Question

if speed of a aeroplane is reduced by 40km per hr,it takes 20 min more to cover 1200km.find the speed of the aeroplane

Answer 1

Solution:-
Let the actual speed of the aeroplane be 'x' km/hr
So, speed after reduction = (x - 40) km/hr
Let the actual time taken by the aeroplane be y hr
and the time taken after the speed reduction = y + 20 min = (y + 20/60)
= (y + 1/3) hr
Total distance = 1200 km
So, 
Distance = speed*time
1200 = x*y 
xy = 1200
y = 1200/x .............(1)
After speed reduction,
1200 = (x - 40) (y + 1/3)
1200 = (x - 40) (3y + 1/3)
1200*3 = (x -40) (3y + 1)
Now, substituting the value of y = 1200/x, we get 
3600 = (x - 40) (3*1200/x + 1)
3600 = (x - 40) (3600/x + 1)
3600x = (x - 40) (3600 + x)
3600x = x² + 3600x - 40x - 144000
x² - 40x - 144000 = 0
x² - 400x +360x - 144000 = 0
x(x - 400) + 360 (x - 400) = 0
(x - 400) (x + 360)
x = 400 and x = -360
Speed can't be negative, so the actual speed of the aeroplane is 400 km/hr
substituting the value of y = 1200/x.
y = 1200/400
y = 3 hours
So, the actual time taken by the aeroplane is 3 hours.
Answer.

Answer 2

speed= v
distance = 1200
time = t..... 20 min =1/3 hr
so, v= 1200/t
v- 40= 1200/ t +1/3
putting the value of v in above equation
1200/t -40 = 1200/t +1/3
t = -3.31 or 2.98
-ve time is impossible so for t= 2.98 we get 
v= 1200/ 2.98
v= 402.01 km/hr