Question:

The total stopping distance for the car traveling at 60 miles per hour is approximately what percent greater than the total stopping distance for the car traveling at 50 miles per hour?

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When comparing two quantities, always take the difference and divide by the smaller value to compute the percentage increase.
Updated On: Oct 7, 2025
  • 22%
  • 30%
  • 38%
  • 45%
  • 52%
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The Correct Option is C

Solution and Explanation

Step 1: Total stopping distance at 50 mph. 
From the graphs: - Distance during reaction time = 55 feet. 
- Distance after brakes = 137 feet. 
\[ \text{Total at 50 mph} = 55 + 137 = 192 \, \text{feet}. \] Step 2: Total stopping distance at 60 mph. 
From the graphs: - Distance during reaction time = 66 feet. 
- Distance after brakes = 198 feet. 
\[ \text{Total at 60 mph} = 66 + 198 = 264 \, \text{feet}. \] Step 3: Find difference. 
\[ 264 - 192 = 72. \] Step 4: Express as percent of shorter distance. 
\[ \frac{72}{192} = 0.375 = 37.5% \approx 38%. \] Step 5: Conclusion. 
Thus, the total stopping distance at 60 mph is approximately 38% greater than at 50 mph. 

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