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 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.

the buses are at 50 mph  and 60 mph

the distance = 210 miles

Step-by-step explanation:

r1 = rate of bus 1

r2 = rate of bus 2

r1 = r2+10   (The speed of one of the buses was 10 mph greater than the speed of the other bus.)

distance = rate * time

for bus 1

distance = r1 * (3.5)

bus 2 was 1/6 short of making it to town b

so it made it 5/6 of the way

5/6 distance = r2 * (3.5)

multiply by 6/5 on each side

distance = 6/5 * r2 * 3.5

set the distances equal

r1 * (3.5)=6/5 * r2 * 3.5

r1 = 6/5  * r2

now set rate 1 equal  (r1 = r2+10)

6/5  * r2 = r2 + 10

solve for r2 by subtracting r2 on each side

6/5 r2 -r2 = 10

6/5 = 1.2

1.2 r2 - r2 =10

.2 r2 = 10

divide by .2 on each side

r2 = 10/.2

r2 = 50

r1 = 50 + 10

r1 =60

to find the distance

distance = r1 * (3.5)

distance = 60 * 3.5

distance = 210 miles

i think the answer would be c