Question

Repartir 153 cards em tres montes de forma que o primeiro contenha 2/3 do segundo. O qual deveria ter 3/4 do terceiro

Answer 1

$p1=\frac{2}{3}.p2;~~~p2=\frac{3}{4}.p3;~~~p3=?;$

Para p3;
$p1+p2+p3=153\\\\(\frac{2}{3}.(\frac{3}{4}.p3))+\frac{3}{4}.p3+p3=153\\\\\frac{6p3}{12}+\frac{3p3}{4}+p3=153\\\\\frac{6p3+9p3+12p3}{12}=153\\\\27p3=153.12\\\\p3=\frac{1836}{27}\\\\p3=68$

Para p2:
$p2=\frac{3}{4}.p3\\\\p2=\frac{3}{4}.68\\\\p2=\frac{3.68}{4}\\\\p2=3.17\\\\p2=51$

Para p1:
$p1=\frac{2}{3}.p2\\\\p1=\frac{2}{3}.51\\\\p1=\frac{2.51}{3}\\\\p1=2.17\\\\p1=34$

Logo, o primeiro recebe 34, o segundo 51 e o terceiro 68.

Answer 2

3°= c
2°=3/4c
1°=(2/3*3/4).c

1°=(2/3*3/4).c
1°=6/12.c

1°=6/12c
2°=3/4.c
3°= c

mmc
12,4  2
6,2    2
3,1    3
1,1    12

1°=6/12.c
2°=9/12.c
3°=12/12.c

153=1°+2°+3°.c
153=(6/12+9/12+12/12)c
153=27/12.c
153=27.c/12
c/12=153/27
c/12=17/3

1°=6.17/3
1°=102/3
1°=34

2°=9*17/3
2°=153/3
2°=51

3°=12*17/3
3°=204/3
3°=68

Espero ter ajudado