Question

The accuracy of a certain stopwatch varies by 5%, meaning that the time it shows may be off as much as 5% from the actual time. If the stopwatch shows a time of 3:26 (3 minutes, 26 seconds) for an athlete’s 800m run, then x is the actual time and 0.05x is the potential amount of error. Use this information to write your equation and determine the range of possible times that the run may actually have taken.

Answer 1

3 minutes, 16.19 s ≤ Actual time ≤ 3 minutes, 36.842 s

Explanation:

The accuracy of the stopwatch varies as much of 5%

This means that the time it shows can be

Time*0.95 ≤ Time ≤ Time*1.05

if x is the actual time

x- 0.05*x ≤ x ≤ x+ 0.05*x

x *0.95 ≤ x ≤ x*1.05

If the stopwatch shows

3 minutes, 26 sec = 206 seconds

There are two possible ranges for the actual run

x *0.95 = 206 s

x = 216.842 s

or

x *1.05 = 206 s

x = 196.190

196.190 s ≤ Actual time ≤ 216.842 s

3 minutes, 16.19 s ≤ Actual time ≤ 3 minutes, 36.842 s