Question

Carol ran a 100 meter race in 13. 4 seconds. But, it was wrongly

measured as 14.1 seconds. What is the percent error involved in the
measurement?

Answer 1

Answer: Approximately 5.22%

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Work Shown:

A = true value = 13.4 seconds

B = measured value (erroneous value) = 14.1 seconds

C = percent error

C = [ (B-A)/A ] * 100%

C = [ (14.1-13.4)/(13.4) ] * 100%

C = 0.0522 * 100%

C = 5.22%

This value is approximate.