[CR][PnR] HSPA posts

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[CR][PnR] HSPA posts

by HSPA » Tue Apr 12, 2011 10:39 pm
Is 1/p > r/(r^2 + 2) ?

(1) p = r
(2) r > 0
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by Stuart@KaplanGMAT » Tue Apr 12, 2011 11:02 pm
HSPA wrote:Is 1/p > r/(r^2 + 2) ?

(1) p = r
(2) r > 0
We see a yes/no inequality question with variables. We think: be careful about multiplying or dividing both sides by a variable - if the variable could be negative, weird things may happen!

(1) we immediately consider p=r=0. In this case, 1/p is undefined, so it's impossible to answer the question: insufficient.

(2) no info about p: insufficient.

Combined: we know that p=r>0, i.e. both p and r are positive. It's now safe to multiply or divide by p and/or q!

Subbing in p=r:

Is 1/r > r/(r^2 + 2)?

Cross multiplying:

Is r^2 + 2 > r^2?

Subtracting r^2 from both sides:

Is 2 > 0?

That's a definite yes - sufficient, choose (C)!
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by manpsingh87 » Wed Apr 13, 2011 1:13 am
HSPA wrote:Is 1/p > r/(r^2 + 2) ?

(1) p = r
(2) r > 0
we have no info about r,p, so r,p can be either integer, rational or irrational no. let's just try to substitute different possible values to get the final result.

1) p=r, 1/r>r/(r^2+2)
let r=sqrt(2); 1/sqrt(2)=sqrt(2)/2
r^2+2=sqrt(2)^2+2=4;
r/r^2+2=sqrt(2)/4; therefore 1/r>r/r^2+2;
let r=1/2; 1/r=2;
r^2+2=(1/2)^2+2=9/2;
r/r^2+2=1/9 therefore 1/r>r^2+2;
let r=-1; 1/r=-1;
r^2+2=(-1)^2+2=3 r/r^2+2=-1/3 now 1/r<r/r^2+2;

hence 1 alone is not sufficient to answer the question.

2) r>0; it didn't provide any information about p so 2 alone is not sufficient to answer the question.
combining 1 and 2 we will get the cases where 1/r>r/r^2+2 hence C
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by force5 » Wed Apr 13, 2011 2:12 am
good one Stuart i agree with you. Good solution.
My answer is also C.