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If x and y are positive, which of the following ...

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Source: — Problem Solving |
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by aditi_bc » Sat Nov 29, 2008 10:52 am
Just use the same cross-multiply approach which you use when you compare numbers.

For e.g to compare which one is greater 2/3 or 3/4, you multiply 2*4 & 3*3.

Use the same approach. Also as soon as you find an option which is greater, just stop. You don't have to test every answer.
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by cramya » Sat Nov 29, 2008 11:11 am
I think I posted my solution before for this.

Since sqrt(x+y),sqrt(x) and sqrt(y) are present take x=16 y=9
Their sum 16+9 = 25 is also a pefect square

This makes the computations simple and fool proof!
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by parallel_chase » Sat Nov 29, 2008 1:42 pm
My approach is algebraic in solving such questions.

important keywords = "must" + "positive"

1/sqrt(x+y) = sqrt(x+y)/(x+y)

I. sqrt(x+y)/2x, if x and y have the same value sqrt(x+y)/(x+y) = sqrt(x+y)/2x - eliminate

II. sqrtx+sqrty/(x+y) III. sqrtx - sqrty/(x+y)

The denominator of two options is same as that of question stem, therefore, option whose numerator is greatest must be greater.

if x and y are positive sqrtx + sqrt y will always be greater than sqrt(x+y)

Hence II only.

Hope this helps.
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by niraj_a » Sat Nov 29, 2008 2:45 pm
i chose numbers which would make a perfect square i.e. X = 2 and Y = 1
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by kranti » Sun May 03, 2009 1:57 pm
Comparing equations is more accurate way of solving, but it is complicated.

I think simpler way is to try with numbers,
- try with 9 and 16, this eliminates III
- try with 36 and 64, this eliminates I
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