A milkman has an 80-liter mixture of milk and water which has 20% of water.…
2025
A milkman has an 80-liter mixture of milk and water which has 20% of water. The milkman sold some mixture and add 8 liters of water to the remaining mixture. Now the ratio of the milk and water is 12:5. Find the sold mixture.
Answer: A. 20 Liter — Solution: Initial amounts: total = 80 L. Water = 20% of 80 = 16 L. Milk = 80 - 16 = 64 L. Let x be the liters of mixture sold. The sold mixture has the same…
- A.
20 Liter
- B.
18 Liter
- C.
22 Liter
- D.
24 Liter
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Correct answer: A
Solution:
Initial amounts: total = 80 L. Water = 20% of 80 = 16 L. Milk = 80 - 16 = 64 L.
Let x be the liters of mixture sold. The sold mixture has the same proportions, so milk removed = 0.8x and water removed = 0.2x.
Remaining milk = 64 - 0.8x. Remaining water before adding = 16 - 0.2x. After adding 8 L water, new water = 16 - 0.2x + 8 = 24 - 0.2x.
Set the required ratio: (64 - 0.8x) / (24 - 0.2x) = 12 / 5. Cross-multiply and solve:
5(64 - 0.8x) = 12(24 - 0.2x) ⇒ 320 - 4x = 288 - 2.4x
Rearrange: 320 - 288 = 4x - 2.4x ⇒ 32 = 1.6x ⇒ x = 20.
Answer: The milkman sold 20 liters of the mixture.
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