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Matematica E Simpla • View topic - Probleme de ciclul primar (3)

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 Post subject: Probleme de ciclul primar (3)
PostPosted: Mon Dec 08, 2008 4:26 pm 
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Joined: Mon Dec 01, 2008 3:20 pm
Posts: 14
Location: Ialomita-Ograda
Enunt:
Bogdan are impreuna cu tatal sau 48 de ani. Acum doi ani, copilul avea de 3 ori mai putin decat tatal.
Sa se afle varsta tatalui si varsta coplului.


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 Post subject: Re: Probleme de ciclul primar (3)
PostPosted: Mon Dec 08, 2008 8:42 pm 
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Joined: Wed Oct 22, 2008 6:27 pm
Posts: 29
Rezolvare:
Notam cu x varsta fiului (Bogdan) si cu y varsta tatalui.

Bogdan are impreuna cu tatal sau 48 de ani.
Adica, x + y = 48.

Acum doi ani, copilul avea de 3 ori mai putin decat tatal.
Cu alte cuvinte, acum doi ani, tatal avea de 3 ori mai mult decat copilul.
Acum doi ani, Bogdan avea x - 2 ani iar tatal sau avea y-2 ani.
Deci, y - 2 = 3 (x - 2)

Din prima ecuatie, avem y = 48 - x
Inlocuim in a doua ecuatie:
48 - x - 2 = 3 (x - 2),
o rezolvam si aflam pe x.
Dupa aceea, din y = 48 - x il aflam pe y.


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