Two taps together can fill a tank in 75/8 hours.the tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can seperately fill the tank.
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Tank capacity = V
Tap 1 filling duration = T => flow rate = V/T
Tap 2 filling duration = T+10 => flow rate = V/(T+10)
When two taps are filling together, flow rate = V (1/T + 1/(T+10) )
= V * (2 T+10) / (T² +10 T)
So Time to fill tank = (T²+10T) / (2T+10) = 75/8
=> 8T² + 80 T = 150 T + 750
8 T² - 70 T - 750 = 0
4 T² - 35 T - 375 = 0
T = [ 35 +- √(35² + 6000) ] / 8
T = 15 hours The bigger tap
Smaller tap takes 25 hours to fill.
Tap 1 filling duration = T => flow rate = V/T
Tap 2 filling duration = T+10 => flow rate = V/(T+10)
When two taps are filling together, flow rate = V (1/T + 1/(T+10) )
= V * (2 T+10) / (T² +10 T)
So Time to fill tank = (T²+10T) / (2T+10) = 75/8
=> 8T² + 80 T = 150 T + 750
8 T² - 70 T - 750 = 0
4 T² - 35 T - 375 = 0
T = [ 35 +- √(35² + 6000) ] / 8
T = 15 hours The bigger tap
Smaller tap takes 25 hours to fill.
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Answer:
Step-by-step explanation:
Tank capacity = V
Tap 1 filling duration = T => flow rate = V/T
Tap 2 filling duration = T+10 => flow rate = V/(T+10)
When two taps are filling together, flow rate = V (1/T + 1/(T+10) )
= V * (2 T+10) / (T² +10 T)
So Time to fill tank = (T²+10T) / (2T+10) = 75/8
=> 8T² + 80 T = 150 T + 750
8 T² - 70 T - 750 = 0
4 T² - 35 T - 375 = 0
T = [ 35 +- √(35² + 6000) ] / 8
T = 15 hours The bigger tap
Smaller tap takes 25 hours to fill
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