In June, a runner ran a mile in 7:45. In September, the runner ran a mile in 5:51.Find the percent of change from June to September. Round to the nearest hundredth of a percent.

1 answer

To find the percent change from June to September, we start with the two times the runner completed a mile:

  • June: 7 minutes and 45 seconds (which is \( 7 \times 60 + 45 = 465 \) seconds)
  • September: 5 minutes and 51 seconds (which is \( 5 \times 60 + 51 = 351 \) seconds)

Next, we use the percent change formula:

\[ \text{Percent Change} = \left(\frac{\text{New Value} - \text{Old Value}}{\text{Old Value}}\right) \times 100% \]

In this case, the "Old Value" is the time from June (465 seconds) and the "New Value" is the time from September (351 seconds).

Substituting the values into the formula:

\[ \text{Percent Change} = \left(\frac{351 - 465}{465}\right) \times 100% \]

Calculating the difference:

\[ 351 - 465 = -114 \]

Now substituting back into the formula:

\[ \text{Percent Change} = \left(\frac{-114}{465}\right) \times 100% \]

Calculating the fraction:

\[ \frac{-114}{465} \approx -0.245161 \]

Now converting to percentage:

\[ -0.245161 \times 100% \approx -24.5161% \]

Rounding to the nearest hundredth of a percent gives us:

\[ \text{Percent Change} \approx -24.52% \]

Since we are looking for the percent of change in the runner's time (which represents an improvement), we can express this as a reduction in time.

Thus, the percent change from June to September is approximately:

\[ \boxed{24.52%} \]

This indicates a decrease of approximately 24.52%, which is an improvement in the time taken to run a mile.