Ages

Ages

Concept Explanation

Age problems are a staple of quantitative aptitude because they test your ability to translate real situations into mathematical equations. The good news: the underlying math is straightforward algebra. The tricky part is reading the wording carefully, because these problems love to play with time — “5 years ago,” “after 10 years,” “twice as old as,” and so on.

The single most important principle in age problems is that everyone ages at the same rate. Time passes for everyone equally. This means if you’re solving a problem involving two or more people, the number of years you add or subtract applies to ALL of them. A common mistake is to apply the time shift to only one person, which breaks the relationship.

Another key concept: age difference never changes. If Rina is 3 years older than Sameer today, she will be 3 years older than Sameer in 10 years, and she was 3 years older than Sameer 20 years ago. The gap is constant because both people get older by the same amount each year. This principle is incredibly useful — if you ever know the age gap and one person’s current age, you immediately know the other’s current age too.

Age ratios work differently. If a mother is 40 and her daughter is 10, the ratio is 4:1. After 20 years, they’ll be 60 and 30 — the ratio is now 2:1. The ratio shrinks over time because both numbers are growing, but the gap (30 years) stays fixed. This is why problems that give you future or past ratios usually involve more algebra — you have to solve for the present ages first.

Key Formulas

Symbol	Meaning
P	Present age of person 1
Q	Present age of person 2
x	Years in the future (positive) or past (negative)
G	Constant age gap between two people

Step-by-Step Example

Q: The sum of the ages of a father and his son is 60 years. Five years ago, the father’s age was four times the son’s age. Find their present ages.

Step 1: Define variables. Let son’s present age = S, father’s present age = F. Given: F + S = 60

Step 2: Translate “5 years ago” condition. Five years ago: father’s age = F – 5, son’s age = S – 5. Given: F – 5 = 4(S – 5)

Step 3: Solve the system of equations. From equation 1: F = 60 – S Substitute: (60 – S) – 5 = 4(S – 5) 55 – S = 4S – 20 55 + 20 = 4S + S 75 = 5S → S = 15

Step 4: Find father’s age. F = 60 – 15 = 45

Answer: Son is 15 years old, father is 45 years old.

Common Mistakes

• Applying time change to only one person → Always apply “n years ago” or “after n years” to EVERY person’s present age in the equation.
• Forgetting that ages must be positive → If you get a negative age, something went wrong with the time direction (likely the “n years ago” subtraction was applied wrong).
• Confusing average age with individual ages → “Sum of ages” problems require knowing how many people are involved.

Quick Test (2 Qs)

1. Q: A mother is 3 times as old as her son. After 12 years, she will be twice as old. Find the son’s present age. Options: A) 6 B) 10 C) 12 D) 15. Ans: C (Reason: Let son’s age = x, mother’s = 3x. After 12 years: 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12)

2. Q: The age difference between two brothers is 8 years. Five years ago, the elder brother’s age was three times the younger’s. Find the elder brother’s age. Options: A) 15 B) 17 C) 20 D) 22. Ans: B (Reason: Let younger = y, elder = y + 8. Five years ago: y + 8 – 5 = 3(y – 5) → y + 3 = 3y – 15 → 18 = 2y → y = 9, elder = 17)