The sum of the present ages of a brother and sister is 21 years. Five years ago, the product of their ages was 28. What are their present ages (brother, sister)?

Introduction / Context:
This is a classic quadratic-type age problem involving the sum and the product of ages at different times. We are told the present sum of the ages of a brother and sister and the product of their ages five years ago. The question asks us to find their present ages. This problem illustrates how to translate real-world conditions into simultaneous equations and solve them logically.

Given Data / Assumptions:

- Let the brother's present age be B years and the sister's present age be S years.
- The sum of their present ages is B + S = 21 years.
- Five years ago, their ages were B − 5 and S − 5 years respectively, and the product of those ages was 28, i.e., (B − 5)(S − 5) = 28.
- Ages are positive whole numbers and B, S > 5, since we talk about ages 5 years ago.

Concept / Approach:
We have two unknowns and two equations: one linear equation (sum of ages) and one quadratic equation (product of ages in the past). We can express one variable in terms of the other using the sum equation and substitute into the product equation, or reason by trying integer pairs that sum to 21 and satisfy the product condition. Both methods ultimately identify the correct pair of ages.

Step-by-Step Solution:
Step 1: From the sum, B + S = 21. Step 2: Five years ago, their ages were B − 5 and S − 5, and their product was 28, so (B − 5)(S − 5) = 28. Step 3: Expand the product: (B − 5)(S − 5) = BS − 5B − 5S + 25 = 28. Step 4: Rearrange to BS − 5B − 5S + 25 − 28 = 0 ⇒ BS − 5B − 5S − 3 = 0. Step 5: From B + S = 21, we can write S = 21 − B and substitute into the product equation. Step 6: Substitute S: B(21 − B) − 5B − 5(21 − B) − 3 = 0. Step 7: Simplify step by step: 21B − B^2 − 5B − 105 + 5B − 3 = 0. Step 8: Combine like terms: −B^2 + 21B − 105 − 3 = −B^2 + 21B − 108 = 0 ⇒ B^2 − 21B + 108 = 0. Step 9: Factor the quadratic: B^2 − 21B + 108 = (B − 9)(B − 12) = 0. Step 10: So B = 9 or B = 12. Accordingly, S = 21 − B gives (B, S) as (9, 12) or (12, 9).

Verification / Alternative check:
Taking the pair (9, 12): Five years ago, their ages were 4 and 7. The product 4 × 7 = 28, which matches the given condition. Taking the pair (12, 9) simply swaps the roles of brother and sister. The problem asks for ages in the order (brother, sister), and the option 9 years, 12 years matches one consistent assignment. Both the sum and product conditions hold, so these are the correct present ages.

Why Other Options Are Wrong:
Pairs like (6, 15), (7, 14), (8, 13) or (10, 11) either do not sum to 21 or, when reduced by 5 years each, do not give a product of 28. For example, (6, 15) sums to 21, but five years ago the ages would be 1 and 10, whose product is only 10, not 28. Similar checks rule out the other pairs.

Common Pitfalls:
Students sometimes forget that both ages must be reduced by 5 when going back in time, or they misinterpret the product condition as applying to present ages. Others may attempt to solve the quadratic but make factoring errors. A quick integer trial of pairs that sum to 21 can also help: (9, 12) gives past ages (4, 7) with product 28, a strong hint that the algebra is consistent.

Final Answer:
The present ages of the brother and sister are 9 years and 12 years respectively.