Question 763148: man A set out from a certain point and travelled at the rate of 6 kph. After A had gone 2 hrs, another man B set out to overtake him and went 4km the 1st hour, 5km for the second hour, 6 km for the 3rd hour and so on gaining 1 km every hour. After how many hours will they be together?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! man A set out from a certain point and travelled at the rate of 6 kph.
After A had gone 2 hrs, another man B set out to overtake him and
went 4km the 1st hour, 5km for the second hour, 6 km for the 3rd hour
and so on gaining 1 km every hour.
After how many hours will they be together?
;
We know from the information given, that Man A has traveled 12 km when
Man B starts after him.
:
Let x = no. of hrs for B to catch A
x also = no. of km added to 4 km by the man B after x hrs
:
B's dist = A's dist
x( +4) = 6x + 12
 +4x = 6x + 12
multiply both sides by 2
x^2 - x + 8x = 12x + 24
combine like terms on the left
x^2 - x + 8x - 12x - 24 = 0
x^2 - 5x - 24 = 0
Factors to:
(x-8)(x+3) = 0
The positive solution
x = 8 hrs from the time B starts,they are together
:
:
Prove this by doing it the long way
Hr_A___B (add one km each hr)
1 18 km, 4 km
2 24 km, 9 km, added 5
3 30 km, 15 km, added 6
4 36 km, 22 km, added 7
5 42 km, 30 km, added 8
6 48 km, 39 km, added 9
7 54 km, 49 km, added 10
8 60 km, 60 km, added 11, same distance after 8 hrs
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