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the harmonic mean of two positive integers p and q is 3, given that [ (6p+1)/(7q-4) ] <(p/q), what could be the value of q?

a)q<=4

b)q>9

c)q<8

d)q>=13

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if harmonic mean of  p and q is 3

then 2pq/(p+q)=3

2pq=3p+3q

2pq-3p=3q

p(2q-3)=3q

p=3q/(2q-3).......................(1)

and

[ (6p+1)/(7q-4) ] <(p/q)

6pq+q<7pq-4p

4p+q<pq

from (1)

{12q/(2q-3) + q  }< 3q2/(2q-3)

(2q2+9q)/2q-3 <3q2/(2q-3)

{(2q2+9q)/2q-3} - 3q2/(2q-3)<0

(-q2+9q)/(2q-3)<0

(q2-9q)/(2q-3)>0

this is true if either both numerator and denominator is positive or both are negative

case 1) if (q2-9q)>0 and (2q-3)>0

ie q(q-9)>0 and (2q-3)>0

from here we get q > 9

case 2) if (q2-9q)<0 and (2q-3)<0

from here we get q belongs to (0,3/2)

so option b) is correct

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