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If xy =  6, what is the value of xy(x+y)?
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18 Feb 2014, 03:52
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Re: If xy =  6, what is the value of xy(x+y)?
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18 Feb 2014, 03:53




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Re: If xy =  6, what is the value of xy(x+y)?
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18 Feb 2014, 05:34
Basically we need to find the value for (x+y).
St1: xy = 5
From stem, y = 6/x. Plugging this in statement 1, we will have x + 6/x = 5. Upon solving, we get x = 2 or 3. This means, y = 3 or 2. (x+y) = 1 or 1. Two different answers. Not sufficient.
St2: xy^2 = 18
This can be written as (xy)*y or 6y. This gives y = 3. We have no information about x. Not sufficient.
St1 + St2: We get x = 2 and y = 3. (x+y) = 1. Hence xy(x+y) = (6)(1) = 6. Sufficient.
Answer (C).



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Re: If xy =  6, what is the value of xy(x+y)?
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18 Feb 2014, 10:00
known xy=6, unknown xy(x+y)
statement 1) xy=5; we know that xy=6, we can substitute x=(6/y) in the given equation, which becomes (6/y)y=5; y^2+5y+6=0; (y+3)(y+2)=0; hence y=3 or y=2, when y=3, x=2. thus x+y=1 and xy(x+y)=(6)(1)=6 when x=2, y=3 thus x+y=1 and xy(x+y)= (6)(1)= 6 hence 1 is not sufficient
statement 2) xy^2=18; can be written as xy(y)=18, 6y=18; y=3; also we know from the question statement that xy=6; thus x=2 hence x+y=23=1, hence xy(x+y) = 6(1)= 6
hence sufficient. Therefore answer should be B



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Re: If xy =  6, what is the value of xy(x+y)?
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18 Feb 2014, 23:19
st1: xy=5 we know that xy=6 i.e. x=6/y this give us 2 solution so not sufficient. st2: xy2=18 i.e. 6y=18 give us y and we can find x. sufficient. Answer is B.
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Re: If xy =  6, what is the value of xy(x+y)?
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19 Feb 2014, 01:45
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionIf xy =  6, what is the value of xy(x+y)? (1) x  y = 5 (2) xy^2 = 18 Data Sufficiency Question: 100 Category: Algebra First and seconddegree equations Page: 160 Difficulty: 650 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you! If xy =  6, what is the value of xy(x+y)? (1) x  y = 5 (2) xy^2 = 18 the question asks what is the value of 6(x+y)? Statement I is insufficient: xy = 6 x = y + 5 6 = y^2 + 5y y^2 + 5y + 6 = 0 y = 3, y = 2 y = 3, x = 2, x + y = 1 y = 2, x = 3, x + y = 1 Since we have two different answers hence statement I is insufficient Statement II x(y)(y) = 18 6(y) = 18 y = 3 and since xy = 6, x = 2. Hence answer is B
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Re: If xy =  6, what is the value of xy(x+y)?
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31 Mar 2015, 23:21
Solving A we get 2 values of x and y hence not sure which one is the value. Solving B we get one set of values for x and y hence definite answer so B.



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Re: If xy =  6, what is the value of xy(x+y)?
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01 Apr 2015, 20:49
Hi All, There's a subtle, but important lesson in how this question is designed: you MUST make sure that you finish the work and answer the question that is asked..... Donnie84 wrote: Basically we need to find the value for (x+y).
St2: xy^2 = 18
This can be written as (xy)*y or 6y. This gives y = 3. We have no information about x. Not sufficient.
Answer (C). This poster correctly figured out the value of Y from Fact 2, but then stopped working....by plugging the value of Y back into either the starting equation or the one in Fact 2, we have the value of X.....AND we can answer the question. Fact 2 is SUFFICIENT. It's important to remember that DS questions have NO "safety net"  the moment you make a little mistake, you're probably going to get the question wrong and you won't even realize it. It's important to make sure that you have PROOF of what you believe. In this case, if we really had no idea about the value of X, then we should be able to quickly come up with 2 different possibilities that fit the given restrictions and PROVE that the answer to the question changes. When Quant scores take a significant drop, it's almost always due to some type of little mistakes. Practice putting ALL of the work that yo do on the pad and you'll be far more likely to remove those silly mistakes from your approaches. GMAT assassins aren't born, they're made, Rich
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Re: If xy =  6, what is the value of xy(x+y)?
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03 Oct 2015, 15:50
Bunuel wrote: SOLUTION
If xy =  6, what is the value of xy(x+y)?
(1) x  y = 5 > \(x=y+5\) > \(xy=(y+5)y=6\) > \(y^2+5y+6=0\) > \(y=3\) or \(y=2\). If \(y=3\) then \(x=y+5=2\) so \(xy(x+y)=6(x+y)=6\). But if \(y=2\) then \(x=y+5=3\) so \(xy(x+y)=6(x+y)=6\). Two different answers, hence not sufficient.
(2) xy^2 = 18 > \(xy*y=6*y=18\) > \(y=3\) > \(x=2\) > \(xy(x+y)=6(x+y)=6\). Sufficient.
Answer: B. In the Option 1 , cant we write X  Y =5 as (X  Y)^2 = 25 ?? as in this way we can get answer by A also.



