A is 20% more efficient than B and takes 40 days less than what B takes to complete a particular job. B is 50% more efficient than C. On the same job, A works for 50 days and quits. And then B takes over and works for 20 days and quits. C completed the remaining work. Find the number of days taken by C to complete the work.

Let the amount of work required to complete the job be represented as \( W \).

### Step 1: Defining efficiency and time for A, B, and C

#### Efficiency of B:
Let B's rate of work (efficiency) be \( r_B \) units of work per day. Thus, B takes \( \frac{W}{r_B} \) days to complete the job.

#### Efficiency of A:
A is 20% more efficient than B, meaning A's rate of work is \( r_A = 1.2r_B \).

Let the time taken by B to complete the job be \( t_B \) days. From the problem, we know that A takes 40 days less than B to complete the job:
\[
t_A = t_B - 40
\]
Since the time to complete the job is the total work divided by the rate of work, we know:
\[
t_A = \frac{W}{r_A} = \frac{W}{1.2r_B}
\]
\[
t_B = \frac{W}{r_B}
\]
Thus, from the given information, we have:
\[
\frac{W}{1.2r_B} = \frac{W}{r_B} - 40
\]
Multiplying through by \( 1.2r_B \) to eliminate the denominators:
\[
W = 1.2W - 40 \times 1.2r_B
\]
\[
0.2W = 40 \times 1.2r_B
\]
\[
r_B = \frac{0.2W}{40 \times 1.2} = \frac{W}{240}
\]
Thus, B's rate of work is \( r_B = \frac{W}{240} \).

#### Efficiency of C:
B is 50% more efficient than C, meaning B’s rate of work is 1.5 times that of C. Hence, C's rate of work is:
\[
r_C = \frac{r_B}{1.5} = \frac{W}{240 \times 1.5} = \frac{W}{360}
\]

### Step 2: Amount of work completed by A, B, and C

#### Work done by A in 50 days:
A works for 50 days, so the work done by A is:
\[
\text{Work by A} = r_A \times 50 = 1.2r_B \times 50 = 1.2 \times \frac{W}{240} \times 50 = \frac{1.2W \times 50}{240} = \frac{W}{4}
\]
Thus, A completes \( \frac{W}{4} \) of the total work.

#### Work done by B in 20 days:
B works for 20 days, so the work done by B is:
\[
\text{Work by B} = r_B \times 20 = \frac{W}{240} \times 20 = \frac{W}{12}
\]
Thus, B completes \( \frac{W}{12} \) of the total work.

#### Remaining work for C:
The total work done by A and B is:
\[
\frac{W}{4} + \frac{W}{12} = \frac{3W}{12} + \frac{W}{12} = \frac{4W}{12} = \frac{W}{3}
\]
Thus, the remaining work for C is:
\[
W - \frac{W}{3} = \frac{2W}{3}
\]

### Step 3: Time taken by C to complete the remaining work
The time taken by C to complete \( \frac{2W}{3} \) of the total work is:
\[
\text{Time by C} = \frac{\frac{2W}{3}}{r_C} = \frac{\frac{2W}{3}}{\frac{W}{360}} = \frac{2W}{3} \times \frac{360}{W} = \frac{2 \times 360}{3} = 240 \text{ days}
\]

### Final Answer:
C takes **240 days** to complete the remaining work.

Let me know if you'd like more details or have any questions.

Here are 5 related questions for further practice:
1. If A worked for 40 days instead of 50, how many days would C take to complete the remaining work?
2. If C were 25% more efficient, how many days would C take to complete the work left after A and B?
3. How long would it take A to complete the entire job on their own?
4. If B worked for 30 days instead of 20, how much work would be left for C to complete?
5. If C worked for the first 60 days, how much work would be left for A and B to complete?

**Tip:** Efficiency comparisons are often relative, so always express one worker's efficiency in terms of another to simplify the calculations.

A is 20% more efficient than B and takes 40 days less than what B takes to complete a particular job. B is 50% more efficient than C. On the same job, A works for 50 days and quits. Then B takes over and works for 20 days and quits. C completed the remaining work. Find the number of days taken by C to complete the work.

Learn how to solve a complex work and efficiency problem involving workers A, B, and C, where A is 20% more efficient than B, and B is 50% more efficient than C. A works for 50 days, B for 20 days, and C completes the remaining job. The problem involves calculating how many days C takes to finish the work. Suitable for students in Grades 9-11.

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