This problem involves calculating the amount of pure milk to add to a mixture to decrease the water percentage.
Initial Mixture Analysis
The initial milk sample has a total volume of 10 L.
- Water percentage = 5%
- Amount of water = $5\%$ of 10 L $= \frac{5}{100} \times 10 = 0.5$ L
- Amount of pure milk = Total Volume - Water Amount $= 10 - 0.5 = 9.5$ L
Adding Pure Milk
Let $x$ represent the quantity (in Liters) of pure milk added to the sample.
- The amount of water remains constant at 0.5 L.
- The amount of pure milk increases to $9.5 + x$ L.
- The new total volume becomes $10 + x$ L.
- The target water percentage is 2%.
Calculating Added Milk Quantity
The concentration of water in the new mixture is expressed as:
$ \text{Water Percentage} = \frac{\text{Amount of Water}}{\text{New Total Volume}} \times 100 $We set this to the desired 2%:
$ \frac{0.5}{10 + x} = \frac{2}{100} $Now, we solve for $x$:
- Simplify the fraction: $\frac{0.5}{10 + x} = 0.02$
- Cross-multiply: $0.5 = 0.02 \times (10 + x)$
- Distribute: $0.5 = 0.2 + 0.02x$
- Isolate the term with $x$: $0.5 - 0.2 = 0.02x \implies 0.3 = 0.02x$
- Solve for $x$: $x = \frac{0.3}{0.02} = \frac{30}{2} = 15$ L
Therefore, 15 L of pure milk must be added.
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If a + b + c = 0, then the value of (a² + b² + 2ab) is equal to:
