Question

16 workers working 8 hours per day can demolish a building in 32 days. In how many days 24 workers working 12 hours per day can demolish the same building?

• A. 128/3 days
• B. 56/3 days
• C. 128/9 days
• D. 56/9 days

Hint:

To avoid calculation errors, do not multiply the numbers out before dividing.
Keep them in factored form, as shown in Step 3. This allows for quick cancellation and prevents dealing with unnecessarily large numbers during the exam.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem involves the concept of Work Equivalence.
Total work is directly proportional to the number of workers (M), the number of days they work (D), and the number of hours they work per day (H).
If the same job (demolishing the building) is done by two different sets of workers, the product of these variables must be equal, provided the work done (W) is constant.
Step 2: Key Formula or Approach:
We use the MDH formula:
\[ M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2 \]
Where:
\( M \) = Men (Workers)
\( D \) = Days
\( H \) = Hours per day.
Step 3: Detailed Explanation:
From the question, we define our values:
Scenario 1: \( M_1 = 16, D_1 = 32, H_1 = 8 \).
Scenario 2: \( M_2 = 24, H_2 = 12 \), and we need to find \( D_2 \).
Set up the equation:
\[ 16 \times 32 \times 8 = 24 \times D_2 \times 12 \]
To isolate \( D_2 \), rearrange the terms:
\[ D_2 = \frac{16 \times 32 \times 8}{24 \times 12} \]
Now, simplify by canceling common factors:
Divide 8 in the numerator and 24 in the denominator by 8:
\[ D_2 = \frac{16 \times 32 \times 1}{3 \times 12} \]
Now, divide 16 and 12 by their common factor of 4:
\[ D_2 = \frac{4 \times 32}{3 \times 3} \]
Multiply the remaining terms:
\[ D_2 = \frac{128}{9} \]
The number of days required is 128/9.
Step 4: Final Answer:
The number of days required by 24 workers working 12 hours per day is 128/9 days.