Average of P and its reciprocal = \(\frac{\frac{1}{P} + P}{2}\)

\(=\frac{\frac{1 + P^2}{P}}{\frac{2}{1}}\)

\(=\frac{1+P^2}{2P}\)

So, we have:
QUANTITY A: \(\frac{1+P^2}{2P}\)

QUANTITY B: \(1\)

We can complete this solution using matching operations

Since \(2P\) is POSITIVE, we can safely multiply both quantities by \(2P\) to get:
QUANTITY A: \(1+P^2\)
QUANTITY B: \(2P\)

If you notice that we have a PERFECT SQUARE hiding among the two quantities, then we can first subtract \(2P\) from both quantities to get:
QUANTITY A: \(1+P^2-2P\)
QUANTITY B: \(0\)

Rearrange to get:
QUANTITY A: \(P^2-2P+1\)
QUANTITY B: \(0\)

Factor to get:
QUANTITY A: \((P-1)^2\)
QUANTITY B: \(0\)

Since we know that P-1 does not equal 0, we know that \((P-1)^2\) is POSITIVE

Answer: A

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Since we know 0<p<1 can't we just take numbers in p/q form?

For example if p=1/2 then A is basically stating 1/2+2 / 2= 5/4. Similarly if we take other fractions the result will be A>B. For eg if we take 1/4+4/1=> 17/8.