Pdgmat2010 wrote:is p + q > r + s ?

stmt 1: p > r + s
stmt 2: q > r + s


pls pick nos while solving and explain!!

OA after a couple of posts

[spoiler]IMO : C[/spoiler]

St1 : p > r + s but we cannot be sure whether p + q > r + s, since q could be -ve as well, we don't know.
St2 : q > r + s but we cannot be sure whether p + q > r + s, since p could be -ve as well, we don't know.

Adding both the eqns we get,
p+q > 2 ( r+s) , hence definitely ( p+q) > ( r+s)

OA ?

i used the same logic.
consider p = q = -4

r = s = -3

stmt a : - 4 > -3 + -3
-4 > -6

stmt b: - 4 > -6

adding a and b
-4 -4 > -6 -6
-8 > -12

So , the answer should be C...

.dunno how Kaplan says its E

I am 100% sure that C should be the answer. Let see if some expert just drop in with something unique otherwise..

When you combine the two statements you get:

p+q>2(r+s)

This does not imply p+q is necessarily > than (r+s)

if both (p+q) and (r+s) are negative

eg (p+q) =-5 and (r+s)=-4

---> (p+q)> 2(r+s)--->-5>-8

BUT (p+q)<(r+s)

therefore (E)

I got it now....i was really confused..thanks PRABHDEEP..

consider p = q = -4

r = s = -3

stmt a : - 4 > -3 + -3
-4 > -6

stmt b: - 4 > -6

adding a and b
-4 -4 > -6 -6

-8 > -12

but, -8 < -6 (r+s) hence, E
[edited my last post]

Thats a nice quistion

thanks Prabhdeep , Suman for your replies..
got it now

so if i take p=-4 , q=-4 then p+q = -8
also if r=s=-3 then r+s =-6
p+q < r+s
although
p+q > 2(r+s)
guess the whole purpose of choosing -ve nos was to show that p+q can be less than r+s
that's what i missed.

Cheers,
Pd

PRABHDEEP gill wrote:When you combine the two statements you get:

p+q>2(r+s)

This does not imply p+q is necessarily > than (r+s)

if both (p+q) and (r+s) are negative

eg (p+q) =-5 and (r+s)=-4

---> (p+q)> 2(r+s)--->-5>-8

BUT (p+q)<(r+s)

therefore (E)