while boarding an aeroplane a passenger got hurt. the pilot showing promptness and concern made arrangements to hospitalise the injured and so plane started late by 30 mintues. to reach the destination,1500 km away in time the pilot increased the speed by 100km/hour.find the original speed/hour of the plane.
Here is the link to your query.
let T be original time and S be original speed .
T = distance / S = 1500/S [ TIME IS IN MIN ] ...........1
as per the condition,
NEW SPEED = S+100
NEW TIME= T + 30
WE GET EQUATION-
NEW TIME = 1500/ (S+30)
FROM 1, WE GET
1500/S - 1/2 = 1500/(S+100)
⇒1500/S - 1500/(S+100) = 1/2
⇒1/S - 1/(S+100) = 1/3000
⇒(S+100-S)3000 = S(S+100)
⇒300000 = S^2 + 100S
⇒S^2 + 100S - 300000 = 0
⇒S^2+600S-500S-300000 = 0
⇒S(S+600)-500(S-600) = 0
⇒(S-500)(S+600) = 0
Therefore, either S = 500 or S = -600. Since speed of the aeroplane cannot be negative, we reject S = -600. Therefore, the usual speed of the aeroplane is 500 kmph.
Let the usual speed of the plane = x km/hr.
Increased speed of the plane = (x +100) km/hr
Time taken to reach the destination at usual speed,
Time taken to reach the destination at increased speed,
t 1 – t 2 = 30 min
Thus, the usual speed of the plane=500 km/hr