Question 29471: Janice is training for a triathlon. She is working on running and cycling. She runs for 45 minutes then bikes for 1.5 hours. If her combined distance running and cycling is 31.5 miles and her cycling is 3 times her speed in running, what is her running speed? (here's how I worked the problem).
45X + 1.5(3) = 31.5
45X + 4.5 = 31.5
4.5 - 4.5 = 31.5 - 4.5
45X = 27
27/45
X=.6 miles per hour
Answer by AnlytcPhil(1806) (Show Source):
You can put this solution on YOUR website! Janice is training for a triathlon. She is working on running and cycling. She
runs for 45 minutes then bikes for 1.5 hours. If her combined distance running
and cycling is 31.5 miles and her cycling is 3 times her speed in running, what
is her running speed?
First of all let's change the 45 minutes to 3/4 hr or .75 hr, so we will not mix
minutes and hours.
Janice is training for a triathlon. She is working on running and cycling. She
runs for .75 hr. then bikes for 1.5 hours. If her combined distance running
and cycling is 31.5 miles and her cycling is 3 times her speed in running, what
is her running speed?
Let her running speed be X, then
her cycling speed = 3X
Then using DISTANCE = SPEED × TIME
Her running distance = X times .75 hr. or .75X miles
Her cycling distance = 3X times 1.5 hr or 1.5(3X) miles
>>...her combined distance running and cycling is 31.5 miles...<<
This says .75X + 1.5(3X) = 31.5
Solve this and get 6 miles per hour.
(here's how I worked the problem).
45X + 1.5(3) = 31.5
No. Your first mistake is that you mixed time units, that is, minutes and
hours. Your second mistake is that you used only 3 miles per hour for her
cycling speed, when you should have used 3X miles per hour. (I don't think a
bicycle can be ridden that slow, as the bike would fall over.)
45X + 4.5 = 31.5
4.5 - 4.5 = 31.5 - 4.5
45X = 27
27/45
X=.6 miles per hour
Was Janice a turtle? [Just kidding.] LOL. You did the algebra correct.
It was just the wrong equation. Solve this one
.75X + 1.5(3X) = 31.5
correctly and you'll get 6 mph.
Edwin
AnlytcPhil@aol.com
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