paresh_patil wrote:If 2 of the 4 expressions x+y, x+5y, x-y & 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 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
