SOLUTION: Tom Quig traveled 280 miles east of St. Louis. For most of the trip he averaged 60 mph, but for one period of time he was slowed to 10 mph due to a major accident. If the total tim

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: Tom Quig traveled 280 miles east of St. Louis. For most of the trip he averaged 60 mph, but for one period of time he was slowed to 10 mph due to a major accident. If the total tim      Log On


   

Question 173626This question is from textbook Introductory Algebra
: Tom Quig traveled 280 miles east of St. Louis. For most of the trip he averaged 60 mph, but for one period of time he was slowed to 10 mph due to a major accident. If the total time of travel was 8 hours, how many miles did he drive at the reduced speed? This question is from textbook Introductory Algebra

Found 2 solutions by nerdybill, stanbon:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Tom Quig traveled 280 miles east of St. Louis. For most of the trip he averaged 60 mph, but for one period of time he was slowed to 10 mph due to a major accident. If the total time of travel was 8 hours, how many miles did he drive at the reduced speed?
.
You will need to apply the "distance formula":
d = rt
where
d is distance
r is rate or speed
t is time
.
Let x = time driving at reduced speed
then
8-x = time driving at 60 mph
.
60(8-x) + 10x = 280
480 - 60x + 10x = 280
480 - 50x = 280
480 = 50x + 280
200 = 50x
4 hours = x
.
This means he spent 4 hours driving at 10 mph, the distance he drove at this rate then is:
4 * 10 = 40 miles

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Tom Quig traveled 280 miles east of St. Louis. For most of the trip he averaged 60 mph, but for one period of time he was slowed to 10 mph due to a major accident. If the total time of travel was 8 hours, how many miles did he drive at the reduced speed?
------------------------------
Most-trip DATA:
distance = x miles ; rate = 60 mph ; time = d/r = x/60 hrs.
----------------------
Slow part DATA:
distance = 280-x miles ; rate = 10 mph ; time = d/r = (280-x)/10 hrs
------------------------
EQUATION:
time + time = 8 hrs.
x/60 + (280-x)/10 = 8
Multiply thru by 60 to get:
x + 6(280-x) = 480
-5x = 480 - 6*280
-5x = -1200
x = 240 miles (distance traveled at 60 mph)
280-x = 40 miles (distance traveled at 10 mph)
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Cheers,
Stan H.