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Re: If xy =  6, what is the value of xy(x+y)?
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03 Oct 2015, 16:18
Hi rahulhkdwivedi, While you could square both sides of that equation, you haven't shown any of the steps that you would do next... so how do you know that it's sufficient? I suggest that you attempt what you've suggested, and then answer try to answer the GIVEN question: What is the value of XY(X+Y)? You should be able to prove that it's insufficient. GMAT assassins aren't born, they're made, Rich
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Re: If xy =  6, what is the value of xy(x+y)?
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03 Oct 2015, 16:28
rahulhkdwivedi wrote: Bunuel wrote: SOLUTION
If xy =  6, what is the value of xy(x+y)?
(1) x  y = 5 > \(x=y+5\) > \(xy=(y+5)y=6\) > \(y^2+5y+6=0\) > \(y=3\) or \(y=2\). If \(y=3\) then \(x=y+5=2\) so \(xy(x+y)=6(x+y)=6\). But if \(y=2\) then \(x=y+5=3\) so \(xy(x+y)=6(x+y)=6\). Two different answers, hence not sufficient.
(2) xy^2 = 18 > \(xy*y=6*y=18\) > \(y=3\) > \(x=2\) > \(xy(x+y)=6(x+y)=6\). Sufficient.
Answer: B. In the Option 1 , cant we write X  Y =5 as (X  Y)^2 = 25 ?? as in this way we can get answer by A also. I agree with EMPOWERgmatRichC. Additionally, if you do square the expression to get \((xy)^2=25\) >\(x^2+y^22xy=25\), you still wont get any answers for \(xy(x+y)\) or \(x^2y+xy^2\) Also, when you square any expression you end up increasing the number of solutions by a factor of 2. For example, you are given xy=5 but when you square the expression, \((xy)^2=25\) > this is quadratic equation, you end up getting, \(xy=\pm 5\), thus giving you an additional equation in \(xy=5\) Do not square or cube any given expression unless absolutely needed to. Even when you do make sure to check whether all solutions given by the squared or the cubed equations satisfy the given requirements. Hope this helps.



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If xy =  6, what is the value of xy(x+y)?
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05 Jan 2016, 21:11
I was actually thinking about A as well to square (xy). xy)^2=25 or x^2+y^22xy=25 x^2+y^2+12=25 x^2+y^2=13
now, (x+y)^2=x^2+y^2+2xy. 1312=1. from this: x+y=1 or x+y=1.
since 2 outcomes are possible, A is not correct.
I answered B, because I solved the way it was explained above, yet this method bothers me as well..



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If xy =  6, what is the value of xy(x+y)?
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21 Apr 2016, 12:18
I got ditched! help pls. If xy =  6, what is the value of xy(x+y)? (1) x  y = 5 \((xy)^2\) = \((x+y)^2\)  4xy > 25 = \((x+y)^2\) + 24 > \((x+y)^2\) = 1 > \((x+y)\)=1 ( Possible to find xy(x+y) ) (2) \(xy^2\) = 18 (xy)(y)=18  the same method everyone did above. ( Possible to find xy(x+y) ) I selected D okk.. I know it isn't D After my analysis, I reckon the mistake I committed was deducing \((x+y)^2\) =1 > \((x+y)\)= 1 ; probably this must've been (+1) or ( 1). Somebody, please confirm is this where I got it wrong
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If xy =  6, what is the value of xy(x+y)?
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21 Apr 2016, 16:14
snorkeler wrote: okk.. I know it isn't D After my analysis, I reckon the mistake I committed was deducing \((x+y)^2\) =1 > \((x+y)\)= 1 ; probably this must've been (+1) or ( 1). Somebody, please confirm is this where I got it wrong Yes. The catch with S1 was that you will end up getting x= \(\pm\) 1, giving you 2 different values for xy(x+y) > Not sufficient. FYI, when you have a quadratic equation , you must have 2 solutions (as per the definition). Even if you dont know this, you can also realize that there are 2 values that will give you a value =1 when squared > when a (any variable)= \(\pm\) 1 Hope this helps.



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