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Actividades extraescolares colegios soluciones

Actividades extraescolares colegios. Soluciones

En este post iremos publicando las soluciones a las adivinanzas que se vayan publicando en los muros de nuestras RRSS. Futuros Talentos-Actividades extraescolares colegios os da las gracias por participar en los retos que os proponemos para todos los niveles y edades.

 

Post 1. 08/12/2015

Actividades extraescolares colegios

 

Solución:

La idea principal para llegar a la solución a este problema es intentar dividir en cada pesada las bolas en grupos más pequeños, para ser capaces de llegar a la solución sea cual sea la bola que sea distinta. Para que quede claro en todo momento, numero las bolas (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12).

En la primera pesada tenemos que pesar cuatro bolas a un lado (1, 2, 3, 4), y otras cuatro bolas a otro lado (5, 6, 7, 8), dejando otras cuatro bolas fuera de la balanza (9, 10, 11, 12). De este modo tendremos tres grupos de cuatro bolas bien diferenciados. Ahora, dependiendo de cuál sea el resultado, serán en consecuencia las siguientes pesadas. A continuación os expongo todos los posibles resultados.

Caso 1) La balanza se equilibra. Con esto podemos deducir que (1, 2, 3, 4, 5, 6, 7, 8) no son la bola que pesa distinto al resto. La bola distinta tiene que estar en el grupo (9, 10, 11, 12). Para la segunda pesada, tomaremos tres bolas de este último grupo (9, 10, 11) y las pondremos en un lado de la balanza, poniendo en el otro a tres bolas que sabemos que pesan lo normal (1, 2, 3).

Caso 1.1) La balanza se equilibra. Con esto nos queda claro que la bola que pesa distinto es (12), así que únicamente tendríamos que hacer una última pesada comparándola con cualquier otra bola, (1) por ejemplo, y determinar si pesa más o menos que el resto.

Caso 1.2) La balanza se inclina para uno de los lados. Como sabemos que (1, 2, 3) no son bolas que pesen distinto, si (9, 10, 11) baja será una de estas la que pese más, y si sube será una de estas la que pese menos. Para la tercera pesada tomaremos las bolas (9) y (10) y las compararemos.

Caso 1.2.1) La balanza se equilibra. Entonces sabremos que la bola (11) pesa más o menos que el resto (dependiendo de lo que hubiera sucedido en la segunda pesada)

Caso 1.2.2) La balanza se inclina hacia uno de los lados. Dependiendo de la información obtenida en la segunda pesada, sabremos que la bola pesa más o menos que el resto, y por lo tanto, la bola que en este caso también suba, o también baje, será la que pese distinto. Pesará menos si su lado de la balanza sube, y pesará más si su lado de la balanza baja.

Caso 2) La balanza se desequilibra hacia (1, 2, 3, 4). Por lo tanto sabemos que una bola de (1, 2, 3, 4) pesa más que el resto o una bola de (5, 6, 7, 8) pesa menos que el resto. Para la siguiente pesada tendremos que dividir, tal y como hemos hecho al principio, estas 8 bolas en tres grupos iguales (o similares) para poder dividir el problema. Así que, ponemos en un lado de la balanza (1, 2, 5) y en el otro lado de la balanza (3, 4, 6), dejando (7) y (8) fuera de la balanza.

Caso 2.1) La balanza se equilibra. En este caso sabemos que (7) u (8) pesa menos que el resto. Bastará con poner una bola en cada lado de la balanza, y la bola que vaya para arriba será la que pese menos que el resto.

Caso 2.2) La balanza se inclina hacia (1, 2, 5). En este caso sabemos que, (1) o (2) pesa más que el resto, o (6) pesa menos que el resto. Bastará con poner en la última pesada (1) a un lado y (2) al otro. Si se equilibran será (6) la que pese menos, y si se inclina para un lado, será la bola en ese lado la que pese más que el resto.

Caso 2.3) La balanza se inclina hacia (3, 4, 6). En este caso sabemos que, (3) o (4) pesa más que el resto, o (5) pesa menos que el resto. Bastará con poner en la última pesada (3) a un lado y (4) al otro. Si se equilibran será (5) la que pese menos, y si se inclina para un lado, será la bola en ese lado la que pese más que el resto.

Caso 3) La balanza se desequilibra hacia (5, 6, 7, 8). Por lo tanto sabemos que una bola de (5, 6, 7, 8) pesa más que el resto o una bola de (1, 2, 3, 4) pesa menos que el resto. Para la siguiente pesada tendremos que dividir, tal y como hemos hecho al principio, estas 8 bolas en tres grupos iguales (o similares) para poder dividir el problema. Así que, ponemos en un lado de la balanza (5, 6, 1) y en el otro lado de la balanza (7, 8, 2), dejando (3) y (4) fuera de la balanza.

Caso 3.1) La balanza se equilibra. En este caso sabemos que (3) u (4) pesa menos que el resto. Bastará con poner una bola en cada lado de la balanza, y la bola que vaya para arriba será la que pese menos que el resto.

Caso 3.2) La balanza se inclina hacia (5, 6, 1). En este caso sabemos que, (5) o (6) pesa más que el resto, o (2) pesa menos que el resto. Bastará con poner en la última pesada (5) a un lado y (6) al otro. Si se equilibran será (6) la que pese menos, y si se inclina para un lado, será la bola en ese lado la que pese más que el resto.

Caso 3.3) La balanza se inclina hacia (7, 8, 2). En este caso sabemos que, (7) o (8) pesa más que el resto, o (1) pesa menos que el resto. Bastará con poner en la última pesada (7) a un lado y (8) al otro. Si se equilibran será (1) la que pese menos, y si se inclina para un lado, será la bola en ese lado la que pese más que el resto.

 

 

Futuros Talentos – Actividades extraescolares colegios os desea un feliz día

Actividades extraescolares colegios es un proyecto dedicado al desarrollo de actividades fuera del horario escolar de los colegios, pero que se realizan en las propias instalaciones de los mismos tras acabar la jornada lectiva de los alumnos.

Actividades extraescolares colegios presenta 4 vertientes principales: actividades extraescolares formativas, actividades extraescolares deportivas, actividades extraescolares artisticas y actividades extraescolares idiomas.

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