Two water taps together can fill a tank in 9 3/8 hours the tap of larger diameter takes 10 hours less than the smaller one fill the tank separately find the time in which each tap can separately fill the tank
Answers
Answer:
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.
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Answer:
Step-by-step explanation:
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.