48 kg of 30% sulfuric acid in a mixture of 10% and 40% sulfuric acid arose. How many kilograms were each of the original solutions?

Let's assume that x kg of 10% sulfuric acid solution is mixed with y kg of 40% sulfuric acid solution to obtain 48 kg of the final mixture.

Step 1: Write the equation based on the amount of sulfuric acid in the mixture:
0.1x + 0.4y = 0.3(48)

Step 2: Simplify the equation:
0.1x + 0.4y = 14.4

Step 3: Use the fact that the total mixture is 48 kg:
x + y = 48

Step 4: Solve the system of equations.
First, multiply the first equation by 10 to eliminate decimals:
10(0.1x + 0.4y) = 10(14.4)
x + 4y = 144

Now, subtract the second equation from the first equation:
x + 4y - x - y = 144 - 48
3y = 96

Divide both sides of the equation by 3:
y = \frac{96}{3}
y = 32

Substitute the value of y into the second equation to find x:
x + 32 = 48
x = 48 - 32
x = 16

Step 5: Answer
Therefore, there were 16 kg of the 10% sulfuric acid solution and 32 kg of the 40% sulfuric acid solution in the original mixture.