SOLUTION: Hindered by a strong wind, a jet flew 15% slower than usual and made a 4000 mile trip in 30 minutes longer than usual. What is the usual speed of the jet?

Question 421537: Hindered by a strong wind, a jet flew 15% slower than usual and made a 4000 mile trip in 30 minutes longer than usual. What is the usual speed of the jet?
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Hindered by a strong wind, a jet flew 15% slower than usual and made a 4000 mile trip in 30 minutes longer than usual. What is the usual speed of the jet?
-------
With wind DATA:
distance = 4000 miles ; time = x hrs ; rate = d/t = 4000/x mph
--------
Against wind DATA:
distance = 4000 miles ; time = x+(1/2) hrs ; rate = d/t = 4000/[(2x+1)/2] mph
-----
Equation:
against rate = 0.85(with rate)
8000/(2x+1) = 0.85(4000/x)
---
800000/(2x+1) = 85(4000/x)
---
800000x = 85*4000(2x+1)
800000x = 680000x + 340000
120000x = 340000
x = 2.833333 hr (with wind time)
----
With rate = 4000/2.833333 = 1411.76 mph
---
Against rate = 8000/(2*2.83333+1) = 1200 mph
-----
Solve:
Plane + wind = 1411.76
Plane - wind = 1200
---
2*plane = 2611.76
plane speed = 1305.88 mph
================================
Cheers,
Stan H.
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = it's usual speed in mi/hr
Let = it's usual time in hrs
given:
(1)
(2)
--------------------
(2)
(2)
By substitution:
(2)
(2)
(2)
(2)
(2) mi/hr
check answer:
(1)
(1)
(1) hrs
and
(2)
(2)
(2)
(2)
close enough

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