C1, C2 and C3 can do a work alone in 10, 12 and 15 days respectively. All three of them began the work together but C2 left 2 days before the completion of the work. In how many days was the work completed?
(14/3) days
Solving a Work and Time Problem with Multiple Workers
This problem involves calculating the total time taken to complete a piece of work when multiple individuals work together, but one of them leaves before the work is finished. We need to use the concept of individual work rates and set up an equation based on the total work done.
Understanding Individual Work Rates
The work rate of a person is the amount of work they can complete in one day. If a person takes \(d\) days to complete a work alone, their daily work rate is \(1/d\) of the work.
- C1 takes 10 days to complete the work. C1's daily work rate is \( \frac{1}{10} \).
- C2 takes 12 days to complete the work. C2's daily work rate is \( \frac{1}{12} \).
- C3 takes 15 days to complete the work. C3's daily work rate is \( \frac{1}{15} \).
Setting up the Problem
Let the total number of days taken to complete the work be \(T\) days. We are given that C2 left 2 days before the completion of the work. This means:
- C1 worked for the entire duration, which is \(T\) days.
- C2 worked for \(T - 2\) days.
- C3 worked for the entire duration, which is \(T\) days.
Calculating Total Work Done
The total work done is the sum of the work done by C1, C2, and C3. The total work completed is represented as 1 (or 100%).
- Work done by C1 in \(T\) days = C1's daily rate \(\times\) Number of days C1 worked = \( \frac{1}{10} \times T = \frac{T}{10} \).
- Work done by C2 in \(T - 2\) days = C2's daily rate \(\times\) Number of days C2 worked = \( \frac{1}{12} \times (T - 2) = \frac{T - 2}{12} \).
- Work done by C3 in \(T\) days = C3's daily rate \(\times\) Number of days C3 worked = \( \frac{1}{15} \times T = \frac{T}{15} \).
The total work done is the sum of these individual contributions:
\( \text{Total Work} = \frac{T}{10} + \frac{T - 2}{12} + \frac{T}{15} \)
Since the work is completed, the total work done equals 1:
\( \frac{T}{10} + \frac{T - 2}{12} + \frac{T}{15} = 1 \)
Solving the Equation for T
To solve this equation for \(T\), we need to find a common denominator for 10, 12, and 15. The least common multiple (LCM) of 10, 12, and 15 is 60.
Multiply the entire equation by 60 to eliminate the denominators:
\( 60 \times \left( \frac{T}{10} + \frac{T - 2}{12} + \frac{T}{15} \right) = 60 \times 1 \)
\( 60 \times \frac{T}{10} + 60 \times \frac{T - 2}{12} + 60 \times \frac{T}{15} = 60 \)
\( 6T + 5(T - 2) + 4T = 60 \)
Now, simplify and solve for \(T\):
\( 6T + 5T - 10 + 4T = 60 \)
\( (6T + 5T + 4T) - 10 = 60 \)
\( 15T - 10 = 60 \)
Add 10 to both sides:
\( 15T = 60 + 10 \)
\( 15T = 70 \)
Divide by 15:
\( T = \frac{70}{15} \)
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 5:
\( T = \frac{70 \div 5}{15 \div 5} = \frac{14}{3} \)
So, the total number of days the work was completed is \( \frac{14}{3} \) days.
Verification (Optional but Recommended)
Let's check if the sum of the work done by each person in the calculated time equals 1.
- C1 works for \( \frac{14}{3} \) days. Work by C1 = \( \frac{1}{10} \times \frac{14}{3} = \frac{14}{30} = \frac{7}{15} \).
- C2 works for \( \frac{14}{3} - 2 \) days. \( \frac{14}{3} - 2 = \frac{14}{3} - \frac{6}{3} = \frac{8}{3} \) days. Work by C2 = \( \frac{1}{12} \times \frac{8}{3} = \frac{8}{36} = \frac{2}{9} \).
- C3 works for \( \frac{14}{3} \) days. Work by C3 = \( \frac{1}{15} \times \frac{14}{3} = \frac{14}{45} \).
