Question

Sa se imparta numarul 78 in parti invers proportionale cu numerele 2,5 ; 1/2 ; 5

Answer 1

{ a, b, c } i.p { 2,5 ; 1/2 ; 5 }
a x 2,5 = b x 1/2 = c x 5 = k
a x 2,5 = k ⇒ a = k/2,5
b x 1/2 = k ⇒ b = 2k
c x 5 = k ⇒ c = k/5
k/2,5 + 2k + k/5 = 78
10k/25 + 2k + k/5 = 78
         5)
2k/5 + 2k + k/5 = 78
2k/5 + 10k/5 + k/5 = 78
13k/5 = 78
13k = 78 x 5
13k = 390
k = 390 : 13
k = 30
a = 30/2,5 = 12
b = 2 x 30 = 60
c = 30/5 = 6