R, S and T can finish a work in 20, 15 and 10 days, respectively. R works on all days and S and T work on alternate days with T starting the work on the first day. In how many days is the work finished?
52/7
Work and Time Problem Solution
This problem involves calculating the total time taken by R, S, and T to complete a piece of work, given their individual efficiencies and a specific work pattern.
Understanding Individual Work Rates
First, we need to determine the amount of work each person can complete in one day. This is their work rate.
- R can finish the work in 20 days. So, R's work rate is $\frac{1}{20}$ of the work per day.
- S can finish the work in 15 days. So, S's work rate is $\frac{1}{15}$ of the work per day.
- T can finish the work in 10 days. So, T's work rate is $\frac{1}{10}$ of the work per day.
Analyzing the Work Pattern and Cycle
R works on all days. S and T work on alternate days, with T starting on the first day.
The work pattern is as follows:
- Day 1: R and T work together.
- Day 2: R and S work together.
- Day 3: R and T work together (repeats Day 1).
- Day 4: R and S work together (repeats Day 2).
This establishes a 2-day cycle involving (R+T) on Day 1 of the cycle and (R+S) on Day 2 of the cycle.
Calculating Work Done in One Cycle
Let's calculate the total work done in one complete 2-day cycle:
- Work done on Day 1 (R + T): $R's\ rate + T's\ rate = \frac{1}{20} + \frac{1}{10}$
- To add these fractions, we find a common denominator, which is 20.
- $\frac{1}{20} + \frac{1}{10} = \frac{1}{20} + \frac{1 \times 2}{10 \times 2} = \frac{1}{20} + \frac{2}{20} = \frac{1+2}{20} = \frac{3}{20}$
- So, R and T together complete $\frac{3}{20}$ of the work on Day 1 of the cycle.
- Work done on Day 2 (R + S): $R's\ rate + S's\ rate = \frac{1}{20} + \frac{1}{15}$
- To add these fractions, we find a common denominator, which is 60.
- $\frac{1}{20} + \frac{1}{15} = \frac{1 \times 3}{20 \times 3} + \frac{1 \times 4}{15 \times 4} = \frac{3}{60} + \frac{4}{60} = \frac{3+4}{60} = \frac{7}{60}$
- So, R and S together complete $\frac{7}{60}$ of the work on Day 2 of the cycle.
Total work done in one 2-day cycle:
- Work per cycle = Work on Day 1 + Work on Day 2
- Work per cycle = $\frac{3}{20} + \frac{7}{60}$
- Common denominator is 60.
- $\frac{3}{20} + \frac{7}{60} = \frac{3 \times 3}{20 \times 3} + \frac{7}{60} = \frac{9}{60} + \frac{7}{60} = \frac{9+7}{60} = \frac{16}{60}$
- Simplify the fraction: $\frac{16}{60} = \frac{4 \times 4}{15 \times 4} = \frac{4}{15}$
- In one 2-day cycle, $\frac{4}{15}$ of the work is completed.
Progress Towards Completing the Work
We need to find how many full cycles are completed before the remaining work can be finished in a fraction of a day or one more day. The total work is 1 (or 100%).
Work done after 'n' cycles is $n \times \frac{4}{15}$. We want this value to be close to 1 but not exceed it for full cycles.
- After 1 cycle (2 days): Work done = $1 \times \frac{4}{15} = \frac{4}{15}$. Remaining work = $1 - \frac{4}{15} = \frac{11}{15}$.
- After 2 cycles (4 days): Work done = $2 \times \frac{4}{15} = \frac{8}{15}$. Remaining work = $1 - \frac{8}{15} = \frac{7}{15}$.
- After 3 cycles (6 days): Work done = $3 \times \frac{4}{15} = \frac{12}{15} = \frac{4}{5}$. Remaining work = $1 - \frac{4}{5} = \frac{1}{5}$.
- After 4 cycles (8 days): Work done = $4 \times \frac{4}{15} = \frac{16}{15}$. This is more than 1, so work finishes before 4 full cycles are completed.
So, 3 full cycles are completed in 6 days. After 6 days, $\frac{4}{5}$ of the work is done, and $\frac{1}{5}$ of the work remains.
Finishing the Remaining Work
After 6 days, the work remaining is $\frac{1}{5}$. The 6 days complete 3 full cycles. The 7th day is the start of the next cycle (Day 1 of the cycle), which means R and T work together.
