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18 January, 16:59

Doomtown is 300 miles due west of Sagebrush and Joshua is due west of Doomtown. At 9 a. m. Mr. Archer leaves Sagebrush for Joshua. At 1 p. m. Mr. Sassoon leaves Doomtown for Joshua. If Mr. Archer travels at an average speed 20 mph faster than Mr. Sassoon and they each reach Joshua at 4 p. m., how fast is each traveling?

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  1. 18 January, 20:14
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    This is an example of distance problem in algebra.

    Ley x be the rate of mr. sassoon

    X + 20 be the rate of mr. archer

    Y = distance from doomtown to joshua

    So the distance travelled by mr. sassoon:

    Y = 4x

    The distance travelled by mr. archer:

    (300 + y) = (x + 20) * 7

    Substitute y = 4x to the second eqation

    300 + 4x = (x + 20) * 7

    Solving for x

    X = 53.33 mph is the rate of mr sassoon

    X + 20 = 73.33 mph the rate of mr. archer
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