A can do a piece of work in 10 days. B can do it in 24 days. If C also works with them then it takes only 6 days to complete the whole work. In how many days C alone can complete the whole work?

Options

A. 25
B. 40
C. 50
D. 75

Answer

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

### Step 1: Determine the total work
We will take the total work to be the least common multiple (LCM) of the days taken by A, B, and the combined work with C. The days are 10, 24, and 6.

**LCM of 10, 24, and 6:**
- The prime factorization of 10 is \(2^1 \times 5^1\)
- The prime factorization of 24 is \(2^3 \times 3^1\)
- The prime factorization of 6 is \(2^1 \times 3^1\)

The LCM will take the highest power of each prime:
- \(2^3\) from 24
- \(3^1\) from 24
- \(5^1\) from 10

So, LCM = \(2^3 \times 3^1 \times 5^1 = 120\).

**Total work = 120 units.**

### Step 2: Calculate the work done by A and B in one day
- A can complete the work in 10 days, so in one day, A does:
\[ \text{Work done by A in 1 day} = \frac{120 \text{ units}}{10 \text{ days}} = 12 \text{ units/day} \]

- B can complete the work in 24 days, so in one day, B does:
\[ \text{Work done by B in 1 day} = \frac{120 \text{ units}}{24 \text{ days}} = 5 \text{ units/day} \]

### Step 3: Calculate the combined work done by A, B, and C in one day
When A, B, and C work together, they complete the work in 6 days. Therefore, the combined work done in one day is:
\[ \text{Work done by A, B, and C in 1 day} = \frac{120 \text{ units}}{6 \text{ days}} = 20 \text{ units/day} \]

### Step 4: Find the work done by C in one day
Now we can set up the equation:
\[ \text{Work done by A} + \text{Work done by B} + \text{Work done by C} = \text{Total work done in 1 day} \]
Substituting the known values:
\[ 12 + 5 + C = 20 \]
Thus,
\[ C = 20 - (12 + 5) = 20 - 17 = 3 \text{ units/day} \]

### Step 5: Calculate the time taken by C to complete the work alone
Now we need to find out how many days C will take to complete the total work of 120 units at the rate of 3 units per day:
\[ \text{Time taken by C} = \frac{\text{Total work}}{\text{C's efficiency}} = \frac{120 \text{ units}}{3 \text{ units/day}} = 40 \text{ days} \]

### Final Answer
C alone can complete the whole work in **40 days**.

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