A model car has a width of 4 inches and a length of 16 inches. If the actual car has a width of 6...

Question:

A model car has a width of 4 inches and a length of 16 inches. If the actual car has a width of 6 feet, how long is the car?

Proportions and Variation:

In some real-life circumstances, we can appreciate situations in which there are problems related to proportions. In particular, we can see them in the scales of the models that represent some objects such as cars. When their width and length dimensions are compared with the real dimensions, a constant value is produced that represents the proportional scale.

Answer and Explanation:

{eq}\eqalign{ & {\text{In this specific case}}{\text{, we have two proportional values }}\,x\,\left( {inches} \right){\text{ and }} \cr & y\,\left( {feet} \right){\text{ that have a variation in directly proportional form}}{\text{. }} \cr & {\text{So we have:}} \cr & {\text{Model car:}} \cr & \,\,\,\,{x_1} = 4\,inches{\text{ (wide)}} \cr & \,\,\,\,{x_2} = 16\,inches{\text{ (length)}} \cr & {\text{Actual car:}} \cr & \,\,\,\,{y_1} = 6\,feet{\text{ (wide)}} \cr & \,\,\,\,{y_2} = ?\,\,feet{\text{ (length)}} \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{{6 \times 16}}{4} = 24\,feet \cr & {\text{Therefore}}{\text{, the length of the car is }}\boxed{24{\text{ }}feet} \cr} {/eq}


Learn more about this topic:

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Ratios and Proportions: Definition and Examples

from Geometry: High School

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