document.write( "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? \n" ); document.write( "

Algebra.Com's Answer #709662 by greenestamps(13200) $\"\"$ $\"About$

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.

\n" ); document.write( "(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.

\n" ); document.write( "(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.

\n" ); document.write( "Answer: 70/3 hours, or 23 1/3 hours. \n" ); document.write( "

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