Math Problem Statement

The can you are holding contains a combination of two different liquids. Seven parts water to 5 parts milk makes up this proportion. The proportion of milk to water in the container is changed to 7:9 when 9 liters of the mixture are removed, and it is then filled with water. How many liters of milk were initially contained inside of the can when it was first opened?

Solution

We are given a problem involving a mixture of water and milk. Initially, the ratio of water to milk is 7:57:5, and after some operations, the proportion of milk to water changes to 7:97:9. Let's find out how much milk was in the container initially.

Step-by-Step Solution

  1. Initial ratio and total amount: The initial ratio of water to milk is 7:57:5. Let the total amount of liquid in the can be xx liters. Thus, the initial amount of milk and water can be expressed as: Amount of water=712x(since water makes up 7 parts of the total 12 parts)\text{Amount of water} = \frac{7}{12}x \quad \text{(since water makes up 7 parts of the total 12 parts)} Amount of milk=512x(since milk makes up 5 parts of the total 12 parts)\text{Amount of milk} = \frac{5}{12}x \quad \text{(since milk makes up 5 parts of the total 12 parts)}

  2. Removal of 9 liters of mixture: When 9 liters of the mixture are removed, it means that the same ratio 7:57:5 is preserved in the removed mixture. Therefore, in the 9 liters removed: Amount of water removed=712×9=5.25 liters\text{Amount of water removed} = \frac{7}{12} \times 9 = 5.25 \text{ liters} Amount of milk removed=512×9=3.75 liters\text{Amount of milk removed} = \frac{5}{12} \times 9 = 3.75 \text{ liters}

  3. Remaining amounts of water and milk: After removing the 9 liters, the remaining amounts of water and milk are: Remaining water=712x5.25\text{Remaining water} = \frac{7}{12}x - 5.25 Remaining milk=512x3.75\text{Remaining milk} = \frac{5}{12}x - 3.75

  4. Filling with water: The 9 liters removed are replaced entirely with water. So, the new amount of water becomes: New amount of water=712x5.25+9=712x+3.75\text{New amount of water} = \frac{7}{12}x - 5.25 + 9 = \frac{7}{12}x + 3.75

  5. New ratio of milk to water: After this operation, the ratio of milk to water becomes 7:97:9. Therefore, we can set up the following equation using the new amounts of milk and water: 512x3.75712x+3.75=79\frac{\frac{5}{12}x - 3.75}{\frac{7}{12}x + 3.75} = \frac{7}{9}

  6. Solving the equation: Cross-multiply to get: 9(512x3.75)=7(712x+3.75)9\left(\frac{5}{12}x - 3.75\right) = 7\left(\frac{7}{12}x + 3.75\right) Expanding both sides: 9×512x9×3.75=7×712x+7×3.759 \times \frac{5}{12}x - 9 \times 3.75 = 7 \times \frac{7}{12}x + 7 \times 3.75 Simplify: 4512x33.75=4912x+26.25\frac{45}{12}x - 33.75 = \frac{49}{12}x + 26.25 Multiply through by 12 to eliminate the fractions: 45x405=49x+31545x - 405 = 49x + 315 Rearranging the terms: 45x49x=315+40545x - 49x = 315 + 405 4x=720-4x = 720 x=7204=180x = \frac{720}{-4} = -180

  7. Check for logical inconsistency: Upon reviewing, I made an algebraic mistake; however, correcting it

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Math Problem Analysis

Mathematical Concepts

Ratios
Algebra
Mixtures
Linear Equations

Formulas

Amount of component in mixture = (ratio part / total parts) * total volume
Cross multiplication for ratios
Linear equation manipulation

Theorems

Ratio and Proportion
Cross Multiplication

Suitable Grade Level

Grades 9-11