SOLUTION: An army officer made the first part of a trip in a plane which flew at the rate of 300 miles an hour. At the landing field he was met by a car which took him the rest of the way t

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Question 766802: An army officer made the first part of a trip in a plane which flew at the rate of 300 miles an hour. At the landing field he was met by a car which took him the rest of the way to his destination at the rate of 55 miles an hour. the trip required 3 hours and 42 minutes. on his return trip the car traveled at the rate of 60 miles an hour and the plane which he took flew at the rate of 250 miles an hour. the return journey required 4 hours and 6 minutes. find the total distance that he flew and the total distance he traveled by car.
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An army officer made the first part of a trip in a plane which flew at the rate of 300 miles an hour.
At the landing field he was met by a car which took him the rest of the way to his destination at the rate of 55 miles an hour.
the trip required 3 hours and 42 minutes. on his return trip the car traveled at the rate of 60 miles an hour and the plane which he took flew at the rate of 250 miles an hour.
the return journey required 4 hours and 6 minutes.
find the total distance that he flew and the total distance he traveled by car.
:
let p = distance traveled by plane
let c = distance traveled by car
:
write a time equation; time = dist/speed
original trip:
p%2F300 + c%2F55 = 3 + 42%2F60 hrs
p%2F300 + c%2F55 = 3.7 hrs
multiply by 3300, resulting in
11p + 60c = 12210
:
Return trip
p%2F250 + c%2F60 = 4 + 6%2F60 hrs
p%2F250 + c%2F60 = 4.1 hrs
multiply by 1500, resulting in:
6p + 25c = 6150
:
Use elimination here, multiply the 1st eq by, 10, the 2nd eq by 24
Subtract the 1st eq from the 2nd
144p + 600c = 147600
110p + 600c = 122100
---------------------------subtraction eliminates c find p
34p = 25500
p = 25500/34
p = 750 mi by plane
Total dist: 1500 mi
find c
6(750) + 25c = 6150
25c = 6150 - 4500
c = 1650/25
6 = 66 mi by car
total dist: 132 mi
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