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Five farmers have 7,9,11,13 & 14 apple trees respectively in their orchards. Last year each of them discovered that every tree in their own orchard yields exactly the same number of apples. Further, if the 3rd farmer gives one apple to the 1st and the 5th gives 3 to each of the 2nd & d 4th, they would all exactly have the same number of apples, what were the yields per tree in the orchards of the 3rd & 4th farmers ?Options 9 & 1117 & 9 9 & 9 11 & 9

Question

Five farmers have 7,9,11,13 & 14 apple trees respectively in their orchards. Last year each of them discovered that every tree in their own orchard yields exactly the same number of apples. Further, if the 3rd farmer gives one apple to the 1st and the 5th gives 3 to each of the 2nd & d 4th, they would all exactly have the same number of apples, what were the yields per tree in the orchards of the 3rd & 4th farmers ?Options 9 & 1117 & 9 9 & 9 11 & 9

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Solution

Vamos a resolver el problema paso a paso.

  1. Denotemos el número de manzanas por árbol en cada huerto como a1,a2,a3,a4a_1, a_2, a_3, a_4 y a5a_5 respectivamente.

  2. Entonces, el número total de manzanas en cada huerto será:

    • Primer agricultor: 7a17a_1
    • Segundo agricultor: 9a29a_2
    • Tercer agricultor: 11a311a_3
    • Cuarto agricultor: 13a413a_4
    • Quinto agricultor: 14a514a_5
  3. Según el problema, si el tercer agricultor da una manzana al primero y el quinto da 3 manzanas a cada uno del segundo y cuarto, todos tendrían el mismo número de manzanas.

  4. Esto se traduce en las siguientes ecuaciones:

    • 7a1+1=9a2+37a_1 + 1 = 9a_2 + 3
    • 7a1+1=11a317a_1 + 1 = 11a_3 - 1
    • 7a1+1=13a4+37a_1 + 1 = 13a_4 + 3
    • 7a1+1=14a567a_1 + 1 = 14a_5 - 6
  5. Simplificamos las ecuaciones:

    • 7a1+1=9a2+3    7a1=9a2+27a_1 + 1 = 9a_2 + 3 \implies 7a_1 = 9a_2 + 2
    • 7a1+1=11a31    7a1=11a327a_1 + 1 = 11a_3 - 1 \implies 7a_1 = 11a_3 - 2
    • 7a1+1=13a4+3    7a1=13a4+27a_1 + 1 = 13a_4 + 3 \implies 7a_1 = 13a_4 + 2
    • 7a1+1=14a56    7a1=14a577a_1 + 1 = 14a_5 - 6 \implies 7a_1 = 14a_5 - 7
  6. Ahora, buscamos valores enteros que satisfagan estas ecuaciones. Probamos con las opciones dadas:

    • Para a3=9a_3 = 9 y a4=11a_4 = 11:

      • 7a1=1192=992=977a_1 = 11 \cdot 9 - 2 = 99 - 2 = 97
      • 7a1=1311+2=143+2=1457a_1 = 13 \cdot 11 + 2 = 143 + 2 = 145
      • No coincide.
    • Para a3=17a_3 = 17 y a4=9a_4 = 9:

      • 7a1=11172=1872=1857a_1 = 11 \cdot 17 - 2 = 187 - 2 = 185
      • 7a1=139+2=117+2=1197a_1 = 13 \cdot 9 + 2 = 117 + 2 = 119
      • No coincide.
    • Para a3=9a_3 = 9 y a4=9a_4 = 9:

      • 7a1=1192=992=977a_1 = 11 \cdot 9 - 2 = 99 - 2 = 97
      • 7a1=139+2=117+2=1197a_1 = 13 \cdot 9 + 2 = 117 + 2 = 119
      • No coincide.
    • Para a3=11a_3 = 11 y a4=9a_4 = 9:

      • 7a1=11112=1212=1197a_1 = 11 \cdot 11 - 2 = 121 - 2 = 119
      • 7a1=139+2=117+2=1197a_1 = 13 \cdot 9 + 2 = 117 + 2 = 119
      • Coincide.

Por lo tanto, los rendimientos por árbol en los huertos del tercer y cuarto agricultor son 1111 y 99 respectivamente.

This problem has been solved

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