In these question two equations numebred I and numberd II are given. You have to solve both the equations and find out the correct option.
Quantity I: X, Y and Z can finish a work alone in 12,18 and 9 days respectively. X started the work and Y and Z assisted him on every 3rd day. Find the total number of days in which the work is completed.
Quantity II: A, B and C can finish a work alone in 20,12 and 15 days respectively. All three start working together. After 2 days B left the work. After 4 more days Cleft and the remaining work is completed by A alone. A worked for how many days in all?

Options

A. `Quantity I gt quantity II`
B. `Quantity I le quantity II`
C. `Quantity I ge quantity II`
D. `Quantity I lt quantity II`

Answer

To solve the given problem, we need to analyze both quantities step by step.

### Quantity I:

**Step 1: Determine the work rates of X, Y, and Z.**

- X can finish the work in 12 days, so his work rate is:
\[ \text{Work rate of X} = \frac{1}{12} \text{ of the work per day} \]

- Y can finish the work in 18 days, so his work rate is:
\[ \text{Work rate of Y} = \frac{1}{18} \text{ of the work per day} \]

- Z can finish the work in 9 days, so his work rate is:
\[ \text{Work rate of Z} = \frac{1}{9} \text{ of the work per day} \]

**Step 2: Find the total work done in one cycle of 3 days.**

- In the first two days, only X works:
\[ \text{Work done by X in 2 days} = 2 \times \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \]

- On the third day, X, Y, and Z work together:
\[ \text{Work done on the 3rd day} = \frac{1}{12} + \frac{1}{18} + \frac{1}{9} \]
To add these fractions, we find a common denominator, which is 36:
\[ = \frac{3}{36} + \frac{2}{36} + \frac{4}{36} = \frac{9}{36} = \frac{1}{4} \]

- Total work done in 3 days:
\[ \text{Total work in 3 days} = \frac{1}{6} + \frac{1}{4} \]
Finding a common denominator (12):
\[ = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \]

**Step 3: Determine how many cycles are needed to complete the work.**

- Total work to be done = 1 (whole work)
- Work done in 3 days = \(\frac{5}{12}\)

Let \(n\) be the number of 3-day cycles:
\[ n \times \frac{5}{12} \leq 1 \]
\[ n \leq \frac{12}{5} = 2.4 \]
Thus, \(n = 2\) full cycles can be completed.

**Step 4: Calculate total work done after 6 days.**

- Work done in 6 days:
\[ 2 \times \frac{5}{12} = \frac{10}{12} = \frac{5}{6} \]

**Step 5: Calculate remaining work and days needed.**

- Remaining work:
\[ 1 - \frac{5}{6} = \frac{1}{6} \]

- On the 7th day, X works alone:
\[ \text{Work done by X on 7th day} = \frac{1}{12} \]

- On the 8th day, X works again:
\[ \text{Work done by X on 8th day} = \frac{1}{12} \]

- Total work done in the last two days:
\[ \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \]

Thus, the total number of days to complete the work is:
\[ 6 + 2 = 8 \text{ days} \]

### Quantity II:

**Step 1: Determine the work rates of A, B, and C.**

- A can finish the work in 20 days, so his work rate is:
\[ \text{Work rate of A} = \frac{1}{20} \]

- B can finish the work in 12 days, so his work rate is:
\[ \text{Work rate of B} = \frac{1}{12} \]

- C can finish the work in 15 days, so his work rate is:
\[ \text{Work rate of C} = \frac{1}{15} \]

**Step 2: Find the total work done in the first 2 days.**

- In 2 days, A, B, and C work together:
\[ \text{Work done in 2 days} = 2 \left(\frac{1}{20} + \frac{1}{12} + \frac{1}{15}\right) \]
Finding a common denominator (60):
\[ = 2 \left(\frac{3}{60} + \frac{5}{60} + \frac{4}{60}\right) = 2 \left(\frac{12}{60}\right) = \frac{24}{60} = \frac{2}{5} \]

**Step 3: Determine remaining work after 2 days.**

- Remaining work:
\[ 1 - \frac{2}{5} = \frac{3}{5} \]

**Step 4: Calculate work done in the next 4 days (A and C only).**

- A and C work together for the next 4 days:
\[ \text{Work done by A and C in 1 day} = \frac{1}{20} + \frac{1}{15} \]
Finding a common denominator (60):
\[ = \frac{3}{60} + \frac{4}{60} = \frac{7}{60} \]

- Work done in 4 days:
\[ 4 \times \frac{7}{60} = \frac{28}{60} = \frac{14}{30} = \frac{7}{15} \]

**Step 5: Calculate remaining work after 6 days.**

- Remaining work:
\[ \frac{3}{5} - \frac{7}{15} \]
Finding a common denominator (15):
\[ = \frac{9}{15} - \frac{7}{15} = \frac{2}{15} \]

**Step 6: Calculate how long A works alone to finish the remaining work.**

- A's work rate is \(\frac{1}{20}\), so time taken to finish \(\frac{2}{15}\) work:
\[ \text{Time} = \frac{\frac{2}{15}}{\frac{1}{20}} = \frac{2}{15} \times 20 = \frac{40}{15} = \frac{8}{3} \text{ days} \]

**Step 7: Total days A worked.**

- Total days A worked:
\[ 2 \text{ (initial days)} + 4 \text{ (next days)} + \frac{8}{3} \text{ (final days)} = 6 + \frac{8}{3} = \frac{18}{3} + \frac{8}{3} = \frac{26}{3} \text{ days} \]

### Final Comparison:

- Quantity I: 8 days
- Quantity II: \(\frac{26}{3} \approx 8.67\) days

### Conclusion:
Since \(8 < \frac{26}{3}\), we conclude that Quantity I is less than Quantity II.