Seven years ago, the average age of A, B, and C was 51 years. Today A is 3 years older than B, and B is 3 years older than C. What are their present ages (in years) for A, B, and C?

Introduction:
Age-averages often require translating averages into totals and then applying given relationships between people. Here, we know a past average and fixed differences among the three persons' ages. We compute the current total and solve using linear relations.

Given Data / Assumptions:
Seven years ago: average of A, B, C = 51 years ⇒ past total = 3 * 51 = 153. Each person aged 7 years since then ⇒ current total adds 3 * 7 = 21. A = B + 3; B = C + 3 ⇒ A = C + 6.

Concept / Approach:
Compute the present total first, then express all ages in terms of the youngest (C). Solve the single-variable equation for C, then back-calculate B and A using the given differences.

Step-by-Step Solution:
Present total = 153 + 21 = 174. Let C = x. Then B = x + 3, A = x + 6. Sum = x + (x + 3) + (x + 6) = 3x + 9. Set 3x + 9 = 174 ⇒ 3x = 165 ⇒ x = 55. So C = 55, B = 58, A = 61.

Verification / Alternative check:
Seven years ago the ages would be 54, 51, and 48. Average then = (54 + 51 + 48) / 3 = 153 / 3 = 51, which matches the given value.

Why Other Options Are Wrong:
54, 51, and 48: These are the ages seven years ago, not the present ages. 55, 58, and 61 or 48, 51, and 54 or 62, 59, and 56: These violate the required order A = B + 3 and B = C + 3 with the correct present total 174.

Common Pitfalls:
Using 51 as the present average, or forgetting to add 7 years to each person before solving. Always update the total when time passes for multiple people.

Final Answer:
The present ages are 61, 58, and 55 years for A, B, and C respectively.