- Work remaining = $\frac{1}{5}$.
- Work rate of R and T together = $\frac{3}{20}$ per day.
Can R and T finish the remaining $\frac{1}{5}$ work in a full day?
- Work R+T can do in 1 day = $\frac{3}{20}$.
- Compare remaining work ($\frac{1}{5} = \frac{4}{20}$) with work R+T can do in 1 day ($\frac{3}{20}$).
- Since $\frac{4}{20} > \frac{3}{20}$, R and T cannot finish the remaining work in one full 7th day. They will complete $\frac{3}{20}$ of the work on Day 7.
Work remaining after 7 days (6 days from cycles + 1 full day of R+T):
- Work done after 6 days = $\frac{4}{5}$.
- Work done on Day 7 (R+T) = $\frac{3}{20}$.
- Total work done after 7 days = $\frac{4}{5} + \frac{3}{20} = \frac{16}{20} + \frac{3}{20} = \frac{19}{20}$.
- Remaining work after 7 days = $1 - \frac{19}{20} = \frac{1}{20}$.
Now, it's the start of Day 8. This is Day 2 of the work cycle, meaning R and S work together.
- Work remaining = $\frac{1}{20}$.
- Work rate of R and S together = $\frac{7}{60}$ per day.
Time taken by R and S to finish the remaining $\frac{1}{20}$ work:
- Time = $\frac{\text{Remaining Work}}{\text{Rate}} = \frac{\frac{1}{20}}{\frac{7}{60}}$
- Time = $\frac{1}{20} \times \frac{60}{7} = \frac{60}{140} = \frac{6}{14} = \frac{3}{7}$ days.
So, on the 8th day, R and S work for $\frac{3}{7}$ of the day to finish the remaining work.
Calculating Total Time
Total time taken to finish the work is the sum of the full days worked and the fraction of the last day.
- Full days worked = 7 days (6 days from 3 cycles + 1 full Day 7)
- Fraction of the 8th day worked = $\frac{3}{7}$ days
- Total time = $7 + \frac{3}{7}$ days
- Total time = $\frac{7 \times 7 + 3}{7} = \frac{49 + 3}{7} = \frac{52}{7}$ days.
The work is finished in $\frac{52}{7}$ days.
Step-by-Step Work Calculation Summary
- R's rate = $1/20$ work/day
- S's rate = $1/15$ work/day
- T's rate = $1/10$ work/day
- Work on Day 1 (R+T) = $1/20 + 1/10 = 3/20$
- Work on Day 2 (R+S) = $1/20 + 1/15 = 7/60$
- Work in 1 cycle (2 days) = $3/20 + 7/60 = 9/60 + 7/60 = 16/60 = 4/15$
- Work after 3 cycles (6 days) = $3 \times 4/15 = 12/15 = 4/5$
- Remaining work after 6 days = $1 - 4/5 = 1/5$
- Day 7 is an R+T day. Work done on Day 7 = $3/20$
- Remaining work after 7 days = $1/5 - 3/20 = 4/20 - 3/20 = 1/20$
- Day 8 is an R+S day. Their rate = $7/60$ work/day
- Time on Day 8 to finish $1/20$ work = $(1/20) / (7/60) = (1/20) \times (60/7) = 3/7$ days
- Total time = 7 full days + $3/7$ days = $7 + 3/7 = 52/7$ days.
| Concept | Formula/Explanation |
|---|---|
| Work Rate | If a person finishes work in 'n' days, their rate is $\frac{1}{n}$ per day. |
| Total Work | Usually considered as 1 unit. |
| Time Taken | $\frac{\text{Total Work}}{\text{Work Rate}}$ (if rate is constant). |
| Work Done | Rate $\times$ Time |
| Combined Rate | Sum of individual rates when working together. |
| Alternate Day Work | Calculate work done over a cycle (usually 2 days) and find how many cycles are needed. Address remaining work separately. |
Additional Information: Alternate Day Work Problems
Alternate day work problems are a common type in time and work. The key is to identify the repeating cycle of workers and calculate the work done in one such cycle. Then, determine how many full cycles are completed and deal with the remaining work by considering the work rate of the individuals scheduled for the day(s) following the last full cycle. Pay close attention to who starts the work, as this determines the sequence of workers in the cycle.
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