Question

Date : 1
An aeroplane left 30 min later than its
Shedualed time and in order to such reach
it destination 300 km in time at had
to increase its speed byy 250 km/hr from it usual speed
​

Answer 1

Answer:

Step-by-step explanation:

Sol:

Let the usual time taken by the aeroplane = x km/hr

Distance to the destination = 1500 km

Case (i)

Speed = Distance / Time = (1500 / x) Hrs

 

Case (iI)

Time taken by the aeroplane = (x - 1/2) Hrs

Distance to the destination = 1500 km

Speed = Distance / Time = 1500 / (x - 1/2) Hrs

 

Increased speed = 250 km/hr

 

⇒ [1500 / (x - 1/2)] - [1500 / x] = 250

⇒ 1/(2x2 - x) = 1/6

⇒ 2x2 - x = 6

⇒ (x - 2)(2x + 3) = 0

⇒ x = 2 or -3/2

Since, the time can not be negative,

The usual time taken by the aeroplane = 2 hrs

and the usual speed = (1500 / 2) = 750 km/hr.

Recommend(9) Comment (0)

person

Syeda , SubjectMatterExpert

Member since Jan 25 2017

Answer,

Let the usual time taken by the aeroplane = x km/hr

Distance to the destination = 1500 km

Case (i)

Speed = Distance / Time = (1500 / x) Hrs

 

Case (iI)

Time taken by the aeroplane = (x - 1/2) Hrs

Distance to the destination = 1500 km

Speed = Distance / Time = 1500 / (x - 1/2) Hrs

 

Increased speed = 250 km/hr

 

⇒ [1500 / (x - 1/2)] - [1500 / x] = 250

⇒ 1/(2x2 - x) = 1/6

⇒ 2x2 - x = 6

⇒ (x - 2)(2x + 3) = 0

⇒ x = 2 or -3/2

Since, the time can not be negative,

The usual time taken by the aeroplane = 2 hrs

and the usual speed = (1500 / 2) = 750 km/hr.