Question

В двух теплицах , каждая из которых имеет прямоугольную форму , собрали 608кг перца . длина каждой теплицы 8 м, ширина одной 2м, а другой 3м . сколько кг перца. собрали в каждой теплице , если с 1м2 соберали одинаковое количество кг перца? объясни , почему эту задачу можно решить двумя способами.

Answer 1

8*2=16(м2)-площадь первой теплицы                                                                         8*3=24(м2)-площадь второй теплицы                                                                          16+24=40(м2)-общая площадь                                                                                  680:40=17(кг)-на 1 м2                                                                                                16*17=272(кг)-с первой теплицы                                                                                24*17=408(кг)-со второй теплицы                                                                               Ответ:272кг,408кг.

Answer 2

Problem Analysis

We have two greenhouses, each with a rectangular shape, and the total weight of peppers collected from both greenhouses is 608 kg. The length of one greenhouse is 8 m, and the width of the other greenhouse is 2 m, while the width of the second greenhouse is 3 m. We need to determine the weight of peppers collected in each greenhouse if the weight of peppers collected per square meter is the same in both greenhouses.

Solution

To solve this problem, we can use two different approaches: one based on the area of each greenhouse and the other based on the weight of peppers collected per square meter.

Approach 1: Based on the Area of Each Greenhouse We can calculate the area of each greenhouse and then determine the weight of peppers collected in each greenhouse based on the proportion of their areas.

1. Calculate the area of the first greenhouse: - Length = 8 m - Width = 2 m - Area = Length * Width = 8 m * 2 m = 16 m²

2. Calculate the area of the second greenhouse: - Length = 8 m - Width = 3 m - Area = Length * Width = 8 m * 3 m = 24 m²

3. Determine the weight of peppers collected in each greenhouse based on the proportion of their areas: - Let's assume the weight of peppers collected per square meter is x kg/m². - Weight of peppers collected in the first greenhouse = Area of the first greenhouse * x kg/m² = 16 m² * x kg/m² = 16x kg - Weight of peppers collected in the second greenhouse = Area of the second greenhouse * x kg/m² = 24 m² * x kg/m² = 24x kg

4. We know that the total weight of peppers collected from both greenhouses is 608 kg. Therefore, we can set up the following equation: - Weight of peppers collected in the first greenhouse + Weight of peppers collected in the second greenhouse = 608 kg - 16x kg + 24x kg = 608 kg - 40x kg = 608 kg - x = 608 kg / 40 kg = 15.2 kg/m²

5. Substitute the value of x back into the equations to find the weight of peppers collected in each greenhouse: - Weight of peppers collected in the first greenhouse = 16x kg = 16 * 15.2 kg = 243.2 kg - Weight of peppers collected in the second greenhouse = 24x kg = 24 * 15.2 kg = 364.8 kg

Therefore, the weight of peppers collected in the first greenhouse is 243.2 kg, and the weight of peppers collected in the second greenhouse is 364.8 kg.

Approach 2: Based on the Weight of Peppers Collected per Square Meter We can directly calculate the weight of peppers collected in each greenhouse based on the weight of peppers collected per square meter.

1. Let's assume the weight of peppers collected per square meter is y kg/m².

2. Calculate the area of the first greenhouse: - Length = 8 m - Width = 2 m - Area = Length * Width = 8 m * 2 m = 16 m²

3. Calculate the area of the second greenhouse: - Length = 8 m - Width = 3 m - Area = Length * Width = 8 m * 3 m = 24 m²

4. Determine the weight of peppers collected in each greenhouse based on the weight of peppers collected per square meter: - Weight of peppers collected in the first greenhouse = Area of the first greenhouse * y kg/m² = 16 m² * y kg/m² = 16y kg - Weight of peppers collected in the second greenhouse = Area of the second greenhouse * y kg/m² = 24 m² * y kg/m² = 24y kg

5. We know that the total weight of peppers collected from both greenhouses is 608 kg. Therefore, we can set up the following equation: - Weight of peppers collected in the first greenhouse + Weight of peppers collected in the second greenhouse = 608 kg - 16y kg + 24y kg = 608 kg - 40y kg = 608 kg - y = 608 kg / 40 kg = 15.2 kg/m²

6. Substitute the value of y back into the equations to find the weight of peppers collected in each greenhouse: - Weight of peppers collected in the first greenhouse = 16y kg = 16 * 15.2 kg = 243.2 kg - Weight of peppers collected in the second greenhouse = 24y kg = 24 * 15.2 kg = 364.8 kg

Therefore, the weight of peppers collected in the first greenhouse is 243.2 kg, and the weight of peppers collected in the second greenhouse is 364.8 kg.

Explanation

This problem can be solved using two different approaches because the weight of peppers collected per square meter is the same in both greenhouses. Therefore, we can either calculate the weight of peppers collected in each greenhouse based on the proportion of their areas or directly calculate the weight of peppers collected in each greenhouse based on the weight of peppers collected per square meter. Both approaches will yield the same result.

Approach 1 calculates the area of each greenhouse and then determines the weight of peppers collected in each greenhouse based on the proportion of their areas. This approach is based on the assumption that the weight of peppers collected per square meter is the same in both greenhouses.

Approach 2 directly calculates the weight of peppers collected in each greenhouse based on the weight of peppers collected per square meter. This approach is also based on the assumption that the weight of peppers collected per square meter is the same in both greenhouses.

Both approaches are valid and will give the same result because they are based on the same assumption. The choice of approach depends on personal preference or the specific requirements of the problem.