A corner shop has \(12\frac{1}{2}\) kg of flour. Each customer order uses \(1\frac{3}{4}\) kg. Work out the greatest number of complete orders that can be f...
Question 1 Report
A corner shop has \(12\frac{1}{2}\) kg of flour. Each customer order uses \(1\frac{3}{4}\) kg.
- Work out the greatest number of complete orders that can be filled. (3)
- Find the mass of flour left over. (1)
Answer Details
Dividing by a mixed number is much safer once both quantities are improper fractions, because the division rule applies directly to those.
(a) Converting:
\[12\frac{1}{2} = \frac{25}{2} \qquad \text{and} \qquad 1\frac{3}{4} = \frac{7}{4}\][1]
Dividing by a fraction is the same as multiplying by its reciprocal, so turn the second fraction upside down:
\[\frac{25}{2} \div \frac{7}{4} = \frac{25}{2} \times \frac{4}{7} = \frac{100}{14} = \frac{50}{7}\][1]
\[\frac{50}{7} = 7\frac{1}{7}\]Only whole orders can be filled, so the answer is rounded down: \(7\) complete orders. [1]
Rounding down rather than to the nearest whole number is essential here, since the eighth order could not be completed with the flour that remains.
(b) The flour used by \(7\) orders is
\[7 \times \frac{7}{4} = \frac{49}{4} \text{ kg}\]Subtracting from the stock, over a common denominator of \(4\):
\[\frac{25}{2} - \frac{49}{4} = \frac{50}{4} - \frac{49}{4} = \frac{1}{4} \text{ kg}\][1]
The leftover of \(\frac{1}{4}\) kg agrees with the \(\frac{1}{7}\) of an order left in part (a), since \(\frac{1}{7} \times \frac{7}{4} = \frac{1}{4}\) kg. Note that \(\frac{1}{7}\) is a fraction of an order, not a mass in kilograms, which is why part (b) needs its own calculation.
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