Question

Сплав содержит меди на 640г больше чем цинка . Когда из сплава выделили 6/7 содержащейся в нём меди и 60% цинка масса сплава стала 200г . Сколько весил сплав первоночально ? СРОЧНО НАДО ! ПОЖАЛУЙСТА ПОМОГИТЕ С ЗАДАЧЕЙ !!!!!!!!!! УМОЛЯЮ

Answer 1

Пусть х г - содержание цинка, а х + 640 г - содержание меди.
Тогда масса сплава - 2х + 640 г.
(х + 640) : 7 · 6  - количество выделенной меди.
60% - 0,6
0,6х - количество выделенного цинка.
2х + 640 - (х + 640) : 7 · 6 - 0,6х = 200   I * 7
14х + 4480 - (х + 640) · 6 - 4.2х = 1400
14х + 4480 - 6х - 3840 - 4,2х = 1400
3,8х + 640 = 1400
3,8х = 1400 - 640
3,8х = 760
Х = 760 : 3,8
Х = 200 (г) - содержание цинка.
200 + 640 = 840 (г) - содержание меди.
840 + 200 = 1040 (г) - масса сплава первоначально.
Ответ: 1040г.

Answer 2

To solve this problem, we can set up an equation based on the given information. Let's denote the initial weight of the alloy as x.

According to the problem, the alloy contains 640g more copper than zinc. This means that the weight of copper in the alloy is x + 640g, and the weight of zinc is x.

When 6/7 of the copper and 60% of the zinc are removed from the alloy, the remaining alloy weighs 200g. This means that the weight of the remaining copper is (6/7)(x + 640g), and the weight of the remaining zinc is (40/100)(x).

We can now set up the equation:

(6/7)(x + 640g) + (40/100)(x) = 200g

To solve this equation, we can simplify it by multiplying through by 7 and 100 to get rid of the fractions:

6(x + 640g) + 40x = 1400g

Simplifying further:

6x + 3840g + 40x = 1400g

Combining like terms:

46x + 3840g = 1400g

Subtracting 3840g from both sides:

46x = 1400g - 3840g

Simplifying:

46x = -2440g

Dividing both sides by 46:

x = -2440g / 46

Simplifying:

x = -53.043g

Therefore, the initial weight of the alloy was approximately -53.043g. However, it is important to note that a negative weight does not make sense in this context. It is possible that there is an error in the problem statement or in the calculations. Please double-check the given information and calculations to ensure accuracy.