Schema:
I cutie- [tex] 18\frac{2}{3} [/tex] kg de mere
II cutie- ? kg de mere, cu [tex] 3\frac{5}{6} [/tex] mai putin decit in prima
III cutie- ? kg de mere
In total avem [tex]50 \frac{1}{9} [/tex] kg
Rezolvare:
1) [tex]18 \frac{2}{3}- 3\frac{5}{6}= 14\frac{5}{6} [/tex] kg sunt in a 2a cutie
2)[tex]18 \frac{2}{3}+14 \frac{5}{6}= 33\frac{1}{2} [/tex] kg sunt in 1a si a 2a cutie
3)[tex] 50\frac{1}{9}-33 \frac{1}{2}= 16\frac{11}{18} [/tex] kg sunt in a 3a cutie
Raspuns: In a 3a cutie sunt [tex]16 \frac{11}{18} [/tex] kg de mere.
Sper ca te-am ajutat....
