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Question

I. 12p27p=1
II. 6q27q+2=0

A
If p<q
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B
If p>q
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C
If pq
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D
If pq
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E
If p=q
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Solution

The correct option is A If p<q
Equation I
12p27p=1
12p27p+1=0
12p24p3p+1=0
4p(3p1)1(3p1)=0
(3p1)(4p1)=0
p=13 or p=14

Equation II
6q27q+2=0
6q24q3q+2=0
2q(3q2)+(3q2)=0
(3q2)(2q1)=0
q=23 or 12
Hence, p<q

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