Question

How much pure antifreeze should be added to 3 gallons of a 30 percent antifreeze solution to get 65 percent antifreeze?

Answer 1

Hint: Here, we will assume the number of required gallons 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 number of gallons of pure antifreeze to be added to 3 gallons of a 30 percent antifreeze solution be $x$
Therefore, the new volume after mixing will be: $\left( {x + 3} \right)$
According to the question, in the new mixture, we have 65 percent antifreeze solution.
Thus, our antifreeze balance becomes:
$\left( {100\% } \right)x + \left( {30\% } \right)\left( 3 \right) = \left( {65\% } \right)\left( {x + 3} \right)$
This is because, when the solution is pure, it means that its percentage of antifreeze solution is $100\% $
Solving this further, we get,
$ \Rightarrow x + \left( {\dfrac{{90}}{{100}}} \right) = \dfrac{{65}}{{100}}\left( {x + 3} \right)$
$ \Rightarrow x + 0.9 = 0.65x + 1.95$
Or, we get,
$ \Rightarrow x - 0.65x = 1.95 - 0.9$
Solving this further,
$ \Rightarrow 0.35x = 1.05$
Dividing both sides by $0.35$
$ \Rightarrow x = \dfrac{{1.05}}{{0.35}} = \dfrac{{105}}{{35}} = 3$
Therefore, we get, $x = 3$

Hence, 3 gallons of pure antifreeze should be added to 3 gallons of a 30 percent antifreeze solution to get 65 percent antifreeze.
Thus, this is the required answer.

Note:
In this question, we have framed a linear equation. A linear equation is defined as an equation that has a highest degree of 1 and has one solution. As here percent of 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.