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saintjingjing
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q^2,p^2

by saintjingjing Fri Nov 11, 2011 6:04 am

Is the sum of p2 and q2 greater than 1?
(1) p is greater than 1/2.
(2) q is greater than 1/2.
prep
I think the answer is E, because p^2>1/4 q^2>1/4 so p^2q^2>1/16, but I am not sure can I adopt my method? because I think when q^2 or p^2, I get the answer like q^2>1/4 actually magnify the result.
is there another idea about this question
Last edited by saintjingjing on Mon Nov 14, 2011 5:19 am, edited 1 time in total.
be.a.true.winner
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Re: q^2,p^2

by be.a.true.winner Sat Nov 12, 2011 2:56 am

saintjingjing Wrote:Is the sum of p2 and q2 greater than 1?
(1) p is greater than 1/2.
(2) q is greater than 1/2.
prep
I think the answer is E, because p^2>1/4 q^2>1/4 so p^2q^2>1/16, but I am not sure can I adopt my method? because I think when I doube q/p, I get the answer like q^2>1/4 actually magnify the result.
is there another idea about this question


Not sure what you mean by "when I double q/p,I get..."

However instead of multiplying you can add both your equations to get the answer as E
P squared > 1/4
q squared > 1/4

=> p squared + q squared > 1/4 + 1/4
=> p squared + q squared > 1/2
=> Insufficient

You can also see by plugging numbers although above shall be quicker!
RonPurewal
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Re: q^2,p^2

by RonPurewal Wed Nov 23, 2011 7:07 am

... or you can just plug numbers. in just about any INEQUALITY situation, PLUGGING EXTREMES is a good way to determine behavior.

here, it's quite obvious that the answer to the question can be "yes", i.e., that the sum can be greater than 1. therefore, the essence of the problem lies in trying to make the sum less than 1.
fortunately, since such specific boundaries are given, this is a straightforward task: just consider the lower extremes of those ranges, i.e., numbers just above 1/2.

if p = a little bit more than 1/2 (like 0.500001) and q = same, then p^2 + q^2 will be a little more than 1/4 + 1/4, or a little more than 1/2.
that's less than 1, so, insufficient.