SOLUTION: On the 24-mile trip to school, Brandon sauntered at a leisurely pace. Thus, he had to double his speed on the way back to complete the trip in 9 hours. How fast did he travel in
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-> SOLUTION: On the 24-mile trip to school, Brandon sauntered at a leisurely pace. Thus, he had to double his speed on the way back to complete the trip in 9 hours. How fast did he travel in
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Question 67215This question is from textbook Advanced mathematics
: On the 24-mile trip to school, Brandon sauntered at a leisurely pace. Thus, he had to double his speed on the way back to complete the trip in 9 hours. How fast did he travel in each direction, and what were the two times? This question is from textbook Advanced mathematics
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Let x=his speed to school
Then 2x=his speed back
We are told that the total trip (to & back) is 48 mi
We are also told that the time to school plus the time back equals 9 hours
We'll use the formula: distance(d) =rate(r) times time (t) or d=rt and
t=d/r
time to school: 24/x
time from school: 24/2x
Our equation to solve is
24/x+24/(2x)=9 multiply through by x
24(x/x)+24(x/2x)=9x simplifying
24+12=9x
9x=36
x=4 miles per hour ---------speed to school
time to school: 24/x=24/4=6 hours-----------time to school
2x=2*4=8 miles per hour---------------rate from school
time from school: 24/2x=24/8=3 hours ----time from school
CK
6 hours + 3 hours = 9 hours
8 miles per hour = 2 times 4 miles per hour
