The efficiencies of A, B and C are in the ratio 2:5:3. Working together, they can compete a task in 9 days. In how many days will C alone complete 40% of that task ?

Options

A. 14
B. 16
C. 15
D. 12

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the total efficiency of A, B, and C.
The efficiencies of A, B, and C are given in the ratio 2:5:3.

Let the efficiencies be:
- Efficiency of A = 2x
- Efficiency of B = 5x
- Efficiency of C = 3x

### Step 2: Calculate the total efficiency.
Total efficiency = Efficiency of A + Efficiency of B + Efficiency of C
= 2x + 5x + 3x
= 10x

### Step 3: Calculate the total work done.
It is given that A, B, and C together can complete the task in 9 days.

Total work = Total efficiency × Number of days
= 10x × 9
= 90x

### Step 4: Calculate the amount of work C can do in one day.
C's efficiency is 3x. Therefore, the work done by C in one day is:
= 3x

### Step 5: Calculate how much work C needs to complete 40% of the total task.
40% of the total work (90x) is:
= 0.40 × 90x
= 36x

### Step 6: Determine the number of days C will take to complete 36x work.
Since C does 3x work in one day, the number of days C will take to complete 36x work is:
= (Total work needed) / (Work done by C in one day)
= 36x / 3x
= 12 days

### Conclusion:
C alone will complete 40% of the task in **12 days**.

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