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How many gallons each of 
15% alcohol and 
10% alcohol should be mixed to obtain 5 gal of 
11% alcohol?







Gallons of


Solution







Percent


(as a decimal)







Gallons of


Pure


Alcohol







x

15%=0.15




y

10%=0.1



5

11%=





How many gallons of 
15% alcohol should be in the mixture? 
◻ gal

How many gallons each of 15% 15 \% alcohol and 10% 10 \% alcohol should be mixed to obtain 55 gal of 11% 11 \% alcohol?\newline\begin{tabular}{|c|c|c|}\newline\hline \begin{tabular}{c} \newlineGallons of \\\newlineSolution\newline\end{tabular} & \begin{tabular}{c} \newlinePercent \\\newline(as a decimal)\newline\end{tabular} & \begin{tabular}{c} \newlineGallons of \\\newlinePure \\\newlineAlcohol\newline\end{tabular} \\\newline\hlinex x & 15%=0.15 15 \%=0.15 & \\\newline\hliney y & 10%=0.1 10 \%=0.1 & \\\newline\hline 55 & 11%= 11 \%= & \\\newline\hline\newline\end{tabular}\newlineHow many gallons of 15% 15 \% alcohol should be in the mixture? \square gal

Full solution

Q. How many gallons each of 15% 15 \% alcohol and 10% 10 \% alcohol should be mixed to obtain 55 gal of 11% 11 \% alcohol?\newline\begin{tabular}{|c|c|c|}\newline\hline \begin{tabular}{c} \newlineGallons of \\\newlineSolution\newline\end{tabular} & \begin{tabular}{c} \newlinePercent \\\newline(as a decimal)\newline\end{tabular} & \begin{tabular}{c} \newlineGallons of \\\newlinePure \\\newlineAlcohol\newline\end{tabular} \\\newline\hlinex x & 15%=0.15 15 \%=0.15 & \\\newline\hliney y & 10%=0.1 10 \%=0.1 & \\\newline\hline 55 & 11%= 11 \%= & \\\newline\hline\newline\end{tabular}\newlineHow many gallons of 15% 15 \% alcohol should be in the mixture? \square gal
  1. Define Variables: Let xx be the gallons of 15%15\% alcohol and yy be the gallons of 10%10\% alcohol. We know that x+y=5x + y = 5 because we want 55 gallons of the final mixture.
  2. Calculate Total Alcohol: We also know that the amount of pure alcohol in the 1515\% solution is 0.15x0.15x and in the 1010\% solution is 0.1y0.1y. The final mixture has 1111\% alcohol, so the total pure alcohol in the final mixture is 0.11×50.11 \times 5.
  3. Set Up Equation: Now we can set up the equation for the total pure alcohol: 0.15x+0.1y=0.11×50.15x + 0.1y = 0.11 \times 5.
  4. Substitute and Simplify: Substitute yy with 5x5 - x (from the first equation) into the alcohol equation: 0.15x+0.1(5x)=0.550.15x + 0.1(5 - x) = 0.55.
  5. Combine Like Terms: Distribute the 0.10.1 into the parentheses: 0.15x+0.50.1x=0.550.15x + 0.5 - 0.1x = 0.55.
  6. Isolate Variable: Combine like terms: 0.05x+0.5=0.550.05x + 0.5 = 0.55.
  7. Solve for x: Subtract 0.50.5 from both sides: 0.05x=0.050.05x = 0.05.
  8. Solve for x: Subtract 0.50.5 from both sides: 0.05x=0.050.05x = 0.05. Divide both sides by 0.050.05 to solve for x: x=0.050.05x = \frac{0.05}{0.05}.

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