If two of the four expressions x+y,x+5y,x-y and 5x-y ar

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varun289: Master | Next Rank: 500 Posts; Posts: 298; Joined: Sun Jun 03, 2012 6:42 amFollowed by: 7 members; Location: New delhi; Thanked: 10 times; GMAT Score:590

by varun289 » Sat Dec 15, 2012 4:08 am

260. 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 x^2-(by)^2, where b is an integer?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

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Source: Beat The GMAT — Data Sufficiency | Sat Dec 15, 2012 4:08 am

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by Anurag@Gurome » Sat Dec 15, 2012 4:33 am

varun289 wrote:260. 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 x^2-(by)^2, where b is an integer?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

(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 [spoiler]1/6[/spoiler].

The correct answer is E.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Dec 15, 2012 7:14 am

varun289 wrote:260. 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 x^2-(by)^2, where b is an integer?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Okay, first recognize that x^2 - (by)^2 is a difference of squares.
Here are some examples of differences of squares:
x^2 - 25y^2
4x^2 - 9y^2
49m^2 - 100k^2
etc.

In general, we can factor differences of squares as follows:
a^2 - b^2 = (a-b)(a+b)

So . . .
x^2 - 25y^2 = (x+5y)(x-5y)
4x^2 - 9y^2 = (2x+3y)(2x-3y)
49m^2 - 100k^2 = (7m+10k)(7m-10k)

So, from the 4 expressions (x+y, x+5y ,x-y and 5x-y), only one pair (x+y and x-y) will result in a difference of squares when multiplied.

So, the question now becomes:
If 2 expressions are randomly selected from the 4 expressions, what is the probability that x+y and x-y are both selected?

P(both selected) = [# of outcomes in which x+y and x-y are both selected]/[total # of outcomes]

As always, we'll begin with the denominator.

total # of outcomes
There are 4 expressions, and we must select 2 of them.
Since the order of the selected expressions does not matter, we can use combinations to answer this.
We can select 2 expressions from 4 expressions in 4C2 ways (= 6 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 4C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

# of outcomes in which x+y and x-y are both selected
There is only 1 way to select both x+y and x-y

So, P(both selected) = 1/6 = E

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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ritind: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Wed Nov 21, 2012 10:40 pm; Thanked: 4 times

by ritind » Wed Dec 19, 2012 12:50 am

we know that
(a+b)(a-b) = a^2 - b^2

Looking at the question there's only one pair that satisfies this eqn
(x+y)(x-y) = x^2 - y^2

Nw the question is how many total outcomes are possible
Answer is 4C2 = 6
It is like selecting 2 items from 4 objects where the order doesn't matter
Eg: (x+y)(x-y) = (x-y)(x+y)

Probability = 1/6

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paml: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Feb 05, 2015 7:14 am; Thanked: 1 times

by paml » Sun Apr 19, 2015 7:15 pm

Does anyone know of any similar problems to this one, preferably from an Official Guide, that we can practice with? Thanks!

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by Brent@GMATPrepNow » Sun Apr 19, 2015 8:03 pm

paml wrote:Does anyone know of any similar problems to this one, preferably from an Official Guide, that we can practice with? Thanks!

It depends what you mean by "similar"
This question involves probability, counting, and algebraic concepts like factoring a quadratic expression and expanding/simplifying products. It's unlikely to find a question will require all of these components.
To find practice questions related to a particular topic, you can use BTG's tagging feature. For example, here are all of the questions tagged as probability questions: https://www.beatthegmat.com/forums/tags/ ... robability
See the left side of that linked page for more tag options.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Sun Apr 19, 2015 11:38 pm

paml wrote:Does anyone know of any similar problems to this one, preferably from an Official Guide, that we can practice with? Thanks!

Suppose n is a randomly selected positive divisor of (12â�´ - 11â�´). What is the probability that n is divisible by 5?

A:: 1/4
B:: 1/3
C:: 3/8
D:: 3/7
E:: 1/2

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paml: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Feb 05, 2015 7:14 am; Thanked: 1 times

by paml » Mon Apr 20, 2015 3:54 pm

Thanks Brent! I mentioned this in another thread as well, but as FYI - my instructor advised that I look for questions that "mirror" questions that I got wrong, in order to start recognizing patterns. That's why I was looking for similar-looking questions, not just DS probability questions in general. I've been having a hard time finding questions that mirror the questions I got wrong, which is why I tried posting in this forum to ask around.

Thank you Matt! Is the answer to your question A?

(12^2+11^2)(12^2-11^2)
=(265)(12+11)(12-11)
=5(53)(23)(1)

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by DavidG@VeritasPrep » Mon Apr 20, 2015 4:42 pm

You're on the right track, Pam. But note that once you break the larger number down, you can assemble factors in a variety of ways. For example, 5*23, or 115, would be a factor; 5*53, or 265 would be another factor; and 5*23*53 would be a factor of itself, etc. Think in terms of # desired/# total. How many total factors are there for 5*53*23? How many of those factors are divisible by 5? (And Kudos to Matt for a clever question!)

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paml: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Thu Feb 05, 2015 7:14 am; Thanked: 1 times

by paml » Mon Apr 20, 2015 4:58 pm

That's helpful, thanks David! Is the answer 1/2?
The divisors are 1, 5, 23, 53, 5*23, 5*53, 5*23*53, 23*53.
4/8 probability.

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by DavidG@VeritasPrep » Mon Apr 20, 2015 5:04 pm

Nicely done

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by Scott@TargetTestPrep » Tue May 12, 2015 5:54 am

varun289 wrote:260. 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 x^2-(by)^2, where b is an integer?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Solution:

First, notice that we are being tested on the difference of squares. We can restate the problem as: What is the probability, when selecting two expressions at random, that the product of those expressions will create a difference of two squares? Remember, the difference of two squares can be written as follows:

a^2 - b^2 = (a + b)(a - b)

So x^2 - (by)^2 can be written as (x + by)(x - by)

Thus, we are looking for two expressions in the form of (x + by)(x - by). Although this problem is attempting to trick us with the expressions provided, the only two expressions that, when multiplied together, will give us a difference of squares are x + y and x - y. When we multiply x + y and x - y the result is x^2 - y^2 or x^2 - (1y)^2

We see that there is just one favorable product, namely (x + y)(x - y). In order to determine the probability of this event, we must determine the total number of possible products. Since we have a total of four expressions and we are selecting two of them to form a product, we have 4C2, which is calculated as follows:

4C2 = 4!/[(4-2)! x 2!] = 4!/(2! x 2!) = (4x3x2x1)/(2x1x2x1) = 24/4 = 6 products

Of these 6 products, we have already determined that only one will be of the form x^2 - (by)^2. Therefore, the probability is 1/6.

The answer is E.

Note: If you don't know how to use the combination formula, here is a method that will work equally well.

We are choosing 2 expressions from a pool of 4 possible expressions. That is, there are 2 decisions being made:

Decision 1: Choosing the first expression
Decision 2: Choosing the second expression

Four different expressions are available to be the first decision.

For the second decision, 3 remaining expressions are available because 1 expression was already chosen. We multiply these two numbers: 4 x 3 = 12.

The final step is to divide by the factorial of the number of decisions because the order in which we multiply the expressions doesn't matter (for example, (x+y)(x-y) = (x-y)(x+y). In this case, we had 2 decisions so we divide by divide by 2!.

(4x3)/2! = 12/2 = 6

Once again the answer would be E.

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