Question

Un automobilist si a propus sa parcurga o distanta in 3 zile. In prima zi a parcurs 2 pe 5 si inca 100km, in a doua zi a parcurs cu 50 km mai putin decat 1 pe 4 din rest si i au ramas inca 425km ptr a treia zi. Care e distanta ?

Answer 1

x+y+z = d
x = 2d/5 + 100
y = (3d/5 - 100)/4 -50
z = 425
2d/5+100 + (3d/5 - 100)/4 -50 + 425 = d
aduci la acelasi numitor comun = 20
8d + 2000 + 3d - 1500 + 8500 = 20d
11d + 9000 = 20d
20d -11d = 9000
9d = 9000
d = 9000 : 9
d = 1000
x = 2 *1000/5 + 100
x = 2000/5 + 100
x = 400 + 100
x = 500
y = (3 * 1000/5 - 100)/4 - 50
y = 500/4 - 50
y = 125 - 50
y = 75

Answer 2

x=500
y=75................................