A and B can finish a work in 10 days while B and C can do it in 18 days. A started the work, worked for 5 days, then B worked for 10 days and the remaining work was finished by C in 15 days. In how many days could C alone have finished the whole work ?
A)30 days
B)15 days
C)45 days
D)24 days

Options

A. 30 days
B. 15 days
C. 45 days
D. 24 days

Answer

To solve the problem step by step, we will first determine the efficiencies of A, B, and C based on the information provided.

### Step 1: Determine the total work
Let the total work be represented in units. Since A and B can finish the work in 10 days, we can assume the total work is 90 units (as the LCM of 10 and 18 is 90).

### Step 2: Calculate the efficiencies of A, B, and C
- **Efficiency of A and B**: They complete the work in 10 days.
\[ \text{Efficiency of A and B} = \frac{90 \text{ units}}{10 \text{ days}} = 9 \text{ units/day} \]

- **Efficiency of B and C**: They complete the work in 18 days.
\[ \text{Efficiency of B and C} = \frac{90 \text{ units}}{18 \text{ days}} = 5 \text{ units/day} \]

### Step 3: Set up equations for individual efficiencies
Let the efficiencies of A, B, and C be represented as \(a\), \(b\), and \(c\) respectively:
- From the efficiency of A and B:
\[ a + b = 9 \quad \text{(1)} \]
- From the efficiency of B and C:
\[ b + c = 5 \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1), we can express \(a\) in terms of \(b\):
\[ a = 9 - b \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ b + c = 5 \implies c = 5 - b \quad \text{(4)} \]

### Step 5: Calculate work done by A, B, and C
Now, we analyze the work done:
- A worked for 5 days:
\[ \text{Work done by A} = 5a \]
- B worked for 10 days:
\[ \text{Work done by B} = 10b \]
- C worked for 15 days:
\[ \text{Work done by C} = 15c \]

### Step 6: Total work equation
The total work done by A, B, and C must equal 90 units:
\[ 5a + 10b + 15c = 90 \]

### Step 7: Substitute values from equations
Substituting \(c\) from equation (4) into the total work equation:
\[ 5a + 10b + 15(5 - b) = 90 \]
Expanding this gives:
\[ 5a + 10b + 75 - 15b = 90 \]
Simplifying:
\[ 5a - 5b + 75 = 90 \]
\[ 5a - 5b = 15 \]
Dividing by 5:
\[ a - b = 3 \quad \text{(5)} \]

### Step 8: Solve for \(a\) and \(b\)
Now we have two equations (1) and (5):
1. \(a + b = 9\)
2. \(a - b = 3\)

Adding these two equations:
\[ 2a = 12 \implies a = 6 \]
Substituting \(a\) back into equation (1):
\[ 6 + b = 9 \implies b = 3 \]

### Step 9: Find \(c\)
Using equation (4):
\[ c = 5 - b = 5 - 3 = 2 \]

### Step 10: Calculate the time taken by C alone
C's efficiency is \(c = 2\) units/day. To find how many days C would take to finish the whole work:
\[ \text{Days taken by C} = \frac{90 \text{ units}}{2 \text{ units/day}} = 45 \text{ days} \]

### Final Answer
C alone could finish the whole work in **45 days**.