Question:
On test tube contains some acid and another test tube contains an equal quantity of water. To prepare a solution, 20 grams of the acid is poured into the second test tube. Then, two thirds of the so-formed solution is poured from the second tube into the first. If the fluid in the first test tube is four times that in the second, what quantity of water was taken initially?
On test tube contains some acid and another test tube contains an equal quantity of water. To prepare a solution, 20 grams of the acid is poured into the second test tube. Then, two thirds of the so-formed solution is poured from the second tube into the first. If the fluid in the first test tube is four times that in the second, what quantity of water was taken initially?
Updated On: Sep 25, 2025
- 80 grams
- 60 grams
- 40 grams
- 100 grams
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The Correct Option is C
Solution and Explanation
Let's assume the initial quantity of water and acid = \(X\) grams.
Second test tube = \(X + 20\) grams (20 g Acid)
= \(\frac{2}{3} × (X + 20) = \frac{2}{3}X + \frac{40}{3}\)
= \(X + \frac{2}{3}X + \frac{40}{3} = 4 × \frac{1}{3}(X + 20)\)
= \(X + \frac{2}{3}X + \frac{40}{3} = \frac{4}{3}(X + 20)\)
On multiplying we get,
= \(3X + 2X + 40 = 4(X + 20)\)
= \(5X + 40 = 4X + 80\)
= \(X = 80 - 40\)
\(X = 40\) grams
The correct option is (C): 40 grams
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