Question

Помогите решить задачу: . У Сережи и Пети всего 69 марок. У Пети марок в раза больше, чем у Сережи. Сколько марок у каждого из мальчиков? Заранее спасибо

Answer 1

С-x
2x+x=69
3x=69:3
x=23(м)-Сережа
23*2=46(м)

Answer 2

Problem Analysis

We are given that Sergey and Petya have a total of 69 marbles, and Petya has marbles that are a multiple of the number of marbles Sergey has. We need to determine how many marbles each boy has.

Solution

Let's assume that Sergey has x marbles. According to the problem, Petya has marbles that are a multiple of x. Let's say Petya has y times the number of marbles Sergey has.

We can set up the following equation based on the given information:

x + yx = 69

Simplifying the equation, we get:

x(1 + y) = 69

Now, we need to find values of x and y that satisfy this equation.

Solving the Equation

To find the values of x and y, we can use trial and error or solve the equation algebraically.

Let's start with trial and error. We know that the total number of marbles is 69, so let's try different values of x and see if we can find a value of y that satisfies the equation.

Let's start with x = 1: 1(1 + y) = 69 1 + y = 69 y = 68

This means that if Sergey has 1 marble, Petya would have 68 marbles. However, this doesn't satisfy the condition that Petya has marbles that are a multiple of Sergey's marbles.

Let's try another value of x. Let's say x = 2: 2(1 + y) = 69 2 + 2y = 69 2y = 69 - 2 2y = 67 y = 67/2 y = 33.5

Since y is not a whole number, this solution doesn't work either.

Let's try another value of x. Let's say x = 3: 3(1 + y) = 69 3 + 3y = 69 3y = 69 - 3 3y = 66 y = 66/3 y = 22

This means that if Sergey has 3 marbles, Petya would have 22 times the number of marbles Sergey has. This satisfies the condition that Petya has marbles that are a multiple of Sergey's marbles.

Therefore, Sergey has 3 marbles and Petya has 22 times the number of marbles Sergey has, which is 66 marbles.

Answer

Sergey has 3 marbles and Petya has 66 marbles.

Note: The solution was obtained through trial and error. There may be other values of x and y that satisfy the equation, but this is one possible solution.