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# If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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04 Aug 2013, 09:52
2
Stiv 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

the result will be of form : $$x^2 - (by)^2$$ only when we choose x+y and x-y...rest all pair will not satisfy.
therefore favorable case=1
total case = we have to select 2 from 4 ....= $$4C2$$ = 6

probability = favorable/total =$$1/6$$
hence E
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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05 Sep 2013, 07:44
How come the answer is not b) 1/3? The way I approached it was: first chose an expression (any one), then out of the three remaining, one is its "partner". Hence, 1/3. Am I double counting with this approach? Thanks!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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05 Sep 2013, 07:49
dumptruck wrote:
How come the answer is not b) 1/3? The way I approached it was: first chose an expression (any one), then out of the three remaining, one is its "partner". Hence, 1/3. Am I double counting with this approach? Thanks!

Not all expressions have their appropriate pairs.

Check here: if-two-of-the-four-expressions-x-y-x-5y-x-y-5x-y-are-92727.html#p713823
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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19 Nov 2013, 11:55
1
Hi Bunuel,

Here is how I did it. Since we can choose either (x+y) or (x-y) at first, the probability is 2/4 and for second pick we have only one of these left so the prob is 1/3. Total prob= 2/4*1/3=1/6
Is that right?
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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26 Nov 2013, 18:09
NvrEvrGvUp wrote:
I also got 1/6 but not really understanding why the equation pair is (x+y)(x-y).

My thoughts were that we needed to factor $x^2-(by)^2$ into (x-by)(x+by) where the factored "b" was the squared root of the original "b", or essentially a placeholder for a perfect square.

This led me to think that the only equation pair possible that would work to satisfy the original ask would be (x+5y) and (x-y) since we need two equations with differing signs but also needed an constant infront of the "y" term.

Thanks,
Rich

I thought the same thing. Bunuel why does this not work?
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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27 Nov 2013, 05:11
1
runningguy wrote:
NvrEvrGvUp wrote:
I also got 1/6 but not really understanding why the equation pair is (x+y)(x-y).

My thoughts were that we needed to factor $x^2-(by)^2$ into (x-by)(x+by) where the factored "b" was the squared root of the original "b", or essentially a placeholder for a perfect square.

This led me to think that the only equation pair possible that would work to satisfy the original ask would be (x+5y) and (x-y) since we need two equations with differing signs but also needed an constant infront of the "y" term.

Thanks,
Rich

I thought the same thing. Bunuel why does this not work?

Not sure I understand what you mean but the pair we need is $$(x+y)(x-y)$$ because $$(x+1*y)(x-1*y)$$ is of the form of $$(x-by)(x+by)$$. How is $$(x+5y)(x-y)$$ of the form of $$x^2 - (by)^2$$?
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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14 Oct 2015, 18:49
ronny13 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

First of all, realize that only two expressions form the given 4, when multiplied will give the form $$x^2 -(by)^2$$. These are $$x+y$$ and $$x-y$$
$$(x+y)(x-y) = x^2 - y^2 = x^2 - (by)^2$$ ; $$b = 1$$

Now, the question becomes: Probability of choosing these 2 expressions from given 4 expressions.

In other words,
What is the probability of choosing a and c from the set {a,b,c,d}??

Solution:
Choose (a AND c) OR (c AND a)
=> $$\frac{1}{4}*\frac{1}{3} + \frac{1}{4}*\frac{1}{3}$$
=> $$2(\frac{1}{4}*\frac{1}{3})$$
=> $$\frac{1}{6}$$
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Oct 2015, 11:08
1
atshy21saraf wrote:
Engr2012 wrote:
atshy21saraf wrote:
Q. If two of the expressions (x+y), (x-y), (x+5y), (5x-y) are chosen then What is the probability that their product will be of the form of x^2-(by)^2, b is an integer.
My doubt: What is the difference between 4c2 and (4c1 x 3c1) in case of total possibilities?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Search for a question before you post. This question has already been discussed before. Follow posting guidelines (link in my signatures).

Topics merged. Refer above for the solution.

My bad, I will keep that in mind. I am not very familiar with forum rules.

Although my doubt is still not cleared.

Why did't we do 4C1*3C1 to select 2 out of 4 expression?

Lets say you have 4 objects A,B,C,D and you are asked to select 2 out of the 4. In this case there is no difference whether you take A and then B or you take B and then A. You still are selecting A and B.

