Concursul "Teodor Topan" - problema 3

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maky
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Concursul "Teodor Topan" - problema 3

Post 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.

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Natalee
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Sa rezolvam:

Post 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: \( x + 3x + 13 = 113 \)

...

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Marius Dragoi
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Post by Marius Dragoi »

\( a+b=113 \) si \( a=3b+13 \Rightarrow 3b+13=113 \Rightarrow b=25 \Rightarrow a=88 \).
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miruna.lazar
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Post 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
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