To solve this problem, let's denote the present ages of A, B, and C as $4x$, $5x$, and $6x$ respectively (since their present ages are in the ratio 4:5:6).

Step-by-Step Solution:

1. 18 Years Ago:

   • 18 years ago, the ages of A, B, and C would have been $4x - 18$, $5x - 18$, and $6x - 18$ respectively.

2. Ratio 18 Years Ago:

   • According to the problem, 18 years ago, the ratio of their ages was 11:16:21.
   • So, we set up the following equations: $\frac{4x - 18}{11} = \frac{5x - 18}{16} = \frac{6x - 18}{21}$

3. Solving the Ratios:

   • Let’s solve for $x$ by equating the first two terms: $\frac{4x - 18}{11} = \frac{5x - 18}{16}$
   • Cross-multiply: $16(4x - 18) = 11(5x - 18)$ $64x - 288 = 55x - 198$ $9x = 90$ $x = 10$

4. Finding Present Ages:

   • Substitute $x = 10$ into the present ages:
     • $A = 4x = 4 \times 10 = 40$
     • $B = 5x = 5 \times 10 = 50$
     • $C = 6x = 6 \times 10 = 60$

5. Sum of Present Ages:

   • The sum of their present ages is: $40 + 50 + 60 = 150$

Answer:

The correct answer is b. 150 years.

Would you like further details on any part of this solution?

Here are some additional related questions you might explore:

1. How would the answer change if the ratio 18 years ago was different?
2. What would happen if the present age ratio was changed to something else, say 3:4:5?
3. Can this method be applied to similar problems with different age gaps?
4. How do ratios help simplify complex age problems?
5. What are some alternative methods to solve age-related ratio problems?

Tip: When dealing with age ratio problems, always start by setting the present ages based on the given ratio and use algebraic equations to work through the conditions provided.