Question

A) aflati x,y,z stiind ca sunt direct proportionale cu numerele 0,1 ; 2/5 ; 1/3 si ca suma lor este 1
B)Aflati x,y,z,t stiind ca sunt invers proportionale cu nr. 1/2 : 1/2 ; 0,4 : 0,2 ; si ca suma lor este 5

Answer 1

$x+y+z=1 \\ \frac{x}{0,1} = \frac{y}{2/5} = \frac{z}{1/3} = \frac{x+y+z}{0,1+2/5+1/3} = k \\ 0,1+\frac{2}{5}+\frac{1}{3}= \frac{1,5}{15} + \frac{6}{15} + \frac{5}{15} = \frac{1,5+11}{15} = \frac{12,5}{15} \\ \frac{x+y+z}{12,5/15} = \frac{1}{1} : \frac{12,5}{15} = \frac{1}{1} \cdot \frac{15}{12,5} = \frac{15}{12,5} \\ k= \frac{15}{12,5} \\ x=0,1 \cdot k=0,1 \cdot \frac{15}{12,5} = \frac{1,5}{12,5} \\ \boxed{\bold{x=0,012}} \\ y= \frac{2}{5} \cdot k=\frac{2}{5} \cdot \frac{15}{12,5}= \frac{30}{62,5}$
$\boxed{\bold{y=0,48}} \\ z= \frac{1}{3} \cdot \frac{15}{12,5}= \frac{15}{37,5} \\ \boxed{\bold{z=0,4}}$

Answer 2

                 d.p
{x, y, z}    ______ > {0,1; 2/5; 1/3}=>x/10=2y/5=z/3=k

x/10=k=>x=10k
2y/5=k=>y=5k/2
   z/3=k=>z=3k
10k+5k/2+3k=1
20k+5k+6k=2
k(20+5+6)=2
k*31=2
k=2/31
=>x=10*2/31=20/31
     y=5*2/31/2=5/31
     z=3*2/31  =6/31


                         i.p
B)  {x, y, z, t}______> {1/2; 1/2; 0,4; 0,2}=>x/2=y/2=2z/5=t/5=k

x/2=k=>x=2k
y/2=k=>y=2k
2z/5=k=>z=5k/2
t/5=k=>t=5k
2k+2k+5k/2+5k=5
k(4+4+5+10)=10
k=10/23
x=2*10/23=20/23
y=2*10/23=20/23
z=5*10/23/2=50/23*1/2=25/23
t=5*10/23=50/23