Question

Sa se imparta nr 1268 in patru parti invers proportionale cu nr 2 supra 3, 3 supra 4, 4 supra 5, 5 supra 6. ( As dori daca se poate si rezolvarea)

Answer 1

2x/3=3y/4=4z/5=5t/6
y=8x/9
z=10z/12=5x/6
t=12x/15=4x/5
x+y+z+t=1268
x+8x/9+5x/6+4x/5=1268
90x+80x+75x+72x=114120
317x=114120
x=360
y=8x/9=8×360/9=320
z=5x/6=5×360/6=300
t=4x/5=4×360/5=288