Question 17944: Mr. X biked 2.5 hrs up a mountain. Rested for 30 minutes. Biked down in 1.5 hrs. How far did he bike, if his rate of ascent was 3 km/hr less than his rate of decent?
Answer by venugopalramana(3286)   (Show Source):
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Susan's communte to work in the morning takes 45 minutes and coming home she is able to drive 10 miles per hour faster, so it only takes 30 minutes, how far is her communte (one-way)?
let her speed of travel going to work be =x
on return she drives 10 mph faster.hence her return speed =x+10
let the distance of home to work =d=distance of work to home
using the formula d=rt...or t=d/r...
time of travel to work =d/x=45/60..or d=x*45/60=3x/4...(1)
time of travel to home =d/(x+10)=30/60..or d=(x+10)30/60=(x+10)/2..(2)
equating d from the 2 eqns.
d=3x/4=(x+10)/2..crossmultiplying
2*3x=4*(x+10)
6x=4x+40
6x-4x=40
2x=40
x=20 mph
d=3x/4=3*20/4=15 miles.
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If Mr. X leaves now and drives 66 km/hr, he will reach Tuscaloosa in time for an appointment. On the other hand, if he has lunch first and leaves in 40 minutes, he will have to drive 90 km/hr to make the appointment. How far away is Tuscaloosa? (thanks for the assistance).
let distance =d
speed of travel in case 1 =66...
time of travel in case 1 ..=d/r=d/66
speed of travel in case 2 =90...
time of travel in case 2 ..=d/r=d/90
difference in time of travel =d/66-d/90=40 minutes or 40/60 hrs=2/3
d[(1/66)-(1/90)]=2/3
d[(90-66)/(90*66)]=2/3
d=2*90*66/(24*3)
d=165
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