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How much water must be added to 12 L (litres) of a $40\% $ solution of alcohol to obtain a $30\% $ solution?

Last updated date: 13th Jun 2024
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Hint: Here, we will assume the required quantity of solution to be some variable. We will then frame a linear equation based on the given information. We will then solve the equation using the basic mathematical operation to get the required answer.

Complete step-by-step answer:
Let the volume of water to be added be $x$ litres.
Now, according to the question,
In a 12 L (litres) solution, we have $40\% $ solution of alcohol.
Hence, the amount of alcohol in the given solution is $\dfrac{{40}}{{100}} \times 12 = \dfrac{{48}}{{10}} = 4.8$ litres
Now, when $x$ litres of water is added to 12 litres of a solution, the total volume obtained is $\left( {12 + x} \right)$ litres.
But, according to the question, it is given that this amount of alcohol in the new solution will be $30\% $ of the new solution.
Thus, we get,
$\dfrac{{30}}{{100}} \times \left( {12 + x} \right) = 4.8$
$ \Rightarrow \dfrac{{360}}{{100}} + \dfrac{{30}}{{100}}x = 4.8$
Converting this to decimals,
$ \Rightarrow 3.6 + 0.3x = 4.8$
Subtracting $3.6$ from both sides, we get,
$ \Rightarrow 0.3x = 1.2$
Dividing both sides by $0.3$, we get
$ \Rightarrow x = \dfrac{{1.2}}{{0.3}} = \dfrac{{12}}{3} = 4$
Hence, the value of $x = 4$

Therefore, 4 litres of water must be added to 12 L (litres) of a $40\% $ solution of alcohol to obtain a $30\% $ solution.
Thus, this is the required answer.

In this question, we have framed a linear equation. A linear equation is defined as an equation that has the highest degree of 1 and has one solution. As here percent of the solution is given so we had to first convert it to the fraction then simplify the equation. These types of problems are known as mixture problems as they have a mixture of two or more things and we are required to find some quantity, percentage, price, etc. of the resulting mixture. Thus, allegations and mixtures is the required method to solve these questions.