An aeroplane covers a distance of 1500 km at a uniform speed and reaches its destination by 30 minutes. To reach at right time, the pilot increases he speed by 250 km/h. Find the original speed of the aeroplane.

Options

A. 1000 kmph
B. 800 kmph
C. 1500 kmph
D. 750 kmph

Answer

To solve the problem, we need to find the original speed of the aeroplane given that it covers a distance of 1500 km and reaches its destination 30 minutes late. The pilot increases the speed by 250 km/h to reach on time.

### Step-by-Step Solution:

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

2. **Calculate Time Taken at Original Speed**:
The time taken to cover 1500 km at speed \( S \) is given by:
\[ \text{Time} = \frac{1500}{S} \text{ hours} \]

3. **Calculate Time Taken at Increased Speed**:
The increased speed of the aeroplane is \( S + 250 \) km/h. The time taken to cover the same distance at this speed is:
\[ \text{Time} = \frac{1500}{S + 250} \text{ hours} \]

4. **Set Up the Equation**:
According to the problem, the aeroplane is 30 minutes late, which is equivalent to \( \frac{1}{2} \) hours. Therefore, we can set up the equation:
\[ \frac{1500}{S} - \frac{1500}{S + 250} = \frac{1}{2} \]

5. **Clear the Fractions**:
To eliminate the fractions, multiply through by \( S(S + 250) \):
\[ 1500(S + 250) - 1500S = \frac{1}{2} S(S + 250) \]
Simplifying this gives:
\[ 1500 \times 250 = \frac{1}{2} S^2 + 125S \]
\[ 375000 = \frac{1}{2} S^2 + 125S \]

6. **Multiply Through by 2**:
To eliminate the fraction, multiply the entire equation by 2:
\[ 750000 = S^2 + 250S \]

7. **Rearrange the Equation**:
Rearranging gives us a standard quadratic equation:
\[ S^2 + 250S - 750000 = 0 \]

8. **Use the Quadratic Formula**:
The quadratic formula is given by:
\[ S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 250 \), and \( c = -750000 \):
\[ S = \frac{-250 \pm \sqrt{250^2 - 4 \times 1 \times (-750000)}}{2 \times 1} \]
\[ S = \frac{-250 \pm \sqrt{62500 + 3000000}}{2} \]
\[ S = \frac{-250 \pm \sqrt{3062500}}{2} \]
\[ S = \frac{-250 \pm 1750}{2} \]

9. **Calculate the Two Possible Values**:
- For the positive root:
\[ S = \frac{1500}{2} = 750 \text{ km/h} \]
- For the negative root:
\[ S = \frac{-2000}{2} = -1000 \text{ km/h} \text{ (not valid)} \]

10. **Conclusion**:
Therefore, the original speed of the aeroplane is:
\[ S = 750 \text{ km/h} \]