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30 April, 07:21

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5.

B. 2.4.

C. 2.3.

D. 2.2.

E. 2.1.

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Answers (1)
  1. 30 April, 07:44
    0
    A) 2.5

    Explanation:

    Extracting vital information from the question;

    speed upstream = (v-3) mile/hr, distance traveled = 90 mile, speed downstream = (v+3), time for downstream in hours = t while time for upstream = t + 0.5 hr since the upstream journey is half hour longer.

    speed = distance / time = 90 / (t + 0.5)

    (v - 3) = 90 / (t + 0.5)

    cross multiply

    (v-3) (t + 0.5) = 90 equation (1) for upstream motion

    (v+3) = 90 / t

    cross multiply

    t (v+3) = 90 equation (2) for downstream motion

    make v subject of the formula in equation 2

    vt + 3t = 90

    vt = 90 - 3t

    divide both side by t

    vt/t = (90 - 3t) / t

    v = (90 - 3t) / t

    substitute for t in equation 1

    ((90 - 3t) / t) - 3) (t + 0.5) = 90

    solve through finding l. c. m (lowest common multiple)

    (90 - 3t - 3t) / t (t + 0.5) = 90

    ((90 - 6t) / t) (t + 0.5) = 90

    open the brackets and cross multiply

    90t + 45 - 6t² - 3t = 90 t

    rearrange and collect the like terms

    - 6t² - 3t + 45 = 90t - 90t

    - 6t² - 3t + 45 = 0 multiply both side by - 1

    6t² + 3t - 45 = 0

    divide both side by 3

    2t² + t - 15 = 0

    factorize the expression by multiplying - 15 by 2t² = - 30t²

    find factors of - 30t² that adds to t = 6t * (-5t)

    replace t with (+6t - 5t) in the equation

    2t² + 6t - 5t - 15 = 0

    factorize

    2t (t + 3) - 5 (t + 3) = 0

    (2t - 5) (t + 3) = 0

    2t - 5 = 0 or t + 3 = 0

    2t = 5 or t = - 3

    divide through 2

    2t / 2 = 5 / 2 = 2.5 or t = - 3 since time cannot be negative

    them t = 2 ... 5 seconds
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