Question

в трех коробках 60 кг конфет. Если из первой высыпать 10 кг конфет, а из второй 14 кг. конфет, то в них остается в 2 раза меньше, чем в третьей. Сколько конфет было в коробках первоначально.

Answer 1

После отсыпки в 3 коробках осталось 36кг
в 1и2 в 2 раза меньше чем в 3,т,е 12 и 24.
в1-й и 2-й (6+10)и(6+14)=36 в 3-й-24.

Answer 2

Problem Analysis

We are given three boxes of candies, and it is stated that if we take out 10 kg of candies from the first box and 14 kg of candies from the second box, the remaining candies in the first two boxes will be half the amount of candies in the third box. We need to determine the initial number of candies in the three boxes.

Solution

Let's assume the initial number of candies in the first box is x, in the second box is y, and in the third box is z.

According to the given information, if we take out 10 kg of candies from the first box and 14 kg of candies from the second box, the remaining candies in the first two boxes will be half the amount of candies in the third box. Mathematically, we can express this as:

(x - 10) + (y - 14) = (z / 2)

Simplifying the equation, we get:

x + y - 24 = z / 2

We also know that the total weight of candies in the three boxes is 60 kg. Mathematically, we can express this as:

x + y + z = 60

We now have a system of two equations with two variables. We can solve this system of equations to find the values of x, y, and z.

Let's solve the system of equations:

Equation 1: x + y - 24 = z / 2 (Equation A)

Equation 2: x + y + z = 60 (Equation B)

To eliminate the fraction in Equation A, we can multiply both sides of the equation by 2:

2(x + y) - 48 = z

Substituting this value of z in Equation B, we get:

x + y + 2(x + y) - 48 = 60

Simplifying the equation, we get:

3x + 3y = 108

Dividing both sides of the equation by 3, we get:

x + y = 36

Now we have a system of two equations with two variables:

Equation 3: x + y = 36 (Equation C)

Equation 4: x + y + z = 60 (Equation D)

Subtracting Equation C from Equation D, we get:

z = 60 - 36

z = 24

Now we know that the initial number of candies in the third box is 24.

Substituting this value of z in Equation C, we get:

x + y = 36 - 24

x + y = 12

We have two equations with two variables:

Equation 5: x + y = 12 (Equation E)

Equation 6: x + y + z = 60 (Equation F)

Subtracting Equation E from Equation F, we get:

z = 60 - 12

z = 48

Now we know that the initial number of candies in the second box is 48.

Substituting this value of z in Equation E, we get:

x + y = 12 - 48

x + y = -36

We have two equations with two variables:

Equation 7: x + y = -36 (Equation G)

Equation 8: x + y + z = 60 (Equation H)

Subtracting Equation G from Equation H, we get:

z = 60 - (-36)

z = 96

Now we know that the initial number of candies in the first box is 96.

Therefore, the initial number of candies in the three boxes are as follows: - First box: 96 candies - Second box: 48 candies - Third box: 24 candies

Answer

The initial number of candies in the three boxes were as follows: - First box: 96 candies - Second box: 48 candies - Third box: 24 candies