Dave drove to his friends house and back. it took one hour longer to go there than it did to come back. the average speed on the trip there was 25mph. the average speed on the way back was 30mph. how many hours did the trip there take? (urgent)

Remark

You know the most about the trip on the way back and the total of the two trips.

Way Back

r = 30 mph

d = distance traveled

t = t - 1 where t = the time to go there which we know nothing about

Total Distance

2d = 25 mph * (t + t - 1)

2d = 25 * (2t - 1)

2d = 50t  - 25                 Divide by 2

2d = 50t/2 - 25/2

d = 25t - 12.5

Equation coming back

d = 30*(t - 1)

d = 30t - 30

Comment

Since the distances are the same, equate them.

Solution

25t - 12.5 = 30t - 30                     Subtract 25t from both sides

25t - 25t -12.5 = 30t - 25t - 30   Combine like terms

-12.5 = 5t - 30                              Add 30 to both sides

-12.5 + 30 = 5t - 30 + 30             Combine

17.5 = 5t                                       Divide by 5

17.5/5 = t                                      Divide and Switch

t = 3.5

t is the time going there

t - 1 is the time coming back

Total time = 2*t - 1 = 2*3.5 - 1 = 6

The total time is 6 hours.   Answer

answer: the answer is 5/22

step-by-step explanation: if you subtract 17 from 22 the answer is 5 therefore the remainder of students are female