Just finished my first GMAC Practice Test and am Trying to figure out the ones I got wrong.
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 x2  (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. I understand the difference of squares for (x+y) and (xy). I thought the answer would be (1/12). 1/4*1/3, but guessed 1/6 since that wasn't an answer choice.
GMAT Prep Practice Test Quant Questions #3
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 ceilidh.erickson
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Probability is always defined this way:
probability = (# of desired outcomes)/(total possible # of outcomes)
You correctly identified that there was only one pairing of terms that would give you a difference of squares: (x + y) and (x  y).
However, you miscounted your total # of possible outcomes. Since order does NOT matter, we don't just want to multiply 4*3. We must divide that by the number of duplicates: 2! duplicates if 2 positions are interchangeable. Or we can use the formula for 4 choose 2: (4!)/(2!2!) = 6.
You could also simply list out the possible combinations:
1. (x + y) & (x + 5y)
2. (x + y) & (x  y)
3. (x + y) & (5x  y)
4. (x + 5y) & (x  y)
5. (x + 5y) & (5x  y)
6. (x  y) & (5x  y)
The pairing in red is the only one that works. Thus, we have a 1/6 chance of getting a difference of squares.
probability = (# of desired outcomes)/(total possible # of outcomes)
You correctly identified that there was only one pairing of terms that would give you a difference of squares: (x + y) and (x  y).
However, you miscounted your total # of possible outcomes. Since order does NOT matter, we don't just want to multiply 4*3. We must divide that by the number of duplicates: 2! duplicates if 2 positions are interchangeable. Or we can use the formula for 4 choose 2: (4!)/(2!2!) = 6.
You could also simply list out the possible combinations:
1. (x + y) & (x + 5y)
2. (x + y) & (x  y)
3. (x + y) & (5x  y)
4. (x + 5y) & (x  y)
5. (x + 5y) & (5x  y)
6. (x  y) & (5x  y)
The pairing in red is the only one that works. Thus, we have a 1/6 chance of getting a difference of squares.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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Harvard Graduate School of Education
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Okay, first recognize that xÂ²  (by)Â² is a DIFFERENCE OF SQUARES.simpm14 wrote:Just finished my first GMAC Practice Test and am Trying to figure out the ones I got wrong.
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 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:
xÂ²  25yÂ²
4xÂ²  9yÂ²
49mÂ²  100kÂ²
In general, we can factor differences of squares as follows:
aÂ²  bÂ² = (ab)(a+b)
So . . .
xÂ²  25yÂ² = (x+5y)(x5y)
4xÂ²  9yÂ² = (2x+3y)(2x3y)
49mÂ²  100kÂ² = (7m+10k)(7m10k)
So, from the 4 expressions (x+y, x+5y ,xy and 5xy), only one pair (x+y and xy) 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 xy are both selected?
P(both selected) = [# of outcomes in which x+y and xy 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/gmatcounting?id=789
# of outcomes in which x+y and xy are both selected
There is only 1 way to select both x+y and xy
So, P(both selected) = 1/6
Answer: E
Cheers,
Brent
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Hi Simpm14,
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), (XY) and (5XY). 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 bruteforcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(XY) = X^2  Y^2
(X+Y)(5XY) = X^2 + 4XY  Y^2
(X+5Y)(XY) = X^2 + 4XY  5Y^2
(X+5Y)(5XY)= 5X^2 +24XY  5Y^2
(XY)(5XY) = 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
GMAT assassins aren't born, they're made,
Rich
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), (XY) and (5XY). 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 bruteforcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(XY) = X^2  Y^2
(X+Y)(5XY) = X^2 + 4XY  Y^2
(X+5Y)(XY) = X^2 + 4XY  5Y^2
(X+5Y)(5XY)= 5X^2 +24XY  5Y^2
(XY)(5XY) = 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
GMAT assassins aren't born, they're made,
Rich

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