The problem provides the current ratio of ages for A, B, and C as $4:5:6$, and the ratio of their ages 18 years ago as $11:16:21$. We are tasked with finding the total of their present ages.

Step-by-Step Solution:

1. Define Variables: Let the present ages of A, B, and C be $4x$, $5x$, and $6x$, respectively. Here $x$ is the common multiplier.

2. Set up the equation for 18 years ago: The ages of A, B, and C 18 years ago were: $(4x - 18), (5x - 18), \text{ and } (6x - 18)$ Their ratio 18 years ago is given as $11:16:21$. Hence, we can write: $\frac{4x - 18}{11} = \frac{5x - 18}{16} = \frac{6x - 18}{21}$

3. Equating the first two ratios: $\frac{4x - 18}{11} = \frac{5x - 18}{16}$ Cross-multiply: $16(4x - 18) = 11(5x - 18)$ Simplify: $64x - 288 = 55x - 198$ $64x - 55x = 288 - 198$ $9x = 90 \implies x = 10$

4. Calculate present ages: Substitute $x = 10$ into the expressions for the present ages: $A = 4x = 40, \; B = 5x = 50, \; C = 6x = 60$

5. Find the total of their present ages: $\text{Total} = 40 + 50 + 60 = 150$

Final Answer:

The sum total of their present ages is 150 years.

Follow-up Questions:

1. How would the solution change if the difference in years was not 18?
2. What is the ratio of their ages in 10 years?
3. How can we verify that $x = 10$ satisfies all the ratios?
4. What mathematical concept underlies the use of ratios and proportions in this problem?
5. Could the problem be solved using a different algebraic approach?

Tip: Always verify your results by substituting back into the original problem to check for consistency.