She averaged 9 mph.
27 / 9 = 3 miles
I need help please.
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?
6 answers
d = r t ... t = d / r
27/60 = (d / 10) + (d / 8)
54/120 = (12 d / 120) + (15 d / 120)
d is the distance to the lake
27/60 = (d / 10) + (d / 8)
54/120 = (12 d / 120) + (15 d / 120)
d is the distance to the lake
the trip times are not the same, so the average speed is not 9 mph
sorry, Ms. Sue
sorry, Ms. Sue
Oops! Thanks for catching that, Scott.
I agree with Scott
time for first leg = d/10
time for return leg = d/8
total time = d/10 + d/8 = 9d/40
avg rate = total distance/total time
= 2d/(9d/40) = 2d(40/9d) = 80/9 mph
distance = (27/60)(80/9) = 4 miles
So she ran 4 miles in total.
check: 2 miles at 10 mph = 2/10 hrs
2 miles at 8 mph = 2/8 miles
total time = 2/10 + 2/8
= 1/5 + 1/4 = 9/20 hrs = 27 minutes.
time for first leg = d/10
time for return leg = d/8
total time = d/10 + d/8 = 9d/40
avg rate = total distance/total time
= 2d/(9d/40) = 2d(40/9d) = 80/9 mph
distance = (27/60)(80/9) = 4 miles
So she ran 4 miles in total.
check: 2 miles at 10 mph = 2/10 hrs
2 miles at 8 mph = 2/8 miles
total time = 2/10 + 2/8
= 1/5 + 1/4 = 9/20 hrs = 27 minutes.
why is it 9d/40??
1 answer