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4 April, 19:46

A 16ft seesaw is pivoted in the center. At what distance from the center would a 200lb person sit to balance a 120lb person on the opposite end?

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  1. 4 April, 22:33
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    9.6 ft

    Explanation:

    Distance is inversely proportional to weight

    distance = k / (weight), where

    k is a constant

    or you could say,

    distance * weight = k

    In this scenario,

    120 * 16 = 200 * distance

    On rearranging, making, distance the subject of formula, we have

    Distance = 120 * 16 / 200

    Distance = 1920 / 200

    Distance = 9.6 ft

    So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw
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