SOLUTION: this is supposed to be a linear equation system problem. I couldn't find the answer. Maybe because it's a bit related to physics.
2 groups of people depart at the same time from
Question 917471: this is supposed to be a linear equation system problem. I couldn't find the answer. Maybe because it's a bit related to physics.
2 groups of people depart at the same time from a lodge in opposite directions. they move in a circular track. When they find each other one group has moved 38 km and the other has moved 57 km. find the average speed of each group if the second one arrived at the lodge 1 hour before the first one. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 2 groups of people depart at the same time from a lodge in opposite directions.
they move in a circular track.
When they find each other one group has moved 38 km and the other has moved 57 km.
find the average speed of each group if the second one arrived at the lodge 1 hour before the first one.
:
Find the distance of the circular track: 38 + 57 = 95 km
:
The relationship of the distance each traveled is the same as the relationship of their speeds.
:
Let x = the multiplier
then
38x = speed of the slower group
and
57x = speed of the faster group
:
Write a time equation; time = dist/speed
slow grp time - fast grp time = 1 hr - = 1
Least common multiple; 38*57 = 2166
multiply equation by 2166x, cancel the denominators and you have
57(95) - 38(95) = 2166x
5415 - 3610 = 2166x
1805 = 2166x
x = 1505/2166
x = .833 is the multiplier
then
38 * .833 = 31.67 km/hr is the speed of the slow group
57 * .833 = 47.5 km/hr is the speed of the faster group
:
:
Confirm this find the time for each to make a circuit
95/31.67 = 3.0 hrs
95/47.50 - 2.0 hrs
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time dif: 1 hr
