Is it really a tough question or it only looks scaring?
Please help to solve
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
answer is E
probability (algebraic expressions)
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- Olga Lapina
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- Brent@GMATPrepNow
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Okay, first recognize that x² - (by)² is a DIFFERENCE OF SQUARES.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
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 = E
Cheers,
Brent
- Olga Lapina
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Hi Olga Lapina,
Brent's properly explained the "math" behind this question, so I won't rehash that here. Instead, I'm going to suggest that you learn the following patterns that are based on this type of Quadratic Algebra. The GMAT is a test built on patterns, so knowing the ones that are likely to show up will be beneficial during your practice and when you take the actual GMAT.
These are the three "classic" quadratics:
(x+y)(x-y) = x^2 - y^2
(x+y)(x+y) = x^2 +2xy + y^2
(x-y)(x-y) = x^2 -2xy + y^2
You're likely to see at least one of these on Test Day, so keep an eye out for them.
GMAT assassins aren't born, they're made,
Rich
Brent's properly explained the "math" behind this question, so I won't rehash that here. Instead, I'm going to suggest that you learn the following patterns that are based on this type of Quadratic Algebra. The GMAT is a test built on patterns, so knowing the ones that are likely to show up will be beneficial during your practice and when you take the actual GMAT.
These are the three "classic" quadratics:
(x+y)(x-y) = x^2 - y^2
(x+y)(x+y) = x^2 +2xy + y^2
(x-y)(x-y) = x^2 -2xy + y^2
You're likely to see at least one of these on Test Day, so keep an eye out for them.
GMAT assassins aren't born, they're made,
Rich
- theCodeToGMAT
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x+y, x+5y, x-y, and 5x-y
Total Ways = 4c2 = 4 x 3 / 2 = 6
Possible Pair = 1 (x+y AND x-y)
Probability = 1/6
[spoiler]{E}[/spoiler]
Total Ways = 4c2 = 4 x 3 / 2 = 6
Possible Pair = 1 (x+y AND x-y)
Probability = 1/6
[spoiler]{E}[/spoiler]
R A H U L