Probability and Diff of Sq

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i_have_no_cool_username: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Sat Oct 22, 2011 9:01 am

by i_have_no_cool_username » Sat Nov 19, 2011 5:37 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 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

Answer is E. Why?

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Source: Beat The GMAT — Problem Solving | Sat Nov 19, 2011 5:37 pm

neelgandham: Community Manager; Posts: 1060; Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members; Location: Utrecht, The Netherlands; Thanked: 318 times

by neelgandham » Sat Nov 19, 2011 6:36 pm

i_have_no_cool_username 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?

Hi i_have_no_cool_username(:D It indeed is a cool name)

Welcome to the forum. Before I answer the question, I would like to request you to conceal the answers of the questions posted.
How ? Look here -> https://www.beatthegmat.com/new-spoilers-t5302.html. Now the solution...

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, their product will be of the form of x^2-(by)^2 only if the coefficient of x in both the expressions is 1 and if the coefficient of y in both the expressions is equal and is of different signs. x-y,x+y is only such pair.

Total number of pairs = 3+2+1 = 6
Total number of pairs that satisfy the condition = 1
Probability = 1/6

Anil Gandham
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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Sat Nov 19, 2011 7:19 pm

i_have_no_cool_username 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

Answer is E. Why?

Solution:
(x + y)(x + 5y) = x^2 + 6xy + 5y^2: Not in the form x^2- (by)^2
(x + y)(5x - y) = 5x^2 + 4xy - y^2: Not in the form
(x + 5y)(x - y) = x^2 + 4xy - 5y^2: Not in the form
(x + 5y)(5x - y) = 5x^2 + 24xy - 5y^2: Not in the form
(x - y)(5x - y) = 5x^2 - 6xy + y^2: Not in the form
(x + y)(x - y) = x^2 - y^2: This is in the required form

Hence, the required probability is 1/6.

Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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i_have_no_cool_username: Junior | Next Rank: 30 Posts; Posts: 20; Joined: Sat Oct 22, 2011 9:01 am

by i_have_no_cool_username » Sat Nov 19, 2011 9:03 pm

neelgandham wrote:
i_have_no_cool_username 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?
Hi i_have_no_cool_username(:D It indeed is a cool name)

Welcome to the forum. Before I answer the question, I would like to request you to conceal the answers of the questions posted.
How ? Look here -> https://www.beatthegmat.com/new-spoilers-t5302.html. Now the solution...

If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, their product will be of the form of x^2-(by)^2 only if the coefficient of x in both the expressions is 1 and if the coefficient of y in both the expressions is equal and is of different signs. x-y,x+y is only such pair.

Total number of pairs = 3+2+1 = 6
Total number of pairs that satisfy the condition = 1
Probability = 1/6

Request noted, Neelgandham! If I'm still posting questions after my test on Monday (highly unlikely), I will keep that request in mind

Thanks for the explanation, didn't quite get the "3+2+1=6" part, but after reading Anurag@Gurome's breakdown, it all makes sense now.

Thank you both!

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