Question

In 15 days, 9 women and 15 girls can reap a field. In what time could 15 women and 16 girls reap the same field, assuming 3 women can do as much work as 4 girls?

• A. 10.5 days
• B. 25 days
• C. 11.25 days
• D. 20 days

Hint:

In work problems with different types of workers, always start by converting all worker types into a single equivalent unit based on their given efficiency ratio. This transforms the problem into a simpler one with a single type of worker.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a Time and Work problem involving different efficiencies. We must convert the labor force into a single common unit (either all women or all girls) to use the work-rate formula.
Step 2: Key Formula or Approach:
1. Efficiency relation: \( 3W = 4G \Rightarrow 1W = \frac{4}{3}G \).
2. Work \( = \text{Rate} \times \text{Time} \). Since work is constant, \( M_1 D_1 = M_2 D_2 \).
Step 3: Detailed Explanation:
Given: Efficiency of 3 women \( = \) Efficiency of 4 girls.
Therefore, 1 woman \( = \frac{4}{3} \) girls.
Case 1: 9 women and 15 girls.
Convert 9 women to girls: \( 9 \times \frac{4}{3} = 12 \) girls.
Total labor in Case 1 \( = 12 + 15 = 27 \) girls.
Time taken \( = 15 \) days.
Total Work \( = 27 \times 15 \) girl-days.
Case 2: 15 women and 16 girls.
Convert 15 women to girls: \( 15 \times \frac{4}{3} = 20 \) girls.
Total labor in Case 2 \( = 20 + 16 = 36 \) girls.
Let the time taken be \( D \).
Using \( M_1 D_1 = M_2 D_2 \):
\[ 27 \times 15 = 36 \times D \] \[ D = \frac{27 \times 15}{36} \] \[ D = \frac{3 \times 15}{4} \] \[ D = \frac{45}{4} = 11.25 \text{ days} \]
Step 4: Final Answer:
15 women and 16 girls can reap the field in 11.25 days.