A plane left 30 min later than its scheduled time to reach its destination 1500 km away. In order to reach in time it increases its speed by 250 km/h. What is its original speed?

Options

A. 1000 km/h
B. 750 km/h
C. 600 km/h
D. 800 km/h

Answer

To solve the problem step by step, we will use the information provided in the question and apply the relevant formulas.

### Step 1: Understand the Problem
The plane is scheduled to travel a distance of 1500 km but leaves 30 minutes late. To make up for the lost time, it increases its speed by 250 km/h. We need to find the original speed of the plane.

### Step 2: Convert Time Delay to Hours
The plane leaves 30 minutes late. We need to convert this time into hours:
- 30 minutes = 30/60 hours = 0.5 hours.

### Step 3: Set Up the Variables
Let:
- \( S \) = original speed of the plane in km/h.
- \( S + 250 \) = increased speed of the plane in km/h.

### Step 4: Use the Distance Formula
The formula for distance is:
\[ \text{Distance} = \text{Speed} \times \text{Time} \]

Using this, we can write two equations for the time taken at the original speed and the increased speed.

1. Time taken at original speed:
\[ \text{Time}_1 = \frac{1500}{S} \]

2. Time taken at increased speed:
\[ \text{Time}_2 = \frac{1500}{S + 250} \]

### Step 5: Set Up the Equation
Since the plane leaves 0.5 hours late, the time taken at the increased speed will be 0.5 hours less than the time taken at the original speed:
\[ \frac{1500}{S} - \frac{1500}{S + 250} = 0.5 \]

### Step 6: Solve the Equation
To solve the equation, we first find a common denominator:
\[ \frac{1500(S + 250) - 1500S}{S(S + 250)} = 0.5 \]

This simplifies to:
\[ \frac{1500 \times 250}{S(S + 250)} = 0.5 \]

Cross-multiplying gives:
\[ 1500 \times 250 = 0.5 \times S(S + 250) \]

### Step 7: Simplify and Rearrange
Now, let's simplify:
\[ 375000 = 0.5S^2 + 125S \]
Multiplying through by 2 to eliminate the fraction:
\[ 750000 = S^2 + 250S \]

Rearranging gives us a quadratic equation:
\[ S^2 + 250S - 750000 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula \( S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 250, c = -750000 \).

Calculating the discriminant:
\[ b^2 - 4ac = 250^2 - 4 \times 1 \times (-750000) \]
\[ = 62500 + 3000000 \]
\[ = 3062500 \]

Now, take the square root:
\[ \sqrt{3062500} = 1750 \]

Now, substitute back into the quadratic formula:
\[ S = \frac{-250 \pm 1750}{2} \]

Calculating the two possible values:
1. \( S = \frac{1500}{2} = 750 \)
2. \( S = \frac{-2000}{2} = -1000 \) (not a valid speed)

Thus, the original speed \( S \) is:
\[ S = 750 \text{ km/h} \]

### Final Answer
The original speed of the plane is **750 km/h**.

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