Fractions — O/L Mathematics

How do I practise Fractions?

3 questions with worked solutions

1. Simplify: $\left(\frac{3}{4} + \frac{1}{6}\right) \div \left(\frac{5}{8} - \frac{1}{4}\right)$
   1. A) $2\frac{4}{5}$
   2. B) $1\frac{4}{9}$
   3. C) $2\frac{2}{9}$
   4. D) $2\frac{4}{9}$
   Check my answer

   Correct answer: D) $2\frac{4}{9}$

   Evaluate brackets first, then divide by multiplying by the reciprocal. The correct simplification gives $\frac{22}{9} = 2\frac{4}{9}$.

   Show worked solution

   First evaluate the brackets. Left bracket: $\frac{3}{4} + \frac{1}{6}$. The LCM of $4$ and $6$ is $12$. So $\frac{3}{4} = \frac{9}{12}$ and $\frac{1}{6} = \frac{2}{12}$. Hence $\frac{9}{12} + \frac{2}{12} = \frac{11}{12}$. Right bracket: $\frac{5}{8} - \frac{1}{4}$. The LCM of $8$ and $4$ is $8$. So $\frac{1}{4} = \frac{2}{8}$. Hence $\frac{5}{8} - \frac{2}{8} = \frac{3}{8}$. Now divide: $\frac{11}{12} \div \frac{3}{8} = \frac{11}{12} \times \frac{8}{3}$. Cancel: $8$ and $12$ have a common factor $4$, giving $\frac{11}{3} \times \frac{2}{3} = \frac{22}{9}$. Convert to a mixed number: $\frac{22}{9} = 2\frac{4}{9}$. Answer: $2\frac{4}{9}$.
   Modelled on: the O/L Mathematics examination pattern
2. A shopkeeper has $360$ oranges. He sells $\frac{2}{5}$ of them in the morning and $\frac{1}{3}$ of the remainder in the afternoon. How many oranges are left unsold?
   1. A) $120$
   2. B) $96$
   3. C) $72$
   4. D) $144$
   Check my answer

   Correct answer: D) $144$

   After selling $\frac{2}{5}$ of $360$, $216$ remain. Then $\frac{1}{3}$ of $216$ is $72$, leaving $216 - 72 = 144$ oranges.

   Show worked solution

   Total oranges $= 360$. Morning sales: $\frac{2}{5} \times 360 = \frac{720}{5} = 144$ oranges. Remainder after morning: $360 - 144 = 216$ oranges. Afternoon sales: $\frac{1}{3}$ of the remainder $= \frac{1}{3} \times 216 = 72$ oranges. Oranges left unsold: $216 - 72 = 144$. Answer: $144$ oranges.
   Modelled on: the O/L Mathematics examination pattern
3. Kamala spends $\frac{1}{4}$ of her salary on rent and $\frac{1}{3}$ of the remainder on food. If she has Rs. $9000$ left, what is her monthly salary?
   1. A) Rs. $12000$
   2. B) Rs. $18000$
   3. C) Rs. $15000$
   4. D) Rs. $20000$
   Check my answer

   Correct answer: B) Rs. $18000$

   Let salary be $x$. After rent, $\frac{3}{4}x$ remains. Food is $\frac{1}{3}$ of that, i.e. $\frac{1}{4}x$. Total spent $\frac{1}{2}x$, so $\frac{1}{2}x = 9000$, giving $x = 18000$.

   Show worked solution

   Let the salary be $x$. Rent: $\frac{1}{4}x$. Remainder after rent: $x - \frac{1}{4}x = \frac{3}{4}x$. Food: $\frac{1}{3}$ of the remainder $= \frac{1}{3} \times \frac{3}{4}x = \frac{1}{4}x$. Total spent: $\frac{1}{4}x + \frac{1}{4}x = \frac{1}{2}x$. Money left: $x - \frac{1}{2}x = \frac{1}{2}x$. Given $\frac{1}{2}x = 9000$, so $x = 9000 \times 2 = 18000$. Answer: Rs. $18000$.
   Modelled on: the O/L Mathematics examination pattern

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Where does Fractions fit in the syllabus?

Grade 10 · Term 1