Work and Time Problem: Calculating C's Days Worked
This problem involves calculating the time taken for a task when some individuals leave before completion. We use the concept of work rate.
1. Calculate Individual Work Rates
First, find the Least Common Multiple (LCM) of the days taken by A, B, and C to find the total units of work.
- Days taken: A = 36, B = 54, C = 72
- LCM(36, 54, 72) = 216 units. Total work = 216 units.
- Work rate of A: $\frac{216}{36} = 6$ units/day
- Work rate of B: $\frac{216}{54} = 4$ units/day
- Work rate of C: $\frac{216}{72} = 3$ units/day
2. Set Up the Equation Based on Work Done
Let '$D$' be the total number of days the task took to complete. C worked for all $D$ days.
A left 8 days before completion, so A worked for $(D - 8)$ days.
B left 12 days before completion, so B worked for $(D - 12)$ days.
The total work done is the sum of the work done by A, B, and C:
$Total Work = (Rate_A \times Days_A) + (Rate_B \times Days_B) + (Rate_C \times Days_C)$ $216 = (6 \times (D - 8)) + (4 \times (D - 12)) + (3 \times D)$3. Solve for Total Days (D)
Now, solve the equation:
$216 = 6D - 48 + 4D - 48 + 3D$ $216 = (6D + 4D + 3D) - (48 + 48)$ $216 = 13D - 96$ $216 + 96 = 13D$ $312 = 13D$ $D = \frac{312}{13}$ $D = 24$4. Determine Days C Worked
The total number of days the task took is $D$, which is 24 days. C worked for the entire duration.
Therefore, C worked for 24 days.
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