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Cargo and Passenger Planes Trivia?
A cargo plane flying at average speed of 750 miles per hour leaves station B East bound at 1 AM local time and reaches station A at 11 AM local time. A passenger plane leaves station B at 6 AM
local time and flying at average speed of 900 miles per hour lands at station A at 7 AM same day.
(a) What is the distance between places A and B
(b) What is the difference in local times between these 2 places.
(c) At what distance from station A the planes would cross each other?
Clue: Time difference between the two stations is 4 hours. Can you solve the problem without clue.
How?
1 Answer
- Tino-sbyLv 59 years agoFavorite Answer
S=distance between A and B
X=time different between A and B, where A is x hour ahead B, if x is negatif, it will means the opposite
So, when B is 1 o'clock, A is 1+x o'clock
Tc: cargo flying time = 11-1-X = 10-x
Tp: passenger flying time = 7-6+X = 1+x
S = Tc x cargospeed = Tp x passenger speed
(10-x) 750 = (1+x) 900
7500 - 750x = 900 + 900x
7500-900 = 900x + 750x
6600 = 1650x
x = 4
Time difference 4 hours
S = (10-x)750 = (10-4) 750 = 6x750 = 4500 miles
Crossing place is n miles from A, so
t = n/900
and also
t = (4500-n)/750
n/900 = (4500-n)/750
750n = 4050000 - 900n
750n + 900n = 4050000
1650n = 4050000
n = 2454 miles from A
