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A and B can complete a piece of work in 20 days and 30 days, respectively. A and B started the work and after 10 days, A left the job. Then, C joined with B and completed the job taking one day more than the estimated time. How many days would C take to complete the work alone?
Most Upvoted Answer
A and B can complete a piece of work in 20 days and 30 days, respectiv...
Given Information:
A can complete a piece of work in 20 days.
B can complete a piece of work in 30 days.
A and B started the work together.
After 10 days, A left the job.
C joined with B and completed the job in one day more than the estimated time.
To Find:
The number of days C would take to complete the work alone.
Assumptions:
The work is considered as 1 unit.
The efficiency of the workers is constant.
Solution:
Step 1: Calculate the work done by A and B in 10 days.
Let's assume the total work is 1 unit.
So, in 10 days, A completes 10/20 = 1/2 of the work.
In 10 days, B completes 10/30 = 1/3 of the work.
Step 2: Calculate the work remaining after 10 days.
The work remaining after 10 days is 1 - (1/2 + 1/3) = 1 - (3/6 + 2/6) = 1 - (5/6) = 1/6.
Step 3: Calculate the time taken by C to complete 1/6 of the work.
Since C and B together took one day more than the estimated time to complete the work, B alone would have taken 1/6 - 1/30 = 1/6 - 1/30 = 5/30 = 1/6 of the work.
So, C alone would take 1 day to complete 1/6 of the work.
Step 4: Calculate the time taken by C to complete the entire work.
Since C alone completes 1/6 of the work in 1 day, C would take 6 times more to complete the entire work.
Therefore, C would take 6 days to complete the work alone.
Final Answer:
C would take 6 days to complete the work alone.
A can complete a piece of work in 20 days.
B can complete a piece of work in 30 days.
A and B started the work together.
After 10 days, A left the job.
C joined with B and completed the job in one day more than the estimated time.
To Find:
The number of days C would take to complete the work alone.
Assumptions:
The work is considered as 1 unit.
The efficiency of the workers is constant.
Solution:
Step 1: Calculate the work done by A and B in 10 days.
Let's assume the total work is 1 unit.
So, in 10 days, A completes 10/20 = 1/2 of the work.
In 10 days, B completes 10/30 = 1/3 of the work.
Step 2: Calculate the work remaining after 10 days.
The work remaining after 10 days is 1 - (1/2 + 1/3) = 1 - (3/6 + 2/6) = 1 - (5/6) = 1/6.
Step 3: Calculate the time taken by C to complete 1/6 of the work.
Since C and B together took one day more than the estimated time to complete the work, B alone would have taken 1/6 - 1/30 = 1/6 - 1/30 = 5/30 = 1/6 of the work.
So, C alone would take 1 day to complete 1/6 of the work.
Step 4: Calculate the time taken by C to complete the entire work.
Since C alone completes 1/6 of the work in 1 day, C would take 6 times more to complete the entire work.
Therefore, C would take 6 days to complete the work alone.
Final Answer:
C would take 6 days to complete the work alone.
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A and B can complete a piece of work in 20 days and 30 days, respectively. A and B started the work and after 10 days, A left the job. Then, C joined with B and completed the job taking one day more than the estimated time. How many days would C take to complete the work alone?
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