AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 3 Medium

A container A holds 90\,\text{mL} of saline solution with a salt concentration of 10.5\%, and container B holds 210\,\text{mL} of saline solution with a salt concentration of 11.7\%. If the same amount of solution is taken from both containers and poured into the other container, then stirred evenly, the resulting salt concentrations in both containers become equal. How many milliliters of solution were poured out from each container?

  • A.

    55

  • B.

    57

  • C.

    59

  • D.

    63

  • E.

    67

Answer:D

After exchanging, the salt concentration becomes: \left(90 \times 10.5\% + 210 \times 11.7\%\right) \div (90 + 210) \times 100\% = 11.34\%

Volume of exchanged solution: \frac{90 \times 11.34\% - 90 \times 10.5\%}{11.7\% - 10.5\%} = 63\,\text{mL}

Each of containers A and B had 63\,\text{mL} of solution exchanged.

Related Problems
Problem 1 AMC 10 Weekly Practice Round 2
Problem 2 AMC 10 Weekly Practice Round 2
Problem 4 AMC 10 Weekly Practice Round 2
Problem 5 AMC 10 Weekly Practice Round 2
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