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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
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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 of the form of x^2 -(by)^2, where b is an integer?

The number of ways we can select 2 out of 4 expressions are 4c2= 6 ways

These are 6 total outcomes possible.
(x+y) (x+5y)
(x+y) (x-y )
(x+y ) (5x-y)
(x+5y) (x-y)
(x+5y) (5x-y)
(x-y) (5x-y)

Out of this outcomes, we should look for the product in form of x^2 - (by)^2 , which is same as (x+by)(x - by), where b is an integer.

We can see that only 1 out of 6 is in the form of (x+by)(x - by) i.e (x+y) (x-y )

So the probability = 1/6
Option E is the answer.

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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
GMATNinja, can you please clarify why the following approach is not correct?
According to probability definition, probability=total # of desirable outcomes/total # of possible outcomes.
In this problem, total # of desirable outcomes=2 expressions. 2 expressions from the list can be used to produce the following outcome- x^2 - (by)^2 ; total # of possible outcomes=4 expressions--> Probability that their product will be of the form of x^2 - (by)^2, where b is an integer = 2/4=1/2
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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
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tkorzhan1995 wrote:
GMATNinja, can you please clarify why the following approach is not correct?
According to probability definition, probability=total # of desirable outcomes/total # of possible outcomes.
In this problem, total # of desirable outcomes=2 expressions. 2 expressions from the list can be used to produce the following outcome- x^2 - (by)^2 ; total # of possible outcomes=4 expressions--> Probability that their product will be of the form of x^2 - (by)^2, where b is an integer = 2/4=1/2


Hi tkorzhan1995,

Your definition of probability is mathematically correct, but you have to follow the instructions in the prompt to make sure that you are correctly applying that formula.

We're told to take the product of 2 of the four given 'expressions.' When multiplying two terms together, it does not matter which one is first - meaning that there are 6 possible pairings of the four expressions (you could list them all out or use the Combination Formula --> 4c2).

From those 6 possible pairings, there is just 1 that is a match for the 'format' defined in the prompt, thus the probability is 1/6.

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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
Bunuel wrote:
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 of 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

Four expressions can be paired in \(C^2_4=6\) ways (choosing 2 out of 4);

\(x^2 - (by)^2=(x-by)(x+by)\) and only one pair is making such expression: \((x+1*y)(x-1*y)\), hence probability is 1/6.

Answer: E.



Dear Bunuel, thank you for the explanation but I have one doubt-

I got answer as A - 1/3

We want : (x + by) ( x - by)

We have x + y , x - y, x + 5y , 5x - y
While (x-y)(x+y) is obvious, can we say that --> x + 5y = x - y + 6y = x-y(1-6) = x - by form?
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If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
Expert Reply
NCC wrote:
Bunuel wrote:
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 of 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

Four expressions can be paired in \(C^2_4=6\) ways (choosing 2 out of 4);

\(x^2 - (by)^2=(x-by)(x+by)\) and only one pair is making such expression: \((x+1*y)(x-1*y)\), hence probability is 1/6.

Answer: E.



Dear Bunuel, thank you for the explanation but I have one doubt-

I got answer as A - 1/3

We want : (x + by) ( x - by)

We have x + y , x - y, x + 5y , 5x - y
While (x-y)(x+y) is obvious, can we say that --> x + 5y = x - y + 6y = x-y(1-6) = x - by form?


If I got you right, you mean that if we write x + 5y as x - (-5)y, then it when multiplied by 5x-y will also be of the form of (x-by)(x+by). But if one of the expression is x - (-5)y, then, the product to be of the form of (x-by)(x+by), the second one must be x + (-5)y = x - 5y, while we have 5x-y.

Does this make sense?
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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
Hi Bunuel or any other expert.

Quote:
It should be 1/4*1/3*2, since you can choose (x+y) first and then (x-y) or vise versa.


Why did you multiply the above the 2? I think I am missing a concept here. Could you please help explain or refer to relatable links for guidance?
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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
Vegita wrote:
Hi Bunuel or any other expert.

Quote:
It should be 1/4*1/3*2, since you can choose (x+y) first and then (x-y) or vise versa.


Why did you multiply the above the 2? I think I am missing a concept here. Could you please help explain or refer to relatable links for guidance?

See this explanation https://gmatclub.com/forum/if-two-of-th ... 92727.html
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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
Quote:
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 of 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


1. The six possible products are:
(x+y)(x+5y) = x^2 + 6xy +5y^2 = doesn't work
(x+y)(x-y) = x^2 - y^2 = x^2 - (1y)^2 = WORKS!
(x+y)(5x-y) = 5x^2 +4xy - y^2 = doesn't work
(x+5y)(x-y) = x^2 + 4xy - 5y^2 = doesn't work
(x+5y)(5x-y) = 5x^2 +24xy -5y^2 = doesn't work
(x-y)(5x-y) = 5x^2 -6xy + y^2 = doesn't work
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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
Asked: 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 of the form of x^2 - (by)^2, where b is an integer?

(x+y)(x-y) = xˆ2 - yˆ2 : Of the form x^2 - (by)^2, where b is an integer
(x+y)(5x-y) = 5xˆ2 +4xy - yˆ2 : NOT Of the form x^2 - (by)^2, where b is an integer
(x+y)(x+5y) = xˆ2 + 6xy + 5yˆ2: NOT Of the form x^2 - (by)^2, where b is an integer
(x-y)(5x-y) = 5xˆ2 - 6xy +yˆ2: NOT Of the form x^2 - (by)^2, where b is an integer
(x-y)(x+5y) = xˆ2 + 4xy - 5yˆ2: NOT Of the form x^2 - (by)^2, where b is an integer
(x+5y)(5x-y)= 5xˆ2 + 9xy - 5yˆ2: NOT Of the form x^2 - (by)^2, where b is an integer

The probability that their product will be of the form of x^2 - (by)^2 = 1/6

IMO E
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Re: If two of the four expressions x + y, x + 5y, x - y, and 5x - y are [#permalink]
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