Diptesh and Tushar working separately can do a piece of work in 9 and 12 days respectively. If they work for a day alternately, Diptesh beginning, in how many days, the work will be completed?
41/4
Solving Work and Time Problems with Alternate Working
This problem involves two individuals, Diptesh and Tushar, working alternately on a task. To solve such problems, we first need to determine their individual work rates and then calculate the work done in each cycle (a cycle includes one day for each person).
Understanding Individual Work Rates
The work rate is the amount of work a person can do in one day. If a person can complete a job in 'n' days, their one day work rate is \( \frac{1}{n} \) of the total work.
- Diptesh can do the work in 9 days.
- Diptesh's one day work rate = \( \frac{1}{9} \) of the total work.
- Tushar can do the work in 12 days.
- Tushar's one day work rate = \( \frac{1}{12} \) of the total work.
Calculating Work Done in One Cycle
They work alternately, with Diptesh beginning. A complete cycle consists of Diptesh working for one day and then Tushar working for one day. So, one cycle takes 2 days.
Work done in 1st day (by Diptesh) = \( \frac{1}{9} \)
Work done in 2nd day (by Tushar) = \( \frac{1}{12} \)
Work done in one cycle (2 days) = Work done by Diptesh + Work done by Tushar
Work done in one cycle = \( \frac{1}{9} + \frac{1}{12} \)
To add these fractions, we find a common denominator, which is the LCM of 9 and 12. LCM(9, 12) = 36.
\( \frac{1}{9} = \frac{1 \times 4}{9 \times 4} = \frac{4}{36} \)
\( \frac{1}{12} = \frac{1 \times 3}{12 \times 3} = \frac{3}{36} \)
Work done in one cycle (2 days) = \( \frac{4}{36} + \frac{3}{36} = \frac{4+3}{36} = \frac{7}{36} \)
So, in every 2-day cycle, \( \frac{7}{36} \) of the total work is completed.
Determining the Number of Full Cycles
We need to find out how many full cycles of 2 days can occur before the work is almost finished. The total work is considered as 1 unit. We want to find the number of cycles such that the work done is close to, but not exceeding, 1.
Let 'n' be the number of cycles. Work done in 'n' cycles = \( n \times \frac{7}{36} \).
We look for the largest 'n' such that \( n \times \frac{7}{36} \le 1 \).
If \( n=5 \), work done = \( 5 \times \frac{7}{36} = \frac{35}{36} \). This is less than 1.
If \( n=6 \), work done = \( 6 \times \frac{7}{36} = \frac{42}{36} \). This is greater than 1, meaning the work is completed within the 6th cycle.
Therefore, 5 full cycles are completed.
Calculating Work Done After Full Cycles and Remaining Work
After 5 full cycles (which take \( 5 \times 2 = 10 \) days), the work completed is \( \frac{35}{36} \).
Remaining work = Total work - Work done in 5 cycles
Remaining work = \( 1 - \frac{35}{36} = \frac{36}{36} - \frac{35}{36} = \frac{1}{36} \).
So, \( \frac{1}{36} \) of the work still needs to be done.
Completing the Remaining Work
After 5 cycles (10 days), the 11th day begins. The turn is for the person who started the work, which is Diptesh.
Diptesh's one day work rate is \( \frac{1}{9} \).
He needs to complete the remaining work of \( \frac{1}{36} \).
Time taken by Diptesh to complete \( \frac{1}{36} \) of the work = \( \frac{\text{Remaining Work}}{\text{Diptesh's One Day Work Rate}} \)
Time taken = \( \frac{\frac{1}{36}}{\frac{1}{9}} = \frac{1}{36} \times \frac{9}{1} = \frac{9}{36} = \frac{1}{4} \) day.
Total Time Taken
Total time = Time for 5 cycles + Time for remaining work
Total time = 10 days + \( \frac{1}{4} \) day
Total time = \( 10 \frac{1}{4} \) days.
In improper fraction form, \( 10 \frac{1}{4} = \frac{(10 \times 4) + 1}{4} = \frac{40 + 1}{4} = \frac{41}{4} \) days.
Thus, the work will be completed in \( \frac{41}{4} \) days.
Revision Table: Work and Time Concepts
| Concept | Explanation | Formula/Calculation |
|---|---|---|
| Work Rate | Amount of work done per unit of time (e.g., per day). | If work takes N days, rate = \( \frac{1}{N} \) work/day. |
| Total Work | Usually represented as 1 unit. | Work Rate \( \times \) Time Taken = Total Work (1) |
| Combined Work Rate | Sum of individual work rates when working together. | Rate1 + Rate2 + ... |
| Alternate Working | Individuals take turns doing the work. Calculate work done in a cycle (usually includes one turn for each). | Work in cycle = Sum of work rates for one turn each. |
Additional Information: Efficiency and Time
Efficiency is inversely proportional to the time taken to complete a task. If a person is more efficient, they take less time, and their work rate is higher.
- Diptesh takes 9 days, Tushar takes 12 days.
- Diptesh is more efficient than Tushar because he takes less time.
- Diptesh's rate (\( \frac{1}{9} \)) is greater than Tushar's rate (\( \frac{1}{12} \)).
When solving alternate working problems, it's crucial to track whose turn it is after each full cycle of turns is completed, as this person will start the next sequence of turns or complete the remaining work.
Important Questions from Work Efficiency
A and B working together can complete a job in 30 days. The ratio of their efficiencies is 3 : 2. In how many days can the faster person complete the job?
A takes 15 days to complete \(\frac{5}{7} \) of a work. With the help of B, they finish the whole work in 12 days. In how many days, B alone will complete the same work
A alone can complete a work in 14 days and B alone can complete the same work in 21 days. A and B start the work together but A leaves the work after 4 days of the starting of work. In how many days B will complete the remaining work?
For completing a certain work, A is 50% less efficient than B and B is 50% more efficient than C. Working together A, B and C can complete the work in 48 days. A alone can complete the same work in:
30 persons can do a piece of work in 24 days. How many more persons are required to complete the work in 20 days?
