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fredd [130]
3 years ago
11

Quiero distribuir 72 sandwiches de queso,54 de jamón y 30 de carne en la mayor cantidad posible de bandejas,y que en todas haya

lo mismo ¿cuantas bandejas usará? ¿cuántos sándwiches de cada gusto habrá en cada bandeja?
Mathematics
1 answer:
Lelu [443]3 years ago
7 0

Answer:

Se necesitan 6 bandejas

6 trays needed

Se pueden colocar 12 sándwiches de queso, 9 de jamón y 5 de carne en 6 bandejas

12 cheese sandwiches, 9 ham and 5 meat sandwiches can be placed in 6 trays

Step-by-step explanation:

Esto se puede averiguar calculando el HCF de 72,54 y 30 para que haya el mismo número de sándwiches de los tres tipos.

HCF se descubre enumerando primero los factores.

Factores de 72 = 2 * 2 * 2 * 3 * 3

Factores de 54 = 3 * 2 * 3 * 3

Factores de 30 = 2 * 5 * 3

Ahora escribe los factores comunes que son 2 y 3

El factor común más alto es 2 * 3 = 6

Por lo tanto, se necesitan 6 bandejas para los bocadillos.

Si se utilizan 6 bandejas, habrá 72/6 = 12 sándwiches de queso.

Si se utilizan 6 bandejas, habrá 54/6 = 9 sándwiches de jamón.

Si se utilizan 6 bandejas, habrá 30/6 = 5 sándwiches de carne.

De esta forma se pueden distribuir por igual para que el número total de los mismos tipos de bocadillos siga siendo el mismo.

This can be found out by calculating the HCF of 72,54 and 30 so that there are equal number of the sandwiches of all the three types.

HCF is found out by first listing the factors.

Factors of 72= 2*2*2*3*3

Factors of 54= 3*2*3*3

Factors of 30= 2*5*3

Now write the common factors which are 2 and 3

Highest Common Factor is 2*3= 6

Therefore 6 trays are needed for the sandwiches .

If 6 trays are used then there will be 72 /6= 12 cheese sandwiches.

If 6 trays are used then there will be 54/6= 9  ham sandwiches.

If 6 trays are used then there will be 30/6= 5 meat sandwiches.

In this way they can distributed equally so the total number of the same types of sandwiches remains the same.

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