Question 156080: I am having difficulty setting up the equation for this problem. Thank you for any help.
An army messenger will travel at a speed of 60 miles per hour on land and in a motorboat whose speed is 20 miles per hour in still water. In delivering a message he will go by land to a dock on a river and then on the river against a current of 4 miles per hour. If he reaches his destination in 4 1/2 hours and then returns to his starting point in 3 1/2 hours, how far did he travel by land and how far by water.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
GOING TO DESTINATION:
Let d1=distance travelled on land (one way)
And let d2=distance travelled by boat (one way)
Then total distance travelled=2d1+2d2
Time travelled on land=d1/60
Time travelled by boat=d2/(20-4)=d2/16 (must subtract speed of current)
Now we are told that the above two times equals 4.5 hr, so:
d1/60 + d2/16=4.5-------------------------eq1
RETURNING TO STARTING POINT:
Time travelled by boat=d2/(20+4)=d2/24-------------(must add speed of current)
Time travelled on land=d1/60
And we are told that the above two times equals 3.5 hrs, so:
d1/60 +d2/24=3.5-------------------------------eq2
subtract eq2 from eq1:
d2/16 -d2/24=1 or
3d2/48-2d2/48=1 and
d2/48=1 multiply each side by 48
d2=48 mi; substitute into eq1
d1/60+48/16=4.5
d1/60+3=4.5 subtract 3 from each side
d1/60=1.5 multiply each side by 60
d1=90 mi
Total distance travelled=2d1+2d2=2*90+2*48=180+96=276 mi
Distance travelled by land=2d1=2*90=180 mi
Distance travelled by water=2d2=2*48=96 mi
CK
90/60+48/16=4.5
1.5+3=4.5
4.5=4.5
and
48/24+90/60=3.5
2+1.5=3.5
3.5=3.5
Hope this helps---ptaylor
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