Humberto has a toy train that is a scale model of an actual train. The length of the actual train is 600 feet and the width is 30 feet. If the scale model measures 10 inches long,... Humberto has a toy train that is a scale model of an actual train. The length of the actual train is 600 feet and the width is 30 feet. If the scale model measures 10 inches long, how wide is the scale model?
Understand the Problem
The question asks to find the width of a scale model train, given the length and width of a real train, and the length of the scale model. It requires setting up a proportion comparing the length and width of the real train to the length and width of the scale model, paying attention to unit conversions (feet to inches).
Answer
The width of the scale model is $0.5$ inches.
Answer for screen readers
C. 0.5 inches
Steps to Solve
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Convert the length of the actual train to inches Since the length of the model train is given in inches, we need to convert the length of the actual train from feet to inches. We know that 1 foot = 12 inches. So, 600 feet = $600 \times 12 = 7200$ inches.
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Convert the width of the actual train to inches Similarly, convert the width of the actual train from feet to inches. 30 feet = $30 \times 12 = 360$ inches.
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Set up a proportion Let $x$ be the width of the model train in inches. We can set up a proportion relating the length and width of the actual train to the length and width of the scale model: $$ \frac{\text{length of actual train}}{\text{width of actual train}} = \frac{\text{length of model train}}{\text{width of model train}} $$ $$ \frac{7200}{360} = \frac{10}{x} $$
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Solve for $x$ To solve for $x$, we can cross-multiply: $$ 7200x = 360 \times 10 $$ $$ 7200x = 3600 $$ $$ x = \frac{3600}{7200} $$ $$ x = \frac{1}{2} = 0.5 $$
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Determine the width of the scale model The width of the scale model is 0.5 inches.
C. 0.5 inches
More Information
The problem involves setting up a proportion to relate the dimensions of an actual object to the dimensions of its scale model. Paying attention to unit conversions is crucial for solving this problem.
Tips
A common mistake is forgetting to convert the feet to inches. If the length of the real train and width of the real train stay in feet, then you would have: $$ \frac{600 \text{ feet}}{30 \text{ feet}} = \frac{10 \text{ inches}}{x \text{ inches}} $$ Cross-multiplying gives $600x = 300$ so $x = 0.5$ inches, giving the correct answer by coincidence. If the length of the model train was in feet, then you would get the wrong answer. So it's important to keep units consistent.
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