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A mixture contains alcohol, water, and glycerin in the ratio 4:3:2. If 18 liters of the mixture is replaced with 18 liters of alcohol, the ratio becomes 5:3:2. What was the original volume of the mixture?

GMAT · 2022 · Quantitative
Question:
A mixture contains alcohol, water, and glycerin in the ratio 4:3:2. If 18 liters of the mixture is replaced with 18 liters of alcohol, the ratio becomes 5:3:2. What was the original volume of the mixture?

Options

(A) 36 liters
(B) 45 liters
(C) 54 liters
(D) 63 liters
(E) 72 liters
Answer: (B) 45 liters

Explanation:
Let original volume = V. Original: 4k, 3k, 2k where 9k = V. When 18 liters removed: (4k)(V-18)/V alcohol removed, similarly for others. After adding 18L alcohol: 4k - 4k(18)/V + 18 alcohol. New ratio 5:3:2 with total V means: (4k - 72k/V + 18)/(5) = (3k - 54k/V)/(3) = (2k - 36k/V)/(2). From water and glycerin ratio unchanged: (3k - 54k/V)/(2k - 36k/V) = 3/2. Solving: 2(3k - 54k/V) = 3(2k - 36k/V), so 6k - 108k/V = 6k - 108k/V. This is always true, so try another approach. Total = 9k = V. After operation, proportions become 5:3:2 with same total V. So new amounts: 5V/10, 3V/10, 2V/10. For alcohol: 4V/9 - (4/9)(18) + 18 = 5V/10. Solving: 4V/9 - 8 + 18 = V/2, so 4V/9 + 10 = V/2. Multiply by 18: 8V + 180 = 9V, so V = 180. This does not match options. Let me reconsider the problem setup.

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