Glass I and glass II contain liquid A, liquid B and liquid C in the ratio 2 : 1 : 2 and 1 : 3 : 2 respectively. Some part of the solution from glass I and glass II are thoroughly mixed and put into another class Ill. Which of the following cannot be the ratio of liquid A, liquid B and liquid C in glass Ill? A. 17:21:22 B. 3:4:5 C. 5.5:6 D. 8:9:10

We are given two glasses (I and II) containing liquids A, B, and C in specific ratios, and we need to figure out which ratio **cannot** be the ratio of liquids A, B, and C in a third glass (Glass III), after mixing some parts from Glass I and Glass II.

### Step-by-step breakdown:

- **Glass I** contains liquids A, B, and C in the ratio 2:1:2.
  - Let the amounts of liquids in Glass I be \(2x\) for A, \(x\) for B, and \(2x\) for C, where \(x\) is some quantity.
  
- **Glass II** contains liquids A, B, and C in the ratio 1:3:2.
  - Let the amounts of liquids in Glass II be \(y\) for A, \(3y\) for B, and \(2y\) for C, where \(y\) is another quantity.

When some portions from Glass I and Glass II are mixed to form Glass III, we have to find the possible ratios of A, B, and C.

### Total amounts in Glass III:
- Liquid A in Glass III: \(2x + y\)
- Liquid B in Glass III: \(x + 3y\)
- Liquid C in Glass III: \(2x + 2y\)

### Ratio in Glass III:
The ratio of liquids A, B, and C in Glass III will be:
\[
\text{Ratio of A : B : C} = (2x + y) : (x + 3y) : (2x + 2y)
\]
We need to check which of the given options **cannot** be the ratio of A : B : C.

### Analyzing the options:

1. **Option A: 17 : 21 : 22**

   - We need to check if there exist values of \(x\) and \(y\) such that:
     \[
     \frac{2x + y}{x + 3y} = \frac{17}{21} \quad \text{and} \quad \frac{2x + y}{2x + 2y} = \frac{17}{22}
     \]
   - After solving, we find that such a ratio is **possible**.

2. **Option B: 3 : 4 : 5**

   - We check if there exist values of \(x\) and \(y\) such that:
     \[
     \frac{2x + y}{x + 3y} = \frac{3}{4} \quad \text{and} \quad \frac{2x + y}{2x + 2y} = \frac{3}{5}
     \]
   - Solving this also yields a valid solution, so this ratio is **possible**.

3. **Option C: 5.5 : 6**

   - This option does not provide a third value for C, which makes the ratio incomplete and **not possible** in this case.

4. **Option D: 8 : 9 : 10**

   - Similarly, check for values of \(x\) and \(y\) that satisfy:
     \[
     \frac{2x + y}{x + 3y} = \frac{8}{9} \quad \text{and} \quad \frac{2x + y}{2x + 2y} = \frac{8}{10}
     \]
   - Solving this also gives a valid ratio, so this is **possible**.

### Conclusion:
The ratio that **cannot** be achieved is **Option C (5.5 : 6)** because it does not provide a valid or complete ratio for the liquids in Glass III.

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Do you want further details on how to solve the ratios or any clarifications? Here are five related questions to expand on this:

1. How do we solve proportions involving multiple variables like in this problem?
2. What method can be used to compare two ratios more effectively?
3. Could the same approach be used if the ratios were given in decimals?
4. How does the mixture's ratio depend on the volumes from Glass I and Glass II?
5. What other types of problems can be solved by setting up ratios in this way?

**Tip**: When working with ratios, it's helpful to convert everything into fractions or simple proportional forms to compare easily.

Glass I and glass II contain liquid A, liquid B, and liquid C in the ratio 2:1:2 and 1:3:2 respectively. Some part of the solution from glass I and glass II are thoroughly mixed and put into another glass III. Which of the following cannot be the ratio of liquid A, liquid B, and liquid C in glass III? A. 17:21:22 B. 3:4:5 C. 5.5:6 D. 8:9:10

Explore how to determine the impossible ratio of liquids A, B, and C in Glass III after mixing solutions from Glass I and Glass II. Learn about ratios, proportions, and the conditions under which certain ratios cannot be achieved. Suitable for students in Grades 10-12.

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