Question 176993
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