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.

Answer 1

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:

$$-0.245161\times 100$$

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:

$$\text{Extra open brace or missing close brace}$$

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