OG 11, DS - I think they are wrong, please check

I'm getting (C), Both statements taken together are sufficient.

I don't have OG11, so I have no way of checking, but here's what I did:

Rewrite the stem: is 1/p > r/(r^2+2) becomes,

is p<(r^2+2)/r (in general 1/a>1/b can be rewritten as a<b)

The question therefore becomes, is p<r+2/r?

(1) If p=r, then you cannot answer the question because r could be +ve or -ve.

Eg. 2<2+2/2, 2<3, but -2>-2+2/-2, ie. -2>-3.

Not Sufficient.

(2)r>0 does not tell us anything about P.



Considering both statements,

if r>0,

then p will always be less than p+anything, or p<r+2/r, since p=r.

Both Statements are sufficient.

lets simplify it

r^2 + 2 > pr

now (1) if p = r

=>

r^2 + 2 > r^2 which is always true. Hence A is suficient

Ans is (A).

Ian,

Thanks for the pointers. I think my approach yielded the correct answer because statement (1) guaranteed that p and r will have the same sign, and statement (2) did not impose any conditions that would cause p and r to take on different signs.

Perhaps a better solution would have been as follows:

stem: is 1/p > r/(r^2+2)

(1) If p=r,

The question in the stem, Is 1/p > r/(r^2+2) can be rewritten as:

is p<(r^2+2)/r

(since 1/a>1/b is equivalent to a<b, whenever a and b have the same sign)

The answer to the question can still not be determined, since p and r could be both +ve or both -ve.

Eg. 2<2+2/2, 2<3, but -2>-2+2/-2, ie. -2>-3.

Not Sufficient.

(2)r>0 does not tell us anything about P.



Considering both statements,

if r>0, and p=r

then, using the simplification from (1) above, p will always be less than p+anything, or p<p+2/p, since p=r.

Both Statements are sufficient.

If we dont feel comfortable wiht algebric manipulations picking numbers would be the best.

Pick something engative and positive and we can see 1) is not sufficient.

Since 2) says r> 0 and using 1) p =r we can definitely say 1/p is indeeed greater

C) its is.

Thanks Ian for you nice explanation!

If we dont feel comfortable with algebric manipulations picking numbers would be the best.

Pick something negative and positive and we can see 1) is not sufficient.

Since 2) says r> 0 and using 1) p =r we can definitely say 1/p is indeeed greater

C) it is.

Thanks Ian for you nice explanation!