NECO 2023 · Paper 1 · Q11

Approximation & error

In a Chemistry class, a student recorded $21.23\text{ cm}^3$ for a titre value of $21.32\text{ cm}^3$ . Find the percentage error, correct to one decimal place.

Worked solution (try it first)

The error is $21.32 - 21.23 = 0.09\text{ cm}^3$ .
Divide by the true titre and multiply by 100: $\frac{0.09}{21.32} \times 100\% = 0.422\ldots\%$ .
To one decimal place this is 0.4%, which is 0.40, option B.

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