<PRE><B>Question 142679</B>
Rachel allows herself  1 hr to reach a sales appointment 50 mi away. After she has driven 30 mi, she realizes that she must increase her speed by 15 mph in order to get there on time. What was her speed for the first 30mi?
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1st leg DATA:
distance = 30 mi ; time = x hrs ; rate = 30/x mph
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2nd leg DATA:
distance = 20 mi : time = (1-x) hrs ; rate = 20/(1-x) mph
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EQUATION:
2nd leg rate - 1st leg rate = 15 mph
20/(1-x) - 30/x = 15
4(1-x) - 6/x = 3
4x -6(1-x) = 3x(1-x)
4x -6 + 6x = 3x - 3x^2
3x^2 +7x -6 = 0
3x^2 + 9x - 2x - 6 = 0
3x(x+3) - 2(x+3) =0
(x+3)(3x-2) = 0
Positive solution:
x = 2/3 hrs.
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speed for the 1st leg = 30/(2/3) = 45 mph
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Cheers,
Stan H.

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