The diluted alcohol contains only 7 litres of alcohol and the rest is water. A new mixture whose concentration is 40%, is to be formed by replacing alcohol. How many litres of mixture shall be replaced with pure alcohol if there was initially 28 litres of water in the mixture?
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$\(\frac{42}{5}\)$
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$\(\frac{35}{4}\)$
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$\(\frac{47}{8}\)$
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$\(\frac{14}{3}\)$
B
Correct answer
Explanation
Original mixture: 7L alcohol + 28L water = 35L total (20% alcohol). Let x litres be replaced. Removed: 0.2x alcohol, 0.8x water. Added: x alcohol. Final alcohol = 7 - 0.2x + x = 7 + 0.8x. For 40% concentration: (7 + 0.8x)/35 = 0.4, giving x = 35/4 = 8.75 litres. The replacement maintains total volume while changing composition.