a plane left 30 minutes later than the schedule time and in order to reach its destination 1500 away in time it has to increase its speed by 250 km / hr from its usual speed. find its usual speed.
Answers
Let x be the usual speed of the plane
Increased speed=x+250
Time taken by plane in usual circumstaces,t
1
=
speed
distance
t
1
=
x
1500
Time taken by plane due to delay, t
2
=
x+250
distance
Difference in times = Delay in departure
x
1500
−
x+250
1500
=
60
30
x
1500
−
x+250
1500
=
2
1
⇒
x
2
+250x
x+250−x
=
3000
1
x
2
+250x−750000=0
Solving:
x
2
+1000x−750x−750000=0
x(x+1000)−750(x+1000)=0
(x−750)(x+1000)=0
x=750
∴ The usual speed of the plane is750 km/h.
hope it helps
Answer:
Let the usual speed of the plane be x km/hr.
Increased speed of the plane = (x + 250) km/hr
Time taken to reach the destination at usual speed,
• t1 = 1500/x hr
Time taken to teach the destination at Increase speed
• t2 = 1500/(x+2) hr
We have,
t1 - t2 = 30 min
(1500/x) - 1500/(x+2) = 30/60
(1500/x) - 1500/(x+2) = 1/2
(x+250−x)/(x²+250x) = 1/3000
x²+250x−750000 = 0
On factorizing, we get:
x²+1000x−750x−750000 = 0
x(x+1000)−750(x+1000) = 0
(x−750) (x+1000) = 0
x = 750 or x = -1000
Speed can't be in -ve.
Thus, Usual speed of the plane is 750 km/hr.
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