Multiple choice

There are three containers A, B and C that contain a mixture of water, acid and alcohol. All the three containers contain equal amount of mixture. The ratio of water, acid and alcohol is (2 : 2 : 1), (1 : 3 : 2) and (3 : 2 : 1) in containers A, B and C, respectively. If the contents of all the three containers are poured into a larger tank, then find the ratio of volume of water, acid and alcohol in this larger tank.

  1. 32 : 37 : 21

  2. 31 : 35 : 21

  3. 33 : 37 : 20

  4. 31 : 17 : 25

  5. None of these
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A Correct answer
Explanation

Equal amounts of mixture mean we can sum the ratios directly. A: 2:2:1 (total 5), B: 1:3:2 (total 6), C: 3:2:1 (total 6). Normalize to common denominator 30: A: 12:12:6, B: 5:15:10, C: 15:10:5. Sum: Water = 12+5+15 = 32, Acid = 12+15+10 = 37, Alcohol = 6+10+5 = 21. Ratio = 32:37:21.

AI explanation

Let the equal amount of mixture in each container be 60 units, which allows us to easily divide by their respective sum of ratios. Container A contains 24 units of water, 24 units of acid, and 12 units of alcohol. Container B contains 10 units of water, 30 units of acid, and 20 units of alcohol. Container C contains 30 units of water, 20 units of acid, and 10 units of alcohol. Adding these together, the tank contains (24 + 10 + 30) = 64 units of water, (24 + 30 + 20) = 74 units of acid, and (12 + 20 + 10) = 42 units of alcohol. Dividing the total values of 64, 74, and 42 by their greatest common divisor of 2 gives the final ratio of 32 to 37 to 21.

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