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A chemist is mixing a 40% salt solution with a 20% salt solution to make 50 L of
a new solution that will contain 25% salt. How much of each of the original
solutions should the chemist use?

Respuesta :

LammettHash

Let [tex]x[/tex] be the amount (in L) of the 40% solution to be used, and [tex]y[/tex] the amount (L) of the 20% solution. The chemist wants a new solution of 50 L, so that

[tex]x+y=50[/tex]

Each L of the 40% solution contributes 0.4 L of salt, while each L of the 20% solution contributes 0.2. The new solution should have a concentration of 25% salt, so that

[tex]0.4x+0.2y=0.25(x+y)=12.5[/tex]

Now

[tex]x+y=50\implies y=50-x[/tex]

[tex]0.4x+0.2y=12.5\implies0.4x+0.2(50-x)=12.5\implies0.2x=2.5[/tex]

[tex]\implies\boxed{x=12.5}\implies\boxed{y=37.5}[/tex]