Question

A jet plane was flying at height of 2000 km, due to same technical problem it was slowed down. It’s average speed is reduced by 200 km/h and time increased by 30 minutes. Find the original duration of flight.
1. 2 hrs
2. 4 hrs
3. 3 hrs
4. 3.5 hrs

Answer 1

Correct Answer - Option 1 : 2 hrs

Given∶

Distance = 200 km

Average speed reduced by 200 km/hr

Time increased by 30 minutes = ½ hr

Formula Used∶

Concept of Quadratic, Speed = Distance/Time

Calculation∶

Let the original be of x hours

Distance = 2000 km

Usual speed = 2000/x km/hr

When time is increased by 30 min

Time = (x +1/2) hrs

Then new speed = 2000/(x + ½) km/hr = 4000/(2x + 1) km/hr

According to question,

2000/x – 4000/(2x + 1) = 200

⇒ [2000(2x + 1) – 4000x]/(2x² + x) = 200

⇒ 4000 x + 2000 – 4000 x = 200 (2x² + x)

⇒ 2000 = 400 x² + 200x

⇒ 400 x² + 200x – 2000 = 0

⇒ 200 (2x² + x – 10) = 0

⇒ 2x² + x – 10 = 0

⇒ 2x²+ 5x – 4x – 10 = 0

⇒ x(2x + 5) – 2(2x + 5) = 0

⇒ (x – 2) (2x +5) = 0

⇒ x - 2 = 0 or 2x + 5 = 0

⇒ x = 2 or x = - 5/2, which is not possible as time cannot be negative.

So, x = 2 hours

∴ The original duration is at 2 hrs.