Father four times son now; seven times five years ago: Ravi’s father is four times Ravi’s current age. Five years back, the father was seven times Ravi’s age then. What is the father’s present age?

Difficulty: Easy

Correct Answer: 40 yr

Explanation:


Introduction / Context:
This is a standard two-equation age problem. The current multiplicative relation (father is 4 times the son) is paired with an earlier-time multiplicative relation (7 times five years ago). Solving the simultaneous linear equations gives the exact ages.


Given Data / Assumptions:

  • Let Ravi’s present age be R, father’s present age be F.
  • F = 4R (now).
  • Five years ago: F − 5 = 7(R − 5).


Concept / Approach:
Substitute F from the first relation into the second and solve for R, then compute F. This avoids introducing extra variables and minimizes algebraic errors.


Step-by-Step Solution:

F = 4RF − 5 = 7(R − 5)4R − 5 = 7R − 35 ⇒ 30 = 3R ⇒ R = 10F = 4R = 40


Verification / Alternative check:
Five years ago: father 35, Ravi 5; 35 is indeed 7 times 5. The data is fully consistent.


Why Other Options Are Wrong:
84/70/35 are inconsistent with both relations; they fail either the current 4x relation or the 7x past relation.


Common Pitfalls:
Reversing who is a multiple of whom, or applying the 5-year shift to only one person. Always subtract the same number of years from both ages when going backwards.


Final Answer:
40 yr

More Questions from Problems on Ages

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion