Question

Кассир разделил имеющиеся в кассе монеты на две кучи. В первой кучку было монет по двадцать Центов, а во второй - монеты по пятьдесят центов. Вычислите, Сколько монет каждого номинала было в кассе, если всего было 203 монеты, А денег в каждой кучку было поровну. Спасибо заранее.

Answer 1

Пусть х монет было 20 центов, тогда по 50 центов (203-х) т.к. всего 203 монеты. По условию денег в каждой кучке было поровну. Составим уравнение:
20х=50(203-х)
20х=10150-50х
70х=10150
х=145 монет по 20 центов.
203-145=58 монет по 50 центов.
Ответ: 145 по 20 центов; 58 по 50 центов.
Проверка:
145×20=58×50
2900=2900 (ВЕРНО)

Answer 2

Problem Analysis

A cashier has divided the coins in the cash register into two piles: one with coins worth 20 cents and the other with coins worth 50 cents. The total number of coins in the cash register is 203, and there is an equal amount of money in each pile. We need to determine how many coins of each denomination are in the cash register.

Solution

Let's assume the number of coins worth 20 cents is x, and the number of coins worth 50 cents is y. We can set up the following equations based on the given information:

1. The total number of coins: x + y = 203. 2. The total value of the coins in each pile is the same: 20x = 50y.

We can solve this system of equations to find the values of x and y.

Solving the Equations

To solve the system of equations, we can use the substitution method. We can solve equation 1 for x and substitute it into equation 2:

From equation 1: x = 203 - y.

Substituting into equation 2: 20(203 - y) = 50y.

Now we can solve for y:

20(203 - y) = 50y 4060 - 20y = 50y 4060 = 70y y = 4060 / 70 y = 58.

Now that we have the value of y, we can substitute it back into equation 1 to find x:

x = 203 - y x = 203 - 58 x = 145.

Answer

Therefore, there were 145 coins worth 20 cents and 58 coins worth 50 cents in the cash register.

Please let me know if you need any further assistance!