Question

VA ROG AJUTATI-MA SI PE MINE.. PRIN METODA MERSULUI INVERS PENTRU CLASA A 4 A... Ionut pleaca in tabara cu o suma de bani. In prima zi cheltuie 2/6 din suma si inca 16 lei, a doua zi un sfert din suma ramasa si inca 12 lei, iar a treia zi 3/5 din rest sinca 15 lei. Ii raman 45 lei. Ce suma a cheltuit in fiecare zi? Cu cati lei a plecat Ionut in tabara?

Answer 1

Răspuns:

Explicație pas cu pas:

Notam cu S - suma de bani

$Prima ZI\\\\\\\frac{2}{6} *s +16 = \frac{2(s+48)}{6} =\frac{s+48}{3}$  cheltuie in prima zi

Cat ramane din suma dupa prima zi (restul)

$s-\frac{s+48}{3} = \frac{3s-(s+48)}{3} =\frac{2s-48}{3}$

$II zi\\\\\frac{1}{4} *\frac{2s-48}{3} + 12=\frac{2s-48}{12}+12=\frac{2s-48+144}{12}=\frac{2s+96}{12}=\frac{s+48}{6}$ cheltuie a doua zi

Cat ramane din suma dupa ce cheltuie a doua zi (restul 2)

$\frac{2s-48}{3}-\frac{(s+48)}{6} =\frac{4s-96-(s+48)}{6} = \frac{3s-144}{6}= \frac{s-48}{2}$

$III zi\\\\\frac{3}{5}*\frac{s-48}{2}+15 = \frac{3s-144}{10}+15=\frac{3s-144+150}{10}=\frac{3s+6}{10}$cheltuie a treia zi

$IV zi \\45 lei$

Facem suma si inmultim ecuatia cu 30

$\frac{s+48}{3} + \frac{s+48}{6} +\frac{3s+6}{10} +45=s \\\\10(s+48) +5(s+48)+3(3s+6)+45*30=30s\\\\10s+480+5s+240+9s+18+1350=30s\\\\24s + 2088=30s\\\\2088=6s\\\\s=348$

348 lei a plecat ionut in tabara

1. zi

$(348+48) : 3 = 132$

    2. zi

$(348+48):6 =66$

    3. zi

$(3*348+6):10= 1050:10= 105$

Verificare: 132+66+105+45 = 348