Banuiesc ca au bara deasupra (x=cifra):
a)
(2x+5x+11x)/3=(20+x+50+x+100+10+x)/3=(180+3x)/3=60+x=6x(cu bara)
b)
(21x+37x+32x)/3=(200+10+x+300+70+x+300+20+x)/3=
(900+3x)/3=300+x=30x(cu bara)
Fara bara (x=factor):
a) (2*x+5*x+11*x)/3=18*x/3=6*x
b) (31*x+37*x+32*x)/3=90*x/3=30*x
