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Let initial Quantity = x litres
Milk percentage initialy = (3/4)*100 = 75%
Milk Quantity initially = 0.75x

Milk removed in 20 litres = 75% of 20 = 15 litres
Milk left in original solution = 0.75x - 15

Water addition = 12 litres

i.e Final Total quantity = x - 20 + 12 = x-8
Milk in final quantity = (2/3)*(x-8)

i.e. 0.75x - 15 = (2/3)*(x-8)
i.e. x = 116

Answer: Option E

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A container has a mixture of milk and water in the ratio 3:1. After removing 20 liters of the mixture and adding 12 liters of water, the new ratio of milk to water becomes 2:1. What is the total volume of the original mixture?

A. 36
B. 54
C. 72
D. 92
E. 116
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Original Ratio ( milk: water ): 3:1
Remove 20L: Remove 15L milk, 5L water (maintaining 3:1)
Add 12L water: Only water increases
Check: Does final milk:water = 2:1?

Use Table and start with C(72)

Original Mixture Milk water After Removal(20l of mixture) After Addition(12l water) Final Ratio Answer
725418M:39, W:13 M:39, W:25 39:25Wrong
92 69 23 M:54, W:18 M:54, W:30 27:15Wrong
116 87 29 M:72, W:24 M:72, W:36 2:1 Correct


Time-Saving Strategy:
After testing C and getting ~ (1.5:1) as final ratio, we know we need larger volumes to reach 2:1. This eliminates A, B,C immediately and focuses testing on D, E only.

Answer: E (116)
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I have posted the solution in the reserved post.

Mitalilodaya24
Explain how to solve this question
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V1C1 - milk removed = V2C2

(3/4)x - 20 * (3/4) = 2/3 * (x-8)
9x - 20*9 = 8(x-8)
9x - 18 = 8x - 64
x = 116
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