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A 10% solution of fertilizer is to be mixed with a 60% solution of fertilizer in order to get 150 gallons of a 50% solution. How many gallons of the 10% solution and 60% solution should be mixed?

Answer :

Let the amount of gallons of 10% solution be "x", and

Let the amount of gallons of 60% solution be "y".

We need 150 gallons, thus we can write:

[tex]x+y=150_{}[/tex]

Both 10% and 60% solution needs to be mixed to produce 150 gallons of 50% solution. We can write:

[tex]\begin{gathered} 0.1x+0.6y=0.5\times150 \\ 0.1x+0.6y=75 \end{gathered}[/tex]

Solving for "x" in the first equation >>>

[tex]\begin{gathered} x+y=150 \\ x=150-y \end{gathered}[/tex]

We will substitute it into the second equation and solve for "y" >>>

[tex]\begin{gathered} 0.1x+0.6y=75 \\ 0.1(150-y)+0.6y=75 \\ 15-0.1y+0.6y=75 \\ 0.5y=60 \\ y=120 \end{gathered}[/tex]

Now, we can solve for "x" using the first equation >>>>

[tex]\begin{gathered} x=150-y \\ x=150-120 \\ x=30 \end{gathered}[/tex]Answer

We need 30 gallons of 10% solution and 120 gallons of 60% solution

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