Question

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?

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Jett

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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.

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.

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