Total work = \( \frac{7}{15} + \frac{2}{9} + \frac{14}{45} \)
Find the LCM of 15, 9, and 45, which is 45.
Total work = \( \frac{7 \times 3}{15 \times 3} + \frac{2 \times 5}{9 \times 5} + \frac{14}{45} = \frac{21}{45} + \frac{10}{45} + \frac{14}{45} \)
Total work = \( \frac{21 + 10 + 14}{45} = \frac{45}{45} = 1 \)
The total work is 1, which means the calculation is correct.
| Person | Time Alone (days) | Daily Work Rate | Time Worked (days) | Work Done |
|---|---|---|---|---|
| C1 | 10 | \(1/10\) | \(T\) | \(T/10\) |
| C2 | 12 | \(1/12\) | \(T-2\) | \((T-2)/12\) |
| C3 | 15 | \(1/15\) | \(T\) | \(T/15\) |
The total time taken for the work to be completed is \( \frac{14}{3} \) days.
Revision Table: Work and Time Concepts
| Concept | Explanation | Formula/Example |
|---|---|---|
| Work Rate | The amount of work done per unit of time (e.g., per day). | If A does work in \(D\) days, rate = \(1/D\) per day. |
| Total Work | Usually represented as 1 unit. | Work = Rate \(\times\) Time |
| Combined Rate | Sum of individual rates when working together. | If A's rate is \(R_A\) and B's rate is \(R_B\), combined rate = \(R_A + R_B\). |
| Work Done | Fraction or amount of work completed. | Work Done by A in \(t\) days = \(R_A \times t\). |
Additional Information: Alternative Approach
Another way to approach this problem is to consider the work done in the last 2 days. In the last 2 days, only C1 and C3 were working (since C2 left 2 days before completion).
- Combined daily rate of C1 and C3 = \( \frac{1}{10} + \frac{1}{15} \).
- LCM of 10 and 15 is 30.
- Combined rate = \( \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \) of the work per day.
- Work done by C1 and C3 in the last 2 days = \( \frac{1}{6} \times 2 = \frac{2}{6} = \frac{1}{3} \) of the work.
This means the remaining \( 1 - \frac{1}{3} = \frac{2}{3} \) of the work was done by all three (C1, C2, and C3) working together at the beginning.
- Combined daily rate of C1, C2, and C3 = \( \frac{1}{10} + \frac{1}{12} + \frac{1}{15} \).
- LCM of 10, 12, 15 is 60.
- Combined rate = \( \frac{6}{60} + \frac{5}{60} + \frac{4}{60} = \frac{15}{60} = \frac{1}{4} \) of the work per day.
Let the number of days all three worked together be \(t_1\). They completed \( \frac{2}{3} \) of the work in \(t_1\) days.
\( \text{Combined Rate} \times t_1 = \text{Work Done} \)
\( \frac{1}{4} \times t_1 = \frac{2}{3} \)
\( t_1 = \frac{2}{3} \times 4 = \frac{8}{3} \) days.
The total time for the work is the time all three worked together plus the time only C1 and C3 worked:
\( \text{Total Time} = t_1 + 2 \text{ days} \)
\( \text{Total Time} = \frac{8}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3} \) days.
Both methods give the same result, \( \frac{14}{3} \) days, confirming the answer.
Important Questions from Work Efficiency
A and B together can complete a certain work in 20 days whereas B and C together can complete it in 24 days. If A is twice as good a workman as C, then in what time will B alone do 40% of the same work?
14 men can complete a work in 15 days. If 21 men are employed, then in how many days will they complete the same work?
A can do a certain work in 15 days, while B can do the same work in 21 days. If they work together, then in how many days will the same work be completed?
To do a certain work, A and B work on alternate days with B beginning the work on the first day. A alone can complete the same work in 24 days. If the work gets completed in \(11 \frac{1}{3}\) days, then B alone can complete \(\rm \frac{7}{9}^{th}\) part of the original work in:
Two men and 7 women can complete a work in 28 days whereas 6 men and 16 women can do the same work in 11 days. In how many days can 7 men complete the same work?
