To answer this question, we have the following situation:
We need to have the time in minutes as follows:
[tex]7\min +\frac{40}{60}=7.66666666667\min [/tex]Now, we have that the unit rate for preparing 24 containers at 7.66666...minutes is:
[tex]\frac{24}{(7+\frac{40}{60})}=3.13043478261[/tex]That is 3.1304...containers per minute. If we inverse this rate, we obtain:
[tex]\frac{1}{3.13043478261}\frac{\min}{cont}=0.319444444444\frac{\min}{cont}[/tex]We can convert this time in seconds as follows:
[tex]0.319444444444\frac{\min}{cont}\cdot60\frac{sec}{\min}=19.1666666667\frac{sec}{cont}[/tex]Then, in seconds, we have that the maximum time per container is about:
[tex]0.319444444444\frac{\min}{cont}[/tex]Or equivalently:
19.16667 sec per container.