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29 July, 14:54

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?

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Answers (2)
  1. 29 July, 14:55
    0
    Original speed: 6 miles per hour

    Ending: 8 miles per hour

    Step-by-step 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
  2. 29 July, 15:13
    0
    The first rate is 6 miles per hour

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

    Step-by-step 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
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