BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

GMAT Prpep

Expert replies
by moneyman » Sun May 04, 2008 2:23 am
Is 1/p>r/r^2 + 2??

(1)p=r
(2)r>0

This is how I worked it out

1/p>r^2 + 2 is equal to Is r^2 + 2 > pr by cross multiplying..

(1) says that p=r so it is proved that r^2 + 2=r^2

but the answer is C

pls explain
Maxx
Join the discussion
Source: — Problem Solving |

by AleksandrM » Sun May 04, 2008 5:57 am
Never multiply or divide variables in inequalities because you do not know the sign of the variable, and therefore do not know whether to flip the sign. The first statement tells you that p is equal to r, but it does not tell you their signs. If the sign is positive, then the answer is yes, if it is negative then the answer is no, since the bigger the negative the smaller it is. The second statement tells you that one of the variables is positive, but you do not know what the other variable is. Taking the two statements together allows you to answer the question. Both are positive variables and p = r.
Join the discussion

by nipunkathuria » Fri Oct 01, 2010 5:19 am
AleksandrM wrote:Never multiply or divide variables in inequalities because you do not know the sign of the variable, and therefore do not know whether to flip the sign. The first statement tells you that p is equal to r, but it does not tell you their signs. If the sign is positive, then the answer is yes, if it is negative then the answer is no, since the bigger the negative the smaller it is. The second statement tells you that one of the variables is positive, but you do not know what the other variable is. Taking the two statements together allows you to answer the question. Both are positive variables and p = r.
As said A has be true...but since we have an expression involving 1/P(or after statement1 its 1/R) , we shd note that R!=0.
What i feel, rather that giving R>0 they should have given R!=0 or P!=0...

still the ambiguity of dividing by 0 is removed using (2) , hence the answer should be C
Join the discussion