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Suppose 3 quarts of pure antifreeze is mixed with 5 quarts of a solution that is 14% antifreeze.Answer the questions below. Do not do any rounding.

Suppose 3 quarts of pure antifreeze is mixed with 5 quarts of a solution that is 14% antifreeze-example-1
User Hiddenbit
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The amount of antifreeze in the second volume is only 14% of it, which means it is equivalent to say that it is only 14% of 5 quarts that is being added to the final solution. From this, we perform the following calculation:


\begin{gathered} (3)/(4)+(5)/(4)*14\text{ \%} \\ \\ (3)/(4)+(5*(14)/(100))/(4)=(3+5*0.14)/(4)=(3+0.7)/(4)=(3.7)/(4) \end{gathered}

From the solution developed above, we are able to say that the number of quarts of antifreeze in the resulting mixture is 3.7

Now, about of the percentage of the resulting mixture is antifreeze, we divide the amount of antifreeze by the amount of total mixing and multiply by 100%, as follows:


((3.7)/(4))/((3)/(4)+(5)/(4))*100\text{ \%}=((3.7)/(4))/((8)/(4))*100\text{ \%}=(3.7)/(8)*100\text{ \%=46.25\%}

From the solution we developed above, we are able to conclude that the percentage of the resulting mixture that is antifreeze is equal to 46.25%

User Kasra
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