Posted by PDF on .
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?

Math 
bobpursley,
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*(50/24timerough)=50
timerough(30 12)=50(11/2)
time rough= 25/18 hrs
distance rough= 25/18 * 12=50/3 km
check all that. 
Math 
Reiny,
Let the distance on good road be x km
then distance on rough road is 50x km
time on good road = x/30
time on rough road = (50x)/12
total time = x/30 + (50x)/12
= (2x + 250  5x)/60
= (250  3x)/60
but total time = 50/24
so (2503x)/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 roundoff
all checks out