Alternative approach:

If x = rate of input and y = rate of output [original].

Then:

(x-y) = 10*(5x/6 - 6y/5) [as the time taken is 10 times in the second case, so the rate in the first case must be 10 times the second]

=> y =(2/3)x

Put this in 6*(x-y)=1, to get x=1/2 and y=1/3

Then to find the required time t, we solve t*(x/3 - y/5) = 1, to get t=10 hours.

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