Problema Solution
Two aviators in private planes flew from detroit to St. louis the first plane arrived 1 hour sooner than the second how long did it take each, if the first plane was two/thirds that of the second
Answer provided by our tutors
let
(2/3)v = the speed of the first plane
v = the speed of the second plane
t + 1 = the time of the first plane
t = the time of the second plane
since both planes traveled the same distance and distance = time*speed we have
(2/3)v(t + 1) = vt divide both sides by v
(2/3)(t + 1) = t
by solving we find:
t = 2 hr
click here to see the step by step solution of the equation:
2 + 1 = 3 hr
the first plane needed 3 hours while the second plane needed 2 hours to arrive in St. Louis.