Sagot :
Si on pose d la distance totale parcourue ! alors :
donc ; la distance totale est parcourue (3/4)d à vélo + (1/5)d en courant + xd
d = (3/4)d + (1/5)d + xd
--> xd = d - (3/4)d - (1/5)d
--> x = [ (d - (3/4)x - (1/5)d ] / d
--> x = d(1 - (3/4) - (1/5)) / d
--> x = (1 - (3/4) - (1/5)
--> x = [(1 * 4 * 5) - (3 * 5) - (1 * 4)] / (4 * 5)
--> x = (20 - 15 - 4) / 20
--> x = 1/20
donc Farid a percouru à la nage l'équivalent des (1/20) du trajet total .
2)
(1/20)d = 100m
d / 20 = 100m
d = 100m * 20
d = 2 000 m
donc :
à vélo :
(3/4)d = (3/4) * 2000 m
= 6000m / 4
= 1500 m
en courant :
(1/5)d = (1/5) * 2000m
= 2000 m / 5
= 400 m
