SOLUTION: Kimani and mwaura live 40km apart.one day kimani left his house at 10:30am and cycled towards mwaura's house at an average speed of 15km/hr.Mwaura left his house at 12:00noon on th

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Question 1195987: Kimani and mwaura live 40km apart.one day kimani left his house at 10:30am and cycled towards mwaura's house at an average speed of 15km/hr.Mwaura left his house at 12:00noon on the same day and cycled towards kimani's houseat an average speed of 25km/hr.determine the distance from kimani's house when the two met
Found 2 solutions by math_helper, greenestamps:
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!



Noting that Kimani will travel (15km/hr)*(1.5hrs) = 22.5km  before Mwaura starts, we can write the following, setting t = 0  to be noon:



1.  Dk_k(t) = 15t + 22.5

2.  Dk_m(t) = 40 - 25t 



Where each Dk(t) is distance from Kimani's house.



When the two meet, their distances from Kimani's will be equal, so set eq 1 equal to eq 2.   Then solve for t to get t = 0.4375hr



At t=0.4375hr,   Dk_k(t) = 15(0.4375) + 22.5 = 6.5625 + 22.5 = 29.0625km



Check using Dk_m(t):

    at t=0.4375hr,  Dk_m(t) = 40 - 25(0.4375) = 29.0625km  

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!
 


The problem asks for the distance from kimani's house when the two meet, so what we need to determine is the distance kimani travels before they meet.


He leaves 1.5 hours before mwaura; traveling 15km/h, he travels 1.5(15) = 22.5km before mwaura starts; that leaves 40-22.5 = 17.5km.


When they are both riding, the ratio of their speeds is 15:25 = 3:5.  That means kimani rides 3/8 of the remaining 17.5km before they meet.


(3/8)(17.5) = (3/8)(35/2) = 105/16 = 6.5625 km


So the total distance kimani rides before they meet is


22.5 + 6.5625 = 29.0625km


ANSWER: 29.0625km


 

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