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

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

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Goreveronicaid Goreveronicaid · Answer 2

notam: x=suma de bani de la inceput

x-[tex] \frac{1}{4} [/tex]=y
y-[tex] \frac{1}{6} [/tex]=z
z-[tex] \frac{1}{3} [/tex]=a
a-120=600

a=600+120
a=720

z-[tex] \frac{1}{3} [/tex]=720
z=720+[tex] \frac{1}{3} [/tex]
z=[tex] \frac{2161}{3} [/tex]

y-[tex] \frac{1}{6} [/tex]=[tex] \frac{2161}{3} [/tex]
y=[tex] \frac{1}{6} [/tex]+[tex] \frac{2161}{3} [/tex]
y=[tex] \frac{4323}{6} [/tex]

x-[tex] \frac{1}{4} = \frac{4323}{6} [/tex]
x=[tex] \frac{4323}{6} + \frac{1}{4} [/tex]
x=[tex]720 \frac{1}{2} [/tex]