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A car radiator needs a 60% antifreeze solution. The radiator now holds 6 liters of a 40% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength? liters

User KBP
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Final answer:

To obtain a 60% antifreeze solution from a 40% solution, 2 liters of the original mixture should be drained and replaced with 100% antifreeze.

Step-by-step explanation:

How to Make a 60% Antifreeze Solution in a Car Radiator

The problem given is that a car radiator currently contains 6 liters of a 40% antifreeze solution. To achieve a 60% antifreeze solution, an amount of the current solution needs to be drained and replaced with 100% antifreeze. Here's a step-by-step guide to calculate how much of the 40% solution should be drained:

  1. Determine the amount of pure antifreeze in the original solution: $0.40 \times 6 \text{L} = 2.4 \text{L}$.
  2. Let $x$ be the amount of the solution to be replaced with 100% antifreeze.
  3. The volume of the new solution after replacement will still be 6 liters since we are draining and replacing.
  4. The amount of pure antifreeze after replacement will be $2.4 \text{L} - 0.40x \text{L} + x \text{L}$ (since $x$ liters of 100% antifreeze are added).
  5. We want the new solution to be 60% antifreeze: $\frac{2.4 \text{L} - 0.40x \text{L} + x \text{L}}{6 \text{L}} = 0.60$.
  6. Solve the equation for $x$: $2.4 \text{L} + 0.60x \text{L} = 3.6 \text{L}$.
  7. Rearrange and solve for $x$: $0.60x \text{L} = 1.2 \text{L}$; hence, $x = 2 \text{L}$.
  8. Therefore, 2 liters of the 40% solution should be drained and replaced with 100% antifreeze to achieve the desired 60% solution.

The calculation above assumes that mixing the solutions will not alter the total volume significantly, which is typical for most liquid solutions unless they have significant volume contraction upon mixing.

User Mbm
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