Math Problem Statement

hey, so i have this mixture, which consists of 59% substance A, 35,7% B and 5,3% C

It was mixed from two other mixtures. one of the other mixtures being 57,5% substance A, 38,9% B and 3,6% C

What were the percentages of the other mixture the initial sample was made from?

Solution

To solve this problem, we can use the concept of weighted averages for mixtures. The idea is that the final mixture is a result of combining two different mixtures with known and unknown compositions. Let's call the first mixture (with known percentages) "Mixture 1" and the second mixture (with unknown percentages) "Mixture 2."

Let:

  • xx be the fraction of the final mixture that comes from Mixture 1.
  • 1x1 - x be the fraction that comes from Mixture 2.

We can now set up a system of equations based on the fact that the final percentages of each substance are weighted averages of the percentages from Mixture 1 and Mixture 2.

For substance A: 59%=x57.5%+(1x)A259\% = x \cdot 57.5\% + (1 - x) \cdot A_2 For substance B: 35.7%=x38.9%+(1x)B235.7\% = x \cdot 38.9\% + (1 - x) \cdot B_2 For substance C: 5.3%=x3.6%+(1x)C25.3\% = x \cdot 3.6\% + (1 - x) \cdot C_2

Here, A2A_2, B2B_2, and C2C_2 are the unknown percentages of substances A, B, and C in Mixture 2, which we need to find.

To solve this, we would first need the value of xx or more information about the ratio of the two mixtures. Could you clarify or provide this missing detail, or should we proceed assuming equal parts from both mixtures?

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Math Problem Analysis

Mathematical Concepts

Weighted Averages
Systems of Linear Equations

Formulas

Final percentage = x * Percentage_Mixture1 + (1 - x) * Percentage_Mixture2
59% = x * 57.5% + (1 - x) * A_2
35.7% = x * 38.9% + (1 - x) * B_2
5.3% = x * 3.6% + (1 - x) * C_2

Theorems

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Suitable Grade Level

Undergraduate level or advanced high school (Grades 11-12)