SOLUTION: A piece of manufacturing equipment can produce enough items to fill an order in 10 hours. If an auxiliary piece of equipment is used also, it will take 7 hours. How long would it t

Question 1095086: A piece of manufacturing equipment can produce enough items to fill an order in 10 hours. If an auxiliary piece of equipment is used also, it will take 7 hours. How long would it take the auxiliary equipment to fill the order working alone?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.

The combined rate of work is of the job per hour. The known individual rate of work is . It implies that the unknown individual rate of work is = = . Hence, it will take = 23 hours and 20 minutes for the auxiliary equipment to fill the order working alone.

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".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The solution supplied by ikleyn is a perfectly good traditional algebraic one. I find many students prefer a different method. Here it is for your problem; give it a try and see if maybe it "works" better for you than the traditional algebraic method.

(1) Take the 10 hours that the one piece of equipment takes to do the job, and the 7 hours that it takes the two pieces together to do the same job.

(2) Consider a period of 70 hours, where 70 is the least common multiple of 10 and 7. In those 70 hours, the first piece of equipment could do this job 70/10 = 7 times; together the two pieces of equipment could do the job 70/7 = 10 times. That means the second piece of equipment could do the job 3 times in those 70 hours. Doing the job 3 times in 70 hours means doing the job once in 70/3 hours.

Answer: 70/3 hours, or 23 1/3 hours.
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