PK 2018 Q19
19. Abdul wrote down four positive numbers. He chose one of them and added it to the mean of the other three. He repeated this for each of the four numbers in turn.
The results were 17, 21, 23 and 29. What was the largest of Abdul's numbers?
- A 12
- B 15
- C 21
- D 24
- E 29
Pink Kangaroo 2018, Question 19. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: C 21
Let the four numbers be \(a\), \(b\), \(c\), \(d\) with \(a \leqslant b \leqslant c \leqslant d\). Abdul chooses the numbers in turn, giving
\[\begin{array}{lll} a + \tfrac{1}{3}(b + c + d) = 17 \quad (1) & \qquad & b + \tfrac{1}{3}(a + c + d) = 21 \quad (2) \\ c + \tfrac{1}{3}(a + b + d) = 23 \quad (3) & \qquad & d + \tfrac{1}{3}(a + b + c) = 29 \quad (4) \end{array}\]
Adding these equations gives:
\[\begin{array}{lr} & a + b + c + d + \tfrac{1}{3}(3a + 3b + 3c + 3d) = 90 \\ \text{that is} & 2a + 2b + 2c + 2d = 90. \end{array}\]
Hence \(a + b + c + d = 45\). Subtracting a third of this from equation (4), we get
\[d + \tfrac{1}{3}(a + b + c) - \tfrac{1}{3}(a + b + c + d) = 29 - 15.\]
Which gives \(\tfrac{2}{3}d = 14\), so \(d = 21\). We can continue in a similar way to find \(a = 3\), \(b = 9\) and \(c = 12\).
Solution © UK Mathematics Trust, from the PK 2018 solutions.