# How long will it take you to drive 180 miles at a speed of 25 miles per hour?

## Question:

How long will it take you to drive 180 miles at a speed of 25 miles per hour?

## Proportions and Variation:

We can establish a relation between two numbers that are proportional by means of a ratio {eq}x {/eq} and {eq}y {/eq}. That is to say, a proportion shows the similarity between two ratios. When two variables are dependent, variations in the magnitude of one variable will have a proportional effect on the other. When there is an increase or decrease of a variable {eq}x {/eq} with respect to another {eq}y {/eq}, for a ratio or constant K, variations are present. In the case that we have a direct variation, it happens that when one variable increases the other increases, which can also be written as: {eq}\frac{{{y_1}}}{{{x_1}}} = \frac{{{y_2}}}{{{x_2}}} {/eq}.

{eq}\eqalign{ & {\text{In this specific case }}{\text{,we have two proportional values }}\,y\,\left( {hours} \right){\text{ }} \cr & {\text{and }}x\,\left( {miles} \right){\text{ that have a variation in directly proportional form}}{\text{. }} \cr & {\text{So we have:}} \cr & \,\,\,\,{x_1} = 25\,miles \cr & \,\,\,\,{y_1} = 1\,hour \cr & \,\,\,\,{x_2} = 180\,miles \cr & \,\,\,\,{y_2} = ?\,\,hours \cr & {\text{Since}}{\text{, }}x{\text{ and }}y{\text{ vary directly}}{\text{, then}}{\text{, when }}x{\text{ increases it also }} \cr & {\text{increases }}y{\text{. For this reason}}{\text{, it must be satisfied that:}} \cr & \,\,\,\,\frac{{{y_2}}}{{{x_2}}} = \frac{{{y_1}}}{{{x_1}}} \cr & {\text{So if we do cross - multiplying:}} \cr & \,\,\,\,{y_2} \cdot {x_1} = {y_1} \cdot {x_2} \cr & {\text{Now}}{\text{, solving for }}\,{y_2}{\text{:}} \cr & \,\,\,\,{y_2} = \frac{{{y_1} \cdot {x_2}}}{{{x_1}}} \cr & {\text{So}}{\text{, substituting the given values:}} \cr & \,\,\,\,{y_2} = \frac{{1 \times 180}}{{25}} = 7.2\,hours \cr & {\text{Therefore}}{\text{, it will take }}\boxed{7.2{\text{ }}hours} \cr} {/eq}