<PRE><B>Question 881211</B>
Jacqui commutes 30 mi to her job each day.
 She finds that if she drives 10 mi/h faster, it takes her 6 min less to get to work.
 Find her new speed.]\
:\
let s = the new speed
then
(s-10) = the old speed
:
Change 6 min to hrs (we&#39;re using mph) 6/60 = .1 hr
write a time equation; time = dist/speed
:
Old speed time - new speed time = 6 min
{{{30/((s-10))}}} - {{{30/s}}} = .1
multiply equation by s(s-10), cancel the denominators and you have:
30s - 30(s-10) = .1s(s-10)
30s - 30s + 300 = .1s^2 - s
Form a quadratic equation on the right
0 = .1s^2 - s - 300
we want s^2 have a coefficient of 1, mult eq by 10
s^2 - 10s - 3000 = 0
You can use the quadratic formula, but this will factor easily to
(s-60)(s+50)  0
The positive solution
s = 60 mph is the new speed
:
:
:
Check this by finding the actual time for each
30/50 = .6 hrs
30/60 = .5 hrs
---------------
Diff: .1 hrs which is 6 min
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