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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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03 Jul 2016, 11:21
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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
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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the
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03 Jul 2016, 11:42
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
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03 Jul 2016, 23:56
B
x should be min and y max
so 1/33/9=0



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Re: If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the
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09 Jul 2016, 19:54
I tackled this question by looking to make the denominators the same. So Y*(1/X)  X*(X/Y)=? >> (YX^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
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16 Oct 2016, 12:46
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
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16 Oct 2016, 12:49
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
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16 Oct 2016, 12:51
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
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16 Oct 2016, 13:50
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 = yx^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 yx^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
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24 Jun 2018, 06:12
Abhishek009 niks18 pushpitkc gmatbusters Skywalker18I 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
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24 Jun 2018, 06:27
adkikani wrote: Abhishek009 niks18 pushpitkc gmatbusters Skywalker18I 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 adkikaniFrom the given ranges of x & y we know that both are positive. Now we need Maximum value of 1/xx/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/xx/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
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24 Jun 2018, 08:53
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/xx/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
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24 Jun 2018, 21:59
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/xx/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 adkikaniI 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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If x and y are integers such that 2 < x ≤ 8 and 2 < y ≤ 9, what is the
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28 Jun 2018, 06:05
Hi niks18By 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
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28 Jun 2018, 09:53
adkikani wrote: Hi niks18By 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 adkikaniif we need to maximize AB, 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




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