Question

A tank can be filled by one tap in 209 minutes and by another in 25 minutes. Both the taps are kept open for 5 minutes and then the second is turned off. In how many minutes more is the tank completely filled?
(a) 6
(b) 11
(c) 12
(d) $162\dfrac{1}{5}$

Answer 1

Hint: We start solving the problem by assigning variables for the capacity of tank. We then find the capacity of the tank that can be filled by first tap, second tap in one minute of their opening. We then find the capacity of the tank that can be filled in five minutes if both taps are opened. We then find the remaining portion of the tank that needs to be filled. We then find the time required in minutes to fill the obtained remaining portion of the tank when only tap 1 is opened to get the required answer.

Complete step-by-step answer:
According to the problem, we are given that a tank can be filled by one tap in 209 minutes and by another in 25 minutes. If both the taps are kept open for 5 minutes and then the second is turned off, then we need the time required to fill the tank completely after turning off the tap.
Let us assume the capacity of the tank is ‘x’ litres.
Now, let us find the amount of tank that the first tap can fill in 1 minute. Let us assume it as ‘y’.
So, we get $209\times y=x\Leftrightarrow y=\dfrac{x}{209}$ litres/min ---(1).
Now, let us find the amount of tank that the second tap can fill in 1 minute. Let us assume it as ‘z’.
So, we get $25\times z=x\Leftrightarrow z=\dfrac{x}{25}$ litres/min ---(2).
Now, let us find the amount of tank that can be filled if both taps are opened in one minute. Let us assume it as ‘p’.
So, we get $p=\dfrac{x}{209}+\dfrac{x}{25}$.
$\Rightarrow p=\dfrac{25x+209x}{209\times 25}$.
$\Rightarrow p=\dfrac{234x}{5225}$ litres.
Now, let us find the amount of tank that is filled in 5 mins if both taps are opened.
So, we get $5p=5\times \dfrac{234x}{5225}=\dfrac{234x}{1045}$.
The remaining part of tank to be filled is $x-\dfrac{234x}{1045}=\dfrac{1045x-234x}{1045}=\dfrac{811x}{1045}$ litres.
Now, let us find the time required to fill the $\dfrac{811x}{1045}$ litres if only tap1 is opened. Let us assume it as ‘t’ minutes.
So, we get $t=\dfrac{\dfrac{811x}{1045}}{\dfrac{x}{209}}=\dfrac{811}{5}=162\dfrac{1}{5}$ minutes.

So, the correct answer is “Option (d)”.

Note: Whenever we get this type of problems, we first assign variables for all the knowns to avoid confusion. We can see that the given problem consists of a huge amount of calculation, so we need to perform each step carefully. We can also find the required time in hours by using the conversion that 1 hour = 60 minutes. Similarly, we can expect problems to find the time required to fill the remaining portion of the tank if the second tap is opened instead of the first tank.