Quantity I : Original duration of flight. In a flight of 3000 km an aircraft was slowed down by bad weather . Its average speed for the trip was reduced by 100 km /hr. and the time increased by one hour .
Quantity II : Usual time of a man who, when walks at ` (3)/(4) th` of his usual pace, reaches his office 20 minutes late .

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 problem step by step, we will analyze both quantities separately.

### Step 1: Calculate Quantity I (Original Duration of Flight)

1. **Define Variables**:
- Let the original speed of the aircraft be \( V \) km/hr.
- The reduced speed due to bad weather is \( V - 100 \) km/hr.
- The distance of the flight is 3000 km.

2. **Set Up the Equation**:
- The time taken at the original speed is \( \frac{3000}{V} \) hours.
- The time taken at the reduced speed is \( \frac{3000}{V - 100} \) hours.
- According to the problem, the time increased by 1 hour due to the reduction in speed:
\[ \frac{3000}{V - 100} = \frac{3000}{V} + 1 \]

3. **Cross Multiply**:
- Rearranging the equation:
\[ 3000 = \frac{3000(V - 100)}{V} + V - 100 \]
- Cross-multiplying gives:
\[ 3000V = 3000(V - 100) + V(V - 100) \]

4. **Simplify the Equation**:
- Expanding the equation:
\[ 3000V = 3000V - 300000 + V^2 - 100V \]
- Cancel \( 3000V \) from both sides:
\[ 0 = V^2 - 100V - 300000 \]

5. **Solve the Quadratic Equation**:
- Rearranging gives:
\[ V^2 - 100V - 300000 = 0 \]
- Using the quadratic formula \( V = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ V = \frac{100 \pm \sqrt{(-100)^2 - 4 \cdot 1 \cdot (-300000)}}{2 \cdot 1} \]
\[ V = \frac{100 \pm \sqrt{10000 + 1200000}}{2} \]
\[ V = \frac{100 \pm \sqrt{1210000}}{2} \]
\[ V = \frac{100 \pm 1100}{2} \]
- This gives two possible values for \( V \):
\[ V = 600 \quad \text{(valid speed)} \quad \text{and} \quad V = -500 \quad \text{(not valid)} \]

6. **Calculate the Original Duration**:
- Now, using \( V = 600 \):
\[ \text{Original time} = \frac{3000}{600} = 5 \text{ hours} \]

### Step 2: Calculate Quantity II (Usual Time of the Man)

1. **Define Variables**:
- Let the usual speed of the man be \( S \) km/hr.
- When he walks at \( \frac{3}{4}S \), he reaches 20 minutes late.

2. **Set Up the Equation**:
- The usual time taken is \( \frac{D}{S} \) and the time taken at reduced speed is \( \frac{D}{\frac{3}{4}S} \).
- The difference in time is 20 minutes (or \( \frac{1}{3} \) hours):
\[ \frac{D}{\frac{3}{4}S} - \frac{D}{S} = \frac{1}{3} \]

3. **Simplify the Equation**:
- This can be rewritten as:
\[ \frac{4D}{3S} - \frac{D}{S} = \frac{1}{3} \]
- Finding a common denominator:
\[ \frac{4D - 3D}{3S} = \frac{1}{3} \]
\[ \frac{D}{3S} = \frac{1}{3} \]

4. **Solve for D**:
- Cross-multiplying gives:
\[ D = S \]

5. **Calculate Usual Time**:
- The usual time taken is:
\[ \text{Usual time} = \frac{D}{S} = \frac{S}{S} = 1 \text{ hour} \]

### Conclusion

- **Quantity I**: 5 hours
- **Quantity II**: 1 hour

### Comparison
- Quantity I (5 hours) is greater than Quantity II (1 hour).

### Final Answer
- **Quantity I is greater than Quantity II**.

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