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Comment on Inequalities - Part II
Hi Brent, regarding this
https://gmatclub.com/forum/if-a-b-c-0-is-c-219680.html
Are we able to use the transitive property here?
S1: Rewritten S1 as 1 < a+b-c. Using given a+b+c < 0 it means that a+b+c < 0 < 1 < a+b-c, thus a+b+c < a+b-c. Subtract a and b on both sides --> c < -c, which means c is a negative number and therefore c<1.
Thank you.
Question link: https:/
Question link: https://gmatclub.com/forum/if-a-b-c-0-is-c-219680.html
Great solution! Perfectly reasoned!!
Nice work.
Question Link:
https://gmatclub.com/forum/is-1-p-r-r-2-2-1-p-r-2-r-86165.html
Where is the problem if we solve like this?
Is 1/p > r/(r^2 + 2)?
The question is asking:
r^2 +2 > pr ??
From stmt 1 we know:
r^2 +2 > r^2 ..so why s the answer not A
Is it the case because we are squaring and missing the negative aspect of r?
Link: https://gmatclub.com
Link: https://gmatclub.com/forum/is-1-p-r-r-2-2-1-p-r-2-r-86165.html
Target question: Is 1/p > r/(r^2 + 2)?
You made an error when rephrasing the target question.
Since r^2 + 2 is always POSITIVE, we can safely multiply both sides of the inequality by (r^2 + 2) to get the rephrased target question: Is (r^2 + 2)/p > r?
IMPORTANT: If we multiply both sides of an inequality by a POSITIVE value, the direction of the inequality symbol STAYS THE SAME. If we multiply both sides of an inequality by a NEGATIVE value, the direction of the inequality symbol IS REVERSED.
So, IF p > 0, we can multiply both sides by p to get: Is r^2 + 2 > pr?
IF p < 0, we can multiply both sides by p to get: Is r^2 + 2 < pr?
Since we don't know whether p is POSITIVE or NEGATIVE, we don't know what the rephrased target question is.
Does that help?
Makes Sense! thanks, Brent.
thanks, Brent.
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