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During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed og 60miles per hour. in terms of x, what was Francine's average speed for the entire trip?
a. 180-x/2
b. x + 60/4
c. 300 - x/ 5
d. 600/ 115-x
e. 12000 / x + 200
Assume total distance is a Distance Average speed a*x/100 40 a(1-x/100) 60
Average speed = 40*a*x/100 + 60a(1-x/100)
a[40x/100 + 60 - 60x/100]
a [300-x/5]
this is how I solved this problem. It is definately wrong, but I cannot find exactly what is wrong or what is missing. Anyone can explain it?
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Re: ps 149 algebra
[#permalink]
18 Dec 2011, 00:58
eybrj2 wrote:
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed og 60miles per hour. in terms of x, what was Francine's average speed for the entire trip?
a. 180-x/2
b. x + 60/4
c. 300 - x/ 5
d. 600/ 115-x
e. 12000 / x + 200
Assume total distance is a Distance Average speed a*x/100 40 a(1-x/100) 60
Average speed = 40*a*x/100 + 60a(1-x/100)
a[40x/100 + 60 - 60x/100]
a [300-x/5]
this is how I solved this problem. It is definately wrong, but I cannot find exactly what is wrong or what is missing. Anyone can explain it?
Let the total distance be 100miles(coz the distance is in percentage).
so avg speed is =total distance/total time total distance = x+(100-x)=x total time =(x/40)+((100-x)/60)=(x+200)/12
So, the avg speed is (12000/(x+200))
E is the answer
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.