The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is:

Question

Accepted Answer

Let the three distinct real numbers be $$x$$, $$y$$, and $$z$$ such that $$x < y < z$$.

1. **Initial Conditions:**  
   - The average of the three numbers is given as 28:  
     $$
     x + y + z = 28 	imes 3 = 84
     $$

2. **Modified Conditions:**  
   - After modifying the smallest and largest numbers:  
     $$
     (x + 7) + y + (z - 10) = 84 + 7 - 10 = 81
     $$  
   - The new arithmetic mean becomes $$27$$:  
     $$
     \frac{81}{3} = 27
     $$  
   - The middle number satisfies:  
     $$
     y + 2 = 27 \implies y = 25
     $$

3. **Substitute $$y = 25$$:**  
   - From $$x + y + z = 84$$:  
     $$
     x + 25 + z = 84 \implies x + z = 59
     $$

4. **Difference Between Largest and Smallest Numbers:**  
   - The difference between modified largest and smallest numbers is given as $$64$$:  
     $$
     (z - 10) - (x + 7) = 64
     $$  
   - Simplify:  
     $$
     z - x - 17 = 64 \implies z - x = 81
     $$

5. **Solve Simultaneous Equations:**  
   - From $$x + z = 59$$ and $$z - x = 81$$:  
     Add both equations:  
     $$
     (x + z) + (z - x) = 59 + 81 \implies 2z = 140 \implies z = 70
     $$

Thus, the largest number in the original set is **70**.