A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
Answers
SOLUTION :
Let x km/h be the usual speed of a plane .
Total Distance to reach the destination = 1600 km
ATQ..
Case 1.
Original Time = Distance /speed = (1600 / x) h
Case 2.
Let( x + 400 )km/h be the new speed of a plane
New Time = Distance / Speed = 1600 / (x + 400) Hrs
Original time - New time = 40 min
[1600 / x] - [1600 / x + 400] = 40 min
[1500 / x] - [1500 / x+100] = 40/60
[ 1 min = 1/60 h]
[1600 / x] - [1600 / x + 400] = 2/3 h
1600 [ 1/x - 1 /(x + 400)] = ⅔
1600 × 3 [ x + 400 - x] / [x(x + 400)] = 2
4800 [ 400/(x² + 400x) ] = 2
1920000/(x² + 400x) = 2
1920000 = 2 (x² + 400x)
1920000 = 2x² + 800x
2x² + 800x - 1920000 = 0
2(x² + 400x - 960000) = 0
x² + 400x - 960000 = 0
x² + 1200x -800x - 960000 = 0
[By middle term splitting]
x(x + 1200) - 800(x + 1200) = 0
(x + 1200 )(x - 800) = 0
x = - 1200 or x = 800
Since, the speed can’t be negative, so x ≠ - 1200. Therefore , x = 800
Hence, the usual speed of a plane = 800 km/h.
HOPE THIS ANSWER WILL HELP YOU…
Let the usual speed be = x km/h
now we know that
speed = distance /time
time = distance / speed
Usual time = 1600/x
Now due to bad weather the speed is increased b 400 km/ hr
which means new speed = 1600/x+400
Now it's given that the plain left 40 minutes late
which means new time = 1600/x+400 + 2/3 ( 40/60 = 2/3 hours)
1600/x = 1600/x+400 +2/3
1600/x-1600/x+400 = 2/3
1600(x+400)-1600x)/x(x+400) = 2/3
1600x + 640000-1600x/x²+400x = 2/3
640000 = 2/3(x²+400x)
640000 * 3/2 = x²+400x
960000 = x² +400x
0 = x²+400x-960000
0 = x² +1200x-800x-960000
0 = x( x+ 1200) - 800(x+1200)
0= (x+1200)(x-800)
x= -1200 or x= 800
Speed cannot be negative
∴ usual speed = 800 km/h
:)