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
OA E
Source: GMAT Prep
If two of the four expressions x+y, x+5y, x-y, and 5x-y are
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Okay, first recognize that x² - (by)² is a DIFFERENCE OF SQUARES.BTGmoderatorDC 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
OA E
Source: GMAT Prep
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)
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)
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
Answer: E
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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:BTGmoderatorDC 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
OA E
Source: GMAT Prep
a^2 - b^2 = (a + b)(a - b)
So, x^2 - (by)^2 can be factored 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 desired form x^2 - (by)^2. Therefore, the probability is 1/6.
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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