IMC 2012 Q15
15. Which of the following has a value that is closest to 0?
- A \(\dfrac{1}{2} + \dfrac{1}{3} \times \dfrac{1}{4}\)
- B \(\dfrac{1}{2} + \dfrac{1}{3} \div \dfrac{1}{4}\)
- C \(\dfrac{1}{2} \times \dfrac{1}{3} \div \dfrac{1}{4}\)
- D \(\dfrac{1}{2} - \dfrac{1}{3} \div \dfrac{1}{4}\)
- E \(\dfrac{1}{2} - \dfrac{1}{3} \times \dfrac{1}{4}\)
Intermediate Mathematical Challenge 2012, Question 15. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E \(\dfrac{1}{2} - \dfrac{1}{3} \times \dfrac{1}{4}\)
\(\dfrac{1}{2} + \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{2} + \dfrac{1}{12} = \dfrac{7}{12}\); \(\dfrac{1}{2} + \dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{1}{2} + \dfrac{1}{3} \times \dfrac{4}{1} = \dfrac{1}{2} + \dfrac{4}{3} = \dfrac{11}{6}\); \(\dfrac{1}{2} \times \dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{1}{2} \times \dfrac{1}{3} \times \dfrac{4}{1} = \dfrac{2}{3}\); \(\dfrac{1}{2} - \dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{1}{2} - \dfrac{1}{3} \times \dfrac{4}{1} = \dfrac{1}{2} - \dfrac{4}{3} = -\dfrac{5}{6}\); \(\dfrac{1}{2} - \dfrac{1}{3} \times \dfrac{1}{4} = \dfrac{1}{2} - \dfrac{1}{12} = \dfrac{5}{12}\).
Of the fractions \(\dfrac{7}{12}\), \(\dfrac{11}{6}\), \(\dfrac{2}{3}\), \(-\dfrac{5}{6}\), \(\dfrac{5}{12}\), the closest to 0 is \(\dfrac{5}{12}\).
Solution © UK Mathematics Trust, from the IMC 2012 solutions. Longer versions with investigations: Solutions and investigations (PDF).