Question

Несколько камней разложены в 5 кучек. Известно, что в пятой кучке камней в шесть раз больше, чем в третьей; во второй кучке камней вдвое больше, чем в третьей и пятой вместе взятых; в первой кучке камней втрое меньше, чем в пятой, и на 15 меньше, чем в четвёртой; в четвёртой кучке камней в два раза меньше, чем во второй. Сколько всего суммарно камней в этих пяти кучках?

Answer 1

Ответ:

90 камней.

Пошаговое объяснение:

Пусть в третьей куче х камней, тогда в пятой 6х, во второй 2(х+6х)=14х, в первой 6х:3=2х; в четвертой 2х+15.

По условию 14х=2(2х+15)

14х=4х+30

10х=30

х=3.

В третьей куче 3 камня, во второй 14*3=42 камня, в первой 2*3=6 камней, в четвертой 2*3+15=21 камень, в пятой 6*3=18 камней.

Всего 6+42+3+21+18=90 камней.

Answer 2

Problem Analysis

We are given information about the number of stones in five piles. The fifth pile has six times more stones than the third pile. The second pile has twice as many stones as the combined total of the third and fifth piles. The first pile has one-third the number of stones as the fifth pile and is 15 stones fewer than the fourth pile. The fourth pile has half the number of stones as the second pile. We need to determine the total number of stones in all five piles.

Solution

Let's assign variables to represent the number of stones in each pile: - Let `x` be the number of stones in the first pile. - Let `y` be the number of stones in the second pile. - Let `z` be the number of stones in the third pile. - Let `w` be the number of stones in the fourth pile. - Let `v` be the number of stones in the fifth pile.

We can now translate the given information into equations:

1. The fifth pile has six times more stones than the third pile: `v = 6z`. 2. The second pile has twice as many stones as the combined total of the third and fifth piles: `y = 2(z + v)`. 3. The first pile has one-third the number of stones as the fifth pile and is 15 stones fewer than the fourth pile: `x = (1/3)v` and `x = w - 15`. 4. The fourth pile has half the number of stones as the second pile: `w = (1/2)y`.

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

Solving the Equations

Let's solve the system of equations step by step:

From equation 1, we have `v = 6z`.

Substituting this value of `v` into equation 2, we get `y = 2(z + 6z) = 2(7z) = 14z`.

From equation 4, we have `w = (1/2)y = (1/2)(14z) = 7z`.

Substituting this value of `w` into equation 3, we get `x = 7z - 15`.

Finally, substituting the value of `v` from equation 1 into equation 3, we get `x = (1/3)(6z) = 2z`.

Now we have two expressions for `x`:

- `x = 7z - 15` - `x = 2z`

Setting these two expressions equal to each other, we can solve for `z`:

`7z - 15 = 2z`

Simplifying the equation, we get:

`5z = 15`

Dividing both sides by 5, we find:

`z = 3`

Now that we have the value of `z`, we can substitute it back into the equations to find the values of `x`, `y`, `w`, and `v`:

- `x = 2z = 2(3) = 6` - `y = 14z = 14(3) = 42` - `w = 7z = 7(3) = 21` - `v = 6z = 6(3) = 18`

Answer

Therefore, the total number of stones in all five piles is the sum of `x`, `y`, `z`, `w`, and `v`:

Total = `x + y + z + w + v = 6 + 42 + 3 + 21 + 18 = 90`.

So, there are a total of 90 stones in these five piles.

Please let me know if you need any further assistance!