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Concursul "Teodor Topan" - problema 3
Posted: Mon Dec 03, 2007 2:37 pm
by maky
Suma a doua numere este \( 113 \). Daca se imparte numarul mai mare la cel mai mic se obtine catul \( 3 \)si restul \( 13 \). Aflati numerele.
Crisan Georgeta, Simleul Silvaniei
Sa rezolvam:
Posted: Sun Mar 02, 2008 9:09 pm
by Natalee
Metoda figurativa:
. numarul mic: |_____|
. numarul mare: |_____|_____|_____|...\( 13 \)...|
. suma numerelor: |_____|_____|_____|_____|...\( 13 \)...| \( = 113 \)
\( 113 - 13 = 100 \)
. numarul mic: \( 100 : 4 = 25 \)
. numarul mare: \( 25\cdot 3 + 13 = 75 + 13 = 88 \)
Proba: \( 25 + 88 = 113 \)
Prin ecuatie:
. numarul mic: \( x \)
. numarul mare: \( 3x + 13 \)
Ecuatia problemei: \( x + 3x + 13 = 113 \)
...
Nathaska
Posted: Thu Mar 06, 2008 11:45 pm
by Marius Dragoi
\( a+b=113 \) si \( a=3b+13 \Rightarrow 3b+13=113 \Rightarrow b=25 \Rightarrow a=88 \).
Posted: Thu Oct 23, 2008 8:27 pm
by miruna.lazar
a + b = 113
a : b = 3 rest 13
a = 3b+ 13 -> inlocuim in prima relatie
3b + 13 + b = 113
4b + 13 = 113 / - 13
4b = 100 -> b = 25
a = 113 - 25
a = 88
proba : 88 : 25 = 3 rest 13
88 + 25 = 113