In a flight of 600km , an aircraft was slowed down due to bad weather . Its average speed for the trip was reduced by 200km / hour and the time increased by 30 minutes. Find the duration of the flight.

speed = distance / time =>distance = speed X time let the normal speed of flight be s, time t and distance is 600km. so,

600= sX t----(1)

avg. speed is reduced by 200km/hour, it means now speed is, s-200, and time increased by 30 mins= half hour. so time= t+0.5 hour. (both speed and time should be in same unit so i converted to hour). distance is same in both cases. new equation will be,

600=(s-200 )X(t+0.5)

=>600 = st+.5s-200t-100

=>600 = 600+.5s-200t-100 (from eq 1, we get value of st)

=>0=.5s-200t-100

=>.5s-200t=100

=>s-400t=200

=> s=200+400t----(2)

Put value of s in eq 1, we will get, 600=(200+400t ) X t

=>3=(1+2t)X t

=>3=t+2tXt

=>2tXt+t -3=0

=> 2tXt+3t – 2t -3=0

=>2t(t-1) +3(t-1)

=>(2t+3)(t-1)=0

=> t=1 or t=-3/2 (minus is not possible) so 1 hour ans. And time of flight with reduced speed is, 1+.5= 1.5 hour ans.

i will re-write in x and y in few mins. and i am not teacher and my answers are correct but may be i am not writing in proper way. so you may not get full marks

speed = distance / time

=>distance = speed X time

let the normal speed of flight be x, time y and distance is 600km. so,

600= x y----(1)

avg. speed is reduced by 200km/hour, it means now speed is, x-200, and time increased by 30 mins= half hour. so time= y+0.5 hour. (both speed and time should be in same unit so i converted to hour). distance travelled by flight is same in both cases. new equation will be,

600=(x-200 )X(y+0.5) (because we know, =>distance = speed X time )

=>600 = xy+.5x-200y-100

=>600 = 600+.5x-200y-100 {substitution of (x y =600) from eq 1 }

=>0=.5x-200y-100

=>.5x-200y=100

=>x-400y=200

=> x=200+400y----(2)

Put value of x in eq 1, we will get,

600= (200+400y )  (y)

=>600= 200 (1+2y) (y) (taking 200 as common)

=>3= (1+2y) y (dividing both LHS and RHS by 200)

=>3=y+2y y

=>2y y+ y -3=0

=> 2y y +3y – 2y -3=0

=> 2y y-2y +3y -3=0

=>2y (y-1) +3(y-1) =0

=>(2y+3)(y-1)=0

=> y=1 or y=-3/2 (minus is not possible) so 1 hour ans. And time of flight with reduced speed is, 1+.5= 1.5 hour ans.