Question

An aeroplane takes 1 hour more for a journey of 1000 km if its speed decreases by 50 km/hr from its usual speed. What is the usual speed of the aeroplane?

• A. 200 km/hr
• B. 240 km/hr
• C. 250 km/hr
• D. 300 km/hr

Hint:

When dealing with speed and time problems, set up an equation based on the difference in time taken and solve using algebraic techniques.

Answer 1

The Correct Option is C

Step 1: Define the variables
Let the usual speed of the aeroplane be \( x \) km/hr. Step 2: Write the time equations
- Time taken to travel 1000 km at speed \( x \): \[ \text{Time} = \frac{1000}{x} \] - Time taken to travel 1000 km at reduced speed \( x - 50 \): \[ \text{Time} = \frac{1000}{x - 50} \] Step 3: Set up the equation based on given condition
According to the problem, the difference in time is 1 hour: \[ \frac{1000}{x - 50} - \frac{1000}{x} = 1 \] Step 4: Solve for \( x \)
Take LCM: \[ \frac{1000(x) - 1000(x - 50)}{x(x - 50)} = 1 \] \[ \frac{1000x - 1000x + 50000}{x(x - 50)} = 1 \] \[ \frac{50000}{x(x - 50)} = 1 \] \[ 50000 = x(x - 50) \] Step 5: Solve the quadratic equation
\[ x^2 - 50x - 50000 = 0 \] Using the quadratic formula: \[ x = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(1)(-50000)}}{2(1)} \] \[ x = \frac{50 \pm \sqrt{2500 + 200000}}{2} \] \[ x = \frac{50 \pm \sqrt{202500}}{2} \] \[ x = \frac{50 \pm 450}{2} \] Step 6: Find the valid value of \( x \)
\[ x = \frac{50 + 450}{2} = \frac{500}{2} = 250 \] Since speed cannot be negative, we take \( x = 250 \) km/hr. Thus, the correct answer is 250 km/hr.