if x and y are positive, which of the following

[GMATTaker]

If x and y are positive, which of the following must be greater than 1/sqrt(x + y)?

1. sqrt(x + t)/2x
2. (sqrt(x) + sqrt(y)) / (x + y)
3. (sqrt(x) - sqrt(y)) / (x + y)

Answers:
1. None
2. 1 only
3. 2 only
4. 1 & 3
5. 2 & 3

I tried solving it using some sample values for x & y, but as you can realize that
took some time. Is there a faster way (or an algebraic way) to solve this?.

Thanks in advance.

[RonPurewal]

http://www.manhattangmat.com/forums/post4838.html deals with some number plugging.

there's little sense in dealing with #3 algebraically: because of the subtraction, it can clearly equal 0 (if x and y are the same number). since 1/√(x + y) is a positive number, the possibility of 0 rules out roman numeral III. (in fact, that expression can even be negative, as nothing prohibits x from being smaller than y.)

--

if you want to compare two fractions, you can use the technique of cross products to perform the comparison.
to use this technique, you take the two 'cross products' (one of the numerators, times the denominator of the other fraction), and associate each of the cross products with whichever fraction donated the numerator.
for instance, if you're comparing 2/3 vs. 11/17, then the cross products are 2 x 17 = 34 (associated with 2/3) and 3 x 11 = 33 (associated with 11/17). because 34 is greater than 33, it follows that 2/3 is greater than 11/17.

notice that this technique only applies to positive fractions... but that's all you really need: if the fractions have opposite signs, then the comparison is trivial (the positive one is bigger!), and if the fractions are both negative, then the comparison is the opposite of whatever it would be if they were positive.

--

find cross products in #(i):
√(x + y)/2x vs. 1/√(x + y)
cross products are (x + y) vs. 2x
subtract one x from both sides --> this comparison is the same as y vs. x
we don't know which is bigger.

find cross products in #(ii):
(√x + √y)/(x + y) vs. 1/√(x + y)
cross products are (√x + √y)√(x + y) vs. (x + y)
divide both sides by √(x + y) to give (√x + √y) vs. √(x + y) --- remember that (quantity) divided by √(quantity) is √(quantity) -- that's the definition of what a square root is.
since both of these quantities are positive, we can square them and compare the squares:
(√x + √y)^2 vs. (√(x + y))^2
x + 2√xy + y vs. x + y
left hand side is bigger
so the original fraction is bigger than 1/√(x + y)

ans = ii only

there may well be a shorter and more elegant way to figure out #(ii), but i can't conjure one at the moment. a beer to anyone who can.

[Guest]

change form of original equation by multiplying 1/sqrt(x+y) by / sqrt(x + y ) by sqrt(x+y) / sqrt (x+y) to get sqrt(x+y) / (x+y)

1) sqrt(x +y) / 2x will equal sqrt(x+y) / (x+y) if x = y, so therefore it does not have to be greater than 1/sqrt(x+y)

2) denominators are the same so it's really a matter of comparing if sqrt(x) + sqrt(y) vs sqrt(x+y), and it seems sqrt(x) + sqrt(y) is always greater. i found this by testing numbers, but i'm not sure if there is a theorem for this.

3) sqrt(x) - sqrt(y) could be negative if y is bigger than x, so therefore it does not have to be greater than 1/sqrt(x+y) because 1/sqrt(x+y) is positive

[RonPurewal]

2) denominators are the same so it's really a matter of comparing if sqrt(x) + sqrt(y) vs sqrt(x+y), and it seems sqrt(x) + sqrt(y) is always greater. i found this by testing numbers, but i'm not sure if there is a theorem for this.

you don't need to memorize this sort of result; you can figure it out by squaring both sides.
the square of the left-hand side is x + 2√x√y + y, and the square of the right-hand side is x + y. since x and y are positive numbers, the left side is bigger.
that's a proof.

btw, at this point you can call it a theorem if you want. anything you've proved is a "theorem", although some theorems are of course more famous than others.

[Guest]

It becomes very simple if you just take x and y =1.
[editor: invalid reasoning. see below.]

[RonPurewal]

It becomes very simple if you just take x and y =1.

well, yeah, but you can't solve "MUST" problems that way.

you're looking for the expressions that MUST be greater than 1/√(x + y), no matter what x and y are. if the expressions work for one choice of plug-in, that doesn't tell you anything at all.
(if any of the expressions doesn't work for any single plug-in, though, you can eliminate it.)

--

stupid analogy:
which of the following MUST be less than 10?
(i) x
(ii) x + 1
(iii) x + 2

obviously, the answer is "none of them" - they can all be huge, if x is huge - but your approach above would mistakenly conclude that the answer is "i, ii, and iii".

