A farm starts with \(18\dfrac{3}{4}\) tonnes of silage. In week 1, \(\dfrac{2}{5}\) of the stock is used. In week 2, a further \(\dfrac{1}{3}\) of what rema...
Question 1 Report
A farm starts with \(18\dfrac{3}{4}\) tonnes of silage. In week 1, \(\dfrac{2}{5}\) of the stock is used. In week 2, a further \(\dfrac{1}{3}\) of what remains is used. In week 3, exactly \(3\dfrac{1}{2}\) tonnes are used.
- Find how much silage remains after week 1, as a mixed number in its simplest form. (2)
- Find how much remains after week 2. (2)
- Find how much remains after week 3. (2)
- The farm needs at least 4 tonnes in reserve for week 4. Determine, with a reason, whether the farm has enough. (2)
Answer Details
This question repeatedly applies a fraction of a remaining quantity, converting between mixed numbers and improper fractions at each stage, and finally compares the final amount with a required minimum.
Starting from \(18\dfrac{3}{4} = \dfrac{75}{4}\) tonnes, using \(\dfrac{2}{5}\) leaves a remaining fraction of \(1-\dfrac{2}{5}=\dfrac{3}{5}\):
\[\dfrac{75}{4} \times \dfrac{3}{5} = \dfrac{225}{20} = \dfrac{45}{4} = 11\dfrac{1}{4} \text{ tonnes}\][2 marks]
Using a further \(\dfrac{1}{3}\) leaves a remaining fraction of \(1-\dfrac{1}{3}=\dfrac{2}{3}\) of the amount from part (a):
\[\dfrac{45}{4} \times \dfrac{2}{3} = \dfrac{90}{12} = \dfrac{15}{2} = 7\dfrac{1}{2} \text{ tonnes}\][2 marks]
Subtracting the \(3\dfrac{1}{2} = \dfrac{7}{2}\) tonnes used in week \(3\):
\[\dfrac{15}{2} - \dfrac{7}{2} = \dfrac{8}{2} = 4 \text{ tonnes}\][2 marks]
The farm has exactly \(4\) tonnes remaining, and \(4\) tonnes meets the requirement of "at least \(4\) tonnes", so the farm has exactly enough reserve for week \(4\) [2 marks].
Each fraction used in this question is taken of the amount remaining after the previous stage, not of the original \(18\dfrac{3}{4}\) tonnes, which is why the calculation must proceed stage by stage rather than combining all the fractions at once.
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