A milk vendor has two cans of milk. The first contains 25% water and the rest…

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A milk vendor has two cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 L of milk such that the ratio of water to milk is 3:5 ?

  1. A.

    3 L

  2. B.

    6 L

  3. C.

    7 L

  4. D.

    8 L

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Correct answer: B

Answer: Take 6 L from the first container and 6 L from the second container.

Solution steps:

  1. Let x be the litres taken from the first container (which is 25% water, 75% milk) and y be the litres taken from the second container (50% water, 50% milk).

  2. Total volume: x + y = 12.

  3. Required water:milk = 3:5, so milk fraction = 5/8 and required milk amount = (5/8) × 12 = 7.5 L.

  4. Milk contributed = (3/4)x + (1/2)y. Set (3/4)x + (1/2)y = 7.5 and use x + y = 12.

  5. Substitute y = 12 − x: (3/4)x + (1/2)(12 − x) = 7.5 ⇒ (3/4)x + 6 − (1/2)x = 7.5 ⇒ (1/4)x = 1.5 ⇒ x = 6.

  6. Then y = 12 − 6 = 6. So take 6 L from each container.

Verification: Water from first = 25% of 6 = 1.5 L; water from second = 50% of 6 = 3 L; total water = 4.5 L. Milk = 12 − 4.5 = 7.5 L. Ratio water:milk = 4.5:7.5 = 3:5, as required.

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