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Gmat Prep 2

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Gmat Prep 2

by abhinav85 » Wed Jun 24, 2009 7:54 pm
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random.What is the probability that their product will be the form of
x^2-(by)^2,where b is an integer?

A.1/2

B.1/3

C.1/4

D.1/5

E.1/6.

E
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by ssmiles08 » Wed Jun 24, 2009 8:24 pm
The way I looked at it was there are only two out of the 4 that make x^2-(by)^2: x+y and x-y

here b = 1

the rest would cause x to have an additional value adjacent to it (5x-y) or it would not produce the form x^2 - (by)^2: x+5y (ex. if you take this with any other number for example, x+y you would get the form x^2 + 6xy + 5y^2)

so that eliminates 2 of the 4 possibilities.

we are left with x+y and x-y as our picking options.

So probability of picking the 2 out of 4 possible numbers is 2/4 = 1/2 for our first pick.

now we only have 3 numbers, and only one number we want. so 1/3

1/2*1/3 = 1/6 (E)
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by Robinmrtha » Wed Jun 24, 2009 8:32 pm
only x+y and x- y satisfy...
So there is only one pair that can satisfy the form...
Total no of pairs that can be formed...
=4!/(4-2)!2!
=6
Therefore the required probability = 1/6
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by abhinav85 » Thu Jun 25, 2009 4:19 am
Hey ssmiles

Can u elaborate this, i did'nt get this???

now we only have 3 numbers, and only one number we want. so 1/3

1/2*1/3 = 1/6 (E)
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clarification pls

by deltaforce » Thu Jun 25, 2009 4:34 am
since there are six elements, i went this way - but obviously the answer is wrong...can someone clarify why

4 elements and only two possible can yield the solution

so 2/4 * 1/3 = 1/2 * 1/3 = 1/6

i went a step further, now we can pick (x+y) (x-y) first or (x-y)(x+y)first, that produces two options

1/6 * 2 = 1/3

what is wrong with my reasoning ...thanks
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by ssmiles08 » Thu Jun 25, 2009 4:34 am
abhinav85 wrote:Hey ssmiles

Can u elaborate this, i did'nt get this???

now we only have 3 numbers, and only one number we want. so 1/3

1/2*1/3 = 1/6 (E)
Sure.

We need x+y and x-y for the product to work out in that given form.

so 2/4 options for the first pick. = 1/2

say we pick (x+y) in our first pick and we need to pick (x-y) to complete our product.

in our remaining pool, for our second pick we have: x+5y,x-y and 5x-y

so only 3 remain and we only need 1 of those 3.

So the probability of picking (x-y) is 1/3

so 1/2*1/3 = 1/6

-hope that helps.
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by ghacker » Thu Jun 25, 2009 8:53 am
Very easy answer is 1/6
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by m&m » Thu Jun 25, 2009 9:31 am
Question is poorly worded, given the answer the question should have asked what is probability of the product of any terms being X^2 - by^2.

Question currently asks, ASSUMING or GIVEN that 2 numbers are picked what is the probability that their product is of the above form.

which is 4/4*1/3 = 1/3

if you KNOW that 2 numbers are picked with certainty then you can't multiply by the probability of picking 2 numbers

If you do a table for every term picked there is exactly one other term that will yield the desired form?

Am I missing something?
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