Posted by **PDF** on Friday, February 3, 2012 at 8:12pm.

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**, Friday, February 3, 2012 at 8:24pm
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.

- Math -
**Reiny**, Friday, February 3, 2012 at 9:07pm
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

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