# 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?

##### 2 Answers
Jun 14, 2018

$160 \text{ Litres at 27%}$

$480 \text{ Litres at 7%}$

#### Explanation:

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: $\to {S}_{7} + {S}_{27} = {S}_{12} = 640$

By % content: $\to \left[\frac{7}{100} \times {S}_{7}\right] + \left[\frac{27}{100} \times {S}_{27}\right] = \frac{12}{100} \times 640$

Two unknowns and two equations. Thus solvable.

${S}_{7} + {S}_{27} = 640 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

$\frac{7 {S}_{7}}{100} + \frac{27 {S}_{27}}{100} = \frac{384}{5} \text{ } \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
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$\textcolor{b l u e}{\text{Determine the value of } {S}_{27}}$

Consider $E q n \left(2\right)$

$\frac{7 {S}_{7} + 27 {S}_{27}}{100} = \frac{384}{5}$

Multiply both sides by 100

$7 {S}_{7} + 27 {S}_{27} = 7680 \text{ } \ldots . E q u a t i o n \left({2}_{a}\right)$

$E q n \left({2}_{a}\right) - 7 E q n \left(1\right)$

$7 {S}_{7} + 27 {S}_{27} = 7680$
ul(7S_7+color(white)(2)7S_(27)=4480larr" Subtract"
$\textcolor{w h i t e}{\text{d")0color(white)("d}} + 20 {S}_{27} = 3200$

Divide both sides by 20

$\textcolor{red}{{S}_{27} = 160}$
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$\textcolor{b l u e}{\text{Determine the value of } {S}_{7}}$

Substitute into $E q n \left(1\right)$

$\textcolor{g r e e n}{{S}_{7} + \textcolor{red}{{S}_{27}} = 640 \textcolor{w h i t e}{\text{dddd")->color(white)("dddd}} {S}_{7} + \textcolor{red}{160} = 640}$

${S}_{7} = 480$
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$\textcolor{b l u e}{\text{Check}}$

$\left[\frac{7}{100} \times 480\right] + \left[\frac{27}{100} \times 160\right] = \frac{12}{100} \times 640$

$\left[33.6\right] + \left[43.2\right] = 76.8$

$76.8 = 76.8$ Thus correct

Jun 14, 2018

Alternative and more efficient approach.

$480 \text{ Litres of 7% solution}$
$160 \text{ Litres of 27% solution}$

#### Explanation:

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 ${S}_{12} = 640$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{red}{{S}_{7} = 640 - {S}_{27}}$

Thus we have by percentage we have:

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{dd")[7/100xxcolor(red)(S_7)]color(white)("ddddd}} + \left[\frac{27}{100} \times {S}_{27}\right] = \frac{12}{100} \times 640}$

By substitution for ${S}_{7}$ we have:

$\textcolor{g r e e n}{\left[\frac{7}{100} \times \left(\textcolor{red}{640 - {S}_{27}}\right)\right] + \left[\frac{27}{100} \times {S}_{27}\right] = \frac{12}{100} \times 640}$

$44.8 - \frac{7 {S}_{27}}{100} + \frac{27 {S}_{27}}{100} = 76.8$

$\frac{20 {S}_{27}}{100} = 32$

${S}_{27} = \frac{32 \times 100}{20} = 160$
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Thus ${S}_{7} = 640 - 160 = 480$