<PRE><B>Question 167160</B>
Roma Sherry drove 330 miles from her hometown to Tucson. During her return trip, she was able to increase her speed by 11 mph. If her return trip took 1 hour less time, find her original speed and her speed returning home.
:
Let s = original speed
then
(s+11) = return speed
:
Write a time equation: Time = {{{distance/speed}}}
:
Original time = return time + 1 hr
{{{330/s}}} = {{{330/((s+11))}}} + 1
:
Multiply equation by s(s+11) and you have:
330(s+11) = 330s + s(s+11)
:
330s + 3630 = 330s + s^2 + 11s
:
0 = 330s - 330s + s^2 + 11s - 3630
:
A quadratic equation:
s^2 + 11s - 3630 = 0
Factor this to:
(s + 66)(s - 55) = 0
Positive solution
s = 55 mph is original speed.
:
Find the time
330/55 = 6 hr, original time
and
330/66 = 5 hrs, faster time; confirms our solution.
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