X can do a piece of work in 20 days. He worked at it for 5 days and th...
- X’s five day work = 5/20 = 1/4. Remaining work = 1 – 1/4 = 3/4.
- This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 × 4/3 = 20 days.
- X & Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.
X can do a piece of work in 20 days. He worked at it for 5 days and th...
Given:
X can do a piece of work in 20 days.
X worked for 5 days.
Y finished it in 15 days.
To find: In how many days can X and Y together finish the work?
Solution:
Let us assume that the total work is 1 unit.
X's efficiency = 1/20 (i.e., he can complete 1/20th of the work in 1 day)
Y's efficiency = 1/15 (i.e., he can complete 1/15th of the work in 1 day)
X worked for 5 days, so the work done by him = (1/20) * 5 = 1/4
The remaining work = 1 - 1/4 = 3/4
Now, Y will finish the remaining work in 15 days.
So, the work done by Y in 1 day = 1/15
Let us assume that X and Y together can finish the work in 'd' days.
So, their combined efficiency = 1/d
Now, we can form an equation based on the work done by X and Y together:
(1/20)*5 + (1/d)*x = 3/4
Simplifying the above equation, we get:
x/d = 9/20
So, d/x = 20/9
Therefore, the number of days required for X and Y together to finish the work = x = (20/9)*d
We know that Y can finish the remaining work in 15 days.
So, (1/15)*d = 3/4
Solving the above equation, we get:
d = 45/4
Therefore, X and Y together can finish the work in:
x = (20/9)*d = (20/9)*(45/4) = 10 days
Hence, the correct option is (c) 10 days.