Veritas Quant 700 + question

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Source: — Data Sufficiency |

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by edge » Sun Jul 31, 2011 12:31 pm
My answer is E.

The two statements are obviously insufficient by themselves.
Combining them, you can either solve the set quadratically or plug in numbers. I chose the latter path and chose (12, 4) and (4, 12) for (X, Y). Plugging them into the equation in the question, you get different answers.

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by naveen451 » Mon Aug 01, 2011 11:31 pm
IMO E

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by tpr-becky » Wed Aug 03, 2011 9:21 pm
You can use the bowtie method to add the two fractions:

1/x + 2/y = y + 2x/(xy)

thus in order to solve we need something that tells us both the value of y + 2x and the value of xy.

Statement 1 does not help for either - it can't be maniuplated to y + 2x and you can't manipulate to xy. Thus you are left with BCE

Statement 2 gives you one piece, but not y + 2x so it is not enough - CE

Together you still don't have y + 2x so it is not enough - and the answer is E
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by gmatboost » Wed Aug 03, 2011 10:46 pm
My guess is that something that can save you time on questions like this is to say to yourself:

"Any time I see algebraic fractions with different denominators, I am going to put them over a common denominator."

It's almost always helpful, and it's usually necessary.

If you did this, but then weren't sure if the combined statements were sufficient, then consider the following about systems of equations in which one equation is XY = ? and the other is X+Y = ?:

1. In these systems, there are usually 2 solutions, and X can be either one as long as Y is the other one.
2. You can often guess what the numbers are by looking at the equations. In this case, it's not unreasonable to say that you could guess that the two numbers are 4 and 12.
3. Once you know that EITHER x = 12, y = 4 OR x = 4, y = 12, you can plug both sets of values into the prompt and see what happens. In this case, different outcomes. In other cases, they could be the same.
4. If all else fails, you can use substitution to set up a quadratic equation and find the two values that way. Here, x = 16 - y, so you can substitute to get (16 - y)*y = 48 and solve the resulting quadratic equation. This is messy though. Ty to guess at the two numbers first.

Hope this helps. Let me know if the time sink was something totally different.[/url]
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by ajaykpat » Wed Aug 03, 2011 11:51 pm
I have approached question in this way:


now from St 1. X+Y=16 ( Not sufficient)
st 2. XY=48 ( Not sufficient)

For scenario St.1 + St.2:

Consider st 1.

X+Y=16
dividing bot side of equation with XY then we have ==> X+Y/XY = 16/XY
==> 1/X + 1/Y = 16/48=1/3 ( XY=48 frm st.2 )


Now from question we have 1/X + 2/Y = 1/X + 1/Y + 1/Y

==> 1/3 + 1/Y

Hence , we cant get the solution of given equation using both statements

ANS= E