integer.
E.g. x = 1 (although I do realize that with additional constraints in Q stem, x can not take value of 1,
but can you validate my reasoning.)


Hi adkikani

I am not sure why are you calling the highlighted statement as a "Universal fact"? can you clarify more on your query?
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New post 28 Jun 2018, 05:05
Hi niks18

By universal fact, I meant: if basic conditions are satisfied, the rule holds true for any value of a variable.
Eg. If I am given that x raised to any unknown power yields an even integer, I know for sure:
x is even.

Similarly highlighted text would hold true if (1)the ratio yielded a decimal (x: any value other than 1) or
(2) an integer (I would have to take x to be 1, just to make sure I understood the logic of subtraction clearly).

I wanted to ask you if highlighted text holds true for both conditions above.
In short, for A - B, I need to maximize A and minimize B irrespective of A and B being decimals or integers.
Let me know if we are on same page.
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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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New post 28 Jun 2018, 08:53
1
adkikani wrote:
Hi niks18

By universal fact, I meant: if basic conditions are satisfied, the rule holds true for any value of a variable.
Eg. If I am given that x raised to any unknown power yields an even integer, I know for sure:
x is even.

Similarly highlighted text would hold true if (1)the ratio yielded a decimal (x: any value other than 1) or
(2) an integer (I would have to take x to be 1, just to make sure I understood the logic of subtraction clearly).

I wanted to ask you if highlighted text holds true for both conditions above.
In short, for A - B, I need to maximize A and minimize B irrespective of A and B being decimals or integers.
Let me know if we are on same page.


Hi adkikani

if we need to maximize A-B, then irrespective of nature of A & B, we will have to maximize A and minimize B. If this is your query, then we are on the same page ;)
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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the &nbs [#permalink] 28 Jun 2018, 08:53
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