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4 April, 19:04

It took Fran 2.7 hours to drive to her mother's house on Monday morning. On her return trip on Tuesday night, traffic was heavier, so the trip took her 3 hours. Her average speed on Tuesday was 6 mph slower than on Monday. What was her average speed on Tuesday.

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  1. 4 April, 19:26
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    Step-by-step explanation:

    Set up a table for a distance = rate * time problem. We are considering the trips taken on Monday and Tuesday.

    d = r * t

    Monday

    Tuesday

    Now let's start filling in what we know. First off, Fran went to her Mother's both days. Unless her Mother moved overnight from Monday to Tuesday, the distance from Fran to her mom is the same both days, even though we don't know how far it is. So we will just call that "d".

    d = r * t

    Monday d =

    Tuesday d =

    Now we are told about the times it took to get there on both days. Monday took 2.7 hours and Tuesday took 3 hours. Filling in that:

    d = r * t

    Monday d = * 2.7

    Tuesday d = * 3

    We're getting there. Now let's look at rates. Again, we don't know her rate (that's what we are solving for!) but we do know that she drove faster on Monday than Tuesday. Tuesday she was 6 miles per hour slower than Monday. Since we don't know Monday's rate, we will call it r. Since we don't know Tuesday's rate, but only that it is 6 mph slower than Monday, we will call it r - 6

    d = r * t

    Monday d = r * 2.7

    Tuesday d = r-6 * 3

    Now we have our equations. We know that d = rt. Since the distances are the same, d = d, by the transitive property of equality, we can set the 2 expressions equal to each other:

    2.7r = 3 (r - 6) and solve for r:

    2.7r = 3r - 18

    -.3r = - 18

    r = 60

    If r = 60, then that r value goes in for Monday. That means that Tuesday is 60 - 6 which is 54
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