Answer
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Hint: In this type of question we have to use the concept of problems on pipes and water tanks. Here, we have given time to fill the tank before hole so first we calculate the part of the tank that gets filled in one minute. Then, we find time to fill the tank after hole , so first we calculate the part of the tank that gets filled in one minute. Finally by subtracting these two terms we can obtain the part of the tank which gets emptied in 1 minute and by reversing it we get the time for the tank to get emptied.
Complete step by step answer:
Now, we have to find the time required to empty the drum into the sump.
Here, we have given that the tap can fill a drum in 60 minutes
\[\Rightarrow \text{60 mins = 1 drum}\Rightarrow \text{20 mins = }\dfrac{1}{3}\text{ part of the drum}\]
After 20 mins, a hole is made in the bottom of the drum and the tap is open. Water starts dripping through the hole. So that dripping and filling go simultaneously.
Now, we have given that, to fill the remaining drum it takes 120 minutes. That means to fill remaining \[{{\dfrac{2}{3}}^{rd}}\] part of the drum it takes 120 minutes.
\[\Rightarrow 120\text{ mins = }\dfrac{2}{3}\text{ drum }\Rightarrow \text{ 120}\times \dfrac{3}{2}=180\text{ mins = 1 drum}\]
Hence, before hole a tap can fill the drum in 60 mins.
\[\Rightarrow \dfrac{1}{60}\text{ part of drum in 1 minute}\]
While after hole the tap can fill the drum in 180 mins.
\[\Rightarrow \dfrac{1}{180}\text{ part of drum in 1 minute}\]
So that, in 1 minute \[\left( \dfrac{1}{60}-\dfrac{1}{180} \right)=\left( \dfrac{1}{90} \right)\] the part of tank get emptied.
Therefore, the tank gets emptied in 90 minutes.
Note: In this type of question, students have to take care in the calculation. Students have to take care of the conversion of \[{{\dfrac{2}{3}}^{rd}}\] part of the tank in 120 minutes then how much part in time for complete drumming. Also students have to remember finally reversing the subtraction to obtain the final result.
Complete step by step answer:
Now, we have to find the time required to empty the drum into the sump.
Here, we have given that the tap can fill a drum in 60 minutes
\[\Rightarrow \text{60 mins = 1 drum}\Rightarrow \text{20 mins = }\dfrac{1}{3}\text{ part of the drum}\]
After 20 mins, a hole is made in the bottom of the drum and the tap is open. Water starts dripping through the hole. So that dripping and filling go simultaneously.
Now, we have given that, to fill the remaining drum it takes 120 minutes. That means to fill remaining \[{{\dfrac{2}{3}}^{rd}}\] part of the drum it takes 120 minutes.
\[\Rightarrow 120\text{ mins = }\dfrac{2}{3}\text{ drum }\Rightarrow \text{ 120}\times \dfrac{3}{2}=180\text{ mins = 1 drum}\]
Hence, before hole a tap can fill the drum in 60 mins.
\[\Rightarrow \dfrac{1}{60}\text{ part of drum in 1 minute}\]
While after hole the tap can fill the drum in 180 mins.
\[\Rightarrow \dfrac{1}{180}\text{ part of drum in 1 minute}\]
So that, in 1 minute \[\left( \dfrac{1}{60}-\dfrac{1}{180} \right)=\left( \dfrac{1}{90} \right)\] the part of tank get emptied.
Therefore, the tank gets emptied in 90 minutes.
Note: In this type of question, students have to take care in the calculation. Students have to take care of the conversion of \[{{\dfrac{2}{3}}^{rd}}\] part of the tank in 120 minutes then how much part in time for complete drumming. Also students have to remember finally reversing the subtraction to obtain the final result.
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