John can do a piece of work in 15 days. John and David can do the same amount work in 6 days. In how many days David alone can do the same amount of work?
There are two ways to do the problem: (1) in your head and (2) by algebra.
The first way is "in your head":
The LCM of 6 days and 15 days is 30 days. So in 30 days John could do 2 pieces
of work. And if they worked together for 30 days they could do 5 pieces of
work. So if they worked together for 30 days, John would do 2 pieces of work
and therefore David would do the other 3 pieces of work. Therefore David can
do 3 pieces of work in 30 days, so he can do 1 piece of work in 10 days.
The second way is "by algebra":
Make this chart:
jobs done time in days rate in jobs/day
John
David
together
John can do 1 job in 15 days, so we fill in 1 for his jobs done and 15
for his time in days:
jobs done time in days rate in jobs/day
John 1 15
David
together
John and David together can do 1 job in 6 days, so we fill in 1 for his jobs
done and 6 for their time in days:
jobs done time in days rate in jobs/day
John 1 15
David
together 1 6
Let x = the number of days it would take David alone to do one piece of
work. So we fill in 1 for David's jobs done and x for his time in days:
jobs done time in days rate in jobs/day
John 1 15
David 1 x
together 1 6
Next we fill in the three rates in jobs/day by putting jobs over days:
jobs done time in days rate in jobs/day
John 1 15 1/15
David 1 x 1/x
together 1 6 1/6
The equation comes from the sum of their rates equals their
combined rate:
+ =
+ =
Get an LCD of 30x and multiply through and the answer will be x=10 days,
the same answer as when we do it in our head.
Edwin