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Question 100929This question is from textbook Intermediate Algebra
: Holly takes 50 minutes to mow the lawn. It takes Erich an hour to mow the same lawn. One day Holly started mowing the lawn by herself and worked for 10 minutes. The Erich joined her and the finished the lawn together. How long did take them to finish mowing the lawn after Erich started to help? How do I find the end of this problem. There is the same problem listed on here, but I don't understand where they got the 4/5 fraction from. Someone please help!
This question is from textbook Intermediate Algebra
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Let x=time it takes both to finish the lawn after Erich started to help
Holly mows the lawn at the rate of 1/50 lawn per minute
Erich mows the lawn at the rate of 1/60 lawn per minute
Together they mow the lawn at the rate of (1/50+1/60 )lawn per minute
1/50+1/60: the LCM is 300 so 1/50=6/300 and 1/60=5/300.
So together they mow the lawn at the rate of (6+5)/300 or 11/300 lawn per minute
Now when Holly mowed the lawn for 10 minutes, she finished (1/50)*10 or 1/5 of the lawn so there's 4/5 of the lawn yet to be mowed:
The question now is: If Holly and Erich can mow at the rate of 11/300 lawn per minute, how long will it take them to mow 4/5 of the lawn. Our equation is:
(11/300)*x=4/5 multiply both sides by 300
11x=240 divide both sides by 11
x=240/11 or 21.8181818181818 minutes
CK
(11/300)*(240/11)=4/5
240/300=4/5
4/5=4/5
Hope this helps----ptaylor
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