A and B together can finish a piece of work in 20 days. B and C can together finish the same piece ofwork in 30 days. If A is twice as efficient as C, in How many days can C alone finish the same piece of work?

Options

A. 36
B. 40
C. 42
D. 60

Answer

To solve the problem step by step, we can follow this approach:

### Step 1: Understand the work done by A, B, and C
- A and B together can finish the work in 20 days.
- B and C can finish the work in 30 days.
- A is twice as efficient as C.

### Step 2: Calculate the total work
Let's assume the total work is represented in units. We can take the total work to be 60 units (the least common multiple of 20 and 30).

### Step 3: Calculate the daily work done by A and B
Since A and B can complete 60 units in 20 days:
- Daily work of A and B together = Total work / Days = 60 units / 20 days = 3 units/day.

### Step 4: Calculate the daily work done by B and C
Since B and C can complete 60 units in 30 days:
- Daily work of B and C together = Total work / Days = 60 units / 30 days = 2 units/day.

### Step 5: Set up the equation to find individual work
From the above, we have:
- A + B = 3 units/day (Equation 1)
- B + C = 2 units/day (Equation 2)

Now, subtract Equation 2 from Equation 1:
(A + B) - (B + C) = 3 - 2
This simplifies to:
A - C = 1 unit/day (Equation 3)

### Step 6: Relate A's work to C's work
We know from the problem that A is twice as efficient as C. If we let C's work be X units/day, then A's work will be 2X units/day.

### Step 7: Substitute A's work in Equation 3
Substituting A's work in Equation 3:
2X - X = 1
This simplifies to:
X = 1 unit/day.

### Step 8: Determine C's work and calculate days to finish
Now, we know C works at a rate of 1 unit/day. To find out how many days C will take to finish the total work of 60 units:
Total days = Total work / C's work = 60 units / 1 unit/day = 60 days.

### Final Answer
C alone can finish the work in **60 days**.

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