Question

un elev are o suma de bani . in prima zi el cheltuieste o treime din suma , a doua zi cheltuieste o doime din rest a treia zi o patrime din noul rest ,iar a patra zi o treime din suma ramasa.dupa aceste cheltuieli ii mai raman 20 de lei . ce suma a avut elevul la inceput

Answer 1

Notez cu s suma de bani.
Prima zi) $\frac{1}{3} * s$

rest : x- $s - \frac{s}{3}$

A doua zi ) $\frac{1}{2} * (s - \frac{s}{3} ) = \frac{1}{2} * ( \frac{3s-s}{3}) = \frac{1}{2} * \frac{2s}{3} = \frac{s}{3}$

Noul rest:  s - $\frac{s}{3} - \frac{s}{3}$ = $\frac{3s-s-s}{3} = \frac{s}{3}$

A treia zi) $\frac{1}{4} * \frac{s}{3} = \frac{s}{12}$

A patra zi ) $\frac{1}{3} * ( s- \frac{s}{3} - \frac{s}{3} - \frac{s}{12} ) = \frac{1}{3} * ( \frac{12s-4s-4s-s}{12} ) = \frac{s}{12}$

Mai raman 20 lei.

Adunam toate cheltuielile + suma ramasa:

s= $\frac{s}{3} + \frac{s}{3} + \frac{s}{12} + \frac{s}{12} +20= \frac{2s}{3} + \frac{2s}{12} +20 = \frac{2s}{3} + \frac{s}{6}+20 = \frac{4s+s}{6} +20 = \frac{5s}{6}$ +20

s= $\frac{5s +20*6}{6}$

6s= 5s + 20 *6
6s - 5s = 20*6
s = 120 lei

Answer 2

fie s-suma
I  s/3 - cheltuieste
 rest r1  (2*s/3)
II r1/2 - cheltuieste
   rest r2 (r1/2)
III r2/4 - cheltuieste
   rest r3 (3*r2/4)
IV r3/3
     rest 20lei=2*r3/3⇒ r3=3*20/2=30 lei
     dar r3=30 lei=3*r2/4⇒ r2=4*30/3=40 lei
    dar r2=40 lei= r1/2⇒ r1=2*40=80 lei
   dar r1=80 lei= 2*s/3 ⇒ s=3*80/2=120 lei
verificare I-120/3=40 lei
              r1=80
            II 80/2=40 lei
              r2=40 lei
           III 40/4=10 lei
             r3=30 lei
           IV 30/3=10 lei
              r4=20 lei