Is 1/p > r/(r^2+2) ? 1: p=r 2:r>0 : DS Archive
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# Is 1/p > r/(r^2+2) ? 1: p=r 2:r>0

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Is 1/p > r/(r^2+2) ? 1: p=r 2:r>0 [#permalink]

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12 May 2007, 09:55
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Is 1/p > r/(r^2+2) ?
1: p=r
2:r>0
Should we solve this problem by simplifying the given inequality i.e. inequality can be simplified to r^2-rp+2>0? Or should we keep the denominator (p) and solve the problem?. Please let me know why/why not?
Thanks
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12 May 2007, 10:03
a insuff If p or r are positive then thsi is correct otherwise not.
b insuff
a+b suff
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12 May 2007, 10:15
My way of solving inequalities DS is as follow:

(as for your question I will try to reach a formula i can work with in according to the given statements)

One of the biggest problems in inequalities DS is the fact that multiplying or dividing an inequality by a negative number reverses the order of the inequality, so people tend to make the following (and quite common) mistake.

1/p > r/(r^2+2) (multiple both sides by p*[r^2+2]

r^2+2 > rp

implementing statement 1 as p=r

and you get

r^2+2 > r^2

which seems right and choose (A) - Wrong !!!!

we have to make sure first that p*[r^2+2] is greater then 0 !

and the only way to know that is when r>0 and p=r

so the answer must be (C)

12 May 2007, 10:15
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