PK 2019 Q11
11. The numbers \(a\), \(b\), \(c\) and \(d\) are distinct positive integers chosen from 1 to 10 inclusive. What is the least possible value \(\dfrac{a}{b} + \dfrac{c}{d}\) could have?
- A \(\dfrac{2}{10}\)
- B \(\dfrac{3}{19}\)
- C \(\dfrac{14}{45}\)
- D \(\dfrac{29}{90}\)
- E \(\dfrac{25}{72}\)
Pink Kangaroo 2019, Question 11. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C \(\dfrac{14}{45}\)
To obtain the smallest value, the numerators should be as small as possible, so 1 or 2, and the denominators should be as large as possible, so 9 or 10. The two possible candidates are \(\tfrac{1}{10} + \tfrac{2}{9}\) and \(\tfrac{1}{9} + \tfrac{2}{10}\). The first gives \(\tfrac{9}{90} + \tfrac{20}{90} = \tfrac{29}{90}\), and the second gives \(\tfrac{10}{90} + \tfrac{18}{90} = \tfrac{28}{90} = \tfrac{14}{45}\), which is smaller.
Solution © UK Mathematics Trust, from the PK 2019 solutions.