A student measured the width of a textbook as 8.2\text{ cm} instead of its actual width of… — JAMB UTME

A student measured the width of a textbook as $8.2\text{ cm}$ instead of its actual width of $8.0\text{ cm}$ . What is the percentage error in the measurement?

A
$0.2\%$
B
$2.44\%$
$2.5\%$ Answer
D
$20\%$

Answer

2.5\%

The correct answer is

2.5\%

. The absolute error is

0.2\text{ cm}

. Dividing the absolute error (

0.2\text{ cm}

) by the actual measurement (

8.0\text{ cm}

) yields

\frac{0.2}{8.0} = 0.025

. Expressing this decimal as a percentage gives

0.025 \times 100\% = 2.5\%

Step-by-Step Solution

Find the absolute error in measurement

\text{Error} = |8.2\text{ cm} - 8.0\text{ cm}| = 0.2\text{ cm}

Percentage error requires knowing the difference between measured value and true value.

Apply the percentage error formula

\text{Percentage Error} = \frac{\text{Error}}{\text{True Value}} \times 100\%

Percentage error is always calculated relative to the true (actual) value.

Substitute values and evaluate

\frac{0.2}{8.0} \times 100\% = \frac{1}{40} \times 100\% = 2.5\%

Simplifying

\frac{0.2}{8.0}

yields

\frac{1}{40}

, which equals

2.5\%

Key Concept

Percentage Error Calculation

Estimated Time:45s

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