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by Brent@GMATPrepNow » Sun Jul 31, 2016 5:52 am
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 x² - (by)², where b is an integer?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6
Important concept tested here
First recognize that x² - (by)² is a DIFFERENCE OF SQUARES.
Here are some examples of differences of squares:
x² - 25y²
4x² - 9y²
49m² - 100k²

In general, we can factor differences of squares as follows:
a² - b² = (a-b)(a+b)

So . . .
x² - 25y² = (x+5y)(x-5y)
4x² - 9y² = (2x+3y)(2x-3y)
49m² - 100k² = (7m+10k)(7m-10k)

--------------------------
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)

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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by [email protected] » Sun Jul 31, 2016 9:32 am
Hi Apoorva@5,

This question is based heavily on algebra patterns. If you can spot the patterns involved, then you can save some time; even if you can't spot it though, a bit of 'brute force' math will still get you the solution.

We're given the terms (X+Y), (X+5Y), (X-Y) and (5X-Y). We're asked for the probability that multiplying any randomly chose pair will give a result that is written in the format: X^2 - (BY)^2.

Since there are only 4 terms, and we're MULTIPLYING, there are only 6 possible outcomes. From the prompt, you should notice that the 'first part' of the result MUST be X^2....and that there should be NO 'middle term'....which limits what the first 'term' can be in each of the parentheses....

By brute-forcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(X-Y) = X^2 - Y^2
(X+Y)(5X-Y) = X^2 + 4XY - Y^2
(X+5Y)(X-Y) = X^2 + 4XY - 5Y^2
(X+5Y)(5X-Y)= 5X^2 +24XY - 5Y^2
(X-Y)(5X-Y) = 5X^2 -6XY + Y^2

Only the second option is in the proper format, so we have one option out of six total options.

Final Answer: E

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by Scott@TargetTestPrep » Mon Aug 01, 2016 3:30 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 x² - (by)², where b is an integer?

A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6
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, the two expressions are only considered as one, so we divide by divide by 2.

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

Once again the answer would be E.

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by Matt@VeritasPrep » Thu Aug 04, 2016 8:42 pm
If we think of this as the difference of squares, we can factor it as so:

(x + y√b) * (x - y√b)

So our two terms must have the same y√b term. (x + y) and (x - y) do (in these cases, √b = 1, so it just disappears).

We don't have any other pairs with the same second term, so that's the only pair that works. (For instance, if we had (x + 2y) and (x - 2y), that'd be fine as well: √b = 2, etc.)

Since we have (4*3)/2 = 6 possible pairs, but only 1 we want, our probability = 1/6.
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