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# If two of the factors of the equation x^2-y^2 and x^2+4xy+4y

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If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  25 Apr 2012, 21:03
00:00

Difficulty:

45% (medium)

Question Stats:

35% (02:24) correct 64% (01:18) wrong based on 77 sessions
If two of the factors of the equation x^2-y^2 and x^2+4xy+4y^2 are chosen at random, what is the probability that their product contains the term 2y^2?

A. 0
B. 1/6
C. 1/3
D. 2/3
E. 1

Searched but could not find this problem on here. Tough one, GL.
[Reveal] Spoiler: OA

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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  26 Apr 2012, 00:00
1
KUDOS
Ok lets start by listing the factors of the 2 expressions

x^2-y^2 = (x+y)*(x-y)
x^2+4xy+4y^2 = (x+2y)*(x+2y)

Now total number of factors are 4 of which 2 are identical, for finding the total number ways of choosing 2, lets list all sets of
{(x+y),(x-y)}, {(x+y),(x+2y)},
{(x-y),(x+2y)},
{(x+2y),(x+2y)}

Ok if we need a term of 2y^2, we need 1 2y and one y. (we cant have 2 '2y's since it would result in 4y^2, and we cant have t 'y's since it'd result in just y^2)
No. of such set of factors out of the above 4 factor sets: 2Hence probability = 2/4 = 0.5

Ok this should be the answer. The designer of the question has it seems ignored the fact that the factors of the 2nd expression are identical.
Anyway since this is not one of the options, Lets think the way the designer must have thought:
Number of ways of choosing 2 factors from 4: 4C2=6
Of these number of combinations which have 1 'y' and one '2y' :4
You can think this way: 6 -(The number of sets have 2 'y's:1 + the number of sets havinf 2 '2y's:1) = 6-2=4
Note: Now the sets {(x+y),(x+2y)},{(x-y),(x+2y)} will be repeated twice to get 6 total ways.
Probability: 4/6 = 2/3

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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  26 Apr 2012, 11:07
Here she goes.

I do not know if this is 100% right because this is from an advanced GMAT math class I took through Princeton Review and the lecturer said not to even handle it he just gave us the answer.

But what I did is make:

The two variables equal to each other.

x^2+4xy+4y^2 = x^2-y^2

Making the Problem to be this:

(x^2+4xy+4y^2)(-x^2+y^2)

Then using the foil method the long way giving us:

-x^4+x^2y^2-4x^3y+4xy^3-4y^2x^2+4y^4

Then simplify like terms to make:

-x^4-3y^2x^2+4y^4

Now we have 3 products to make our denominator, making 4y^4 = (2y^2)(2y^2)

Giving us 2 possible ways to make 2y^2 out of a total of three products. = 2/3
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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  11 Jul 2012, 21:30
hfbamafan wrote:
If two of the factors of the equation x^2-y^2 and x^2+4xy+4y^2 are chosen at random, what is the probability that their product contains the term 2y^2?

A. 0
B. 1/6
C. 1/3
D. 2/3
E. 1

Searched but could not find this problem on here. Tough one, GL.

This is my method. Not the most economical though :

Factorizing the equations we get (x+y)(x-y)(x+2y)(x+2y)

To select two from these there are 4C2 ways = 6

Since this is a small number we can list out the combinations for which 2(y^2) will appear. For 2(y^2) to appear we need the pair of factors selcted to have one and only one "(x+2y)"

We get four such pairs ie : (x+y)(x+2y), (x+y)(x+2y), (x-y)(x+2y), (x-y)(x+2y),

So probability = 4/6 = 2/3
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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  11 Jul 2012, 23:36
MacFauz wrote:
hfbamafan wrote:
If two of the factors of the equation x^2-y^2 and x^2+4xy+4y^2 are chosen at random, what is the probability that their product contains the term 2y^2?

A. 0
B. 1/6
C. 1/3
D. 2/3
E. 1

Searched but could not find this problem on here. Tough one, GL.

