Father and two daughters – Father is three times the elder daughter now; five years ago he was eight times the younger daughter. If the daughters differ by 5 years, what is the father’s present age?

Introduction / Context:We are given two conditions involving the father's age relative to each of two daughters at different times, plus a fixed age gap between daughters. This leads to a compact system of equations.

Given Data / Assumptions:

• Let elder daughter = E, younger daughter = Y (present ages).
• Father = F (present age).
• F = 3E (now).
• Five years ago: F − 5 = 8(Y − 5).
• E − Y = 5 (elder is older by 5).

Concept / Approach:Express E as Y + 5, then F as 3(Y + 5). Substitute into the “five years ago” relation to solve for Y. Back-substitute to get E and then F.

Step-by-Step Solution:E = Y + 5; F = 3E = 3(Y + 5) = 3Y + 15.Five years ago: F − 5 = 8(Y − 5) ⇒ (3Y + 15) − 5 = 8Y − 40.3Y + 10 = 8Y − 40 ⇒ 50 = 5Y ⇒ Y = 10.E = Y + 5 = 15; F = 3E = 45.

Verification / Alternative check:Five years ago: F − 5 = 40 and Y − 5 = 5, and 40 = 8 × 5, confirmed.

Why Other Options Are Wrong:Other numbers fail either the “now 3× elder” or the “5 years ago 8× younger” condition.

Common Pitfalls:Mixing which girl is referenced in each condition, or using the present ages inside the “five years ago” condition without subtracting 5.

Final Answer:45