Question

Un elev are o suma de bani. Determinati suma de bani stiind ca dupa ce a cheltuit $\frac{1}{4}$ din ea, apoi $\frac{1}{6}$ din rest , apoi $\frac{1}{3}$ din noul rest si inca 120 de lei , i-au ramas 600 de lei.

Answer 1

x-suma initiala

1/4 din x= 1/4 * x=  x/4 -prima data

x-x/4=  (4x-x)/4=  3x/4 -restul 

1/6 din 3x/4=  1/6 * 3x/4= 1/2 *x/4= x/8 -a doua oara

3x/4- x/8= (6x-x)/8=  5x/8- noul rest 

1/3 din 5x/8 +120 = 1/3 * 5x/8= 5x/24 +120  -a treia oara

5x/8 -5x/24 -120= (15x-5x-2880)/24= (10x-2880)/24 -noul nou rest

(10x-2880)/24=600

10x-2880=600*24

10x-2880=14400

10x=14400+2880

10x=17280

x=1728

Answer 2

notam: x=suma de bani de la inceput
           
x-$\frac{1}{4}$=y
y-$\frac{1}{6}$=z
z-$\frac{1}{3}$=a
a-120=600

a=600+120
a=720

z-$\frac{1}{3}$=720
z=720+$\frac{1}{3}$
z=$\frac{2161}{3}$

y-$\frac{1}{6}$=$\frac{2161}{3}$
y=$\frac{1}{6}$+$\frac{2161}{3}$
y=$\frac{4323}{6}$

x-$\frac{1}{4} = \frac{4323}{6}$
x=$\frac{4323}{6} + \frac{1}{4}$
x=$720 \frac{1}{2}$