avgspeed=distance/time=50km/time=24km/hr
time=50/24 hrs
time rough+timesmooth= 50/24
sum of distances equals 50km or
30*timerough+12*(50/24-timerough)=50
timerough(30 -12)=50(1-1/2)
time rough= 25/18 hrs
distance rough= 25/18 * 12=50/3 km
check all that.
A motorist averages 30 kph on ordinary loads and 12 kph on roads under repair. His average speed for a distance of 50 km is 24 km/h. What length of the road is under repair?
2 answers
Let the distance on good road be x km
then distance on rough road is 50-x km
time on good road = x/30
time on rough road = (50-x)/12
total time = x/30 + (50-x)/12
= (2x + 250 - 5x)/60
= (250 - 3x)/60
but total time = 50/24
so (250-3x)/60 = 50/24
3000 = 6000 - 72x
72x = 3000
x = 3000/72 = 125/3 or 41 .67 km
so 42.67 km were good road and
50 - 41.7 or 8.3 km were under repair.
check: total time = 50/24 = 2.083 hrs
time on good road = 41.67/30 = 1.389
time on rought road = 8.3/12 = .692
total time = 2.081 , close enough allowing for round-off
all checks out
then distance on rough road is 50-x km
time on good road = x/30
time on rough road = (50-x)/12
total time = x/30 + (50-x)/12
= (2x + 250 - 5x)/60
= (250 - 3x)/60
but total time = 50/24
so (250-3x)/60 = 50/24
3000 = 6000 - 72x
72x = 3000
x = 3000/72 = 125/3 or 41 .67 km
so 42.67 km were good road and
50 - 41.7 or 8.3 km were under repair.
check: total time = 50/24 = 2.083 hrs
time on good road = 41.67/30 = 1.389
time on rought road = 8.3/12 = .692
total time = 2.081 , close enough allowing for round-off
all checks out