Question

Tony went on an 8-mile run. For

the first 1/3 of an hour he ran at a
constant speed. For the last 3⁄4 of an
hour he ran at a constant speed that
was 2 miles per hour faster than his
original speed. What was his original
speed, and what was his ending scene on the run?

Answer 1

Original speed: 6 miles per hour

Ending: 8 miles per hour

Explanation:

Speed = distance/time

Distance = speed × time

Let S be the original speed

(S × ⅓) + [(S + 2) × ¾] = 8

S/3 + 3S/4 + 3/2 = 8

Lcm: 12

(S×4 + 3S×3)/12 = 8 - 3/2

13S/12 = 13/2

S = 13/2 × 12/13

S = 6

Answer 2

The first rate is 6 miles per hour

The final rate is 6+2 or 8 miles per hour

Explanation:

We know d = rt where distance is equal to rate times time

First part

d1 = r * 1/3

Second part

d2 = (r+2)(3/4)

We know the total distance is 8 miles and is

d1+d2 = 8

r * 1/3 + (r+2)(3/4) = 8

Multiply each side by 12 to get rid of the fractions

12( r * 1/3 + (r+2)(3/4)) = 8*12

4r +(r+2) 9 = 96

Distribute

4r+ 9r +18 = 96

Combine like terms

13r +18 = 96

Subtract 18 from each side

13r+18-18 = 96-18

13r = 78

Divide each side by 13

13r/13 = 78/13

r=6

The first rate is 6 miles per hour

The final rate is 6+2 or 8 miles per hour