This is my method. Not the most economical though :

Factorizing the equations we get (x+y)(x-y)(x+2y)(x+2y)

To select two from these there are 4C2 ways = 6

Since this is a small number we can list out the combinations for which 2(y^2) will appear. For 2(y^2) to appear we need the pair of factors selcted to have one and only one "(x+2y)"

We get four such pairs ie : (x+y)(x+2y), (x+y)(x+2y), (x-y)(x+2y), (x-y)(x+2y),

So probability = 4/6 = 2/3

This is the most economical methough, it seems.....
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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  12 Jul 2012, 01:06
I think the second equation should be x2+2y2-4xy
if we need to satisfy the answer set!!
can you please check if thr is a typo
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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  31 Mar 2013, 09:40
MacFauz wrote:
hfbamafan wrote:
If two of the factors of the equation x^2-y^2 and x^2+4xy+4y^2 are chosen at random, what is the probability that their product contains the term 2y^2?

A. 0
B. 1/6
C. 1/3
D. 2/3
E. 1

Searched but could not find this problem on here. Tough one, GL.

This is my method. Not the most economical though :

Factorizing the equations we get (x+y)(x-y)(x+2y)(x+2y)

To select two from these there are 4C2 ways = 6

Since this is a small number we can list out the combinations for which 2(y^2) will appear. For 2(y^2) to appear we need the pair of factors selcted to have one and only one "(x+2y)"

We get four such pairs ie : (x+y)(x+2y), (x+y)(x+2y), (x-y)(x+2y), (x-y)(x+2y),

So probability = 4/6 = 2/3

Sorry if my question is stupid, but shouldn't we also consider 1as well as the x^2-y^2 and x^2+4xy+4y^2 themselves as factors?
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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  31 Mar 2013, 10:45
MacFauz wrote:
hfbamafan wrote:
If two of the factors of the equation x^2-y^2 and x^2+4xy+4y^2 are chosen at random, what is the probability that their product contains the term 2y^2?

A. 0
B. 1/6
C. 1/3
D. 2/3
E. 1

Searched but could not find this problem on here. Tough one, GL.

This is my method. Not the most economical though :

Factorizing the equations we get (x+y)(x-y)(x+2y)(x+2y)

To select two from these there are 4C2 ways = 6

Since this is a small number we can list out the combinations for which 2(y^2) will appear. For 2(y^2) to appear we need the pair of factors selcted to have one and only one "(x+2y)"

We get four such pairs ie : (x+y)(x+2y), (x+y)(x+2y), (x-y)(x+2y), (x-y)(x+2y),

So probability = 4/6 = 2/3

We get four such pairs ie : (x+y)(x+2y), (x+y)(x+2y), (x-y)(x+2y), (x-y)(x+2y)
Isn't (x+y)(x+2y) and (x+y)(x+2y) the same thing and likewise (x-y)(x+2y) and (x-y)(x+2y) the same??
How can we calculate the number of ways of selecting 2 things out of 4 when 2 of them are similar??
Request somebody to please shed some light on this
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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  31 Mar 2013, 23:34
There are 4 factors in total: (x+y), (x-y), (x+2y), (x+2y)

Note: even if the (x+2y) is repeated each is considered as a separate factor of the equation x^2 + 4xy + 4y^2

Number of ways of selecting a pair of factors out of 4 is 4C2 = 6

Now the only two pairs which will not contain 2y^2 in their product are the pairs (x+y)(x-y) & (x+2y)(x+2y)

So remaining 4 pairs lead to 2y^2 in its product.

So probability is 4/6 i.e 2/3
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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y [#permalink]  17 Apr 2013, 12:56
denominator: selecting 2 factors out of 4 where 2 of the 4 factors are same is: 3C2 = 3

Note: it cannot be 4C2 (similar to total number of combinations made by selecting any 2 balls from 4 balls which are colored - red, blue, green, green)

Numerator: selecting 2 factors from 4 factors where one of the factor (x + 2y) is mandatory and this factor repeats in the 4 factors is - 2C1 = 2

(similar to total number of combinations made by selecting any 2 balls from 4 balls which are colored - red, blue, green, green and green ball is mandatorily selected)

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Re: If two of the factors of the equation x^2-y^2 and x^2+4xy+4y   [#permalink] 17 Apr 2013, 12:56
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