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A 5-gallon radiator is full and contains a 40% solution of antifreeze. how much needs to be drained out and replaced with pure antifreeze to obtain a 70% solution?

Respuesta :

Solution:

Let x be the number of antifreeze solution. Thus;

[tex](0.4)x+1(5-x)=0.7x[/tex]

We would solve for x;

[tex]\begin{gathered} 0.4x+5-x=0.7x \\ \\ 5=x-0.4x+0.7x \\ \\ 5=1.3x \\ \\ x=\frac{5}{1.3} \\ \\ x=3.85 \end{gathered}[/tex]

Hence, the amount needs to be drained out is;

[tex]\begin{gathered} 5-x=5-3.85 \\ \\ =1.15 \end{gathered}[/tex]

Answer: 1.15 gallon