Mason A Alone Work Calculation
This problem involves calculating the time taken by Mason A to complete a construction work alone, given information about combined work time and the relative speeds of Mason A and Mason B.
Problem Setup
- Let $d_A$ be the number of days Mason A takes to complete the work alone.
- Let $d_B$ be the number of days Mason B takes to complete the work alone.
- Given: A and B together complete the work in 22.5 days.
- Given: Mason A alone completes the work in 24 days less than Mason B alone. This translates to the equation: $d_A = d_B - 24$, or $d_B = d_A + 24$.
Work Rate Equation
The rate at which work is done is the reciprocal of the time taken. The combined rate of A and B is the sum of their individual rates. The equation is:
$ \frac{1}{d_A} + \frac{1}{d_B} = \frac{1}{22.5} $Solving for $d_A$
- Substitute $d_B = d_A + 24$ into the work rate equation: $ \frac{1}{d_A} + \frac{1}{d_A + 24} = \frac{1}{22.5} $
- Combine the fractions on the left side: $ \frac{(d_A + 24) + d_A}{d_A(d_A + 24)} = \frac{1}{22.5} $ $ \frac{2d_A + 24}{d_A^2 + 24d_A} = \frac{1}{22.5} $
- Cross-multiply: $ 22.5 \times (2d_A + 24) = d_A^2 + 24d_A $ $ 45d_A + 540 = d_A^2 + 24d_A $
- Rearrange into a quadratic equation: $ d_A^2 + 24d_A - 45d_A - 540 = 0 $ $ d_A^2 - 21d_A - 540 = 0 $
- Factor the quadratic equation. We need two numbers that multiply to -540 and add to -21. These numbers are -36 and 15. $ (d_A - 36)(d_A + 15) = 0 $
- Solve for $d_A$: The possible values for $d_A$ are $36$ or $-15$. Since time cannot be negative, we discard $-15$.
Conclusion
Therefore, Mason A alone will complete the construction work in 36 days.
Verification
If $d_A = 36$ days, then $d_B = d_A + 24 = 36 + 24 = 60$ days.
Combined work rate = $\frac{1}{36} + \frac{1}{60} = \frac{5}{180} + \frac{3}{180} = \frac{8}{180} = \frac{2}{45}$.
Time taken together = $\frac{1}{\text{Combined rate}} = \frac{1}{2/45} = \frac{45}{2} = 22.5$ days. This matches the given information.
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