--

remember that gmat problems are rarely as easy as you may think at first glance - and almost never, if users are posting them here. the problems posted on this forum tend to be drawn disproportionately from the pool of difficult problems (although there's the occasional easier one here and there on the forum).

[gkumar]

Thanks for a great post guys. I tried plugging in x=9 and y=16 for the sake of elimination. I got stuck between None, I, and II since I got 5/18 for I and a fraction greater than 7/25 for II. I cross multipled with 1/5 from the control value from the question stem to figure out the greater fraction. The "must be" clause made things a lot trickier.

Also when I tried plugging y=16 and x=9, I was left with None and II. I got a 50-50 chance so I picked II, which luckily was right. If I had more time to work on this problem, I would cross multiply equations and stuff, but I resorted to educated guesses. I'm not sure if this is the best approach for tackling this type of problem within 2 minutes.

[Ben Ku]

With "MUST" questions, plugging in values can help to eliminate answer choices, but it doesn't necessarily lead you to the correct answer choice. I would use the "plugging in values" approach to eliminate answers if get get stuck with or don't know how to apply a theoretical approach, or if you're running out of time.

It's not a bad approach; it just doesn't always lead you to the answer.

[aps_asks]

Hi Instructors ,

I take a much time solving this problems using the substitution approach.( around 4 mins which is not acceptable).

Can U please suggest an elegant approach to solve such problems within 2 mins?

[RonPurewal]

Hi Instructors ,

I take a much time solving this problems using the substitution approach.( around 4 mins which is not acceptable).

Can U please suggest an elegant approach to solve such problems within 2 mins?

this whole thread is already full of solutions (and it links to another thread, too).

also, remember that you don't have to be able to solve every problem in 2 minutes or less!
remember that 2 minutes is an AVERAGE. that's an average, not an upper limit! as with basically all other averages, this means that you will have some results above the average and others below it.
since this problem has three parts (designated by the three roman numerals), not only is it perfectly ok for you to take more than 2 minutes, but you should actually expect to take more than 2 minutes on it.

[bineeshnair4u]

Hi Instructors ,

I take a much time solving this problems using the substitution approach.( around 4 mins which is not acceptable).

Can U please suggest an elegant approach to solve such problems within 2 mins?

this whole thread is already full of solutions (and it links to another thread, too).

also, remember that you don't have to be able to solve every problem in 2 minutes or less!
remember that 2 minutes is an AVERAGE. that's an average, not an upper limit! as with basically all other averages, this means that you will have some results above the average and others below it.
since this problem has three parts (designated by the three roman numerals), not only is it perfectly ok for you to take more than 2 minutes, but you should actually expect to take more than 2 minutes on it.

Ron,

When something like this comes for eg: you have to take sqrt(x), sqrt(y) and sqrt(x+y).. always go for perfect pythagorean squares.. here i took x = 64, y = 36 and x = 144, y = 25.. solved the problem can be solved in 30sec to 1 min.. As i got the answer as (ii) in both the cases i selected Option B as my answer.. Is this generalization correct? Am i missing something?

Regards,
Bineesh.

[RonPurewal]

When something like this comes for eg: you have to take sqrt(x), sqrt(y) and sqrt(x+y).. always go for perfect pythagorean squares..

this is a pretty deep insight -- i.e., pythagorean triples can ease a lot of the computational nuisance in problems like this. however, you should be careful not to over-generalize; there could be problems where the sole consideration of pythagorean triples would land you on the wrong answer.

for instance, consider the following data sufficiency problem (which i'm making up right now):
If ABC is a right triangle with right angle at C, what is the perimeter of triangle ABC?
(1) AB = 5
(2) The area of triangle ABC is 12.5

in this problem, if you think only in terms of pythagorean triples, you will mistakenly think that statement 1 is sufficient -- because you'll think that the 3-4-5 triangle is the only valid possibility, when in fact there are plenty of other triangles that also work (such as 1-√24-5, etc.)

here i took x = 64, y = 36 and x = 144, y = 25.. solved the problem can be solved in 30sec to 1 min.. As i got the answer as (ii) in both the cases i selected Option B as my answer.. Is this generalization correct? Am i missing something?

Regards,
Bineesh.

well, this kind of approach isn't 100 percent trustworthy. i.e., if you find a couple of random cases that give you "yes" answers, you can't necessarily conclude that a statement MUST be true -- there could always be an exception lurking around the next corner.
(by contrast, if you plug in numbers in a definite, exhaustive pattern and keep getting "yes" answers, you can be sure that the statement is always true.)

[pravin22jd]

That's awesome solution. Thanks Ron & Ben.
@Ron
I don't think anybody will be claiming that beer, unless there is new mathematical theory in Math world which simplifies all of our problems ;)

[RonPurewal]

Glad to help.