# How many quarts of olive oil did they use? (see below) Thanks!

Jul 3, 2018

Solved for one of the volumes. I will let you solve for the other by subtraction.

#### Explanation:

$\textcolor{b l u e}{\text{Building the model}}$

To maintain a high level of precision I will stick with fractions.
Rounding decimals can lead to accumulated errors. So if you wish to use decimals convert the fractions at the very end.

Let the volume of 30% oil be ${v}_{30}$
Let the volume of 79% oil be ${v}_{79}$
Let the volume of the blended be ${v}_{b} = \frac{19}{20} \text{ }$quarts

For now ignore the units of quarts.

When blended for total volume we have:

${v}_{30} + {v}_{79} = \frac{19}{20} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

When blended for % of oil volume content we have (in quarts):

$\left[\frac{30}{100} \times {v}_{30}\right] + \left[\frac{79}{100} \times {v}_{79}\right] = \left[\frac{40 \frac{6}{19}}{100} \times \frac{19}{20}\right] \ldots . E q u a t i o n \left(2\right)$
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$\textcolor{b l u e}{\text{Answering the question}}$

In $E q n \left(2\right)$ we have 2 unknowns. If we can express one of these in terms of the other we end up with just 1 unknown. Thus solvable.

I choose to substitute for ${v}_{79}$

From $E q u a t i o n \left(1\right)$

${v}_{79} = \frac{19}{20} - {v}_{30} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(3\right)$

Using $E q u a t i o n \left(3\right)$ substitute for ${v}_{79}$ in $E q u a t i o n \left(2\right)$

$\left[\frac{30}{100} \times {v}_{30}\right] + \left[\frac{79}{100} \times \left(\frac{19}{20} - {v}_{30}\right)\right] = \left[\frac{40 \frac{6}{19}}{100} \times \frac{19}{20}\right]$

$\textcolor{w h i t e}{d \text{d")(30v_30)/100color(white)("dd.d")+color(white)("ddd")1501/2000-(79v_30)/100 color(white)("ddd")= color(white)("ddd}} \frac{766}{2000}$

Having a common denominator of 2000 gives:

$\frac{600 {v}_{30}}{2000} + \frac{1501}{2000} - \frac{1580 {v}_{30}}{2000} = \frac{766}{2000}$

Multiply both sides by 2000

$600 {v}_{30} + 1501 - 1580 {v}_{30} = 766$

$- 980 {v}_{30} = - 2267$

$\textcolor{g r e e n}{{v}_{30} = \frac{- 2267}{- 980} = 2 \frac{307}{980} \text{quarts } \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(3\right)}$
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Using $E q u a t i o n \left(3\right)$ substitute for ${v}_{30} \text{ in } E q u a t i o n \left(1\right)$

I will let you finish that off.