Concentrations 82563

In the hospital laboratory, two containers contained different concentrations of alcohol. After mixing 10 liters of alcohol from the first container with 14 liters from the second container, 75% of the alcohol was formed. If 40 liters from the first container were mixed with 8 liters from the second container, 70% alcohol would be formed. What percentage of alcohol was in the containers?

Correct answer:

a =  68 %
b =  80 %

Step-by-step explanation:

$$\begin{gathered} r_1=75\%=\frac{75}{100}=\frac{3}{4}=0.75 \\ {r}_2=70\%=\frac{70}{100}=\frac{7}{10}=0.7 \\ 10\cdot a/100+14b/100=75/100(10+14) \\ 40\cdot a/100+8b/100=70/100\cdot (40+8) \\ 10\cdot a+14b=75(10+14) \\ 40\cdot a+8b=70\cdot (40+8) \\ 10\cdot a+14\cdot b=75\cdot (10+14) \\ 40\cdot a+8\cdot b=70\cdot (40+8) \\ 10a+14b=1800 \\ 40a+8b=3360 \\ Pivot:Row1\leftrightarrow Row2 \\ 40a+8b=3360 \\ 10a+14b=1800 \\ Row2-\frac{10}{40}\cdot Row1\to Row2 \\ 40a+8b=3360 \\ 12b=960 \\ b=\frac{960}{12}=80 \\ a=\frac{3360-8b}{40}=\frac{3360-8\cdot 80}{40}=68=68\% \\ a=68 \\ b=80 \end{gathered}$$

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