Question

an aeroplane left 50 minutes later than its scheduled time and in order to reach the destination 1250 km away in time had to increase its speed by 250km/h from its usual time. find its usual time

Answer 1

Let the normal time be t hours and the normal speed of the plane be x kmph.

As the distance is 1250 km, x = 1250/t.

Since the plane leaves 50 minutes late, it has to travel at a speed of x+250 in order to reach at the same time. Thus, we get

x + 250 = 1250 / (t-50/60)

             = 1250 / (t-5/6)
             = 1250 * 6 / (6t-5)
             = 7500 / (6t-5)

(x+250) (6t-5) = 7500

Expanding the LHS, we get

6xt -5x + 1500t-1250 = 7500

But xt = 1250

Substituting the value of xt = 1250 in the above equation, we get

6*1250 - 5x + 1500t - 1250 = 7500

7500 - 5x + 1500t - 1250 = 7500

7500 gets cancelled, and hence, we get

-5x + 1500t - 1250 = 0

Substituting the value of x = 1250/t, and multiplying the entire equation by t, we get

-5 * 1250/t + 1500t - 1250 = 0

-6250 + 1500t^2 - 1250t = 0

Rearranging the equation, we get

1500t^2 - 1250t - 6250 = 0

Dividing the equation by 250, we get

6t^2 - 5t - 25 = 0

Therefore t = [5 +/- sqrt (5^2-4*6*25)]/2*6

t = [5+/- sqrt (25+600)] / 12 = [5+/- sqrt 625]/12

  = (5 +/- 25) / 12 

Since time cannot be negative, only the positive value of 25 is considered. Thus, the normal time is

t = (5+25) / 12 = 30 / 12 = 2.5

Thus, the normal time is 2.5 hours or 2 hours and 30 minutes.

Answer 2

speed=v
time=T

vT = 1250 km 

The speed is increased in delation to (v + 250)
the time is then (T - 5/6) because we are talking in hours. 

(v + 250) (T - 5/6) = 1250 km 

vT = (v + 250) (T - 5/6) 

vT = vT - (5/6)v + 250T - 250(5/6) 

(5/6)v - 250T + 250(5/6) = 0 

5v - 1500T + 1250 = 0 

v - 300T + 250 = 0 

Putting T=1250/v

v - 300(1250/v) + 250 = 0 

v² + 250v - 375000 = 0 

(v - 500) (v + 750) = 0 

So v = 500
and v=- 750