Hamid walked from town A to town B in 6 hours. On his return journey, he increased his average speed by 1km/h and covered the distance in 5 hours. Find the distance between the two towns.
Answers
Answered by
10
here in the first journey let the velocity be v1 and time is 6hours. so distance traveled is v1 multiplied by time (6)=6v1
in return journey time is 5 hours and velocity is addition of v1 and 1. and as distance are equal 6v1=5(v1➕1) solving it you get v1=5km/hr and by that distance is 30 km
in return journey time is 5 hours and velocity is addition of v1 and 1. and as distance are equal 6v1=5(v1➕1) solving it you get v1=5km/hr and by that distance is 30 km
Answered by
32
Let the speed Hamid travelled be x
Distance = Speed *Time
On first journey, speed =x and time = 6
So distance 6x
On second journey, speed increased = 1km/h
Therefore, distance = 5(x+1)
From above statements, we can understand that distance equals to the below equation :
6x = 5(x+1)
6x = 5x+5
6x-5x = 5
x = 5
Therefore, speed Hamid travelled first second journeys are 5km/h and (5+1)km/h=6km/h respectively
And
Distance = 5hrs*6km/h = 6hrs*5km/h = 30 km
Distance = Speed *Time
On first journey, speed =x and time = 6
So distance 6x
On second journey, speed increased = 1km/h
Therefore, distance = 5(x+1)
From above statements, we can understand that distance equals to the below equation :
6x = 5(x+1)
6x = 5x+5
6x-5x = 5
x = 5
Therefore, speed Hamid travelled first second journeys are 5km/h and (5+1)km/h=6km/h respectively
And
Distance = 5hrs*6km/h = 6hrs*5km/h = 30 km
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