While boarding an aeroplane a passenger got hurt.The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 50 minutes in order to reach the destination, 1250km away in time it had to increase its speed by 250km/hr from its usual speed. Find its usual speed.
Answers
Answer:
let the speed be x km/hr
distance given = 1500 km
time= distance/speed
T1 (time 1) =1500/x
T2 (time 2) = 1500/x+100
acc to the ques
T1-(30/60)hr (late by 30 mins) = T2
=1500/x -1/2 = 1500/x+100
=1500/x - 1500/x+100 = 1/2
=1500x + 1,50,000 -1500x = 1/2
x²+100x
=1,50,000= 1/2 X ( x²+ 100x )
2 X 1,50,000 = x² + 100x
3,00,000 = x² + 100x
x² + 100x - 3,00,000 = 0
splitting the middle term
x² + 600x -500x -3,00,000 = 0
x ( x + 600 ) - 500 ( x + 600 ) = 0
( x - 500 ) ( x + 600 ) = 0
x = 500 or x = - 600
since speed cannot be negative hence we consider x = 500
therefore x = original speed = 500 km/hr
Let the usual speed of aeroplane be x km/hr,
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hr
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hr
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hr
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hrAcc. To the question,
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hrAcc. To the question,t1 – t2 = 30 min.
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hrAcc. To the question,t1 – t2 = 30 min.1500x–1500(x+100)=3060
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hrAcc. To the question,t1 – t2 = 30 min.1500x–1500(x+100)=30601500(x+100)–1500xx(x+100)=12
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hrAcc. To the question,t1 – t2 = 30 min.1500x–1500(x+100)=30601500(x+100)–1500xx(x+100)=121500x+150000–1500xx2+100x=12
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hrAcc. To the question,t1 – t2 = 30 min.1500x–1500(x+100)=30601500(x+100)–1500xx(x+100)=121500x+150000–1500xx2+100x=12Since, the speed of plane can never be negative
Let the usual speed of aeroplane be x km/hr,Then the increased speed of the aeroplane is = (x + 100) km/hrDistance to be travelled = 1500 km.Time taken to reach the destination at original speed, t1 = 1500/x hrTime taken to reach the destination at increasing speed, t2 = 1500/(x + 100)hrAcc. To the question,t1 – t2 = 30 min.1500x–1500(x+100)=30601500(x+100)–1500xx(x+100)=121500x+150000–1500xx2+100x=12Since, the speed of plane can never be negativeTherefore, the original speed of the plane is 500 km/hr.