The average of three distinct real numbers is 50. If the smallest number is increased by 16 and the largest is reduced by 7, the order of the numbers stays the same, the new arithmetic mean becomes 9 more than the middle number, and the difference between the smallest and largest numbers becomes 55. What is the largest number in the original set?
Answer and solution
Correct answer: 92
Solution
Let the numbers be $A < B < C$. $A + B + C = 150$.
New sum $= (A + 16) + B + (C - 7) = 159$, new mean $= 53$
Mean is 9 more than the middle number: $B = 53 - 9 = 44$
$A + C = 106 \Rightarrow A = 106 - C$
$(C - 7) - (A + 16) = 55 \Rightarrow C - 7 - 106 + C - 16 = 55$
$2C = 184 \Rightarrow C = 92$
New sum $= (A + 16) + B + (C - 7) = 159$, new mean $= 53$
Mean is 9 more than the middle number: $B = 53 - 9 = 44$
$A + C = 106 \Rightarrow A = 106 - C$
$(C - 7) - (A + 16) = 55 \Rightarrow C - 7 - 106 + C - 16 = 55$
$2C = 184 \Rightarrow C = 92$