Linear equations in ages – A is 16 years older than B, and half of B’s age equals one-third of A’s age. Find their present ages.

Introduction / Context:
Here we mix an additive relation (“older by 16”) with a fractional comparison (“half equals one-third”). Such systems are best handled with two equations in two variables, then solved by substitution or elimination.

Given Data / Assumptions:

• A = B + 16.
• (1/2) * B = (1/3) * A.
• All ages are present ages.

Concept / Approach:
Convert the fractional statement to an equation without fractions. From (1/2)B = (1/3)A, multiply both sides by 6 to get 3B = 2A. Substitute A = B + 16 into 3B = 2A and solve for B, then back-solve for A.

Step-by-Step Solution:
3B = 2A.With A = B + 16 ⇒ 3B = 2(B + 16) = 2B + 32.Thus B = 32, and A = B + 16 = 48.

Verification / Alternative check:
(1/2)B = 16 and (1/3)A = 16, so the fractional condition holds. The 16-year difference also holds: 48 − 32 = 16.

Why Other Options Are Wrong:

• 40/32 does not satisfy 3B = 2A.
• 46/30 and 32/16 violate one or both constraints.
• “None of these” is not needed because 48/32 is consistent.

Common Pitfalls:
Misreading “half B equals one-third A,” or subtracting 16 in the wrong direction. Keep the algebraic mapping faithful to the words.

Final Answer:
A = 48, B = 32