Question

В лесу на разных кустах висят 50 шнурков.Сова утверждает,что в среднем три шнурка из пяти,которые можно найти в лесу,ей не подходят,поскольку они слишком длинные для дверного звонка.Ослика Иа утверждает,что в среднем семь из десяти шнурков из леса ему не подходят,поскольку они слишком короткие,чтобы сделать из них хваост.Оба правы.Сколько шнурков,висящих на кустах,не подходят ни сове,ни иа? Найди наименьшее возможное число..

Answer 1

Вот     так  ????????????????

Answer 2

Сове не подходят 3 из 5, а из 50 это равно 30
Иа не подходит 7 из 10, а из 50 это равно 35
Наименьшее число никому неподходящих шнурков равно 30

Answer 3

Analysis and Solution

To find the minimum number of strings that do not fit either the owl or the donkey, we need to calculate the number of strings that do not fit each animal individually and then find the maximum of those two numbers.

According to the information given, the owl claims that on average, three out of five strings found in the forest do not fit her because they are too long for a doorbell. The donkey claims that on average, seven out of ten strings from the forest do not fit him because they are too short to make a tail.

Let's calculate the number of strings that do not fit each animal individually:

For the owl: - Out of every 5 strings, 3 do not fit the owl. - So, the percentage of strings that do not fit the owl is 3/5 = 60%.

For the donkey: - Out of every 10 strings, 7 do not fit the donkey. - So, the percentage of strings that do not fit the donkey is 7/10 = 70%.

Now, we need to find the maximum of these two percentages to determine the minimum number of strings that do not fit either the owl or the donkey.

The maximum percentage is 70%, which means that at least 70% of the strings do not fit either the owl or the donkey.

To find the minimum number of strings, we need to calculate 70% of the total number of strings.

According to the question, there are 50 strings hanging on different bushes in the forest.

So, the minimum number of strings that do not fit either the owl or the donkey is 70% of 50, which is 35 strings.

Therefore, the minimum possible number of strings that do not fit either the owl or the donkey is 35.