Question

ЗМЕЮ Горынычу подарили на день рождения 48 кг. конфет. Первая голова съела в 3 раза больше конфет, чем третья, а третья-на 3 кг. меньше, чем вторая. Сколько кг. конфет съела каждая голова Змея Горыныча? Решить задачу подбором.

Answer 1

Problem Analysis

We are given that Zmey Gorunych received 48 kg of candy for his birthday. The first head ate three times more candy than the third head, and the third head ate 3 kg less than the second head. We need to determine how much candy each head of Zmey Gorunych ate.

Solution

Let's assume the amount of candy eaten by the third head is x kg.

According to the problem, the first head ate three times more candy than the third head, so the first head ate 3x kg of candy.

The third head ate 3 kg less than the second head, so the third head ate (3x + 3) kg of candy.

The sum of the candy eaten by all three heads is equal to 48 kg:

x + 3x + (3x + 3) = 48

Simplifying the equation:

7x + 3 = 48

Subtracting 3 from both sides:

7x = 45

Dividing both sides by 7:

x = 45/7

Therefore, the third head ate approximately 6.43 kg of candy.

The first head ate three times more candy than the third head, so the first head ate approximately 19.29 kg of candy.

The second head ate 3 kg more than the third head, so the second head ate approximately 9.43 kg of candy.

Answer

Based on the calculations, each head of Zmey Gorunych ate the following amounts of candy: - The first head ate approximately 19.29 kg of candy. - The second head ate approximately 9.43 kg of candy. - The third head ate approximately 6.43 kg of candy.

Please note that these values are approximate due to rounding.