Quantity I: Days after which A and B meet. A and B set out to meet each other from two places 165 km apart. A travels 15 km the first day, 14 km second day, 13 km the third day and so on, B travels 10 km the first, 12 km the second day, 14 km the third day and so on.
Quantity II: Number of day required to complete the whole work if A, B and C can complete a piece of work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left.

Options

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

Answer

To solve the given problem, we will break it down into two parts: Quantity I and Quantity II.

### Quantity I: Days after which A and B meet

1. **Understanding the distances traveled by A and B:**
- A starts at 15 km on the first day and decreases his distance by 1 km each day.
- B starts at 10 km on the first day and increases his distance by 2 km each day.

2. **Setting up the equations:**
- The distance A travels in N days can be expressed as:
\[ D_A = 15 + 14 + 13 + ... + (15 - (N - 1)) = \frac{N}{2} \times (15 + (15 - N + 1)) = \frac{N}{2} \times (30 - N + 1) = \frac{N(31 - N)}{2} \]
- The distance B travels in N days can be expressed as:
\[ D_B = 10 + 12 + 14 + ... + (10 + 2(N - 1)) = \frac{N}{2} \times (10 + (10 + 2(N - 1))) = \frac{N}{2} \times (20 + 2(N - 1)) = \frac{N(18 + 2N)}{2} \]

3. **Setting up the total distance equation:**
- The total distance covered by A and B together should equal 165 km:
\[ D_A + D_B = 165 \]
- Substituting the expressions for \(D_A\) and \(D_B\):
\[ \frac{N(31 - N)}{2} + \frac{N(18 + 2N)}{2} = 165 \]
- Simplifying this:
\[ N(31 - N + 18 + 2N) = 330 \]
\[ N(49 - N) = 330 \]
- Rearranging gives us:
\[ N^2 - 49N + 330 = 0 \]

4. **Solving the quadratic equation:**
- Using the quadratic formula:
\[ N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = -49\), and \(c = 330\):
\[ N = \frac{49 \pm \sqrt{(-49)^2 - 4 \cdot 1 \cdot 330}}{2 \cdot 1} \]
\[ N = \frac{49 \pm \sqrt{2401 - 1320}}{2} \]
\[ N = \frac{49 \pm \sqrt{1071}}{2} \]
- Approximating \(\sqrt{1071} \approx 32.7\):
\[ N \approx \frac{49 \pm 32.7}{2} \]
- Taking the positive root:
\[ N \approx \frac{81.7}{2} \approx 40.85 \]
- Rounding gives \(N \approx 41\) days.

### Quantity II: Number of days required to complete the whole work

1. **Understanding the work done by A, B, and C:**
- A can complete the work in 10 days, so his work rate is \( \frac{1}{10} \) of the work per day.
- B can complete the work in 12 days, so his work rate is \( \frac{1}{12} \) of the work per day.
- C can complete the work in 15 days, so his work rate is \( \frac{1}{15} \) of the work per day.

2. **Setting up the equation for total work:**
- Let \(x\) be the total number of days to complete the work.
- A works for \(x - 5\) days, B works for \(x - 3\) days, and C works for \(x\) days.
- The equation for the total work done is:
\[ \frac{x - 5}{10} + \frac{x - 3}{12} + \frac{x}{15} = 1 \]

3. **Finding a common denominator (60):**
- Multiplying through by 60 gives:
\[ 6(x - 5) + 5(x - 3) + 4x = 60 \]
- Expanding this:
\[ 6x - 30 + 5x - 15 + 4x = 60 \]
- Combining like terms:
\[ 15x - 45 = 60 \]
- Solving for \(x\):
\[ 15x = 105 \implies x = 7 \]

### Conclusion

- **Quantity I:** Days after which A and B meet is approximately 41 days.
- **Quantity II:** Number of days required to complete the whole work is 7 days.

### Comparison

- Quantity I (41 days) is greater than Quantity II (7 days).

### Final Answer

**Quantity I > Quantity II.**