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.
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OG 11, DS - I think they are wrong, please check
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jackcrystal
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bbaah
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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.
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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Ian Stewart
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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.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)
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.
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jackcrystal
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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).
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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Ian Stewart
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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.jackcrystal wrote:lets simplify it
r^2 + 2 > pr
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bbaah
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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.
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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cramya
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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!
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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cramya
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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!
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!
