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, twothirds 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?

• A. 80 grams
• B. 60 grams
• C. 40 grams
• D. 100 grams

Answer 1

The Correct Option is C

Let's use algebra to solve this problem:
Let "x" be the initial quantity of water in the second test tube. 
Initially, the first test tube contains some acid and the second test tube contains "x" grams of water. 
After pouring 20 grams of acid into the second test tube, the second test tube contains "x + 20" grams of a solution consisting of both water and acid. 
Now, two-thirds of this solution is poured from the second test tube into the first test tube. 
So, the first test tube contains \((\frac{2}{3})\)(x + 20) grams of the solution. 
According to the problem, the fluid in the first test tube is four times that in the second. 
Therefore, we can set up the equation: 
\((\frac{2}{3})\)(x + 20) = 4x 
Now, let's solve for "x": 
Multiply both sides of the equation by 3 to get rid of the fraction: 
2(x + 20) = 12x 
Distribute on the left side: 
2x + 40 = 12x 
Subtract 2x from both sides: 
40 = 10x 
Now, divide by 10: 
x = 40 
So, the initial quantity of water in the second test tube was 40 grams.
The correct option is (C): 40 grams