SOLUTION: A cyclist rode 40 mi. before having a flat tire and then walking 5 mi to a service station. The cycling rate was four times the walking rate. The time spent cycling and walking was
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Question 149371This question is from textbook Intermediate Algebra with Applications
: A cyclist rode 40 mi. before having a flat tire and then walking 5 mi to a service station. The cycling rate was four times the walking rate. The time spent cycling and walking was 5 h. Find the rate at which the cyclist was riding.
This question is from textbook Intermediate Algebra with Applications
Found 2 solutions by ankor@dixie-net.com, mangopeeler07:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A cyclist rode 40 mi. before having a flat tire and then walking 5 mi to a service station. The cycling rate was four times the walking rate. The time spent cycling and walking was 5 h. Find the rate at which the cyclist was riding.
:
Let x = the walking rate
then
4x = the riding rate
:
Write a time equation: Time =
:
Riding time + walking time = 5 hrs
+ = 5
Reduce the fraction:
+ = 5
= 5
Multiply both sides by x
5x = 15
x =
x = 3 mph is the walking rate
then
4(3) = 12 mph is the riding rate
;
:
Check solution
+ =
3 + 1 = 5 hrs
Answer by mangopeeler07(462) (Show Source): You can put this solution on YOUR website!
A cyclist rode 40 mi. before having a flat tire and then walking 5 mi to a service station. The cycling rate was four times the walking rate. The time spent cycling and walking was 5 h. Find the rate at which the cyclist was riding.
total miles=45 miles
total time=5 hrs
walking speed=x mph
cycling speed=4x mph
40/4x+5/x=5-----------------------because miles over mph gives you hours
Common denominator=4x
40/4x+20/4x=5
Add the numerators
60/4x=5
Multiply both sides by 4x
60=20x
Divide by 20
3=x
walking speed=4 mph
cycling speed=12 mph
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