Question

Для комплектования угощений участникам праздничного факультетского капустника в гипермаркете приобрели 5,5 кг чипсов. Купленные чипсы были упакованы в 46 пакетов трех различных видов, вместимость которых составляла соответственно 70, 80 и 150 г. Сколько пакетов каждого вида было приобретено? Сколько различных решений имеет данная задача? В ответе укажите все возможные решения.

Answer 1

Это система 2 уравнений с 3 неизвестными
{ x+y+z = 46
{ 70x+80y+150z = 5500
2 уравнение делим на 10
{ z = 46-x-y
{ 7x+8y+15(46-x-y) = 550
7x+8y+690-15x-15y = 550
690-550 = 15x+15y-7x-8y
8x+7y = 140
Так как числа 140 и 7y оба делятся на 7, то и 8x кратно 7. Значит, x делится на 7.
Варианты решения:
x = 0; y = 20; z = 26
x = 7; y = 12; z = 27
x = 14; y = 4; z = 28
Других вариантов быть не может быть, иначе будет y < 0.

Answer 2

To determine the number of packets of each type that were purchased, we need to solve a system of equations based on the given information.

Let's assume: - x represents the number of packets of the first type (with a capacity of 70g), - y represents the number of packets of the second type (with a capacity of 80g), and - z represents the number of packets of the third type (with a capacity of 150g).

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

1. The total weight of the chips is 5.5kg, which is equal to 5500g: - 70x + 80y + 150z = 5500

2. The total number of packets is 46: - x + y + z = 46

To solve this system of equations, we can use various methods such as substitution, elimination, or matrix methods. Let's use the substitution method in this case.

From equation 2, we can express x in terms of y and z: - x = 46 - y - z

Substituting this value of x into equation 1, we get: - 70(46 - y - z) + 80y + 150z = 5500

Simplifying the equation: - 3220 - 70y - 70z + 80y + 150z = 5500 - 80y - 70y + 150z - 70z = 5500 - 3220 - 10y + 80z = 2280

Now, let's find the possible values of y and z that satisfy this equation.

Using integer values for y and z, we can find the following solutions:

1. y = 0, z = 28 - This means that no packets of the second type (80g) were purchased, and 28 packets of the third type (150g) were purchased.

2. y = 8, z = 20 - This means that 8 packets of the second type (80g) were purchased, and 20 packets of the third type (150g) were purchased.

3. y = 16, z = 12 - This means that 16 packets of the second type (80g) were purchased, and 12 packets of the third type (150g) were purchased.

4. y = 24, z = 4 - This means that 24 packets of the second type (80g) were purchased, and 4 packets of the third type (150g) were purchased.

5. y = 32, z = -4 - This solution is not valid since the number of packets cannot be negative.

Therefore, there are four possible solutions to this problem, as listed above.