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

[jackcrystal]

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

1. p = r
2. r > 0

you can see the solution on page # 336. But, I think they are wrong.

[bbaah]

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.

[Ian Stewart]

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 answer is indeed C, but the logic above is not quite right. It is not universally true that 1/a > 1/b can be rewritten as a < b. You can only do this if a and b have the same sign (both positive or both negative). This is easy to see with an example: take a = 2, and b = -2. Then clearly 1/a > 1/b, but a is also greater than b.

When you have an inequality like:

1/a > 1/b

and you want an inequality comparing a and b, really what you're doing is multiplying by ab on both sides. If ab > 0 (that is, if a and b have the same sign), we do not need to reverse the inequality, and we get:

b > a

If, on the other hand, ab < 0 (i.e. if a and b have opposite signs), we do need to reverse the inequality; we get:

b < a

Jackcrystal, perhaps you could explain what you think is wrong about the explanation in the OG? The answer should indeed be C.

[jackcrystal]

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 Stewart]

lets simplify it

r^2 + 2 > pr

You've multiplied both sides of the inequality by r^2 + 2, which is certain to be positive, and also by p. You don't know if p is positive or negative. If p is negative you would need to reverse the inequality after you multiply both sides by p. So you can't 'simplify' as you've done above, at least not without knowing whether p is positive or negative.

[bbaah]

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.

[cramya]

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!

[cramya]

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!