SOLUTION: Chuck and Dana agree to meet in Chicago for the weekend. Chuck travel 124 miles in the time it takes Dana to travel 108 miles. If Chucks rate is 4mph more than Danas, and they trav

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Chuck and Dana agree to meet in Chicago for the weekend. Chuck travel 124 miles in the time it takes Dana to travel 108 miles. If Chucks rate is 4mph more than Danas, and they trav      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 345356: Chuck and Dana agree to meet in Chicago for the weekend. Chuck travel 124 miles in the time it takes Dana to travel 108 miles. If Chucks rate is 4mph more than Danas, and they travel the same length in time, what is chucks speed.
2)Airplane flies from Metro to Gotham with a tailwind that increases is 90mph and on the way back decreases the speed 90 mph. with wind on their back it takes 2.05 hours compared to with wind in face 3.46 hours. Whats the distance between the two cities?
Thank you if you get to this. These stump me like crazy!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Chuck travel 124 miles in the time it takes Dana to travel 108 miles.
If Chucks rate is 4mph more than Danas, and they travel the same length in time, what is chucks speed.
:
Let s = Chuck's speed
then
(s-4) = Dana's speed
;
The times are equal, write a time equation: time = dist/speed
124%2Fs = 108%2F%28%28s-4%29%29
Cross multiply
124(s-4) = 108s
124s - 496 = 108s
124s - 108s = +496
16s = 496
s = 496%2F16
s = 31 mph is Chuck's speed
:
:
Check solution by finding the times, they should be equal
(27 mph is D.s speed)
124/31 = 4 hrs
108/27 = 4 hrs also, confirms our solution of x = 31
:
:
2)Airplane flies from Metro to Gotham with a tailwind that increases is
90mph and on the way back decreases the speed 90 mph.
with wind on their back it takes 2.05 hours compared to with wind in face 3.46 hours.
What's the distance between the two cities?
:
Find the airspeed with out the wind
Let s = speed in still air
then
(s+90) = effective speed with the wind
and
(s-90) = effective speed against the wind
:
The distances are the same each way, write a distance equation; d = speed * time
3.46(s-90) = 2.05(s+90)
3.46s - 311.4 = 2.05s + 184.5
3.46s - 2.05s = 184.5 + 311.4
1.41s = 495.9
s = 495.9%2F1.41
s = 351.7 mph air speed
:
Find the dist
3.46(351.7-90) = 905.5 miles between the two cities
:
Confirm this on the return trip
2.05(351.7+90) = 905.5 miles also