Correction - Exercice 26 page 149 - Activités numériques I
Trouvons le nombre d'écoliers, de collégiens et de lycéens. En écrivons toutes les solutions.
Soient \(l\) le nombre des lycéens, \(c\) celui des collégiens et \(e\) celui des écoliers.
On a \(10\)\(l\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\).
On a aussi \(l\) \(+\) \(c\) \(+\) \(e\) \(=\) \(100\), d'où \(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\)
Alors
\(10\)\(l\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\)
\(\Rightarrow\) \(10\)\((\)\(100\) \(-\) \(c\) \(-\) \(e\)\()\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\)
\(\Rightarrow\) \(1000\) \(-\) \(10c\) \(-\) \(10e\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\)
\(\Rightarrow\) \(-\) \(10c\) \(+\) \(2\)\(c\)\(-\)\(10e\) \(+\) \(0,5\)\(e\) \(=\) \(200\) \(-\) \(1000\)
\(\Rightarrow\) \(-\) \(8\) \(c\) \(-\) \(9,5\) \(e\) \(=\) \(-800\)
\(\Rightarrow\) \(8\)\(c\) \(+\) \(9,5\)\(e\) \(=\) \(800\)
\(\Rightarrow\) \(80\)\(c\) \(+\) \(95\)\(e\) \(=\) \(8000\)
\(\Rightarrow\) \(c\) \(=\) \(\frac{8000\color{Red}{- 95e}}{80}\)
\(\Rightarrow\) \(c\) \(=\) \(\frac{8000}{80}\) \(-\) \(\frac{\color{Red}{95e}}{\color{Red}{80}}\)
\(\Rightarrow\) \(c\) \(=\) \(100\) \(-\) \(\frac{\color{Red}{95e}}{\color{Red}{80}}\)
\(\Rightarrow\) \(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\)
Soient \(l\) le nombre des lycéens, \(c\) celui des collégiens et \(e\) celui des écoliers.
On a \(10\)\(l\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\).
On a aussi \(l\) \(+\) \(c\) \(+\) \(e\) \(=\) \(100\), d'où \(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\)
Alors
\(10\)\(l\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\)
\(\Rightarrow\) \(10\)\((\)\(100\) \(-\) \(c\) \(-\) \(e\)\()\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\)
\(\Rightarrow\) \(1000\) \(-\) \(10c\) \(-\) \(10e\) \(+\) \(2\)\(c\) \(+\) \(0,5\)\(e\) \(=\) \(200\)
\(\Rightarrow\) \(-\) \(10c\) \(+\) \(2\)\(c\)\(-\)\(10e\) \(+\) \(0,5\)\(e\) \(=\) \(200\) \(-\) \(1000\)
\(\Rightarrow\) \(-\) \(8\) \(c\) \(-\) \(9,5\) \(e\) \(=\) \(-800\)
\(\Rightarrow\) \(8\)\(c\) \(+\) \(9,5\)\(e\) \(=\) \(800\)
\(\Rightarrow\) \(80\)\(c\) \(+\) \(95\)\(e\) \(=\) \(8000\)
\(\Rightarrow\) \(c\) \(=\) \(\frac{8000\color{Red}{- 95e}}{80}\)
\(\Rightarrow\) \(c\) \(=\) \(\frac{8000}{80}\) \(-\) \(\frac{\color{Red}{95e}}{\color{Red}{80}}\)
\(\Rightarrow\) \(c\) \(=\) \(100\) \(-\) \(\frac{\color{Red}{95e}}{\color{Red}{80}}\)
\(\Rightarrow\) \(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\)
Bien entendu \(c\) représente le nombre d'écoliers qui doit être un entier naturel, ce qui nous donne \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) un entier naturel, et par la suite \(\color{Red}{\frac{19e}{16}}\) est aussi un entier naturel.
Pour que \(\color{Red}{\frac{19e}{16}}\) soit un entier naturel il faut que \(e\) soit un multiple de \(16\). (on multiplions \(\frac{19}{16}\) par un multiple de 16 nous donne un entier naturel).
Trouvons toutes les solutions des nombres d'écoliers, des collégiens et des lycéens.
Pour que \(\color{Red}{\frac{19e}{16}}\) soit un entier naturel il faut que \(e\) soit un multiple de \(16\). (on multiplions \(\frac{19}{16}\) par un multiple de 16 nous donne un entier naturel).
Trouvons toutes les solutions des nombres d'écoliers, des collégiens et des lycéens.
Calculons \(c\) et \(l\) à partir de \(e\) :
* Si \(e=0\)
\(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19\times0}{16}}\) \(=\) \(100\) \(-\) \(0\) \(=\) \(100\). alors \(c=100\).
\(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\) \(=\) \(100\) \(-\) \(100\) \(-\) \(0\) \(=\) \(0\). alors \(l=0\).
* Si \(e=16\)
\(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19\times16}{16}}\) \(=\) \(100\) \(-\) \(19\) \(=\) \(81\). alors \(c=81\).
\(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\) \(=\) \(100\) \(-\) \(81\) \(-\) \(16\) \(=\) \(3\). alors \(l=3\).
* Si \(e=32\)
\(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19\times32}{16}}\) \(=\) \(100\) \(-\) \(38\) \(=\) \(62\). alors \(c=62\).
\(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\) \(=\) \(100\) \(-\) \(62\) \(-\) \(32\) \(=\) \(6\). alors \(l=6\).
* Si \(e=48\)
\(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19\times48}{16}}\) \(=\) \(100\) \(-\) \(57\) \(=\) \(43\). alors \(c=43\).
\(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\) \(=\) \(100\) \(-\) \(43\) \(-\) \(48\) \(=\) \(9\). alors \(l=9\).
* Si \(e=64\)
\(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19\times64}{16}}\) \(=\) \(100\) \(-\) \(76\) \(=\) \(24\). alors \(c=24\).
\(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\) \(=\) \(100\) \(-\) \(24\) \(-\) \(64\) \(=\) \(12\). alors \(l=12\).
* Si \(e=80\)
\(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19\times80}{16}}\) \(=\) \(100\) \(-\) \(95\) \(=\) \(5\). alors \(c=5\).
\(l\) \(=\) \(100\) \(-\) \(c\) \(-\) \(e\) \(=\) \(100\) \(-\) \(5\) \(-\) \(80\) \(=\) \(15\). alors \(l=15\).
* Si \(e=96\)
\(c\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19e}{16}}\) \(=\) \(100\) \(-\) \(\color{Red}{\frac{19\times96}{16}}\) \(=\) \(100\) \(-\) \(114\) \(=\) \(-5\). le nombre des écoliers ne peut pas être négatif, et par la suite \(e\) ne peut pas être \(96\).
Conclusion :
Les solutions possible sont :
\(l=0\) ; \(c=100\) ; \(e=0\)
\(l=3\) ; \(c=81\) ; \(e=16\)
\(l=6\) ; \(c=62\) ; \(e=32\)
\(l=9\) ; \(c=43\) ; \(e=48\)
\(l=12\) ; \(c=24\) ; \(e=64\)
\(l=15\) ; \(c=5\) ; \(e=80\)
Libellés:
1ère année secondaire
Activités numériques I
Correction
Corrigées
exercice
Le Mathématicien
manuel scolaire
Math
Mathématiques
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