Răspuns :
a+b+c=61
b=a/4-2
c=a/2
a+a/4-2+a/2=61
a+a/4+a/2=63
7a/4=63 => 7a=252 => a=36(primul cupon)
b=36/4 -2=9-2= 7 => b=7 (al doilea cupon)
c=36/2 =>c=18 (al treilea cupon)
PROBLEMA II. 36/4 +18 = 9+18=27
27/9=3
PROBLEMA III
44-24=20
20/2=10 (FEMEI)
10/2= 5 FETE
b=a/4-2
c=a/2
a+a/4-2+a/2=61
a+a/4+a/2=63
7a/4=63 => 7a=252 => a=36(primul cupon)
b=36/4 -2=9-2= 7 => b=7 (al doilea cupon)
c=36/2 =>c=18 (al treilea cupon)
PROBLEMA II. 36/4 +18 = 9+18=27
27/9=3
PROBLEMA III
44-24=20
20/2=10 (FEMEI)
10/2= 5 FETE
1.
a + b + c = 61
b + 2 = a/4
b = a/4 - 2
c = a/2
a + a/4 - 2 + a/2 = 61
4a + a - 8 + 2a = 244
7a = 252
a = 252 : 7
a = 36 m
b = 36/4 - 2
b = 7 m
c = 36/2
c = 18 m
2.
36 : 4 = 9 sfert
18 : 2 = 9 o jumatate
9 + 2 * 9 = 27
27 : 9 = 3
3. 44 - 24 = 20 rest
20 : 2 = 10 femei
44 - 24 - 10 = 10 noul rest
10 : 2 = 5 fete
si 5 baieti
a + b + c = 61
b + 2 = a/4
b = a/4 - 2
c = a/2
a + a/4 - 2 + a/2 = 61
4a + a - 8 + 2a = 244
7a = 252
a = 252 : 7
a = 36 m
b = 36/4 - 2
b = 7 m
c = 36/2
c = 18 m
2.
36 : 4 = 9 sfert
18 : 2 = 9 o jumatate
9 + 2 * 9 = 27
27 : 9 = 3
3. 44 - 24 = 20 rest
20 : 2 = 10 femei
44 - 24 - 10 = 10 noul rest
10 : 2 = 5 fete
si 5 baieti