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A Cashier working alone serves 20 clients in 1 hour. A second Cashier serves the same number of clients in 40 minutes.
What time will they need serving the 20 clients if they work both?
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In this problem, let us call this work, serving 20 clients, as "one job".
Then the first cashier makes of the job per minute. It is his rate of work.
The second cashier makes of the job per minute. It is his rate of work.
When they work together, their combined rate of work is the sum of individual rates, i.e. = = = .
Thus we get that the two cashiers, working together, make of the job per minute.
Now it is clear to you that it will take 24 minutes for both to serve 20 clients.
The problem is solved.
It is a typical joint work problem.
There is a wide variety of similar solved joint-work problems with detailed explanations in this site. See the lessons
- Using Fractions to solve word problems on joint work
- Solving more complicated word problems on joint work
- Selected joint-work word problems from the archive
Read them and get be trained in solving joint-work problems.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic
"Rate of work and joint work problems" of the section "Word problems".
Do not forget to send your "thanks" to me after reading this solution.
(I mean: after you get understanding of the solution).