162 inequalities

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162 inequalities

by phoenix9801 » Tue Jul 13, 2010 4:04 pm
162- If n and k are positive integers, is √n+k > √2 ?

(1) k > 3n

(2) n + k > 3n


Note: the Answer is A. But this there any simple way to solve this question like picking numbers. Please explain in depth and step by step.

I apologize if I miss post the question in the wrong posting.
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by Rahul@gurome » Tue Jul 13, 2010 8:35 pm
phoenix9801 wrote:162- If n and k are positive integers, is √n+k > √2 ?

(1) k > 3n

(2) n + k > 3n


Note: the Answer is A. But this there any simple way to solve this question like picking numbers. Please explain in depth and step by step.

I apologize if I miss post the question in the wrong posting.
Is it (√n) + k > √2 or √(n + k) > √2?
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by this_time_i_will » Tue Jul 13, 2010 9:08 pm
This is a DS prob. plz ask in appropriate forum.

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by phoenix9801 » Tue Jul 13, 2010 10:48 pm
Rahul@gurome wrote:
phoenix9801 wrote:162- If n and k are positive integers, is √n+k > √2 ?

(1) k > 3n

(2) n + k > 3n


Note: the Answer is A. But this there any simple way to solve this question like picking numbers. Please explain in depth and step by step.

I apologize if I miss post the question in the wrong posting.
Is it (√n) + k > √2 or √(n + k) > √2?

Thanks. It should be is sqrt(n+k) > 2*sqrt(n)

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by kvcpk » Tue Jul 13, 2010 11:50 pm
phoenix9801 wrote:
Rahul@gurome wrote:
phoenix9801 wrote:162- If n and k are positive integers, is √n+k > √2 ?

(1) k > 3n

(2) n + k > 3n


Note: the Answer is A. But this there any simple way to solve this question like picking numbers. Please explain in depth and step by step.

I apologize if I miss post the question in the wrong posting.
Is it (√n) + k > √2 or √(n + k) > √2?

Thanks. It should be is sqrt(n+k) > 2*sqrt(n)
Again confused.. 2*sqrt(n) or sqrt(2) ??

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by Rahul@gurome » Fri Jul 16, 2010 6:45 pm
phoenix9801 wrote:162- If n and k are positive integers, is √n+k > √2 ?

(1) k > 3n

(2) n + k > 3n


Note: the Answer is A. But this there any simple way to solve this question like picking numbers. Please explain in depth and step by step.

I apologize if I miss post the question in the wrong posting.
I am assuming the question is: If n and k are positive integers, is √(n+k) > 2√n?

(1) k > 3n

(2) n + k > 3n

Explanation:

We can rewrite the question as √(n+k) > √(4n) or is (n + k) > 4n?

(1) If we add n on both sides of k > 3n, we get (n + k) > 4n, which is the same as the question asked.
So, (1) is SUFFICIENT.

(2) Given n + k > 3n does not imply for sure that (n + k) > 4n. We can take a counter example for the same.
If n = 1 and k = 3 then n + k = 4 > 3, but 4 > 4 is not true.
So, (2) is NOT SUFFICIENT.

The correct answer is (A).
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by shashank.ism » Sat Jul 17, 2010 1:41 am
If the question is √n+k > √2
then also we can have a solution ..
(1) k>3n and since n and k are n integers i.e. k>3 so obviously √n+k > √2 Hence sufficient.
(2) n+k>3n --> k>2n and since n and k are n integers i.e. k>2 so obviously √n+k > √2 Hence sufficient.

so Answer D.
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by outreach » Sat Jul 17, 2010 2:11 am
√(n+k) > 2√n?

A

option 1
since k and n are positive integers let assume n=1
since k>3n, k must be greater than 4
if n=1,k>=4 cond satisfied
if n=4,k=>=12 cond satisfied

with the smallest positive integer cond is satisfied

suff


option 2
n + k > 3n
k>2n

if n=1,k=3, then left and right hand side of Q are equal. Cond not satisfied
if n=1,k=4, Cond satisfied

insuff
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