A container holds a mixture of milk and water in the ratio 3:2. When 75% of…

2023

A container holds a mixture of milk and water in the ratio 3:2. When 75% of the mixture is removed and 40 litres of another mixture (in which 40% is milk) is added, the milk in the resulting mixture becomes 44 4/9 % of the total. Find the quantity of milk in the initial mixture.

  1. A.

    196/7 liters

  2. B.

    None of these

  3. C.

    160/7 liters

  4. D.

    192/7 liters

  5. E.

    320/7 liters

Attempted by 1 students.

Show answer & explanation

Correct answer: D

Concept

Removing a fixed fraction of a uniform mixture does not change its internal ratio: if a fraction f is removed, the same fraction (1 - f) of EACH component remains. After that, any added quantity changes the totals, and the new concentration is found by the alligation/percentage relation: (milk after)/(total after) = required fraction.

Application

  1. Let the initial total mixture be T litres. With milk : water = 3 : 2, initial milk = (3/5)T and initial water = (2/5)T.

  2. 75% is removed, so 25% remains. The ratio is unchanged, so remaining milk = 0.25 x (3/5)T = 0.15T and remaining total = 0.25T.

  3. Add 40 litres that is 40% milk: this adds 0.40 x 40 = 16 litres of milk and increases the total by 40 litres.

  4. Resultant milk = 0.15T + 16 ; resultant total = 0.25T + 40.

  5. Set the milk fraction equal to 44 4/9 % = 400/9 % = 4/9 : (0.15T + 16)/(0.25T + 40) = 4/9.

  6. Cross-multiply: 9(0.15T + 16) = 4(0.25T + 40) => 1.35T + 144 = T + 160 => 0.35T = 16 => T = 320/7 litres.

  7. Initial milk = (3/5)T = (3/5)(320/7) = 192/7 litres.

Cross-check

With T = 320/7: remaining milk = 0.15 x 320/7 = 48/7, plus 16 = 160/7. Resultant total = 0.25 x 320/7 + 40 = 80/7 + 280/7 = 360/7. Ratio = (160/7)/(360/7) = 160/360 = 4/9 = 44 4/9 %, which matches. So the initial milk is 192/7 litres.

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