How many liters of a 7% solution of salt should be added to a 27% solution in order to obtain 640 liters of a 12% solution?

Let the volume of 7% solution be `S_7`

Let the volume of 27% solution be `S_(27)`

Let the target volume of 12% solution be `S_(12)=640`

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In this situation we have two separate condition that link the proportions.

Relationship by total volume: `->S_7+S_(27)=S_(12)=640`

By % content: `->[7/100xxS_7]+[27/100xxS_(27)]=12/100xx640`

Two unknowns and two equations. Thus solvable.

`S_7+S_(27)=640" "......................Equation(1)`

`(7S_7)/100+(27S_27)/100=384/5" ".............Equation(2)`
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`color(blue)("Determine the value of "S_(27))`

Consider `Eqn(2)`

`(7S_7+27S_27)/100=384/5`

Multiply both sides by 100

`7S_7+27S_(27)=7680" "....Equation(2_a)`

`Eqn(2_a)-7Eqn(1)`

`7S_7+27S_(27)=7680`
`ul(7S_7+color(white)(2)7S_(27)=4480larr" Subtract"`
`color(white)("d")0color(white)("d") +20S_(27)=3200`

Divide both sides by 20

`color(red)(S_(27)=160)`
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`color(blue)("Determine the value of "S_7)`

Substitute into `Eqn(1)`

`color(green)(S_7+color(red)(S_(27))=640 color(white)("dddd")->color(white)("dddd") S_7+color(red)(160)=640)`

`S_7=480`
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`color(blue)("Check")`

`[7/100xx480]+[27/100xx160]=12/100xx640`

`[ 33.6]+[43.2 ]=76.8`

`76.8=76.8` Thus correct

Let the volume of 7% solution be `S_7`

Let the volume of 27% solution be `S_(27)`

Let the target volume of 12% solution be `S_(12)=640`
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`color(red)(S_7=640-S_(27))`

Thus we have by percentage we have:

`color(green)(color(white)("dd")[7/100xxcolor(red)(S_7)]color(white)("ddddd")+[27/100xxS_(27)]=12/100xx640)`

By substitution for `S_7` we have:

`color(green)([7/100xx(color(red)(640-S_(27)))]+[27/100xxS_(27)]=12/100xx640)`

`44.8-(7S_27)/100+(27S_27)/100 = 76.8`

`(20S_(27))/100 =32`

`S_27=(32xx100)/20 = 160`
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Thus `S_7=640-160=480`