An aeroplane takes off 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time , it has to increse its speed by 250km/h from its usual speed . Find its usual speed .

Options

A. 1000 km/h
B. 750 km/h
C. 850 km/h
D. 650 km/h

Explanation

To solve the problem step by step, we will define the variables, set up the equations based on the information given, and solve for the usual speed of the aeroplane.

### Step-by-Step Solution:

1. **Define the Variables**:
Let the usual speed of the aeroplane be \( x \) km/h.

2. **Scheduled Time Calculation**:
The distance to the destination is 1500 km. The time taken to cover this distance at the usual speed \( x \) is given by:
\[ \text{Time} = \frac{1500}{x} \text{ hours} \]

3. **Adjusting for Delay**:
Since the aeroplane takes off 30 minutes late, which is \( \frac{30}{60} = \frac{1}{2} \) hours, the effective time available to reach the destination becomes:
\[ \text{Effective Time} = \frac{1500}{x} - \frac{1}{2} \text{ hours} \]

4. **Increased Speed Calculation**:
To reach the destination on time, the aeroplane increases its speed by 250 km/h. Therefore, the new speed is:
\[ \text{New Speed} = x + 250 \text{ km/h} \]

5. **Time at Increased Speed**:
The time taken to cover the distance at the increased speed is:
\[ \text{Time at Increased Speed} = \frac{1500}{x + 250} \text{ hours} \]

6. **Setting Up the Equation**:
Since both times (effective time and time at increased speed) must be equal for the aeroplane to arrive on time, we can set up the equation:
\[ \frac{1500}{x} - \frac{1}{2} = \frac{1500}{x + 250} \]

7. **Clearing the Fractions**:
To eliminate the fractions, multiply through by \( 2x(x + 250) \):
\[ 2x(x + 250) \left( \frac{1500}{x} - \frac{1}{2} \right) = 2x(x + 250) \left( \frac{1500}{x + 250} \right) \]
Simplifying gives:
\[ 3000(x + 250) - x(x + 250) = 3000x \]

8. **Expanding and Rearranging**:
Expanding the left side:
\[ 3000x + 750000 - x^2 - 250x = 3000x \]
Rearranging gives:
\[ -x^2 - 250x + 750000 = 0 \]
Multiplying through by -1:
\[ x^2 + 250x - 750000 = 0 \]

9. **Using the Quadratic Formula**:
We can now use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 250, c = -750000 \):
\[ x = \frac{-250 \pm \sqrt{250^2 - 4 \cdot 1 \cdot (-750000)}}{2 \cdot 1} \]
\[ x = \frac{-250 \pm \sqrt{62500 + 3000000}}{2} \]
\[ x = \frac{-250 \pm \sqrt{3062500}}{2} \]
\[ x = \frac{-250 \pm 1750}{2} \]

10. **Calculating the Usual Speed**:
Solving gives two potential solutions:
\[ x = \frac{1500}{2} = 750 \quad \text{(valid speed)} \]
\[ x = \frac{-2000}{2} = -1000 \quad \text{(not valid)} \]
Therefore, the usual speed of the aeroplane is:
\[ \boxed{750 \text{ km/h}} \]