Asked by Helen
Two buses left town A for town B at the same time. The speed of one of the buses was 10 mph greater than the speed of the other bus. In 3 1/2 hours one bus reached town B, while the other bus was still away from town B at a distance equal 1/6 of the distance between A and B. Find the speed of the buses and the distance between A and B.
Answers
Answered by
henry2,
Bus #1: X mi./h.
Bus #2: x-10 mi,/h.
d1 = X*T = X*3.5 = 3.5x.
d1/6 = (x-10)*3.5 = 3.5x-35.
d1/6 = 3.5x-35,
d1 = 21x-210.
d1 = 21x-210 = 3.5x.
X = 12 m1/h.
x-10 = 2 m1./h.
d1 = X*T = 12 * 3.5 = 42 miles. = Distance from A to B.
Bus #2: x-10 mi,/h.
d1 = X*T = X*3.5 = 3.5x.
d1/6 = (x-10)*3.5 = 3.5x-35.
d1/6 = 3.5x-35,
d1 = 21x-210.
d1 = 21x-210 = 3.5x.
X = 12 m1/h.
x-10 = 2 m1./h.
d1 = X*T = 12 * 3.5 = 42 miles. = Distance from A to B.
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