Pipe A fills the tank in 36 minutes and Pipe B in 48 minutes. If both pipes are opened simultaneously, when should Pipe B be closed so that the tank fills up in 24 minutes?
Answer
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Hint: We will first assume the volume of the tank to be the LCM of time taken by both the pipes. Then, we will find out how much pipe A fills up the tank in 24 minutes and then we will fill the rest by B.
Complete step-by-step solution:
Let us assume that the tank has the volume of LCM (36, 48).
Since, we know that: \[36 = 2 \times 2 \times 3 \times 3\] and \[48 = 2 \times 2 \times 2 \times 2 \times 3\].
So, LCM (36, 48) = \[2 \times 2 \times 2 \times 2 \times 3 \times 3\]
$ \Rightarrow $ LCM (36, 48) = 144
Let the volume of the tank be 144 units.
Since, Pipe A can fill the tank in 36 minutes, we see that Pipe A runs with the speed of 4 units per minute and similarly, if we look at Pipe B, it fills the tank in 48 minutes, so Pipe B runs with the speed of 3 units per minute.
Now, since both are opened together, we are given that they fill up the tank in 24 minutes. During this Pipe B is closed in between.
So, that means Pipe A was open throughout.
So, Pipe A will fill up = $4 \times 24$ units
$ \Rightarrow $ Pipe A fills up 96 units of the tank.
Since, the tank has the volume of 144 units among which 96 are filled using Pipe A only. So, we are left to fill up 144 – 96 = 48 units using Pipe B.
Since, Pipe B runs with a speed of 3 units per minute and it needs to fill up 48 units.
$ \Rightarrow $ Time taken = $\dfrac{{48}}{3}$ minutes = 16 minutes.
$\therefore $ Pipe B should be closed after 16 minutes so that the tank fills up in 24 minutes when both pipes A and B are opened simultaneously.
Note: The students must wonder how we could assume the volume of the tank to be something. But you must notice that we did not actually assume the volume of the tank, we just termed it in units of LCM because it must be multiple of the time it is actually filled in, otherwise the tank may overflow or underflow. So, we just assumed it in terms of random units which can be equal to any multiple volume.
Complete step-by-step solution:
Let us assume that the tank has the volume of LCM (36, 48).
Since, we know that: \[36 = 2 \times 2 \times 3 \times 3\] and \[48 = 2 \times 2 \times 2 \times 2 \times 3\].
So, LCM (36, 48) = \[2 \times 2 \times 2 \times 2 \times 3 \times 3\]
$ \Rightarrow $ LCM (36, 48) = 144
Let the volume of the tank be 144 units.
Since, Pipe A can fill the tank in 36 minutes, we see that Pipe A runs with the speed of 4 units per minute and similarly, if we look at Pipe B, it fills the tank in 48 minutes, so Pipe B runs with the speed of 3 units per minute.
Now, since both are opened together, we are given that they fill up the tank in 24 minutes. During this Pipe B is closed in between.
So, that means Pipe A was open throughout.
So, Pipe A will fill up = $4 \times 24$ units
$ \Rightarrow $ Pipe A fills up 96 units of the tank.
Since, the tank has the volume of 144 units among which 96 are filled using Pipe A only. So, we are left to fill up 144 – 96 = 48 units using Pipe B.
Since, Pipe B runs with a speed of 3 units per minute and it needs to fill up 48 units.
$ \Rightarrow $ Time taken = $\dfrac{{48}}{3}$ minutes = 16 minutes.
$\therefore $ Pipe B should be closed after 16 minutes so that the tank fills up in 24 minutes when both pipes A and B are opened simultaneously.
Note: The students must wonder how we could assume the volume of the tank to be something. But you must notice that we did not actually assume the volume of the tank, we just termed it in units of LCM because it must be multiple of the time it is actually filled in, otherwise the tank may overflow or underflow. So, we just assumed it in terms of random units which can be equal to any multiple volume.
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