GMAT Prep Practice Test Quant Questions #3

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GMAT Prep Practice Test Quant Questions #3

by simpm14 » Mon Apr 01, 2019 4:02 pm
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 (x-y). I thought the answer would be (1/12). 1/4*1/3, but guessed 1/6 since that wasn't an answer choice.

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by ceilidh.erickson » Thu Apr 04, 2019 11:25 am
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.
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by [email protected] » Sat Nov 16, 2019 9:29 am
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
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)

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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by [email protected] » Sat Nov 16, 2019 10:09 am
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), (X-Y) and (5X-Y). 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 brute-forcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(X-Y) = X^2 - Y^2
(X+Y)(5X-Y) = X^2 + 4XY - Y^2
(X+5Y)(X-Y) = X^2 + 4XY - 5Y^2
(X+5Y)(5X-Y)= 5X^2 +24XY - 5Y^2
(X-Y)(5X-Y) = 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

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by whiteutkarsh » Sat Dec 14, 2019 1:04 am
The only way to get x^2 - (by)^2 is if you multiply (x+by) with (x-by). Looking at the expressions given, the only pair that will satisfy this is (x+y) & (x-y)

No of ways to pick pairs = C(4,2) = 6

No of pairs satisfying condition = 1

Probability = 1/6

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simpm14 wrote:
Mon Apr 01, 2019 4:02 pm
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 (x-y). I thought the answer would be (1/12). 1/4*1/3, but guessed 1/6 since that wasn't an answer choice.
Solution:

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:

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