InterviewSolution
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Quantity B: A bar tender stole beer from a bottle that contained 40% of the spirit and he replaced what he had stolen with beer having 20% spirit. The bottle then contained only 35% spirit. How much percentage of the bottle content did he stole? 1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B |
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Answer» Quantity A: ⇒ Water in the MIXTURE = 20% of 50 = 50 × 0.2 = 10 litres ⇒ MILK in the mixture = 50 – 10 = 40 litres ⇒ Water in 10 litres of the mixture = 20% of 10 = 10 × 0.2 = 2 litres ⇒ Milk in 10 litres of the mixture = 10 – 2 = 8 litres After adding 10 litre of water, ⇒ Water in new mixture = 10 – 2 + 10 = 18 litres ⇒ Milk in the new mixture = 40 – 8 = 32 litres ⇒ Percentage of water in the new mixture = {18/(18 + 32)} × 100 = (18/50) × 100 = 36% Quantity B: LET the initial quantity of mixture is x litres and y litre is replaced; According to the question; 0.4x + 0.2y = 0.35x ⇒ 0.2y = -0.05x ⇒ -4y = x ⇒ y/x = -¼ ∴ Percentage of quantity replaced = -¼ = -25%(minus sign implies removal) ∴ Quantity A > Quantity B |
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