Question 896505: A motorcycle messenger left the rear of a motorized troop 7 miles long and rode to the front of the troop, returning at once to the rear. How far did he ride if the troop traveled 24 miles during this time and each traveled at a constant rate?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i was able to figure it out if i made some assumptions about the time.
i assumed that the time it took for the messenger to get to the front of the line was equal to 1 hour.
in that 1 hour, the messenger traveled 31 miles.
in that same hour, the troop traveled 24 miles.
this makes sense because the messenger started at the back of the troop line which was 7 miles long and then went to the front of the troop line.
since the troop line went 24 miles in the time it took for the messenger to get from the back of the troop line to the front of the troop line, the messenger has to travel that same 24 miles plus an additional 7 miles to get to the front.
the troop line traveled 24 miles and the messenger traveled 31 miles.
assuming that the time it took was 1 hour, then the messenger was traveling at 31 miles per hour and the troop line was traveling at 24 miles per hour.
now the messenger is at the front and delivers his message and then heads to the back of the troop line at the same rate of speed that he took to get to the front of the troop line.
the troop line is traveling at 24 miles per hour.
the messenger is traveling at 31 miles per hour.
the back ot the troop line is moving in the opposite direction as the messenger now.
the messenger will be traveling for x miles and, in the same time, the troop line will be traveling for (7-x) miles.
they will meet somewhere in the middle, that point being x miles from the front of the troop line and 7-x miles from the back of the troop line.
the total distance covered by both the messenger and the troop line will be x + 7 - x which will be equal to 7.
the messenger and the troop line will both be traveling for the same amount of time when they meet.
since rate * time = distance.
the messenger formula will be 31 * T = x
the troop line formula will be 24 * T = 7-x
solve both equations for T an you get:
T = x/31
T = (7-x)/24
since T = T, you get x/31 = (7-x)/24
solve for x in this equation.
start with x/31 = (7-x)/24
multiply both sides of the eqution by 31*24 to get:
24x = 31(7-x)
simplify to get:
24x = 217 - 31x
add 31x to both sides of the equation to get:
55x = 217
divide both sides of the equation by 55 to get:
x = 217/55 = 3.9454545....
7-x is therefore equal to 3.05454545.....
x is the distance that the messenger traveled to get back to the end of the troop line.
the total distance traveled by the messenger is therefore 31 + 3.05454545..... which is equal to 34.05454545..... miles.
the actual time that it took is not important.
the original time sets the rate which leads to the miles.
a different time would set a different rate but lead to the same miles.
assume the first time chosen was y hours instead of 1 hour.
the rate of the troop line would be 24/y miles per hour.
the rate of the messenger would be 31/y miles per hour.
that rate stays constant throughout.
when we get to the end, we get:
24/y*x = 31/y*(7-x)
we simplify to get:
24/y*x = 31/y*7 - 31/y*x
we simplify further to get:
24x/y = 217/y - 31x/y
we add 31x/y to both sides of the equation to get:
55x/y = 217/y
we multiply both sides of the equation by y to get:
55x = 217
we divide both sides of the equation by 55 to get:
x = 217/55
this is the same answer as we got before.
the time and rate of speed were changed to be consistent with each other, but the miles remained the same.
the answer will be the same regardless of the time it took for the messenger to get to the front of the line.
the answer, if everything i did is correct, will be:
31 + 3.9454545... which is equal to 34.9454545 miles.
in fractional form this would be equal to 31 + 3 + 52/55 which is equal to 34 + 52/55 miles.
i get the same answer every time.
the messenger traveled 34 + 52/55 miles in total.
i'm reasonably sure it's correct.
|
|
|
