SOLUTION: A person drives 390 miles on a stretch of road. Half the distance is driven traveling 5 miles per hour below the speed limit, and half the distance is driven traveling 5 miles p

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A person drives 390 miles on a stretch of road. Half the distance is driven traveling 5 miles per hour below the speed limit, and half the distance is driven traveling 5 miles p      Log On


   

Question 631628: A person drives 390 miles on a stretch
of road. Half the distance is driven traveling 5
miles per hour below the speed limit, and half the
distance is driven traveling 5 miles per hour above
the speed limit. If the time spent traveling at the slower
speed exceeds the time spent traveling at the faster
speed by 24 minutes, find the speed limit.
Found 2 solutions by mananth, josmiceli:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Let the speed limit be x mph
First half speed = (x-5) mph
Second half speed = (x+5) mph
Half distance = 390/2 = 195 miles
24 minutes = 24/60 = 2/5 hours
time = distance/speed
First half time = 195/(x-5)
Seond half time = 195/(x+5)
195%2F%28x-5%29+-195%2F%28x%2B5%29+=+2%2F5
LCD = 5(x+5)(x-5)
Multiply equation by the LCD
195%2A5%28x%2B5%29+-195%2A5%28x-5%29=+2%28x-5%29%28x%2B5%29
975x+4875-975x+4875 = 2(x^2-25)
9750 = 2x^2-50
2x^2= 9800
/2
x^2= 4900
take the square root
x= 70 mph
Speed limit is 70 mph
m.ananth@hotmail.ca

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = the speed limit
+s+-+5+ = 5 mi/hr below speed limit
+s+%2B+5+ = 5 mi/hr above the speed limit
---------
Let +t+ = time spent driving at +s+%2B+5+ mi/hr
+t+%2B+24%2F60+ = time spent driving at +s+-+5+
------------
(1) +390%2F2+=+%28+s+%2B+5+%29%2At+
(2) +390%2F2+=+%28+s+-+5+%29%2A%28+t+%2B+24%2F60+%29+
------------------------------
(1) +195+=+s%2At+%2B+5t+
(2) +195+=+%28+s+-+5+%29%2A%28+t+%2B+2%2F5+%29+
--------------------------
(2) +195+=+s%2At+-+5t+%2B+%282%2F5%29%2As+-+2+
(1) +s%2At+=+195+-+5t+
(1) +s+=+195%2Ft+-+5+
Substitute (1) into (2)
(2) +195+=+%28+195%2Ft+-+5+%29%2At+-+5t+%2B+%282%2F5%29%2A%28+195%2Ft+-+5+%29+-+2+
(2) +195+=+195+-+5t+-+5t+%2B+78%2Ft+-+2+-+2+
(2) +0+=+-10t+%2B+78%2Ft+-+4+
(2) +10t+=+78%2Ft+-+4+
(2) +10t%5E2+=+78+-+4t+
(2) +10t%5E2+%2B+4t+-+78+=+0+
(2) +5t%5E2+%2B+2t+-+39+=+0+
Use the quadratic formula

+b+=+2+
+c+=+-39+

+t+=+13%2F5+ hrs
and
(1) +195+=+%28+s+%2B+5+%29%2At+
(1) +195+=+%28++s+%2B+5+%29%2A%2813%2F5%29+
(1) +195+=+%2813%2F5%29%2As+%2B+13+
(1) +%2813%2F5%29%2As+=+195+-+13+
(1) +s+=+%285%2F13%29%2A182+
(1) +s+=+70+
The speed limit is 70 mi/hr