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If two of the four expressions x+y, x+5y, x-y, and 5x-y are
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31 Aug 2020, 10:33
31 Aug 2020, 10:33
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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
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6
Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are
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31 Aug 2020, 10:34
31 Aug 2020, 10:34
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are
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31 Aug 2020, 10:59
31 Aug 2020, 10:59
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Carcass 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
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6
Okay, 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)
----OKAY, ONTO THE QUESTION------------------
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, I have added a video (below) on calculating combinations (like 4C2) in your head
# 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
. this question is like we have a group of 4 marbles. 2 red and 1 blue and 1 green. we want to select two marbles, what is the probability that the first one is red.
