SOLUTION: Junior drove his rig on the Interstate 10 from San Antonio to El Paso. At the halfway point he noticed that he had been averaging 80mph, while his company requires his average spee

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Question 177594This question is from textbook College Algebra
: Junior drove his rig on the Interstate 10 from San Antonio to El Paso. At the halfway point he noticed that he had been averaging 80mph, while his company requires his average speed to be 60mph. What must be his speed for the last half of the trip so that he will average 60mph for the entire trip.
The book also gives this hint: The distance from San Antonio to El Paso is irrelevant. Use D or simply make up a distance. This question is from textbook College Algebra

Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Junior drove his rig on the Interstate 10 from San Antonio to El Paso. At the halfway point he noticed that he had been averaging 80mph, while his company requires his average speed to be 60mph. What must be his speed for the last half of the trip so that he will average 60mph for the entire trip.
The book also gives this hint: The distance from San Antonio to El Paso is irrelevant. Use D or simply make up a distance.
----------------------
1st half DATA:
distance = x miles ; rate = 80 mph; time = d/r = x/80 hrs
------------------
Complete trip DATA:
distance = 2x miles ; rate = 60 mph; time = d/r = 2x/60 = x/30 hrs
------------------
2nd half DATA:
distance = x miles ;
time = total time - 1st half time = (x/30)-(x/80) = 50x/2400 = (1/48)x
-------------------------------------
Therefore rate on the 2nd half = x/[(1/48)x] = 48 mph
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Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Junior drove his rig on the Interstate 10 from San Antonio to 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 his speed for the last
half of the trip so that he will average 60mph for the entire trip.
;
let s = speed required on the last half of the trip
let d = half the distance of the trip
:
Write a time equation: Time = dist%2Fspeed
:
d%2F80 + d%2Fs = %282d%29%2F60
:
%28sd%2B80d%29%2F%2880s%29 = %282d%29%2F60
cross multiply
60(sd + 80d) = 80s(2d)
:
60sd + 4800d = 160sd
:
divide eq by d
60s + 4800 = 160s
:
4800 = 160s - 60s
:
4800 = 100s
s = 4800%2F100
s = 48 mph to average out to 60 mph
:
;
check solution using 600 mi from SA to EP (2d)
300%2F80 + 300%2F48 = %28600%29%2F60
3.75 + 6.25 = 10