During rush hour, Fernando can drive 20 miles using the side roads in the same time that it takes to travel 15

Question

During rush hour, Fernando can drive 20 miles using the side roads in the same time that it takes to travel 15 miles on the freeway. If Fernando's rate on the side roads is 9 mi/h faster than his rate on the freeway, find his rate on the side roads.

Answer 1

Rate on the free way = V1 say

Rate on the side roads = V1 + 9

Time taken for 15 miles on the freeway = 15/V1

Time taken for 20 miles on the side roads = 20 / (V1 + 9)

These two times are same

15/V1 = 20/(V1 + 9)

15V1 + 135 = 20V1

5V1 = 135

V1= 27 mph

Rate on the side road is 36 mph.

Answer 2

the time it takes to force 20 miles utilising the element roads would be represented via the variable 't'. observe that this additionally equals the time it takes to circulate fifteen miles on the line. His velocity on the line would be 'h' and his velocity on element roads 's'. because of the fact He is going 9mi/h swifter on the element roads, s=h+9. velocity circumstances time equals distance, so ht=15 and st=20 (this implies that t(h+9)=20 because of the fact s=h+9.) for this reason, ht+9t=20 and ht=15. in case you isolate the 'ht' words, and set the equations equivalent to a minimum of one yet another, you get 15=20-9t, or 9t=5. So, t=5/9 hours. Distance/time=velocity, so 20/(5/9)=36 mi/h=his velocity on the element roads. 36 mi/h

Answer 3

Vs t = 20
Vf t = 15
then Vs/Vf = 4/3
also Vs=Vf +9 , then Vs/Vs-9 = 4/3
then Vs=36 mi/h