[DS][numbers] HSPA posts

[HSPA]

is n<s<n^2?
(1) s>n
(2) s=(n+1)n/2

[vineeshp]

(1) s>n
NS
If S = 30 and n = 5, n^2 = 25

(2) s=(n+1)n/2

n=1
S = 1
n^2 = 1

n=2
S=3
n ^2 =4.

N.S

1 AND 2)

n = -1 S = 0 n^2 = 1

n= -5, S = 10, n^2 = 25

For all values apart from n=1, equation holds. And n=1 nullified by Stmt 1.

So. C?

[anshumishra]

is n<s<n^2?
(1) s>n
(2) s=(n+1)n/2

Is n<s<n^2 ?

Statement 1: s>n
can we conclude s<n^2 ?
If say, n=2, s=2.1, n^2 =4 => s<n^2
However, if n=2, s=5, n^2 = 4, then s>n^2
Hence, it is Insufficient

Statement 2:
s=n(n+1)/2
If, n=0.5, s = 0.5*0.6/2 = 0.15, n^2 = 0.25
then s<n and s<n^2
But if, n=2, s=2*3/2 = 3, n^2 = 4
then n<s<n^2, So not sufficient.

Combining 1 and 2 :s=(n+1)n/2 and s>n
when is s>n ? => n(n+1)/2 > n => n^2-n>0 => n<0 or n>1
when is s<n^2 ? => n(n+1)/2 < n^2 => 2n^2-n^2-n>0 => n^2-n>0 => n<0 or n>1

So, that means whenever s>n (and hence n lies in the range n<0 or n>1), s<n^2 - Sufficient C

[pankajks2010]

Lets check out an another approach to evaluate second statement in this question:

s=n(n+1)/2; 2s=n^2+n; s+s=n^2+n;
Now, three possibilities arise:
a) s=n, in that case for the equation (s+s=n^2+n) to hold true, s should be equal to n^2
b) s<n, in this case for the equation (s+s=n^2+n) to hold true, s>n^2
c) s>n, in this case for the equation (s+s=n^2+n) to hold true, s<n^2

Now, three conditions and three different answers..thus, statement 2 alone is not sufficient, however, if we consider the third possibility discussed above, then n<s<n^2 and thus, the answer C

Would be great to know your thoughts about this method.

[force5]

IMO C

[HSPA]

I was actually looking for below.. I learned this approach from 'a' poster by name pankajks...

n(n+1) / 2 = s => 2s-n = n^2

Now quesiton is if n^2 > S => 2s-n > s => S>n which is true using statement 1..

Thus C