# Question #97f62

May 4, 2017

For 1 gallon you have $\frac{1}{2}$ of the something else

#### Explanation:

Assumption: the given ratio is of 2 gallons : 1 of something else
and that you require to determine how much there is 'something else for 1 gallon.

Let 'something else be represent by $s$

Express initial condition ratio in fractional form $\frac{2 g}{1 s}$

Let the unknown count or volume of $s$ be $x$

$\frac{2 g}{1 s} \equiv \frac{1 g}{x s}$

Turn the whole thing upside down

$\frac{1 s}{2 g} \equiv \frac{x s}{1 g}$

Multiply both sides by $1 g \leftarrow$ gets the $x s$ on its own

$\textcolor{b r o w n}{\frac{1 \textcolor{g r e e n}{s}}{2 \textcolor{g r e e n}{g}} \times 1 \textcolor{g r e e n}{g} \equiv x \textcolor{g r e e n}{s}}$

Separating numbers from units of count/measurement:

$\textcolor{b r o w n}{\frac{1}{2} \times 1 \text{ }} \textcolor{g r e e n}{\frac{s}{\cancel{g}} \times \cancel{g}}$

Numbers left $\frac{1}{2} \text{ units left} \to s$

So for 1 gallon you have $\frac{1}{2}$ of the something else