Solve the following systems of inequalities for all `xinR`
A milkman has 80% milk in this stock of 800 litres of adultered milk. How much 100% pure milk is to be added to it so that purity is between 90% and 95%?

Answer

To solve the problem step by step, we need to find out how much 100% pure milk should be added to the existing 800 liters of 80% adulterated milk so that the final mixture has a purity between 90% and 95%.

### Step 1: Calculate the amount of pure milk in the existing solution
The existing solution is 800 liters of 80% milk.

\[ \text{Amount of pure milk} = 80\% \text{ of } 800 = \frac{80}{100} \times 800 = 640 \text{ liters} \]

**Hint:** To find the amount of pure milk, multiply the total volume by the percentage of pure milk.

### Step 2: Let \( x \) be the amount of 100% pure milk to be added
We will add \( x \) liters of pure milk to the existing 640 liters of pure milk.

### Step 3: Calculate the total volume of the new mixture
The total volume of the new mixture will be:

\[ \text{Total volume} = 800 + x \text{ liters} \]

### Step 4: Set up the inequality for purity
We want the purity of the new mixture to be between 90% and 95%.

The percentage of pure milk in the new mixture can be expressed as:

\[ \text{Purity} = \frac{\text{Amount of pure milk}}{\text{Total volume}} \times 100 = \frac{640 + x}{800 + x} \times 100 \]

We need to set up the inequalities:

\[ 90 < \frac{640 + x}{800 + x} \times 100 < 95 \]

### Step 5: Solve the left inequality
First, we solve the left part of the inequality:

\[ 90 < \frac{640 + x}{800 + x} \times 100 \]

Dividing both sides by 100:

\[ 0.9 < \frac{640 + x}{800 + x} \]

Cross-multiplying gives:

\[ 0.9(800 + x) < 640 + x \]

Expanding and simplifying:

\[ 720 + 0.9x < 640 + x \]

Rearranging terms:

\[ 720 - 640 < x - 0.9x \]

\[ 80 < 0.1x \]

Dividing by 0.1:

\[ 800 < x \]

### Step 6: Solve the right inequality
Now we solve the right part of the inequality:

\[ \frac{640 + x}{800 + x} \times 100 < 95 \]

Dividing both sides by 100:

\[ \frac{640 + x}{800 + x} < 0.95 \]

Cross-multiplying gives:

\[ 640 + x < 0.95(800 + x) \]

Expanding and simplifying:

\[ 640 + x < 760 + 0.95x \]

Rearranging terms:

\[ 640 - 760 < 0.95x - x \]

\[ -120 < -0.05x \]

Dividing by -0.05 (remember to reverse the inequality):

\[ 2400 > x \]

### Step 7: Combine the results
From the two inequalities, we have:

\[ 800 < x < 2400 \]

### Conclusion
The amount of 100% pure milk that should be added to the existing solution should be between 800 liters and 2400 liters.

**Final Answer:** \( x \in (800, 2400) \)

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