As per the problem,
The larger pump takes 40 minutes to fill the pool.
The smaller one takes 60 minutes.
When both the pumps are working , let the time taken be x minutes , then we can write
[tex] \frac{1}{x}=\frac{1}{40}+\frac{1}{60}\\
\\
\text{Make the denominator equal we get}\\
\\
\frac{1}{x}=\frac{1}{40}*\frac{60}{60}+\frac{1}{60}*\frac{40}{40}\\
\\
\frac{1}{x}=\frac{60}{2400}+\frac{40}{24000}\\
\\
\text{Simplify we get}\\
\\
\frac{1}{x}=\frac{60+40}{2400}\\
\\
\frac{1}{x}=\frac{100}{2400}\\
\\
\frac{1}{x}=\frac{1}{24}\\
\\
x=24\\ [/tex]
Hence, when both the pumps works together, Time taken =24 minutes.
