Question

Two taps A and B can together fill a swimming pool in 15 days. Taps A and B are kept open for 12 days and then tap B is closed. It takes another 8 days for the pool to be filled. How many days goes each tap required to fill the pool?

Answer 1

Hint- To solve this question, we need to use the basic concepts of chapter time and work. First, we assume the rate of both tap A job/day and B job/day respectively. And then make equations such that it satisfies a given condition. i.e. Two taps A and B can together fill a swimming pool in 15 days and also Taps A and B are kept open for 12 days and then tap B is closed. It takes another 8 days for the pool to be filled.

Complete step-by-step answer:
Let A and B be the rate of two taps
According to the question
15A+15B=1 (job)..........(i)
20A+12B=1 (job).........(ii)
Multiply (i) by 12 and (ii) by 15
$\Rightarrow$ 180A+180B=12...........(iii)
$\Rightarrow$ 300A+180B=15.........(iv)
Subtract (iii) from (iv) we get
$\Rightarrow$ 120A=3
$\Rightarrow$ A=$\dfrac{1}{{40}}$job/day (A's rate)
A would take 40 days to fill the pool alone.
Now,
$\Rightarrow$ 20A+12B=1
$\Rightarrow$ $\dfrac{1}{2}$+12B=1
$\Rightarrow$ 12B=\[\dfrac{1}{2}\]
$\Rightarrow$ B=$\dfrac{1}{{24}}$job/day (B's rate)
B would take 24 days to fill the pool alone.
Therefore,
A would take 40 days to fill the pool alone and B would take 24 days to fill the pool alone.

Note- Work is defined as something which has an effect or outcome; often the one desired or expected. The basic concept of Time and Work is similar to that across all Arithmetic topics, i.e. the concept of Proportionality. Efficiency is inversely proportional to the Time taken when the amount of work done is constant.
Efficiency = $\dfrac{K}{\text{Time taken}}$, where K is constant.