Question

За 5 конвертов заплатили 45р. Сколько таких конвертов можно купить за 72р? Решить разными способами

Answer 1

45/5=9р - цена одного конверта

72/9=8 конвертов.........................

Answer 2

Узнаем, сколько стоит один конверт 

45:5=9 рублей

72:9=8 конвертов можно купить на 72 рубля:) другого способа решения не знаю:(

Answer 3

Problem Statement

You have been given 72 rubles to buy envelopes. Each envelope costs 9 rubles. You need to determine how many envelopes you can buy with the given amount.

Solution 1: Division

To solve this problem, you can divide the total amount of money by the cost of each envelope. This will give you the number of envelopes you can buy.

Let's calculate it step by step:

1. Divide the total amount of money (72 rubles) by the cost of each envelope (9 rubles): - 72 / 9 = 8

So, you can buy 8 envelopes with 72 rubles.

Solution 2: Multiplication

Another way to solve this problem is by using multiplication. You can find out how many envelopes you can buy by dividing the total amount of money by the cost of each envelope.

Let's calculate it step by step:

1. Divide the total amount of money (72 rubles) by the cost of each envelope (9 rubles): - 72 / 9 = 8

So, you can buy 8 envelopes with 72 rubles.

Solution 3: Subtraction

You can also solve this problem by subtracting the cost of each envelope from the total amount of money until you can no longer afford another envelope.

Let's calculate it step by step:

1. Subtract the cost of the first envelope (9 rubles) from the total amount of money (72 rubles): - 72 - 9 = 63

2. Subtract the cost of the second envelope (9 rubles) from the remaining amount of money (63 rubles): - 63 - 9 = 54

3. Repeat this process until you can no longer afford another envelope.

- 54 - 9 = 45 - 45 - 9 = 36 - 36 - 9 = 27 - 27 - 9 = 18 - 18 - 9 = 9

So, you can buy 8 envelopes with 72 rubles.

Conclusion

In conclusion, you can buy 8 envelopes with 72 rubles using any of the three methods mentioned above: division, multiplication, or subtraction.