An aircraft was to take off from a certain airport at 8 a.m., but it was delayed by 30 min. To make up for the lost time, it was to increase its speed by 250 km/hour from the normal speed to reach its destination 1500 km on time. What was the normal speed of the aircraft?

Options

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

Answer

To find the normal speed of the aircraft, we can follow these steps:

### Step 1: Define the Variables
Let the normal speed of the aircraft be \( x \) km/hour.

### Step 2: Determine the Delayed Time
The aircraft was delayed by 30 minutes, which is equivalent to \( \frac{1}{2} \) hour.

### Step 3: Calculate the Time Taken at Normal Speed
The time taken to cover the distance of 1500 km at the normal speed \( x \) is given by the formula:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1500}{x} \text{ hours} \]

### Step 4: Calculate the Time Taken at Increased Speed
The new speed of the aircraft after increasing by 250 km/hour is \( x + 250 \) km/hour. The time taken to cover the same distance at this new speed is:
\[ \text{Time} = \frac{1500}{x + 250} \text{ hours} \]

### Step 5: Set Up the Equation
Since the aircraft needs to make up for the 30-minute delay, the time taken at normal speed minus the time taken at increased speed should equal \( \frac{1}{2} \) hour:
\[ \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \]

### Step 6: Clear the Fractions
To eliminate the fractions, multiply through by \( 2x(x + 250) \):
\[ 2x(x + 250) \left( \frac{1500}{x} - \frac{1500}{x + 250} \right) = 2x(x + 250) \cdot \frac{1}{2} \]
This simplifies to:
\[ 3000(x + 250) - 3000x = x(x + 250) \]

### Step 7: Simplify the Equation
Expanding both sides:
\[ 3000x + 750000 - 3000x = x^2 + 250x \]
This simplifies to:
\[ 750000 = x^2 + 250x \]

### Step 8: Rearrange into Standard Quadratic Form
Rearranging gives us:
\[ x^2 + 250x - 750000 = 0 \]

### Step 9: Factor the Quadratic Equation
To factor, we look for two numbers that multiply to \(-750000\) and add to \(250\). The factors are \(1000\) and \(-750\):
\[ (x + 1000)(x - 750) = 0 \]

### Step 10: Solve for \( x \)
Setting each factor to zero gives:
1. \( x + 1000 = 0 \) → \( x = -1000 \) (not valid as speed cannot be negative)
2. \( x - 750 = 0 \) → \( x = 750 \)

### Conclusion
The normal speed of the aircraft is \( 750 \) km/hour.

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