Pregatindu-se pt, concursul de mate ,Marta a rezolvat ,in 3 zile,un nr. de probleme.
In prima zi a rezolvat cu 6 mai mult decat 1/3(unu pe trei) din nr. total ,a 2 -a zi cu 8 mai putin decat 1/4 din rest,iar a 2-a zi 3/5 din noul rest plus ultimele 20 de probleme.
Cate probleme a rezolvat Marta in fiecare din cele 3 zile?
Dar in total?

o pot rezolva folosind ecuatii
I x/3+6
rest r1 (2x/3-6)
II r1/4-8
rest r2 (3r1/4+8)
III 3/5*r2+20
20=2/5*r2 ⇒ r2=5*20/2=50
50=3r1/4+8⇒ 3r1/4=42⇒ r1=42*4/3=56
2x/3-6=56 ⇒ 2x/3=62⇒ x=62*3/2=93 probleme
verificare I 93/3+6=37
r1=93-37=56
II 56/4-8=14-8=6
r2=56-6=50
III 3*50/5+20=30+20=50

Answer Link

Miky93id Miky93id · Answer 2

x-nr de probleme

1/3 din x +6= 1/3 *x +6= x/3 +6 -a rezolvat in prima zi

x-(x/3+6)= (3x-x-18)/3= (2x-18)/3 -restul

1/4 din (2x-18)/3 -8 = 1/4 * 2(x-9)/3 -8= (x-9)/2 -8= (x-9-16)/2= (x-25)/2 - a rezolvat in a2a zi

(2x-18)/3 -(x-25)/2= (4x-36-3x+75)/6= (x+39)/6 -noul rest

3/5 din (x+39)/6 +20= 3/5 * (x+39)/6 +20= (x+39)/10 +20= (x+39+200)/10= (x+239)/10 -a rezolvat in a3a zi

x/3 +6 + (x-25)/2 + (x+239)/10 =x aducem la acelasi numitor

10x+180 +15x-375 +3x +717= 30x

28x +897 -375 =30x

28x + 522=30x

28x-30x= -522
-2x=- 522
2x=522
x=261

261/3 +6= 87+6= 93 probleme a rezolvat in prima zi

(261-25)/2= 236/2= 118 probleme a rezolvat in a2a zi

(261+239)/10= 500/10=50 probleme a rezolvat in a3a zi