SOLUTION: A student walks and jogs to college each day. The student averages 5kmp walking and 9kph jogging. The distance from home to college is 8k, and the student makes the trip in 1 hour.
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Question 76449This question is from textbook Algebra 1
: A student walks and jogs to college each day. The student averages 5kmp walking and 9kph jogging. The distance from home to college is 8k, and the student makes the trip in 1 hour. How far does the student jog? This question is from textbook Algebra 1
You can put this solution on YOUR website! The student averages 5 kmh walking and 9 kmh jogging. The distance from home to college is 8 k, and the student makes the trip in 1 hour. How far does the student jog?
:
Let x = jogging distance; then (8-x) = walking distance
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Write a time equation; Time = distance/speed
:
Jogging time + Walking time = 1 hr + = 1
:
Multiply equation by 45 to get rid of the denominators:
5x + 9(8-x) = 1
5x + 72 - 9x = 45
5x - 9x = 45 - 72
-4x = -27
x = -27/-4
x = 6.75 km jogging
:
8 - 6.75 = 1.25 km walking
:
:
Check using time:
6.75/9 + 1.25/5 =
.75 + .25 = 1 hr
