Algebra 1, round trip

Question

Jed rode his scooter at 20 miles per hour and then jogged back at 8 miles per hour. If the round trip took him 7 hours, how far was it to school?

Answer 1

Let's use the formula distance = rate x time to solve the problem.

Let d be the distance to school.

The time it took Jed to ride to school is:

t1 = d / 20

The time it took Jed to jog back from school is:

t2 = d / 8

We know that the total time for the round trip was 7 hours, so we can write the equation:

t1 + t2 = 7

Substituting the expressions we derived earlier, we get:

d / 20 + d / 8 = 7

To solve for d, we can multiply both sides by the least common multiple of the two denominators, which is 40:

2d + 5d = 280

7d = 280

d = 40

Therefore, the distance to school is 40 miles.

Answer 2

Here's another way to do it.

distance = rate x time, so time = distance/rate (or d/r)

For the scooter, t_s = d/20

For the jogging leg, t_j = d/8 (Note that both t's are equal) We know that t_s + t_j = 7 hours.

So the total time was d/20 + d/8 = 7 The common denominator is 40 (multiply both sides by 40 to eliminate the fractions).

2d + 5d = 280

7d = 280 -> d = 40 miles each way

Check: Scooter t = d/r so t = 40/20 = 2 hrs

Jogging: 40/8 = 5 hrs; total 7 hrs.

Answer 3

40 miles to school

20 mph one way for x hours

8 mph back for 7-x hours

distance = speed times time

20/8 = 10/4 =5/2 = 2.5

2.5 times as long coming back as going

2.5= g/c where g+c=7, c =7-g

(7-c)/c = 2.5

2.5c = 17.5 -c

3.5c = 17.5

c = 17.5/3.5= 35/7 =5 hours coming back

g= 2 hours going to school

20 mph times 2 = 40 miles to school

8 mph times 5 hours = 40 miles coming back

20x = 8(7-x) = 56-8x

28x = 56

x = 56/28 =2 hours going

7-x = 7-2=5 hours coming back

20 times 2 = 40 miles going

8 times 5 = 40 miles coming back