Question

Two liquids, A and B, are mixed together. A is 5.00% solids, the rest water. B is 11.0% solids, the rest water. To the mixture is added 18.5 kg of bone-dry solids resulting in a mixture of 2,150 kg containing 8.25% solids. What was the original amount of the liquid A?

Answer 1

To find the original amount of liquid A, we can set up an equation based on the amounts of solids in A and B, and solve for x.

Step-by-step explanation:

Let's assume the original amount of liquid A is x kg.

Since A is 5.00% solids, the rest of it is water. Therefore, the amount of solids in A is 0.05x kg and the amount of water in A is (1 - 0.05)x kg.

Similarly, B is 11.0% solids, so the amount of solids in B is 0.11(2150 - x) kg and the amount of water in B is (1 - 0.11)(2150 - x) kg.

When the two liquids are mixed together, the total amount of solids in the mixture is 0.05x + 0.11(2150 - x) kg, which is equal to 18.5 kg.

This can be represented in an equation as:

0.05x + 0.11(2150 - x) = 18.5

Solving this equation, we can find the value of x, which represents the original amount of liquid A.