If two of the four expression x+y, x+5y...

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If two of the four expression x+y, x+5y...

by shubh425 » Sun Jun 26, 2016 10:35 am
Please explain


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by 800_or_bust » Sun Jun 26, 2016 10:45 am
shubh425 wrote:Please explain


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The expressions can be selected in 4C2 = 6 different ways. Only one of the six possible pairs will result in a product with a quadratic equation in the form x^2 - (by)^2, namely (x - y)(x + y) = x^2 - y^2, where b=1.
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by Brent@GMATPrepNow » Sun Jun 26, 2016 11:21 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

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Brent
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by [email protected] » Sun Jun 26, 2016 4:48 pm
Hi shubh425,

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 hoppycat » Sun Jun 04, 2017 8:47 am
I got one! I think.
I did 4C2 for the bottom and 1 (just the x+y and x-y pair) for the top.
prob = 1/6
That the correct way right?
Also, would the answer be different if they didn't have that part about b is an integer? I don't know why they tell us that. Also, what's the difficulty level for this?

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by [email protected] » Sun Jun 04, 2017 9:48 am
Hi hoppycat,

Yes - the approach that you used is correct (as you might have noticed from the other responses in the thread). The 'restriction' that B must be an integer actually has no impact on the question (so the answer wouldn't change even if that information wasn't there).

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by Matt@VeritasPrep » Mon Jun 05, 2017 11:13 pm
hoppycat wrote:Also, what's the difficulty level for this?
On the modern (2017) GMAT, maybe 65th-70th percentile?

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by hoppycat » Wed Jun 14, 2017 7:22 am
Matt@VeritasPrep wrote:
hoppycat wrote:Also, what's the difficulty level for this?
On the modern (2017) GMAT, maybe 65th-70th percentile?
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by Scott@TargetTestPrep » Thu Jun 15, 2017 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 x 3)/(2!) = 12/2 = 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 (2! = 2) 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 need to divide 12 by 2.

12/2 = 6

Once again the answer is E.

Alternate Solution:

One other way to solve this problem is to use probability.

Once again, we have determined that the only two expressions that when multiplied together will give us a difference of squares are x + y and x - y. If we select either of those expressions first, since there are 2 favorable expressions and 4 total expressions, there is a 2/4 = 1/2 chance that either x + y or x - y will be selected. Next, since there is 1 favorable expression left and 3 total expressions, there is a 1/3 chance that the final favorable expression will be selected.

Thus, the probability of selecting x - y and x + y is 1/2 x 1/3 = 1/6.

Answer: E

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