Question

Two water taps together can fill a tank in 9⅜ hours the top of a larger diameter tanks 10 hours less than the smaller one to fill the tank separately find the time in which each top can separately fill tank

Answer 1

Let the time taken by the smaller diameter tap = x

larger = x-10

total time taken = 75 /8

portion filled in one hour by smaller diameter tap = 1/x

and by larger diamter tap = 1/x-10

1/x + 1/x-10 = 8/75

x-10+x/x(x-10) = 8/75

2x+10/x²-10x = 8/75

8(x²-10x) = 75 ×2 (x-5)

8/2 (x²-10x) = 75 (x-5)

4x²-40x = 75x-375

4x² -40x-75x +375 = 0

4x²-115x + 375 = 0

4x²-100x-15x +375 = 0

4x(x-25)-15(x-25)=0

(x-25)(4x-15)

x= 25

x= 15/4

If x= 25 

the x-10 = 25-10 = 15

if x = 15/4

x-10 = 15/4 - 10 = 15-40/4 = -25/4

since time cannot be negative therefore x = 25

Answer 2

The answer is in the attachment.
Time taken by the small tap= 25 hrs.
Time taken by the big tap=15hrs.