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Question 176993: It's late here and I can't figure out how to start this. I did the crane off loading problem OK but somehow the formula isn't working. I would greatly appreciate your help.
Show an equation and a solution for the problem. At the coal mine, it takes the older conveyor belt 20 hours to move one day's coal output. The new belt can do it in 15 hours. If the old belt began to move the day's output and the new belt started as well an hour later, how long would it take to move it?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! The old belt moves coal at the rate of 1/20 day's output per hour
The new belt moves coal at the rate of 1/15 day's output per hour
Together the two belts move coal at the rate of 1/20 +1/15=3/60 + 4/60=
7/60 day's output per hour
In one hour the old belt can move 1/20 day's output leaving 19/20 day's output yet to be moved
Let t(total)=total time needed to move the coal, then
t(total)=time required by older belt working alone plus time required by both belts working together to finish the job, or:
t(total)=t(older) + t(older+newer) now we know that
t(older)=1
and
(7/60)*t(older+newer)=19/20 or
t(older+newer)=(19/20)(60/7)=57/7=8 1/7 day
So t(total)=1+8 1/7=9 1/7 days
CK
(1/20)*1+(57/7)(7/60)=1 (1 day's output)
1/20 + 57/60=1
3/60 + 57/60=1
1=1
Does this help???-----ptaylor
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