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

a+b+c+d=1268
2a/3=3b/4=4c/5=5d/6=k
a=3k/2
b=4k/3
c=5k/4
d=6k/5

3k/2+4k/3+5k/4+6k/5=1268
90k+80k+75k+72k=1268*60
317k=1268*60
k=4*60=240

a=3*240/2=360
b=4*240/3=320
c=5*240/4=300
d=6*240/5=288
----------------------
total  =      1268