Question:

A mixture of 40 L of alcohol and water contains 10% water. How much water should be added to this mixture so that the new mixture contains 20% water?

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Always express concentration problems as \(\frac{\text{desired quantity}}{\text{total quantity}}\).
Updated On: Aug 11, 2025
  • 9 L
  • 5 L
  • 7 L
  • 6 L
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The Correct Option is B

Solution and Explanation

Initial water = \( 10% \) of 40 = \( 4 \ \text{L} \). Let \( x \) be the amount of water added. New total mixture = \( 40 + x \) litres. New water content = \( 4 + x \) litres. Given: \[ \frac{4 + x}{40 + x} = 0.2 \] \[ 4 + x = 0.2(40 + x) \Rightarrow 4 + x = 8 + 0.2x \] \[ x - 0.2x = 8 - 4 \Rightarrow 0.8x = 4 \] \[ x = 5 \ \text{L} \]
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