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Students of a class are made to stand in rows. If 4 students are extra in each row, then there would be 2 rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is

Question

Students of a class are made to stand in rows. If 4 students are extra in each row, then there would be 2 rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is

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Solution

Vamos resolver o problema passo a passo.

  1. Vamos definir algumas variáveis:

    • n n é o número total de alunos na classe.
    • r r é o número original de filas.
    • s s é o número original de alunos por fila.
  2. De acordo com a primeira condição:

    • Se houverem 4 alunos a mais em cada fila, então o número de filas será r2 r - 2 .
    • Portanto, podemos escrever a equação: n=(s+4)(r2) n = (s + 4)(r - 2) .
  3. De acordo com a segunda condição:

    • Se houverem 4 alunos a menos em cada fila, então o número de filas será r+4 r + 4 .
    • Portanto, podemos escrever a equação: n=(s4)(r+4) n = (s - 4)(r + 4) .
  4. Agora temos duas equações:

    • n=(s+4)(r2) n = (s + 4)(r - 2)
    • n=(s4)(r+4) n = (s - 4)(r + 4)
  5. Vamos expandir ambas as equações:

    • n=sr+4r2s8 n = sr + 4r - 2s - 8
    • n=sr4r+4s16 n = sr - 4r + 4s - 16
  6. Como ambas as expressões são iguais a n n , podemos igualá-las:

    • sr+4r2s8=sr4r+4s16 sr + 4r - 2s - 8 = sr - 4r + 4s - 16
  7. Simplificando a equação:

    • 4r2s8=4r+4s16 4r - 2s - 8 = -4r + 4s - 16
    • 4r+4r2s4s=16+8 4r + 4r - 2s - 4s = -16 + 8
    • 8r6s=8 8r - 6s = -8
    • 4r3s=4 4r - 3s = -4
    • 4r=3s4 4r = 3s - 4
    • r=3s44 r = \frac{3s - 4}{4}
  8. Substituindo r r na primeira equação:

    • n=(s+4)(3s442) n = (s + 4)\left(\frac{3s - 4}{4} - 2\right)
    • n=(s+4)(3s484) n = (s + 4)\left(\frac{3s - 4 - 8}{4}\right)
    • n=(s+4)(3s124) n = (s + 4)\left(\frac{3s - 12}{4}\right)
    • n=(s+4)(3s12)4 n = \frac{(s + 4)(3s - 12)}{4}
    • n=3s212s+12s484 n = \frac{3s^2 - 12s + 12s - 48}{4}
    • n=3s2484 n = \frac{3s^2 - 48}{4}
    • n=3(s216)4 n = \frac{3(s^2 - 16)}{4}
    • n=3(s4)(s+4)4 n = \frac{3(s - 4)(s + 4)}{4}
  9. Substituindo r r na segunda equação:

    • n=(s4)(3s44+4) n = (s - 4)\left(\frac{3s - 4}{4} + 4\right)
    • n=(s4)(3s4+164) n = (s - 4)\left(\frac{3s - 4 + 16}{4}\right)
    • n=(s4)(3s+124) n = (s - 4)\left(\frac{3s + 12}{4}\right)
    • n=(s4)(3s+12)4 n = \frac{(s - 4)(3s + 12)}{4}
    • n=3s2+12s12s484 n = \frac{3s^2 + 12s - 12s - 48}{4}
    • n=3s2484 n = \frac{3s^2 - 48}{4}
    • n=3(s216)4 n = \frac{3(s^2 - 16)}{4}
    • n=3(s4)(s+4)4 n = \frac{3(s - 4)(s + 4)}{4}
  10. Portanto, o número de alunos na classe é n=3(s4)(s+4)4 n = \frac{3(s - 4)(s + 4)}{4} .

Para encontrar o valor exato de n n , precisamos resolver a equação 4r3s=4 4r - 3s = -4 para valores inteiros de r r e s s . Vamos tentar alguns valores:

  • Se s=8 s = 8 :
    • r=3(8)44=2444=204=5 r = \frac{3(8) - 4}{4} = \frac{24 - 4}{4} = \frac{20}{4} = 5
    • n=8×5=40 n = 8 \times 5 = 40

Verificando com as condições:

  • Se houverem 4 alunos a mais em cada fila: 8+4=12 8 + 4 = 12 alunos por fila, 52=3 5 - 2 = 3 filas.

    • 12×3=36 12 \times 3 = 36 (não é igual a 40, então não é a solução correta).
  • Se s=12 s = 12 :

    • r=3(12)44=3644=324=8 r = \frac{3(12) - 4}{4} = \frac{36 - 4}{4} = \frac{32}{4} = 8
    • n=12×8=96 n = 12 \times 8 = 96

Verificando com as condições:

  • Se houverem 4 alunos a mais em cada fila: 12+4=16 12 + 4 = 16 alunos por fila, 82=6 8 - 2 = 6 filas.

    • 16×6=96 16 \times 6 = 96 (condição satisfeita).
  • Se houverem 4 alunos a menos em cada fila: 124=8 12 - 4 = 8 alunos por fila, 8+4=12 8 + 4 = 12 filas.

    • 8×12=96 8 \times 12 = 96 (condição satisfeita).

Portanto, o número de alunos na classe é 96 96 .

This problem has been solved

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