SOLUTION: Laurie left home and ran to the lake at 10 mi/h. She ran back home at 8 mi/h. If the entire trip took 27 minutes, how far did she run in all?

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Laurie left home and ran to the lake at 10 mi/h. She ran back home at 8 mi/h. If the entire trip took 27 minutes, how far did she run in all?      Log On


   



Question 65413This question is from textbook algebra sctructure and method book 1
: Laurie left home and ran to the lake at 10 mi/h. She ran back home at 8 mi/h. If the entire trip took 27 minutes, how far did she run in all? This question is from textbook algebra sctructure and method book 1

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
We'll use the formula distance (d) = rate(r) times time (t) or d=rt
Let d1=distance to the lake
r1=her rate to the lake (10mi/hr)
t1=her time to the lake
Let d2=distance from the lake
r2=her rate from the lake (8 mi/hr)
t2=her time from the lake
Now we know the following:
(a.) t1+t2= 27 mim
(b.) d1=d2
(c.) Total distance is 2d1 or 2d2 or d1+d2
(d.) d1=r1t1=10t1
(e.) d2=r2t2=8t2
Since d1=d2, we know that:
10t1=8t2
t1=(8/10)t2
t1=(4/5)t2 substitute for t1 in (a.) above and we have
(4/5)t2+t2=27min multiply both sides by 5 to get rid of the fraction
4t2+5t2=135min
9t2=135
t2=15min or (1/4) hr
t1=27-15=12min or (1/5)hr
Now substitute the times in (d.) and/or (e.) to get half the total distance that he ran.
From (d.)
d1=10(1/5)=2 miles
From (e.)
d2=8(1/4)=2 miles
Total distance =
d1+d2=4 miles
Happy Holidays----ptaylor