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# if wx=y, what is the value of xy? 1) wx^2 = 16 2) y = 4

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if wx=y, what is the value of xy? 1) wx^2 = 16 2) y = 4 [#permalink]  05 Mar 2009, 16:14
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if wx=y, what is the value of xy?

1) wx^2 = 16
2) y = 4
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Re: DS: exponents [#permalink]  05 Mar 2009, 16:49
vksunder wrote:
if wx=y, what is the value of xy?

1) wx^2 = 16
2) y = 4

IMHO A

wx = y, value of xy? --> value of wx^2?

1) xy = wx^2 = 16 --> sufficient

2) y = 4 --> wx = 4, don't know the value of w --> can't find value of x and thus of xy --> insufficient
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Re: DS: exponents [#permalink]  05 Mar 2009, 22:04
vksunder wrote:
if wx=y, what is the value of xy?

1) wx^2 = 16
2) y = 4

When you see this type of questions, you have to be aware that most of the time you will not be able to solve x, y individually. So try to see if you can form the term (xy) grouped together with the given conditions.
For example, if i tell you 5xy = 10, can you solve for xy? That's the basic idea.

Back to the problem.
1) How can you form (xy) from wx=y and wx^2 = 16. Well, you see can get rid of w by dividing the two equations. Why not just try that. You'll get wx/(w x^2) = y/16. Then 1/x = y/16. Remember you're trying to form the grouped term (xy). So move the 2 variables to the same side and see what you get. 1/(xy) = 1/16.Hence xy = 16.

2) y = 4. Assume x = 2 or x =1, you'll get different answers.
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Re: DS: exponents [#permalink]  11 Mar 2009, 07:05
Thanks for the great input guys!

Well, here is how I approached the stat 1:

wx^2 = 16
Take square root of both sides
sqrt(wx^2) = sqrt (16 )

sqrt(w) x = 4

And then I had no clue how to proceed. Is it even possible to determine the answer following the above route?

Thanks!
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Re: DS: exponents [#permalink]  11 Mar 2009, 08:18
vksunder wrote:
if wx=y, what is the value of xy?

1) wx^2 = 16
2) y = 4

1. wx^2=16
x(wx)=16
wx=16/x => y=16/x => 16=xy
sufficient

2. y=wx=4
w &x could be the same or different digits
insufficient

Re: DS: exponents   [#permalink] 11 Mar 2009, 08:18
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