Question

В ящике находится 153 карандаша ,которые разложены по трём коробкам причём известно, что в какой-то из коробок в 2 раза больше карандашей , чем в какой-то другой коробке , и в какой-то из коробок в 3 раза больше карандашей , чем в какой-то другой коробке. Определите число карандашей в каждой коробке.

Answer 1

Х-коробка, где меньше всего
х+2х+3х=150
6х=150
х=25 (1 коробка)
2×25=50 (2 коробка)
3×25=75 (3 коробка)

Answer 2

X-- -1 каробка
2х-- 2 каробка
3х 3 каробка

Answer 3

Problem Analysis

We are given that there are 153 pencils in total, distributed among three boxes. It is known that one of the boxes contains twice as many pencils as another box, and one of the boxes contains three times as many pencils as another box. We need to determine the number of pencils in each box.

Solution

Let's assume the number of pencils in the boxes are represented by variables `x`, `y`, and `z`. We can set up the following equations based on the given information:

1. The total number of pencils is 153: `x + y + z = 153`. 2. One box contains twice as many pencils as another box: `x = 2y` or `y = x/2`. 3. One box contains three times as many pencils as another box: `z = 3y` or `y = z/3`.

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

Solving the Equations

Substituting the value of `y` from equation 2 into equation 1, we get: `x + (x/2) + z = 153` Simplifying the equation, we have: `3x/2 + z = 153`

Substituting the value of `y` from equation 3 into equation 1, we get: `x + z/3 + z = 153` Simplifying the equation, we have: `x + 4z/3 = 153`

Now we have a system of two equations with two variables: ``` 3x/2 + z = 153 x + 4z/3 = 153 ```

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

Solving the System of Equations

To solve the system of equations, we can use substitution or elimination method. Let's use the elimination method.

Multiplying the first equation by 4/3, we get: ``` 2x + (4/3)z = 204 x + (4/3)z = 153 ```

Subtracting the second equation from the first equation, we eliminate `x`: ``` (2x - x) + (4/3)z - (4/3)z = 204 - 153 x = 51 ```

Substituting the value of `x` back into the second equation, we can solve for `z`: ``` 51 + (4/3)z = 153 (4/3)z = 102 z = (3/4) * 102 z = 76.5 ```

Since the number of pencils must be a whole number, we can round `z` to the nearest whole number: ``` z ≈ 77 ```

Substituting the values of `x` and `z` back into equation 2, we can solve for `y`: ``` y = x/2 y = 51/2 y = 25.5 ```

Again, since the number of pencils must be a whole number, we can round `y` to the nearest whole number: ``` y ≈ 26 ```

Therefore, the number of pencils in each box is approximately: - Box 1: 51 pencils - Box 2: 26 pencils - Box 3: 77 pencils

Please note that the numbers are approximate due to rounding.

Answer

The number of pencils in each box is approximately: - Box 1: 51 pencils - Box 2: 26 pencils - Box 3: 77 pencils