Question

. Смешав 40 % и 15 % растворы кислоты, добавили 3 кг чистой воды и получили 20 % раствор кислоты. Если бы вместо 3 кг воды добавили 3 кг 80 % раствора той же кислоты, то получили бы 50 %-ый раствор кислоты. Сколько килограммов 40 % -го и 15 % растворов кислоты было смешано?

Answer 1

Системка: (0,4x+0.15y)/(x+y+3)=0.2 ;   (0,4x+0.15y+3*0.8)/(x+y+3)=0.5 ;    РЕШЕНИЕ:  (0.4x+0.15y)=0.2*(x+y+3);   0.4x+0.15y=0.2x+0.2y+0.6;    0.4x-0.2x+0.15y-0.2y=0.6;   0.2x-0.05y=0.6;  0.2x=0.05y+0.6;   x=(0.05y+0.6)/0.2;    x=0.25y+3;   (0.4*(0.25y+3)+0.15y+2.4)/(0.25y+3+y+3)=0.5;   (0.1y+1.2+0.15y+2.4)=0.5*(1.25y+6);    0.25y+3.6=0.625y+3;                                0.25y-0.625y=3-3.6;   -0.375y=-0.6;   y=-0.6/(-0.375) ;    y=1.6 кг ;       x=0.25*1.6+3=3.4 кг ;                           ответ:                                                                   3,4 кг 40% ;
1.6 кг 15%






 

Answer 2

Problem Analysis

We are given that a 40% solution and a 15% solution of acid were mixed together. Then, 3 kg of pure water was added to the mixture, resulting in a 20% solution of acid. If instead of adding 3 kg of water, 3 kg of an 80% solution of the same acid was added, a 50% solution of acid would have been obtained. We need to find the amount of the 40% and 15% solutions that were mixed.

Solution

Let's assume that x kg of the 40% solution and y kg of the 15% solution were mixed.

When 3 kg of pure water was added to the mixture, the total weight of the solution became (x + y + 3) kg.

According to the given information, the resulting solution was 20% acid. This means that the amount of acid in the resulting solution was 20% of the total weight of the solution.

Similarly, when 3 kg of an 80% solution of the same acid was added, the total weight of the solution became (x + y + 3) + 3 = (x + y + 6) kg.

According to the given information, the resulting solution was 50% acid. This means that the amount of acid in the resulting solution was 50% of the total weight of the solution.

Now we can set up the following equations:

Equation 1: Amount of acid in the resulting 20% solution = 20% of (x + y + 3) kg Equation 2: Amount of acid in the resulting 50% solution = 50% of (x + y + 6) kg

We can solve these equations to find the values of x and y.

Solving the Equations

Let's solve the equations step by step.

Equation 1: Amount of acid in the resulting 20% solution = 20% of (x + y + 3) kg 0.2(x + y + 3) = 0.2x + 0.2y + 0.6

Equation 2: Amount of acid in the resulting 50% solution = 50% of (x + y + 6) kg 0.5(x + y + 6) = 0.5x + 0.5y + 3

Expanding the equations:

0.2x + 0.2y + 0.6 = 0.2x + 0.2y + 0.6 0.5x + 0.5y + 3 = 0.5x + 0.5y + 3

The equations are identical, which means that the values of x and y can be any real numbers. There are infinitely many solutions to this problem.

Therefore, we cannot determine the exact amount of the 40% and 15% solutions of acid that were mixed based on the given information.

Conclusion

Based on the given information, we cannot determine the exact amount of the 40% and 15% solutions of acid that were mixed. The problem has infinitely many solutions.