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

Expert replies
Source: — Data Sufficiency |

Is the Answer A?

by bbaah » Fri Sep 19, 2008 7:54 am
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.
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Re: Is the Answer A?

by Ian Stewart » Fri Sep 19, 2008 10:28 am
bbaah wrote:
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.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: Is the Answer A?

by jackcrystal » Fri Sep 19, 2008 10:55 am
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).
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Re: Is the Answer A?

by Ian Stewart » Fri Sep 19, 2008 11:27 am
jackcrystal wrote: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.
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OG 11, DS

by bbaah » Mon Sep 22, 2008 2:23 am
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.
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by cramya » Wed Sep 24, 2008 8:14 pm
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!
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by cramya » Wed Sep 24, 2008 8:15 pm
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!
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