Question 824528: Junior drove his rig on Interstate 10 from San Antonio and El Paso. At the halfway point he noticed that he had been averaging 80 mph, while his company requires his average speed to be 60 mph. What must be speed for the last half of the trip so that he will average 60 mph for the entire trip. Distance doesn't matter.
Found 2 solutions by TimothyLamb, MathTherapy:
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
s = d / t
t = d / s
D = half trip distance
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first half (a):
t = d / s
ta = D/80
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second half (b):
t = d / s
tb = D/x
where x is the unknown speed for the second half
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whole trip:
s = d / t
60 = 2D / (ta + tb)
60 = 2D / (D/80 + D/x)
60(D/80 + D/x) = 2D
60(xD/80x + 80D/80x) = 2D
(60D/80x)(x + 80) = 2D
(60/80x)(x + 80) = 2
x + 80 = 2*(80/60)x
x + 80 = (2.666666)x
(2.666666)x - x = 80
(1.666666)x = 80
x = 80/1.666666
x = 48.0
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answer:
the speed for the last half of the trip must be: 48 mph
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Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
Junior drove his rig on Interstate 10 from San Antonio and El Paso. At the halfway point he noticed that he had been averaging 80 mph, while his company requires his average speed to be 60 mph. What must be speed for the last half of the trip so that he will average 60 mph for the entire trip. Distance doesn't matter.
Let distance be D, and speed on last ˝ of trip, S
Then time he took to travel 1st ˝ of the distance = ˝D/80, or 
Time 2nd ˝ of trip should take = ˝D/S, or  ÷ S, or 
Average speed = 
Therefore, 60 = D ÷ 
60 = D ÷  ____ =  * 
60(DS + 80D) = 160DS ------ Cross-multiplying
60DS + 4,800D = 160DS
D(60S + 4,800) = D(160S) ------ Factoring out GCF, D
60S + 4,800 = 160S
60S – 160S = - 4,800
– 100S = - 4,800
S, or speed on 2nd ˝ of journey should be:  , or  mph
You can do the check!!
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