To solve the problem step by step, we will first determine the efficiencies of A, B, and C, then calculate the remaining work after A and B have worked, and finally find the value of x. After that, we will analyze the work rates of P, Q, and R to determine who should start the job to minimize the total time.
### Step 1: Determine the efficiencies of A, B, and C
1. **A, B, and C together can complete the work in 16 days.**
- Work done by A, B, and C in one day = \( \frac{1}{16} \) of the work.
2. **B and C together can complete the work in 32 days.**
- Work done by B and C in one day = \( \frac{1}{32} \) of the work.
3. **C can complete the work in 80 days.**
- Work done by C in one day = \( \frac{1}{80} \) of the work.
Let’s denote:
- Work done by A in one day = \( A \)
- Work done by B in one day = \( B \)
- Work done by C in one day = \( C \)
From the information:
- \( A + B + C = \frac{1}{16} \)
- \( B + C = \frac{1}{32} \)
- \( C = \frac{1}{80} \)
### Step 2: Calculate A and B's efficiencies
From \( B + C = \frac{1}{32} \):
- Substituting \( C \):
\[
B + \frac{1}{80} = \frac{1}{32}
\]
\[
B = \frac{1}{32} - \frac{1}{80}
\]
To subtract these fractions, find a common denominator (which is 160):
\[
B = \frac{5}{160} - \frac{2}{160} = \frac{3}{160}
\]
Now substituting \( B \) back to find \( A \):
\[
A + \frac{3}{160} + \frac{1}{80} = \frac{1}{16}
\]
Substituting \( \frac{1}{80} = \frac{2}{160} \):
\[
A + \frac{3}{160} + \frac{2}{160} = \frac{1}{16}
\]
\[
A + \frac{5}{160} = \frac{10}{160}
\]
\[
A = \frac{10}{160} - \frac{5}{160} = \frac{5}{160} = \frac{1}{32}
\]
### Step 3: Calculate the total work done after A and B have worked
1. **A works for 4 days:**
- Work done by A in 4 days = \( 4 \times \frac{1}{32} = \frac{4}{32} = \frac{1}{8} \)
2. **B works for 12 days:**
- Work done by B in 12 days = \( 12 \times \frac{3}{160} = \frac{36}{160} = \frac{9}{40} \)
3. **Total work done by A and B:**
\[
\text{Total work done} = \frac{1}{8} + \frac{9}{40}
\]
Converting \( \frac{1}{8} \) to a denominator of 40:
\[
\frac{1}{8} = \frac{5}{40}
\]
\[
\text{Total work done} = \frac{5}{40} + \frac{9}{40} = \frac{14}{40} = \frac{7}{20}
\]
4. **Remaining work:**
\[
\text{Remaining work} = 1 - \frac{7}{20} = \frac{13}{20}
\]
### Step 4: Calculate the time taken by C to complete the remaining work
Since C can do \( \frac{1}{80} \) of the work in one day:
- Time taken by C to complete the remaining work:
\[
x = \frac{\text{Remaining work}}{\text{C's work rate}} = \frac{\frac{13}{20}}{\frac{1}{80}} = \frac{13}{20} \times 80 = 52 \text{ days}
\]
### Step 5: Determine the work rates of P, Q, and R
1. **P takes \( x - 28 \) days:**
\[
P = 52 - 28 = 24 \text{ days}
\]
2. **Q takes \( x - 18 \) days:**
\[
Q = 52 - 18 = 34 \text{ days}
\]
3. **R takes \( x - 8 \) days:**
\[
R = 52 - 8 = 44 \text{ days}
\]
### Step 6: Calculate the work done by P, Q, and R in one day
- Work done by P in one day = \( \frac{1}{24} \)
- Work done by Q in one day = \( \frac{1}{34} \)
- Work done by R in one day = \( \frac{1}{44} \)
### Step 7: Determine who should start the job
To minimize the total time, we need to find out who can complete the most work in the least time. The person with the highest efficiency (lowest time) should start.
- P's efficiency: \( \frac{1}{24} \)
- Q's efficiency: \( \frac{1}{34} \)
- R's efficiency: \( \frac{1}{44} \)
Since \( \frac{1}{24} > \frac{1}{34} > \frac{1}{44} \), P is the most efficient.
### Conclusion
**P should start the job to complete it in the least possible time.**
