Answer the following question.
An airplane travels 96 km with a tailwind in a duration that is 25% less than the duration it requires to fly the same distance against a headwind. If the airplane flies 36 km against the headwind in 1.5 hours, calculate the speed of the airplane in still air.

The correct option is 28 km/hr.

Step 1: Understand the given information

Let the speed of the airplane in still air be _x km/hr and the speed of the wind be _y km/hr.

When traveling with a tailwind (downstream / with the wind):

Effective speed = $x+y$

When traveling against a headwind (upstream / against the wind):

Effective speed = $x-y$

Step 2: Calculate the speed against the headwind

The problem states that the airplane flies 36 km against the headwind in 1.5 hours.

Speed against headwind ($x-y$) = $\frac{36}{1.5}$ = 24 km/hr.

So, we have:

$x-y=24$    --- (Equation 1)

Step 3: Calculate the duration to travel 96 km against the headwind

Time taken against the headwind for 96 km (t_against) = $\frac{96}{x-y}=\frac{96}{24}=4\text{ hours}$

Step 4: Determine the time taken to travel 96 km with the tailwind

The duration with a tailwind is 25% less than the duration against the headwind.

Time taken with tailwind (t_tailwind) = 4 - (25% of 4) = 4 - 1 = 3 hours.

Step 5: Calculate the speed with the tailwind

Speed with tailwind ($x+y$) = $\frac{96}{3}=32\text{ km/hr}$

So, we have:

$x+y=32$    --- (Equation 2)

Step 6: Solve for the speed of the airplane in still air (_x)

Adding Equation 1 and Equation 2:

$(x-y)+(x+y)=24+32$

$2x=56$

$x=28\text{ km/hr}$

Therefore, the speed of the airplane in still air is 28 km/hr.