⚡ AZUCATION SHORTCUT:
Total Increase = $5\% \times (30 \times 60000) = 90000$.
This increase comes from $20\%$ of managers' total salary ($5M$):
$0.20 \times 5M = 90000 \implies M = 90000$.
Using Average Eq: $5(90000) + 25(E) = 30(60000) \implies 4.5L + 25E = 18L \implies 25E = 13.5L$
$E = \frac{1350000}{25} = \mathbf{54000}$.

Step 1: Calculate the total increase in salary pool
Number of employees = $30$. Original average = $60,000$.
Increase in average = $5\%$ of $60,000 = 3,000$.
Total increase in salary = $30 \times 3,000 = 90,000$.

Step 2: Find the average salary of a manager (M)
The increase only comes from managers. Let $M$ be the manager's average salary.
Increase $= 20\% \text{ of } (5 \times M) = 90,000$
$0.2 \times 5 \times M = 90,000 \implies 1 \times M = 90,000$.
The average salary of a manager is $90,000$.

Step 3: Solve for the engineer's average salary (E)
Total Salary $= (5 \times 90,000) + (25 \times E) = 30 \times 60,000$
$450,000 + 25E = 1,800,000$
$25E = 1,350,000$
$E = \frac{1,350,000}{25} = \mathbf{54,000}$.

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