The ratio of efficiency of A is to C is `5:3`. The ratio of number of days taken by B is to C is `2 : 3`. A takes 6 days less than C, when A and C completes the work individually. B and C started the work and left after 2 days. The number of days taken by A to finish the remaining work is :

Options

A. 4.5
B. 5
C. 6
D. `9(1)/(3)`

Answer

To solve the problem step by step, let's break down the information provided and calculate the required values.

### Step 1: Understand the efficiency ratios
- The efficiency ratio of A to C is given as \(5:3\).
- This means if A's efficiency is \(5x\), then C's efficiency is \(3x\).

### Step 2: Determine the time ratios
- The time taken by A and C will be the inverse of their efficiencies. Therefore, the time ratio of A to C is:
\[ \text{Time ratio of A to C} = \frac{3}{5} \]
- Let the time taken by C be \(5y\) days. Then, the time taken by A will be \(3y\) days.

### Step 3: Relate A's and C's time
- It is given that A takes 6 days less than C:
\[ 5y - 3y = 6 \implies 2y = 6 \implies y = 3 \]
- Therefore, the time taken by C is:
\[ 5y = 5 \times 3 = 15 \text{ days} \]
- And the time taken by A is:
\[ 3y = 3 \times 3 = 9 \text{ days} \]

### Step 4: Determine the efficiency of B
- The ratio of the number of days taken by B to C is \(2:3\). Thus, if C takes \(15\) days, then B takes:
\[ \text{Time taken by B} = \frac{2}{3} \times 15 = 10 \text{ days} \]
- The efficiency of B will then be:
\[ \text{Efficiency of B} = \frac{1}{10} \text{ of the work per day} \]
- Since C takes 15 days, C's efficiency is:
\[ \text{Efficiency of C} = \frac{1}{15} \text{ of the work per day} \]

### Step 5: Calculate total work
- The total work can be calculated using A's efficiency:
\[ \text{Total work} = \text{Efficiency of A} \times \text{Time taken by A} = 5 \times 9 = 45 \text{ units} \]

### Step 6: Work done by B and C in 2 days
- B and C work together for 2 days:
\[ \text{Work done in 2 days} = (B's \text{ efficiency} + C's \text{ efficiency}) \times 2 \]
- B's efficiency is \( \frac{1}{10} \) and C's efficiency is \( \frac{1}{15} \):
\[ \text{Work done in 2 days} = \left(\frac{1}{10} + \frac{1}{15}\right) \times 2 \]
- To add these fractions, find a common denominator (30):
\[ = \left(\frac{3}{30} + \frac{2}{30}\right) \times 2 = \left(\frac{5}{30}\right) \times 2 = \frac{1}{6} \times 2 = \frac{1}{3} \text{ of the total work} \]
- Thus, the work done in 2 days is:
\[ = \frac{1}{3} \times 45 = 15 \text{ units} \]

### Step 7: Remaining work
- The remaining work after 2 days is:
\[ \text{Remaining work} = 45 - 15 = 30 \text{ units} \]

### Step 8: Time taken by A to finish remaining work
- A's efficiency is \(5\) units per day, so the time taken by A to finish the remaining work is:
\[ \text{Time taken by A} = \frac{\text{Remaining work}}{\text{Efficiency of A}} = \frac{30}{5} = 6 \text{ days} \]

### Final Answer
The number of days taken by A to finish the remaining work is **6 days**.

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