But if I ask you that the order of selection matters, i.e. a selection process in which A and then B will be different from B and then A then you can only do 4C1*3C2

Coming back to the original question, a^2-b^2 or (a-b)(a+b) can be formed either by choosing a-b and then a+b or by choosing a+b and then a-b. It does not matter which order you choose a+b and a-b. This is the reason why the number of ways = 4C2.

Hope this helps.
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If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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Updated on: 01 Jun 2017, 09:31
1
Bunuel wrote:
spc11 wrote:
If two of the four 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 x [square] - (by) square, where b is a constant.
A.1/2
B.1/3
C. 1/4
D.1/5
E.1/6

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 written 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 form x^2 - (by)^2. Therefore, the probability is 1/6.

Note: If you don’t know how to use the combination formula, here is a method that will work equally well:

We are choosing 2 expressions from a pool of 4 possible expressions. That is, there are 2 decisions being made:

Decision 1: Choosing the first expression

Decision 2: Choosing the second expression

Four different expressions are available to be the first decision.

For the second decision, 3 remaining expressions are available because 1 expression was already chosen. We multiply these two numbers: 4 x 3 = 12.

The final step is to divide by the factorial of the number of decisions (2! = 2) because the order in which we multiply the expressions doesn’t matter (for example, (x+y)(x-y) = (x-y)(x+y)). In this case, the two expressions are only considered as one, so we need to divide 12 by 2.

12/2 = 6

Once again the answer would be E.

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.

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Originally posted by JeffTargetTestPrep on 13 Feb 2017, 07:41.
Last edited by JeffTargetTestPrep on 01 Jun 2017, 09:31, edited 1 time in total.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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13 Feb 2017, 11:56
1
Hi All,

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.

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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28 Mar 2017, 09:18
only 2 combination helps us get the answer
we can solve it in 2 ways:
1)
a) we need x+y or x-y out of 4, then 2/4
b) then we need either of them, so 1/3
c)2/4*1/3=1/6

Or let's us combinatorics
there are totally 2C4 combinations of the 4 pairs, when 2 are chose. 2C4=6
there is only one combination that suits us, then 1/6

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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31 May 2017, 18:23
ronny13 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

1. The six possible products are:
(x+y)(x+5y)
(x+y)(x-y)
(x+y)(5x-y)
(x+5y)(x-y)
(x+5y)(5x-y)
(x-y)(5x-y)

2. Of the above, we see only the second gives the form required.
Probability is therefore 1/6
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Jun 2017, 17:46
ronny13 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

Only (x+y)(x-y) or one combination satisfies the condition. There are 4!/(2!)*(2!) total possibilities, since two equations are selected and the other two are not selected (YYNN as an anagram).

The odds are 1/6.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Jun 2017, 21:26
mbaapp1234 wrote:
ronny13 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

Only (x+y)(x-y) or one combination satisfies the condition. There are 4!/(2!)*(2!) total possibilities, since two equations are selected and the other two are not selected (YYNN as an anagram).

The odds are 1/6.

Minor correction: it should be should "probability" instead of "odds" (these two things are not the same and GMAT always asks about probability, not odds).
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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25 Jun 2017, 22:45
2 expression can be chosen 4C2 ways=6
only one pair will form the required product so probability =1/6
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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11 Aug 2017, 06:32
Why is the proper approach for this problem to set up a combination 4C2 where as for other problems such as choosing a team you may do 4 * 3. When I solved this problem I almost thought of the four expressions as players for a team. There are four to choose from and I have two slots to put players in. For the first slot, I have four to choose from and in the second slot, I have 3 to choose from so there are 4* 3 = 12 combinations. If there were replacement, that would yield 4 * 4 = 16 combinations.

I guess my question is: would someone be able to explain to me the flaw in my logic between these two different methods of arriving at the number of combinations. Thanks!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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11 Aug 2017, 11:37
Hi brandmanrocks,

To start, you have to always be sure that you are answering the question that is ASKED (sometimes the question is so specific that there's really only one way to approach it). That having been said, most prompts can be approached in more than one way, so you could certainly approach this prompt in the way that you described. However, you have to make sure to 'adjust' your 'math' to account for how you approached the work.

You can certainly treat this question as a permutation, BUT you then have to keep track of every potential option that 'fits' what is asked for....

(4 choices)(3 choices) = 12 possible options in which the order matters. Of the various 'pairs', there would be TWO that 'fit' what we're looking for:

(X+Y)(X-Y)
(X-Y)(X+Y)

Again, we would have to count that as two options because 'your way' is focused on treating this as a permutation.

2/12 = 1/6 so you end up with the same correct answer as you would if you treated this as a 'Combination' question.

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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24 Apr 2018, 10:01
Top Contributor
ronny13 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

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

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