two tap water together can fill a tank in 9 9 3 1 18 hours the tap of larger diameter text 10 hour less than the smaller one to fill the tank separately find the time in which each step can separately fill the tank
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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
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
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