Question

Suma a trei numere este 1.Stiind ca ele sunt direct proportionale cu 0,1 ; 2/5 ; 1/3 aflati numerele

Answer 1

a + b + c = 1

( a,b,c ) d.p ( 0,1; 2/5; 1/3 ) ⇒ a = 0,1k
                                                b = 2k/5
                                                c = k/3

0,1k + 2k/5 + k/3 = 1
1/10k + 2k/5 + k/3 = 1
3k + 12k + 10k = 30
25k = 30 ⇒ k = 6/5


⇒ a = 0,1k = 0,1 × 6/5 ⇒ a = 1/10 × 6/5 ⇒ a = 1/5 × 3/5 ⇒ a = 3/25 
    b = 2k/5 = 2×6/5/5 = 12/5/5 ⇒ b = 12/25 
    c = k/3 = 6/5/3 = 6/5/3/1 = 6×1/5×3 = 6/5×3 ⇒ c = 2/5



a + b + c = 3/25 + 12/25 + 2/5 
a + b + c = ( 3 + 12 + 10 )/25 
a + b + c = 25/25
a + b + c = 1 

Answer 2

a+b+c =1
{a, b, c} d. p {0,1;2/5;1/3}
a/0,1= b/2/5=c/1/3=k
a=0,1k =1/10 k
b= 2/5k
c=1/3k
1/10k +2/5k +1/3k =1
3k+12k +10k/30=1
25k/30=1
5k /6 =1
k = 1÷5/6
k=1×6.5
k=6/5
a= 1/10k =1/10×6/5=3/25
b=2/5k= 2/5×6/5= 12/25
c= 1/3k = 1/3×6/5=2/5