A,B and C can do a work in 20,45 and 120 days respectively.They started the work. A left 10 days before and B left 5 days before the completion of work. In how many days is the total work completed ?

Answer

To solve the problem step by step, we will first determine the work done by A, B, and C, and then set up an equation based on the information given in the question.

### Step 1: Determine the work done by A, B, and C in one day.
- A can complete the work in 20 days, so A's work in one day = 1/20 of the work.
- B can complete the work in 45 days, so B's work in one day = 1/45 of the work.
- C can complete the work in 120 days, so C's work in one day = 1/120 of the work.

### Step 2: Find the Least Common Multiple (LCM) of the days.
The LCM of 20, 45, and 120 is 360. This means the total work can be considered as 360 units.

### Step 3: Calculate the efficiency (work done in one day) of A, B, and C.
- A's efficiency = 360 / 20 = 18 units/day.
- B's efficiency = 360 / 45 = 8 units/day.
- C's efficiency = 360 / 120 = 3 units/day.

### Step 4: Set up the equation based on the work done.
Let \( X \) be the total number of days taken to complete the work.

- A worked for \( X - 10 \) days (since A left 10 days before completion).
- B worked for \( X - 5 \) days (since B left 5 days before completion).
- C worked for \( X \) days.

### Step 5: Write the equation for the total work done.
The total work done by A, B, and C can be expressed as:
\[ 18(X - 10) + 8(X - 5) + 3X = 360 \]

### Step 6: Expand and simplify the equation.
Expanding the equation:
\[ 18X - 180 + 8X - 40 + 3X = 360 \]

Combine like terms:
\[ (18X + 8X + 3X) - 220 = 360 \]
\[ 29X - 220 = 360 \]

### Step 7: Solve for \( X \).
Add 220 to both sides:
\[ 29X = 580 \]

Now, divide by 29:
\[ X = \frac{580}{29} \]
\[ X = 20 \]

### Conclusion:
The total work is completed in **20 days**.