IMC 2014 Q15
15. Zac halves a certain number and then adds 8 to the result. He finds that he obtains the same answer if he doubles his original number and then subtracts 8 from the result.
What is Zac's original number?
- A \(8\tfrac{2}{3}\)
- B \(9\tfrac{1}{3}\)
- C \(9\tfrac{2}{3}\)
- D \(10\tfrac{1}{3}\)
- E \(10\tfrac{2}{3}\)
Intermediate Mathematical Challenge 2014, Question 15. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: E \(10\tfrac{2}{3}\)
Let Zac's number be \(x\). Then \(\tfrac{1}{2}x + 8 = 2x - 8\). So \(x + 16 = 4x - 16\).
Therefore \(32 = 3x\), that is \(x = 10\tfrac{2}{3}\).
Solution © UK Mathematics Trust, from the IMC 2014 solutions. Longer versions with investigations: Solutions and investigations (PDF).