Calculate the exact volume of the container. Give your answer as a fraction.

Jim estimates the dimensions of the container as 15 m, 2 m and 3 m and uses these to estimate the volume of the container.

Calculate the percentage error in Jim’s estimated volume of the container.

\(V = 16 \times 1\frac{3}{4} \times 2\frac{2}{3}\)     (M1)

Note: Award (M1) for correct substitution in volume formula. Accept decimal substitution of \(2.66\) or better.

\( = 74.6666{\text{ }} \ldots \)     (A1)
\( = {\text{74}}\frac{{\text{2}}}{{\text{3}}}{\text{ }}{{\text{m}}^{\text{3}}}{\text{ }}\left( {\frac{{{\text{224}}}}{{\text{3}}}{\text{ }}{{\text{m}}^{\text{3}}}} \right)\)     (A1)    (C3)

Note: Correct answer only.

[3 marks]

\({\text{%  error}} = \frac{{\left( {90 - 74\frac{2}{3}} \right) \times 100}}{{74\frac{2}{3}}}\)     (A1)(M1)

Note: Award (A1) for \(90\) seen, or inferred in numerator, (M1) for correct substitution into percentage error formula.

\( = 20.5\)     (A1)(ft)     (C3)

Note: Accept \( - 20.5\).

[3 marks]

This question was well answered by the majority of candidates. Candidates encountered difficulty in part (a) with using fractions finding the exact volume. Nearly all candidates could use the formula for volume and most could achieve at least 2 marks in this first part. Most candidates could find the percentage error correctly using the formula once they found the estimate for the volume. Very few candidates substituted the formula incorrectly, or had an incorrect denominator.

This question was well answered by the majority of candidates. Candidates encountered difficulty in part (a) with using fractions finding the exact volume. Nearly all candidates could use the formula for volume and most could achieve at least 2 marks in this first part. Most candidates could find the percentage error correctly using the formula once they found the estimate for the volume. Very few candidates substituted the formula incorrectly, or had an incorrect denominator.