Pump A, working alone, can fill a tank in 3 hours, and pump B can fill the same tank in 2 hrs. If the tank is empty to start and pump A is switched on for one hour, after which pump B is also switched on and the two work together, how many minutes will pump B have been working by the time the pool is filled?
a)48
b)50
c)54
d)60
e)64
pump A, 1/3 tank /hr
pump B, 1/2 tank/hr
in first hour pump A fills 1/3 tank
so together they fill 2/3 tank
x hours (1/3 tank/hr + 1/2 tank/hr) = 2/3 tank
x (5/6) = 2/3 = 4/6
5 x = 4
x = 4/5 hr = (4/5)60 = 48 minutes
Well, it's a classic case of teamwork! Let's break it down. Pump A can fill the tank in 3 hours, so in 1 hour, it can fill 1/3 of the tank. Then, when pump B joins the party, together they fill the tank in 2 hours. So, pump B is twice as fast as pump A.
Since pump A filled 1/3 of the tank in 1 hour, pump B can fill 2/3 of the tank in the same amount of time. That means that pump B does the remaining 2/3 of the work.
So, pump B is doing 2/3 of the work while pump A does 1/3. That means pump B will work for 2 times longer than pump A did. Since pump A worked for 1 hour, pump B will work for 2 hours.
But we need to convert that into minutes. There are 60 minutes in an hour, so 2 hours is 2 * 60 = 120 minutes.
So, pump B will have been working for 120 minutes by the time the tank is filled.
Therefore, the answer is not on the list!
First, let's find out how much of the tank is filled by pump A in one hour. Since pump A can fill the tank in 3 hours, it can fill 1/3 of the tank in one hour.
Next, let's find out how much of the tank is filled by pump B in one hour. Since pump B can fill the tank in 2 hours, it can fill 1/2 of the tank in one hour.
Now, let's calculate how much of the tank is already filled after pump A has been working for one hour: pump A fills 1/3 of the tank.
Let's denote the remaining volume of the tank after one hour as V.
V = 1 - 1/3
V = 2/3
Now, both pump A and pump B are working together to fill the remaining 2/3 of the tank.
Since pump A fills 1/3 of the tank in one hour and pump B fills 1/2 of the tank in one hour, together they fill 1/3 + 1/2 = 5/6 of the remaining tank in one hour.
Now, let's calculate how many hours it takes for pump A and pump B to fill the remaining 2/3 of the tank when working together:
2/3 = (5/6) * t
2/3 = 5t/6
8/9 = 5t/6
t = (8/9) * (6/5)
t = 8/15
Since there are 60 minutes in an hour, pump B would have been working for (8/15) * 60 = 32 minutes.
However, this is only the time taken to fill the remaining 2/3 of the tank. We must also account for the initial one hour when pump A was working alone.
Therefore, pump B will have been working for a total of 32 + 60 = 92 minutes.
None of the given options matches the calculated result, so the correct answer is not available in the options.
To solve this problem, we need to find the rate at which each pump fills the tank, and then calculate how long it will take for both pumps to fill the tank working together.
Given that pump A can fill the tank in 3 hours, we can determine its fill rate as 1 tank / 3 hours = 1/3 tank per hour. Similarly, pump B can fill the tank in 2 hours, so its fill rate is 1 tank / 2 hours = 1/2 tank per hour.
Now, if pump A works alone for 1 hour, it will fill 1/3 of the tank because its fill rate is 1/3 tank per hour. This means that 1 - 1/3 = 2/3 of the tank is left to be filled.
When both pumps A and B work together, their combined fill rate is 1/3 + 1/2 = 5/6 tank per hour.
To find out how long it will take for pumps A and B to fill the remaining 2/3 of the tank working together, we can set up the equation:
(5/6) tank/hour * t hours = 2/3 tank
We can solve for t by multiplying both sides of the equation by (6/5):
t = (2/3 tank) * (6/5) / (5/6 tank/hour) = 4/5 * 6 = 24/5 = 4.8 hours.
Since 1 hour is equal to 60 minutes, pump B will have been working for 4.8 * 60 = 288 minutes.
Therefore, the answer is not given in the options provided.