Solution:
Given:
P alone can complete a work in 16 days
Q alone can complete the same work in 20 days
P and Q start the work together but Q leaves the work 7 days before the completion of work
Formula used:
Number of days = Total work/1 day's work
Calculation:
P → 16 days
Q → 20 days
Total work = LCM (20, 16) = 80 unit
1 day work of P = 80/16 = 5 unit
1 day work of Q = 80/20 = 4 unit
Now,
P and Q start the work together but Q leaves the work 7 days before the completion of work
(P + Q)'s one day's work = 5 + 4 = 9 unit
Let number of days to complete the Total work be x
⇒ (9 × (x - 7)) + (5 × 7) = 80
⇒ (9x - 63) + 35 = 80
⇒ 9x = 108
∴ x= 108/9 = 12
In 12 days the total work will be completed.
Alternate method:
According to the question,
P and Q start the work together but Q leaves the work 7 days before the completion of work
Q work complete in 7 days = 4 × 7 = 28 unit
It means 28 unit work by P
Total work done by P and Q
⇒ Total work/efficiency per day both
⇒ (80 + 28)/(5 + 4)
⇒ 108/9 = 12 days
∴ In 12 days the total work will be completed.
