When 5 inlet taps and 4 outlet taps are opened in a cistern, the cistern is filled in 12 hrs. Had there been 8 inlet taps and 6 outlet taps, it would have taken just 3 hrs. Find the time taken for the cistern to be filled if there are 10 inlet and 8 outlet taps. (Assume all inlet taps fill at the same rate, and all outlet taps drain at the same rate?

Let the inlet be x and outlet be y
5x-4 .....(given)
Capacity of tank =hours ✖️ (inlet-outlet) 1st tanks capacity =12[5x-4y] .......1
8x-6y.....(given)
Capacity of 2nd tank=3[8x-6y] ......2
We know that capacity is the same
Therefore 12[5x-4y]=3[8x-6y].......3
Y= 1.2x........(from1,2,3)
Substituting value of y in eqn 1 we get,
Initial Capacity =2.4x
THE THIRD ONEs capacity
Final capacity=10x-8y
=10x-8(1.2)
=6.4x.
Time taken by the tank to fill is
Initial Capacity/final capacity
=2.4x/0.4x
=6 hours

option a

What will be the answer?? It's been two years but no reply 😑