Question

Tap A can fill the tank in two hours while taps B and C can empty it in six and eight hours respectively. All the taps remained open initially. After 2 hours tap C was closed and after one more hour tap B was also closed. In how much time now would the remaining tank get filled?
A) 12min
B) 15min
C) 30min
D) 20min

Answer 1

Hint:
We can take the volume of the tank as V. Then we can find the volume tank filled or emptied in one hour by each tap. Then we can find the volume of water in the tank after 2 hours when all the taps are open. Then we can find the volume filled in the next one hour, when one tap is closed. Then we can find the remaining volume to be filled and divide it by the volume filled per hour by the opened tap to get the required time.

Complete step by step solution:
Let V be the total volume of the tank.
We are given that tap A can fill the tank in two hours.
So, the volume of the tank filled in 1 hour by tap A is given by,
 ${V_A} = \dfrac{V}{2}$
We are given that tap B can empty the tank in 6 hours.
So, the volume of the tank emptied in 1 hour by tap B is given by,
 ${V_B} = \dfrac{V}{6}$
We are also given that tap C can empty the tank in 8 hours.
So, the volume of the tank emptied in 1 hour by tap C is given by,
 ${V_C} = \dfrac{V}{8}$
It is given that all the taps were open for 2 hours. So, the volume of the tank filled in 2 hours is given by,
 $ \Rightarrow {V_{2hr}} = 2{V_A} - 2{V_B} - 2{V_C}$
On substituting the values, we get
 $ \Rightarrow {V_{2hr}} = 2\dfrac{V}{2} - 2\dfrac{V}{6} - 2\dfrac{V}{8}$
On cancelling the common terms, we get
 $ \Rightarrow {V_{2hr}} = V - \dfrac{V}{3} - \dfrac{V}{4}$
On taking the LCM, we get,
 $ \Rightarrow {V_{2hr}} = \dfrac{{12V - 4V - 3V}}{{12}}$
On simplification we get
 $ \Rightarrow {V_{2hr}} = \dfrac{{5V}}{{12}}$
It is given that the tap C was closed after 2 hours and tap B has closed after 1 hour.
So, the volume of the tank filled in the next 1 hr is given by,
 $ \Rightarrow {V_{1hr}} = {V_A} - {V_B}$
On substituting the values, we get
 $ \Rightarrow {V_{1hr}} = \dfrac{V}{2} - \dfrac{V}{6}$
On taking the LCM, we get
 $ \Rightarrow {V_{1hr}} = \dfrac{{3V - V}}{6}$
On simplification we get,
 $ \Rightarrow {V_{1hr}} = \dfrac{{2V}}{6}$
Now the remaining volume to be filled is given by subtracting the volumes from the total volume.
 $ \Rightarrow v = V - {V_{2hr}} - {V_{1hr}}$
On substituting the values, we get
 $ \Rightarrow v = V - \dfrac{{5V}}{{12}} - \dfrac{{2V}}{6}$
On taking the LCM, we get
 $ \Rightarrow v = \dfrac{{12V - 5V - 4V}}{{12}}$
On simplification we get
 $ \Rightarrow v = \dfrac{{3V}}{{12}}$
Hence, we have
 $ \Rightarrow v = \dfrac{V}{4}$
Now only tap A is open. So, the time required to fill the remaining volume is given by,
 $ \Rightarrow t = \dfrac{v}{{{V_A}}}$
On substituting the values, we get
 \[ \Rightarrow t = \dfrac{{\dfrac{V}{4}}}{{\dfrac{V}{2}}}\]
On simplification, we get
 \[ \Rightarrow t = \dfrac{1}{2}hr\]
On converting to minutes, we get
 \[ \Rightarrow t = \dfrac{{60}}{2}\min \]
So, we have
 \[ \Rightarrow t = 30\min \]

Therefore, the required solution is 30 min.
So, the correct answer is option C.

Note:
Alternate method to solve this problem is given by,
Let V be the total volume of the tank.
We are given that tap A can fill the tank in two hours.
So, the volume of the tank filled in 1 hour by tap A is given by,
 ${V_A} = \dfrac{V}{2}$
We are given that tap B can empty the tank in 6 hours.
So, the volume of the tank emptied in 1 hour by tap B is given by,
 ${V_B} = \dfrac{V}{6}$
We are also given that tap C can empty the tank in 8 hours.
So, the volume of the tank emptied in 1 hour by tap C is given by,
 ${V_C} = \dfrac{V}{8}$
Let t be the required time to fill the tank after closing tap B and C.
From the question, we can say that, Tap A is open for $2 + 1 + t = 3 + t\,hr$, tap B is open for $2 + 1 = 3hr$and tap C is open only for 2 hrs.
So, we can write the total volume as
 $V = \left( {3 + t} \right)\dfrac{V}{2} - 3\dfrac{V}{6} - 2\dfrac{V}{8}$
On cancelling the common factors, we get
 $V = \left( {3 + t} \right)\dfrac{V}{2} - \dfrac{V}{2} - \dfrac{V}{4}$
On taking the LCM and dividing throughout with V, we get
 $ \Rightarrow 1 = \dfrac{{2\left( {3 + t} \right) - 2 - 1}}{4}$
On cross multiplying and expanding the bracket, we get
 $ \Rightarrow 4 = 6 + 2t - 2 - 1$
On simplification, we get,
 $ \Rightarrow 2t = 1$
 \[ \Rightarrow t = \dfrac{1}{2}hr\]
On converting to minutes, we get
 \[ \Rightarrow t = \dfrac{{60}}{2}\min \]
 \[ \Rightarrow t = 30\min \]
Therefore, the required solution is 30 min.