Question

Un elev are o suma de bani. In prima zi cheltuieste 1/3 din suma si inca 10 lei, a doua zi 2/3 din rest si inca 5 lei, a treia zi cu 7 lei mai mult decat 1/9 din noul rest si ii mai raman 33 lei. Ce suma a avut elevul si cat a cheltuit in fiecare zi?

Answer 1

x = suma elevului
x/3+10 = suma cheltuita in prima zi
[x-(x/3+10)] *2/3 +5 = (x-x/3-10)*2/3+5=[(3x-x-30)/3] * 2/3+5 =
=[(2x-30)/3] * 2/3 +5 =( 4x-60)/9+5=(4x-60+45)/9=(4x-15)/9 suma cheltuita a 2-a zi
[x-(x/3+10)-(4x-15)/9]*1/9+7 = [x-x/3-10-(4x-15)/9] *1/9+7 =
=(9x-3x-90-4x +15)/9 *1/9+7 =
=(2x-75)/9*1/9+7= (2x-75)/81+7=(2x-75-7*81)/81=(2x-75+567)/81=(2x+492)/81 suma cheltuita in a 3-a zi
x-(x/3+10)-(4x-15)/9-(2x+492)/81 =33 lei
x-x/3-10-(4x-15)/9-(2x+492)/81=33 se amplifica cu 81
81x-27x-810-36x-135-2x-492=33*81=2673
16x=2673+810+135-492=3840
x=3840/16=240 lei este suma elevului
1 zi x/3+10=240/3+10=80+10 =90lei
a 2-a zi  (4x-15)/9 = (4*240-15)/9=(960-15)/9=945/9=105 lei
a 3-a zi  (2x+492)/81 =(2x240+492)/81=(480+492)/81=972/81=12
verificare
240-90-105-12=33