Three bottles of equal capacity contain milk and water mixtures:\n• Bottle 1 has milk : water = 5 : 7\n• Bottle 2 has milk : water = 7 : 9\n• Bottle 3 has milk : water = 2 : 1\nAll three bottles are emptied into one large container.\nWhat is the percentage of milk in the final combined mixture?

Introduction:
This question tests combining mixtures of equal volumes. Since all bottles have equal capacity, the final milk fraction is simply the average of the milk fractions of each bottle. The correct method is to convert each milk:water ratio to a milk fraction and then average them because volumes are equal.

Given Data / Assumptions:

• All three bottles have equal capacity (same volume)
• Bottle 1 milk fraction = 5/(5+7)
• Bottle 2 milk fraction = 7/(7+9)
• Bottle 3 milk fraction = 2/(2+1)

Concept / Approach:
If each bottle has volume V, total milk = V*(f1 + f2 + f3) and total volume = 3V.\nSo final milk fraction = (f1 + f2 + f3) / 3.\nConvert fraction to percentage by multiplying by 100.

Step-by-Step Solution:
Bottle 1 milk fraction = 5/12Bottle 2 milk fraction = 7/16Bottle 3 milk fraction = 2/3Sum of fractions = 5/12 + 7/16 + 2/3Common denominator 48: 5/12=20/48, 7/16=21/48, 2/3=32/48Sum = (20 + 21 + 32)/48 = 73/48Average milk fraction = (73/48) / 3 = 73/144Milk percentage = (73/144) * 100 ≈ 50.694...% ≈ 50.7%

Verification / Alternative Check:
Assume each bottle is 144 ml for easy calculation.\nMilk from bottle 1 = 144*(5/12)=60.\nMilk from bottle 2 = 144*(7/16)=63.\nMilk from bottle 3 = 144*(2/3)=96.\nTotal milk = 219 out of total 432.\nMilk% = 219/432 * 100 ≈ 50.7%, confirming the result.

Why Other Options Are Wrong:
49.6%, 48.9%: too low, usually from averaging ratios directly.51.2% or 52.3%: too high, often from giving extra weight to bottle 3 without justification.

Common Pitfalls:
Averaging 5:7, 7:9, and 2:1 as ratios instead of converting to fractions.Forgetting that equal capacity allows simple averaging of fractions.Rounding too early and losing accuracy.

Final Answer:
50.7%