A chef is going to use a mixture of two brands of Italian dressing. The first brand contains 7% vinegar, and the second brand contains 12% vinegar. The chef wants to make 200 mL of a dressing that is 11% vinegar. How much of each brand should she use?

Feb 20, 2017

brand 1 =40ml
brand 2=160ml

Explanation:

Let the amount of brand 1 be ${b}_{1}$
Let the amount of brand 2 be ${b}_{2}$

The target is to have 1 equation with just 1 unknown, thus solvable.

Using ${b}_{1}$ as the basis of reference combined with a fixed total of 200ml.

$\textcolor{b r o w n}{\text{We have two conditions.}}$

$\textcolor{b r o w n}{\text{Condition 1:}}$

${b}_{1} + {b}_{2} = 200 m l$

${b}_{2} = 200 m l - {b}_{1} \text{ } \ldots \ldots \ldots . . E q u a t i o n \left(1\right)$

$\textcolor{b r o w n}{\text{Condition 2:}}$

$\frac{7}{100} {b}_{1} + \frac{12}{100} {b}_{2} \text{ "=" } \frac{11}{100} \times 200 m l$

Multiply both sides by 100

$\text{ "7b_1+" "12b_2 " "=" " 2200ml" } \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(2\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for } {b}_{1}}$

Using Equation (1) substitute for ${b}_{2}$ in Equation(2)

$7 {b}_{1} + \text{ "12(200ml-b_1) " "=" " 2200ml" } \ldots E q u a t i o n \left({2}_{a}\right)$

$7 {b}_{1} + \text{ "2400ml-12b_1" "=" } 2200 m l$

$- 5 {b}_{1} = - 200 m l$

Multiply both sides by (-1)

$+ 5 {b}_{1} = + 200 m l$

Divide both sides by 5

$\textcolor{b l u e}{{b}_{1} = 40 m l}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for } {b}_{2}}$

${b}_{1} + {b}_{2} = 200 m l$

$40 + {b}_{2} = 200 m l$

$\textcolor{b l u e}{{b}_{2} = 160 m l}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Check}}$

$\left[\frac{7}{100} \times 40\right] + \left[\frac{12}{100} \times 160\right] = \frac{11}{100} \times 200$

$2 \frac{4}{5} + 19 \frac{1}{5} = 22$

$22 = 22 \leftarrow \text{ True}$ so solution is correct

Jan 23, 2018

A different approach - we can represent this using a graph
The actual explanation is a bit long. However, once you understand the principle the calculation is very fast. May be 4 or 5 lines.

Explanation:

$\textcolor{b l u e}{\text{Setting up the condition}}$

The final blend will always be 200 ml no matter what the proportion of each of the two constituents.

Let the first dressing containing 7% vinegar be ${d}_{7}$
Let the second dressing containing 12% vinegar be ${d}_{12}$
Let the unknown volume of the 12% dressing be $x$

${d}_{7} + {d}_{12} = 200 m l$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Important point 1}}$

Suppose the 200ml contained only ${d}_{7}$ then the amount of vinegar in the 200ml would be 7%

Suppose the 200ml contained only ${d}_{12}$ then the amount of vinegar in the 200ml would be 12%

As you blend them in different ratios then the final vinegar content of that blend would be directly related to how much of each type you use. This final vinegar content will be somewhere between the two extremes of 7% and 12%
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Important point 2}}$

If you know how much of say the 7% dressing there is then the amount of the 12% dressing is: 200ml - the volume of the 7% dressing. $\to {d}_{12} = 200 - {d}_{7}$

So by considering just one of them there is an indirect link to the other. Thus by considering the change in the amount of one of them and observing the change in blended vinegar content we are indirectly observing everything
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

$\textcolor{b r o w n}{\text{Using ratio in fraction format}}$

$\left(\text{the amount of along")/("the amount of up}\right) \to \frac{200}{12 - 7} = \frac{x}{11 - 7}$

$\left(\text{the amount of along")/("the amount of up")->color(white)("d")200/5color(white)("d")=color(white)("dd}\right) \frac{x}{4}$

Multiply both sides by $\textcolor{red}{4}$

color(green)(200/5color(white)("d")=color(white)("d")x/4color(white)("d")->color(white)("ddd")200/5color(red)(xx4)color(white)("d")=color(white)("d")x/4color(red)(xx4)

color(green)(color(white)("ddddddddddd")->color(white)("dddd")40color(white)("d")color(red)(xx4)color(white)("d")=color(white)("d")x xx (color(red)(4))/4

But $\frac{4}{4} \times x \text{ is the same as "1xx x " which is the same as just } x$

color(green)(color(white)("ddddddddddd")->color(white)("dddddd")160color(white)("dd")=color(white)("d")x