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Let's denote the usual speed of the plane as "x" km/hr.

When the plane departs 30 minutes (0.5 hours) late, it has less time to cover the same distance. To reach the destination in time, it needs to make up for this delay by increasing its speed.

The distance to be covered is 1500 km. Let's first calculate the time it would take to cover this distance at the usual speed "x" km/hr:

Time = Distance / Speed

Time = 1500 / x

Now, since the plane left 30 minutes late, it has to cover the same distance in a shorter amount of time, which is:

Time = Total time - Delayed time

Time = 1500 / x - 0.5

The plane increases its speed by 100 km/hr from the usual speed, so its new speed is "x + 100" km/hr.

Now, the time it takes to cover the same distance at the increased speed is:

Time = Distance / Speed

Time = 1500 / (x + 100)

According to the given information, the time taken at the increased speed is equal to the time taken with the delay:

1500 / (x + 100) = 1500 / x - 0.5

To solve this equation, we can cross-multiply:

1500x = 1500(x + 100) - 0.5x(x + 100)

Simplify and solve for "x":

1500x = 1500x + 150000 - 0.5x^2 - 50x

0.5x^2 + 50x - 150000 = 0

Divide the equation by 0.5 to simplify:

x^2 + 100x - 300000 = 0

Now, we can use the quadratic formula to solve for "x":

x = (-b ± √(b² - 4ac)) / 2a

For the quadratic equation x² + 100x - 300000 = 0:

a = 1

b = 100

c = -300000

Calculate the discriminant:

D = b² - 4ac

D = 100² - 4(1)(-300000)

D = 10000 + 1200000

D = 1210000

Now, calculate the two possible values of "x":

x = (-100 ± √(1210000)) / (2 * 1)

x = (-100 ± 1100) / 2

This gives us two possible solutions for "x":

1. x = (1000) / 2 = 500

2. x = (-1200) / 2 = -600 (discard this solution as speed can't be negative)

So, the usual speed of the plane is indeed 500 km/hr.

Answer:

Let's denote the usual speed of the plane as "x" km/hr.

The plane left 30 minutes (0.5 hours) late, so the time available for the journey is less than the scheduled time.

Let's consider the time it takes to cover the distance of 1500 km:

Usual time (T1) = Distance / Usual Speed = 1500 / x hours

Now, due to the delay, the available time becomes (T1 - 0.5) hours.

In order to cover the same distance in this reduced time, the plane has to increase its speed by 100 km/hr. So, the new speed is (x + 100) km/hr.

Time taken with increased speed (T2) = Distance / Increased Speed = 1500 / (x + 100) hours

Given that the available time is 30 minutes less than the usual time, we can set up the equation:

T1 - 0.5 = T2

Substitute the values of T1 and T2:

1500 / x - 0.5 = 1500 / (x + 100)

Now, solve for x:

1500(x + 100) = 1500x - 0.5x(x + 100)

150000 = 1500x - 0.5x^2 - 50x

Combine like terms:

0.5x^2 - 1450x + 150000 = 0

Divide the equation by 0.5 to simplify:

x^2 - 2900x + 300000 = 0

Now we can solve this quadratic equation using the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / 2a

where a = 1, b = -2900, and c = 300000.

Calculating the roots, we find:

x ≈ 350 and x ≈ 850

The usual speed of the plane should be positive, so the answer is x = 350 km/hr.

Step-by-step explanation: