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The greatest possible value of

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Carcass
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The greatest possible value of [#permalink] New post   Updated on: 03 Nov 2018, 22:14
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Quantity A
Quantity B
The greatest possible value of \(\frac{2}{x-y}\) where x and y are integers and \(9 \leq x \leq 12\) and \(-2 \leq y \leq 8\)
2


A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.
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C

Originally posted by Carcass on 07 Dec 2017, 03:06.
Last edited by chetan2u on 03 Nov 2018, 22:14, edited 1 time in total.
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wongpcla
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Re: The greatest possible value of [#permalink] New post   08 Dec 2017, 00:06
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the answer is C?
that x could be 9, y could be 8
2/1=2 ?
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Re: The greatest possible value of [#permalink] New post   08 Dec 2017, 02:17
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OA added

regards
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GREMasterBlaster
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Re: The greatest possible value of [#permalink] New post   17 Feb 2018, 01:32
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Maximum value of Quantity A will be obtained if we put least possible value of x-y in denominator. That will be minimum value of x and maximum of y.
Hence, Choice C is correct.
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Re: The greatest possible value of [#permalink] New post   17 Feb 2018, 04:04
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Yes, I have C as well. Because of the negative in x-y, plug in the least value of x (=9) with the greatest value of y (=8).
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Re: The greatest possible value of [#permalink] New post   03 Nov 2018, 17:30
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I'm sorry to revive an old topic but I disagree with the answer.
What if X = 8.5 and Y = 8?
The difference would be 1/2 = 0.5, which would duplicate the value of A, making it bigger than B.
Is there something Im missing?
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chetan2u
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Re: The greatest possible value of [#permalink] New post   03 Nov 2018, 22:12
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RCDMC wrote:
I'm sorry to revive an old topic but I disagree with the answer.
What if X = 8.5 and Y = 8?
The difference would be 1/2 = 0.5, which would duplicate the value of A, making it bigger than B.
Is there something Im missing?


Yes you are correct.
A could be actually infinity in the present state as it is NOT given that x and y are integers..
Carcass wrote:
Quantity A
Quantity B
The greatest possible value of \(\frac{2}{x-y}\) where \(9 \leq x \leq 12\) and \(-2 \leq y \leq 8\)
2



say x is 9 and y is 8.999999, \(\frac{2}{x-y}=\frac{2}{9-8.999999}=\frac{2}{0.000001}=2*10^6\), way bigger than B

So editing to make it correct..
Now when we take them as integers we take minimum value of x as 9 and maximum value of y as 8
\(A=\frac{2}{x-y}=\frac{2}{9-8}=\frac{2}{1}=2=B\)
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Re: The greatest possible value of [#permalink] New post   20 Dec 2018, 15:25
chetan2u wrote:
RCDMC wrote:
I'm sorry to revive an old topic but I disagree with the answer.
What if X = 8.5 and Y = 8?
The difference would be 1/2 = 0.5, which would duplicate the value of A, making it bigger than B.
Is there something Im missing?


Yes you are correct.
A could be actually infinity in the present state as it is NOT given that x and y are integers..
Carcass wrote:
Quantity A
Quantity B
The greatest possible value of \(\frac{2}{x-y}\) where \(9 \leq x \leq 12\) and \(-2 \leq y \leq 8\)
2



say x is 9 and y is 8.999999, \(\frac{2}{x-y}=\frac{2}{9-8.999999}=\frac{2}{0.000001}=2*10^6\), way bigger than B

So editing to make it correct..
Now when we take them as integers we take minimum value of x as 9 and maximum value of y as 8
\(A=\frac{2}{x-y}=\frac{2}{9-8}=\frac{2}{1}=2=B\)


Please check that −2≤y≤8 means max value for y is 8 not even 8.000.......001, let alone 8.9999999, and
again for x, min vale is 9 because 9≤x≤12.

As a result, 2/(9-8)=2/1=2.
Answer is C.
_______________________________________________________________
Please let me know if I am wrong.
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chetan2u
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Re: The greatest possible value of [#permalink] New post   20 Dec 2018, 18:04
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AE wrote:
chetan2u wrote:
RCDMC wrote:
I'm sorry to revive an old topic but I disagree with the answer.
What if X = 8.5 and Y = 8?
The difference would be 1/2 = 0.5, which would duplicate the value of A, making it bigger than B.
Is there something Im missing?


Yes you are correct.
A could be actually infinity in the present state as it is NOT given that x and y are integers..
Carcass wrote:
Quantity A
Quantity B
The greatest possible value of \(\frac{2}{x-y}\) where \(9 \leq x \leq 12\) and \(-2 \leq y \leq 8\)
2



say x is 9 and y is 8.999999, \(\frac{2}{x-y}=\frac{2}{9-8.999999}=\frac{2}{0.000001}=2*10^6\), way bigger than B

So editing to make it correct..
Now when we take them as integers we take minimum value of x as 9 and maximum value of y as 8
\(A=\frac{2}{x-y}=\frac{2}{9-8}=\frac{2}{1}=2=B\)


Please check that −2≤y≤8 means max value for y is 8 not even 8.000.......001, let alone 8.9999999, and
again for x, min vale is 9 because 9≤x≤12.

As a result, 2/(9-8)=2/1=2.
Answer is C.
_______________________________________________________________
Please let me know if I am wrong.


Yes , that is correct,
For some reason I took −2≤y≤8 as −2≤y≤9..
If it is −2≤y≤8, we have to take minimum value of x-y to get maximum of 2/(x-y)... Ans least value of x-y is 1 when x is 9 and y is 8
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Re: The greatest possible value of [#permalink] New post   21 Dec 2018, 10:25
C
if we try all the border cases we can't seem to get x-y less than 1 in order to get A to its highest value with the lowest denominator
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Re: The greatest possible value of [#permalink] New post   05 Apr 2023, 09:05
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