SOLUTION: A 6000 m relay race is run by two male runners in 18.5 minutes. \the first runner averaged 300m/min and the second 350 m/min. For how many minutes did each hold the baton?

Algebra ->  Linear-equations -> SOLUTION: A 6000 m relay race is run by two male runners in 18.5 minutes. \the first runner averaged 300m/min and the second 350 m/min. For how many minutes did each hold the baton?      Log On


   



Question 179739: A 6000 m relay race is run by two male runners in 18.5 minutes. \the first runner averaged 300m/min and the second 350 m/min. For how many minutes did each hold the baton?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
To do this problem, you have to know what a
relay race is. The runners runners run one
at a time. One after the other.
For 1st runner
(1) d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
For 2nd runner
(2) d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
given:
d%5B1%5D+%2B+d%5B2%5D+=+6000m
d%5B2%5D+=+6000+-+d%5B1%5D
t%5B1%5D+%2B+t%5B2%5D+=+18.5min
t%5B2%5D+=+18.5+-+t%5B1%5Dmin
r%5B1%5D+=+300m/min
r%5B2%5D+=+350m/min
-------------------------
Rewriting (1) and (2)
(1) d%5B1%5D+=+300t%5B1%5D
(2) 6000+-+d%5B1%5D+=+350%2A%2818.5+-+t%5B1%5D%29
This is 2 equations and 2 unknowns, so it's solvable
(1) d%5B1%5D+=+300%2At%5B1%5D
(2) 6000+-+d%5B1%5D+=+6475+-+350t%5B1%5D
(2) -d%5B1%5D+=+-350t%5B1%5D+%2B+475
Now add (1) and (2)
(2) -d%5B1%5D+=+-350t%5B1%5D+%2B+475
(1) d%5B1%5D+=+300t%5B1%5D
0+=+-50t%5B1%5D+%2B+475
50t%5B1%5D+=+475
t%5B1%5D+=+9.5min
And since
t%5B2%5D+=+18.5+-+t%5B1%5Dmin
t%5B2%5D+=+18.5+-+9.5
t%5B2%5D+=+9min
The 1st runner held the baton for 9.5 min, the 2nd for 9 min
check:
(1) d%5B1%5D+=+300%2A9.5
d%5B1%5D+=+2850
(2) d%5B2%5D+=+350%2A%2818.5+-+9.5%29
d%5B2%5D+=+3150
d%5B1%5D+%2B+d%5B2%5D+=+6000m
2850+%2B+3150+=+6000
6000+=+6000OK