Question

Un nr este impartit in parti direct proportionale cu 1/6; 1/4 si 1/3. In acest fel primul nr este cu 100 mai mare decat daca am fi impartit acelasi nr direct proportional cu 1/6; 1/3 si 1/2. Aflati cele 3 nr.

Answer 1

a+b+c=x+y+z
a+b+c=k/6+k/4+k/3 = (2k+3k+4k)/12=9k/12
x+y+z=m/6+m/3+m/2
a-100=x
k/6-100=m/6    k-600=m    m=k-600
x+y+z=m/6+m/3+m/2=(k-600)/6+(k-600)/3+(k-600)/2=k/6+k/3+k/2-600=
=(2k+4k+6k)/12 -600 =k-600
9k/12 =k-600
9k=12k-7200
3k=7200
k=2400

a=k/6 =2400/6=400
b=k/4 =2400/4=600
c=k/3 =2400/3=800
nr este 400+600+800=1800

m=k-600=2400-600=1800
x=m/6=1800/6=300
y=m/3=1800/3=600
z=m/2=1800/2=900
se verifica nr este 300+600+900=1800
si a-100=400-100=300=x