'A' can finish a work in 10 days. The efficiency of 'A' is 20% less than 'B'. In how many days B will finish the same work?

Options

A. 8.5 days
B. 12.3 days
C. 8 days
D. 7.5 days

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Determine A's Efficiency
A can complete the work in 10 days. Therefore, A's efficiency can be calculated as:
\[ \text{Efficiency of A} = \frac{\text{Total Work}}{\text{Time taken by A}} = \frac{W}{10} \]
Let’s denote the total work as \( W \).

### Step 2: Determine B's Efficiency
According to the problem, the efficiency of A is 20% less than that of B. This means:
\[ \text{Efficiency of A} = \text{Efficiency of B} - 20\% \text{ of Efficiency of B} \]
Let the efficiency of B be \( E_B \). Therefore:
\[ \frac{W}{10} = E_B - 0.2E_B = 0.8E_B \]
From this, we can express \( E_B \):
\[ E_B = \frac{W}{10 \times 0.8} = \frac{W}{8} \]

### Step 3: Calculate Total Work
Now we know that the total work \( W \) can be expressed in terms of A's efficiency:
\[ W = \text{Efficiency of A} \times \text{Time taken by A} = \frac{W}{10} \times 10 = W \]
For calculation purposes, let's assume \( W = 40 \) units (as derived from the efficiency calculations).

### Step 4: Calculate Time Taken by B
Now, we can find out how many days B will take to finish the same work:
\[ \text{Time taken by B} = \frac{\text{Total Work}}{\text{Efficiency of B}} = \frac{W}{E_B} = \frac{40}{\frac{W}{8}} = 8 \text{ days} \]

### Final Answer
B will finish the work in **8 days**.

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