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II) must be true.
first lets show this:
(*) sqrt(x)+sqrt(y)>sqrt(x+y)
to do this we square both sides to get:
(sqrt(x)+sqrt(y))^2 = x+ 2sqrt(x)*sqrt(y) + y > x+y = (sqrt(x+y))^2
since both sides when squared hold the inequality, and since both sides are positive then (*) must be true as well.
now we use (*) to continue with original claim:
(sqrt(x)+sqrt(y))/(x+y) > sqrt(x+y)/(x+y) = 1/sqrt(x+y)
III) not necessarily true. consider x=y=anything. then the fraction in III is 0 and definitely not greater than stem.
If x and y are positive, which of the following must be greater than 1/sqrt(x+y)? I) sqrt(x+y)/2x II) (sqrt x+sqrt y)/(x+y) III) (sqrt x-sqrt y)/(x+y)
a) None b) I only c) II only d) I and III e) II and III
I also got C. The way I did it is like this:
1. sqrt(x+y)/2x = x+y/(2x*(sqrt(x+y)))
=> will be true when x+y/2x > 1 or y > x, but we don't know that that is true
2. (sqrt x+sqrt y)/(x+y) can be rewritten as: (sqrt x+sqrt y)/(sqrt(x+y)*sqrt(x+y))
=> will be true when (sqrt x+sqrt y)/(sqrt(x+y) > 1
=> when (sqrt x+sqrt y) > sqrt(x+y)
squaring both sides 2sqrt(xy) > 0 which we know is true because x and y are positive
So will be true when (sqrt x-sqrt y)/sqrt(x+y) > 1
will be true when (sqrt x-sqrt y) > sqrt(x+y)
squaring both sides, when -2sqrt(xy) > 0
=> 2sqrt(xy) < 0 which we know cannot be true
I took time to understand the answer. Am I missing any basics? I tried to pick nos. It was time consuming.
No, I don't think you are. The strategy I used was, I changed each of the provided options to have the same denominator. sqrt(x+y). then compared all the "numerator" portions. Does this explain it better?
I took time to understand the answer. Am I missing any basics? I tried to pick nos. It was time consuming.
No, I don't think you are. The strategy I used was, I changed each of the provided options to have the same denominator. sqrt(x+y). then compared all the "numerator" portions. Does this explain it better?
Thank you. I understood your explanation. But that strategy didn't strike me quickly.
However, I was wondering if anyone could do this quickly picking nos. I don't practice that much.
you can rule out I and III very quicly picking numbers (see my initial response for examples).
after that you remain with options A and C as possible answers.
there is no way whatsoever (!!!!) to know that II is necessarily true by picking numbers.
you can try few pairs, see that II holds but who promise you that the next pair of numbers won't break that?
in these kind of questions and also in DS, picking numbers is a great strategy to refute a claim by giving a counter example. picking numbers should never be used to show that a statement is always true.
you can "get a feel" by trying one or two examples, but if you want to be certain in your answer you need to find another way.
beware of "picking numbers" it is a great method if it is used correctly but will be your greatest enemy if you use it in the wrong places' leading you to many wrong answers or just wasting you time.
You are right about picking numbers. Trying few pairs to make the statement untrue is what I don't like. Maybe I should just practice, it is a backup method.
Yeah, I don't either, and I bet it will be a big time saver on the test. I am trying to force myself to use it at least in some cases.
Sumithra wrote:
hsampath wrote:
Sumithra wrote:
Thank you guys!
I took time to understand the answer. Am I missing any basics? I tried to pick nos. It was time consuming.
No, I don't think you are. The strategy I used was, I changed each of the provided options to have the same denominator. sqrt(x+y). then compared all the "numerator" portions. Does this explain it better?
Thank you. I understood your explanation. But that strategy didn't strike me quickly.
However, I was wondering if anyone could do this quickly picking nos. I don't practice that much.
As per my calculation the answer is D (Both I & II are greater than 1/sqrt(x+y).
I used picking a number strategy.
Let x=9 and y=16 (why 9 & 16 because both of them are squares and there sum is also a square so it will be easier to calculate)
Hence 1/sqrt(x+y)=1/(sqrt(25)=1/5
I. sqrt(x+y)/2x=sqrt(25)/2*9=5/18 > 5/25 > 1/5
II (sqrt(x) + sqrt(y))/(x+y)=(sqrt(9)+sqrt(16))/(9+16)=(3+4)/25=7/25 > 7/35 > 1/5
III (sqrt(x) - sqrt(y))/(x+y)=(sqrt(9)-sqrt(16))/(9+16)=(3-4)/25=-1/25 or 1/25 (if x=16 & y=9) which is less than 1/5
Hence D is the answer.
The Q asks for an answer that MUST be greater than 1/sqrt(x+y)
There are some numbers which answer NO and some other numbers which answer YES for the same statement. When Q says MUST the answer should be same for ay number picked.
It is better to pick numbers that gives the answer NO or check if any number makes the statement untrue.
If x and y are positive, which of the following must be greater than 1/sqrt(x+y)? I) sqrt(x+y)/2x II) (sqrt x+sqrt y)/(x+y) III) (sqrt x-sqrt y)/(x+y)
a) None b) I only c) II only d) I and III e) II and III
---------------------------------------------
x, y are +ve:
I) (x+y)^3/2 > 2x ???...we can't say..
II)(x+y)^1/2 <x^1/2+ y^1/2 ??..sqr both sides ..YES
III)same as II but a -ve term on RHS ..NO
TF, C) II only
Personally, I feel picking numbers here can be confusing..
I would pick numbers when signs are involved ; here x, y both are +ve and I feel better off doing the algebra.. looks intimidating at first glance , but took less time once I started the calcn
Personally, I pick numbers only if timeÂ´s short or the problem is rather straightforward (= easy ). I think working on the theoretical basics of the problems is safer. In this case, I and III are out rather quickly. We are left with II. LetÂ´s see:
Thanks for pointing out about being careful about picking nos for such problems. Need to take care of it in future. I agree the answer should be C. I have to learn using a combination of picking no and algebra in solving such questions. Important thing to bear in mind is that we use picking nos in such questions to try and prove that an answer is wrong and not otherwise.
in these kind of questions and also in DS, picking numbers is a great strategy to refute a claim by giving a counter example. picking numbers should never be used to show that a statement is always true.
beware of "picking numbers" it is a great method if it is used correctly but will be your greatest enemy if you use it in the wrong places' leading you to many wrong answers or just wasting you time. _________________