A group of men decided to do a job in 13 days but 9 men left the work after each day. The work, as a result, got completed in 16 days. How many men were initially in the group?
360
Understanding the Work and Time Problem
This problem involves calculating the initial number of men required for a job when the workforce changes over time. We are given a planned duration and an actual duration with a decreasing number of workers each day. The total amount of work remains the same in both scenarios.
Setting up the Problem Variables
Let's define the variables:
- Let the initial number of men in the group be \(N\).
- The planned time to complete the job was 13 days.
- The work was actually completed in 16 days.
- After each day, 9 men left the work.
Calculating Total Work in the Planned Scenario
If \(N\) men worked for the planned 13 days, the total work done can be calculated as the product of the number of men and the number of days. We can measure work in 'man-days'.
Total Work (Planned) = Number of men \(\times\) Planned days
Total Work (Planned) = \(N \times 13\) man-days
Total Work (Planned) = \(13N\) man-days
Calculating Total Work in the Actual Scenario
In the actual scenario, the number of men working decreased each day for 16 days. Let's list the number of men working on each day:
- Day 1: \(N\) men
- Day 2: \(N - 9\) men
- Day 3: \(N - 2 \times 9\) men
- Day 4: \(N - 3 \times 9\) men
- ...
- Day 16: \(N - 15 \times 9\) men
The total work done in the actual scenario is the sum of the man-days worked over these 16 days.
Total Work (Actual) = Sum of (Number of men on day \(k\)) for \(k\) from 1 to 16.
Total Work (Actual) = \(\sum_{k=1}^{16} (N - 9(k-1))\)
This sum can be expanded as:
Total Work (Actual) = \((N - 9 \times 0) + (N - 9 \times 1) + (N - 9 \times 2) + ... + (N - 9 \times 15)\)
This is the sum of an arithmetic series. We can rewrite the sum as:
Total Work (Actual) = \(\sum_{k=1}^{16} N - \sum_{k=1}^{16} 9(k-1)\)
Total Work (Actual) = \(16N - 9 \sum_{k=1}^{16} (k-1)\)
The summation \(\sum_{k=1}^{16} (k-1)\) is the sum of the first 15 non-negative integers (0, 1, 2, ..., 15). The sum of the first \(m\) integers (starting from 1) is \(\frac{m(m+1)}{2}\). Here, we sum from 0 to 15, which is the same as summing from 1 to 15, adding 0. So, the sum is \(\frac{15(15+1)}{2} = \frac{15 \times 16}{2} = 15 \times 8 = 120\).
So, Total Work (Actual) = \(16N - 9 \times 120\)
Total Work (Actual) = \(16N - 1080\) man-days
Equating Total Work and Solving for the Initial Number of Men
Since the same job was completed in both scenarios, the total work done must be equal.
Total Work (Planned) = Total Work (Actual)
\(13N = 16N - 1080\)
Now, we need to solve this equation for \(N\).
Subtract \(13N\) from both sides:
\(0 = 16N - 13N - 1080\)
\(0 = 3N - 1080\)
Add 1080 to both sides:
\(1080 = 3N\)
Divide by 3:
\(N = \frac{1080}{3}\)
\(N = 360\)
Conclusion
The initial number of men in the group was 360.
Let's quickly check:
- Planned work: 360 men \(\times\) 13 days = 4680 man-days.
- Actual work: Sum of men for 16 days. Men on day 1 = 360, Day 2 = 351, ..., Day 16 = \(360 - 15 \times 9 = 225\). This is an arithmetic series with first term 360, last term 225, and 16 terms. Sum = \(\frac{16}{2}(360 + 225) = 8 \times 585 = 4680\) man-days.
The total work matches, confirming the initial number of men is correct.
| Scenario | Number of Men | Number of Days | Total Work (man-days) |
|---|---|---|---|
| Planned | \(N\) | 13 | \(13N\) |
| Actual | Decreasing by 9 each day (Arithmetic Series) | 16 | \(16N - 1080\) |
Revision Table: Key Concepts in Work and Time Problems
| Concept | Explanation | Formula/Idea |
|---|---|---|
| Total Work | Often measured in man-days (or units of workers \(\times\) time). Represents the total effort needed to complete the job. | Work = Number of Workers \(\times\) Time |
| Constant Work | When the same job is done under different conditions (varying workers/time), the total work remains the same. | Work\(_{Scenario 1}\) = Work\(_{Scenario 2}\) |
| Varying Workforce | If the number of workers changes over time, the total work is the sum of work done each day (or time unit) with the respective number of workers. | Total Work = \(\sum\) (Workers on day \(i\) \(\times\) 1 day) |
| Arithmetic Progression (AP) | A sequence where the difference between consecutive terms is constant. Useful for sum of workers when the number changes uniformly. | Sum of AP = \(\frac{n}{2} (a + l)\) or \(\frac{n}{2} (2a + (n-1)d)\), where \(n\) is number of terms, \(a\) is first term, \(l\) is last term, \(d\) is common difference. |
Additional Information: Solving Work Problems with Arithmetic Progression
When the number of workers increases or decreases by a fixed amount each day, the total work calculation over several days involves the sum of an arithmetic progression. In this specific problem, the number of men on consecutive days formed an arithmetic progression with the first term \(N\) and a common difference of \(-9\). The terms were \(N, N-9, N-18, ..., N-9(16-1)\). To find the total work, we summed these terms over 16 days.
The sum of an arithmetic progression is given by the formula \(S_n = \frac{n}{2}(2a + (n-1)d)\), where \(S_n\) is the sum of the first \(n\) terms, \(a\) is the first term, and \(d\) is the common difference.
In our case:
- \(n = 16\) (number of days)
- \(a = N\) (initial number of men)
- \(d = -9\) (decrease in men each day)
So, the total work (actual) is \(S_{16} = \frac{16}{2}(2N + (16-1)(-9)) = 8(2N + 15(-9)) = 8(2N - 135) = 16N - 1080\). This matches the sum calculated earlier by summing 0 to 15 and multiplying by 9, then subtracting from 16N.
This method provides a structured way to calculate the total work done by a workforce that changes consistently over time.
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Important Questions from Work Efficiency
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