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sk818020
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If xy=1, what is the value of 2(x+y)^2/2(x-y)^2. ( basically 2(x+y)sq/2(x-y)sq.
1. 2
2. 4
3. 8
4. 16
5. 32
I, like the person who replied first, think there is missing information in this problem. If x and y can be any integer or non-integer, then x and y can be any possible set of non-integers. For example, if x = 4 and y = .25, then:
4 x .25 = 1
If x = .25 and y = 4, then:
.25 x 4 = 1
Other possiblities would include, but are not limited to (also include the vice versa of each example), x = 5 and y = .2, x = 10 and y = .1, x = 20 and y = .05 etc...
All of these examples satisfy the first part of the conditional statement, yet yield varying results for the second part of the conditional statment. In fact, to even begin to answer the question you would have to determine two things; if x and y aree integers and if they are positive or negative because:
1 x 1 =1
and
-1 x -1 = 1
1. 2
2. 4
3. 8
4. 16
5. 32
I, like the person who replied first, think there is missing information in this problem. If x and y can be any integer or non-integer, then x and y can be any possible set of non-integers. For example, if x = 4 and y = .25, then:
4 x .25 = 1
If x = .25 and y = 4, then:
.25 x 4 = 1
Other possiblities would include, but are not limited to (also include the vice versa of each example), x = 5 and y = .2, x = 10 and y = .1, x = 20 and y = .05 etc...
All of these examples satisfy the first part of the conditional statement, yet yield varying results for the second part of the conditional statment. In fact, to even begin to answer the question you would have to determine two things; if x and y aree integers and if they are positive or negative because:
1 x 1 =1
and
-1 x -1 = 1
Are you sure that there was no limes involved at some stage?ajas wrote:Actually no. There was no other infomartion and the answer was in absolute value! This was one of the questions in a practice Gmat Prep test from mba.com.
'cause if you start playing with the expression you will get that the constant that you are looking for can be brought to the following:
c = 1 + (2 / (x - 1/x))^2
where the latter tends to 0 if x approaches +- infinity or 0.
Bests,
J
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Stuart@KaplanGMAT
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The question is definitely off.ajas wrote:If xy=1, what is the value of 2(x+y)^2/2(x-y)^2. ( basically 2(x+y)sq/2(x-y)sq.
1. 2
2. 4
3. 8
4. 16
5. 32
Pls let me know..
Thanks!
First, we can cancel out the 2 on top and bottom of the fraction to get:
(x+y)^2/(x-y)^2
simplifying from there:
(x^2 + 2xy + y^2)/(x^2 -2xy + y^2)
= (x^2 + y^2 + 2)/(x^2 + y^2 - 2)
and at this point the solution is "well, it depends on the values of x and y", i.e. there is no numerical solution.
Are you sure the question isn't:
2^(x+y)^2/2^(x-y)^2?
Solving that, we have:
2^(x^2 + 2xy + y^2) / 2^(x^2 -2xy + y^2)
2^((x^2 + 2xy + y^2)-(x^2 -2xy + y^2))
2^(4xy)
(xy=1)
2^4
16... choose (D)!
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