Problems on Ages: Comprehensive Guide for CSAT

1. Introduction and Importance in CSAT

The topic of “Problems on Ages” is a staple in the quantitative and reasoning sections of the CSAT (Civil Services Aptitude Test) and MPSC exams. It is essentially an application of Linear Equations and Ratios. Candidates can expect 1-2 questions from this area every year. These questions are highly scoring if approached systematically. Mastering this topic improves algebraic thinking, enabling you to translate word problems into mathematical equations effortlessly, a skill which is crucial for overall success in the quantitative aptitude section.

2. Core Concepts, Formulas, and Tricks

Problems on ages generally revolve around finding the present, past, or future ages of individuals based on given conditions and ratios.

Key Rules and Equations

• If the current age is x, then age n years ago was x – n.
• If the current age is x, then age n years from now (hence) will be x + n.
• The difference in ages between two individuals always remains constant, regardless of past or future. If A is 5 years older than B today, A will always be 5 years older than B.
• If the ratio of present ages of A and B is x : y, assume their ages are kx and ky.

Shortcut Tricks for Quick Calculation

• Cross Multiplication Method: When ratio of ages in the past (or present) is a:b, and after/before some years it becomes c:d, you can easily find the value of one ratio unit without making long equations. Calculate: (Difference of new ratio × Years) / (Cross product difference of ratios).
• Options Elimination: Age problems can often be solved quickly by plugging in the options. Look for multiples corresponding to ratios given in the question.

3. Solved Examples with Step-by-Step Explanations

Example 1 (Basic Equation)

Question: Ten years ago, A was half of B in age. If the ratio of their present ages is 3:4, what will be the total of their present ages?

Step-by-step Explanation:

1. Let their present ages be 3x (for A) and 4x (for B).
2. Ten years ago, A’s age was (3x – 10) and B’s age was (4x – 10).
3. According to the condition: (3x – 10) = 1/2 × (4x – 10).
4. Solving the equation: 2 × (3x – 10) = 4x – 10
5. 6x – 20 = 4x – 10
6. 2x = 10, which gives x = 5.
7. Present age of A = 3 × 5 = 15 years.
8. Present age of B = 4 × 5 = 20 years.
9. Total of present ages = 15 + 20 = 35 years.

Answer: 35 years

Example 2 (Ratio and Cross Multiplication Method)

Question: The ratio of ages of two persons is 5:9 and the age of one of them is greater than the other by 40 years. The sum of their ages in years is?

Step-by-step Explanation:

1. Let the ages be 5x and 9x.
2. The difference between their ages is given as 40 years.
3. So, 9x – 5x = 40.
4. 4x = 40. Therefore, x = 10.
5. The ages are 5 × 10 = 50 years and 9 × 10 = 90 years.
6. Sum of their ages = 50 + 90 = 140 years.

Answer: 140 years

Example 3 (Future and Past Relations)

Question: Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?

Step-by-step Explanation:

1. “Three times more” means Father’s age is Son’s age + 3 × Son’s age = 4 times Son’s age. (This is a common trap!). Let Ronit’s present age be x. Father’s present age = x + 3x = 4x.
2. After 8 years: Ronit = x + 8, Father = 4x + 8.
3. Condition: (4x + 8) = 2.5 × (x + 8).
4. 4x + 8 = 2.5x + 20.
5. 1.5x = 12, so x = 8.
6. Present ages: Ronit = 8, Father = 32.
7. After further 8 years (Total 16 years from present): Ronit = 8 + 16 = 24. Father = 32 + 16 = 48.
8. Ratio of Father’s age to Ronit’s age = 48 / 24 = 2 times.

Answer: 2 times

Example 4 (Constant Age Difference)

Question: My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, then what was the age of my father and mother respectively when my brother was born?

Step-by-step Explanation:

1. Let’s establish a timeline based on “When Brother was born”.
2. When Brother was born: Brother’s age = 0.
3. Sister was 4 years old when Brother was born. So, Sister’s age = 4.
4. Father was 28 when Sister was born. So when Brother was born (4 years later), Father’s age = 28 + 4 = 32.
5. Brother is 3 years elder to me. This means I was born 3 years AFTER my brother.
6. Mother was 26 when I was born.
7. Therefore, 3 years BEFORE I was born (which is when my Brother was born), Mother’s age = 26 – 3 = 23.
8. Age of Father = 32, Age of Mother = 23.

Answer: Father 32, Mother 23

4. Pro-tips to Avoid Common Mistakes

• “Times more” vs “Times”: As seen in Example 3, “A is 3 times older than B” or “A is 3 times more than B” means A = B + 3B = 4B. But “A is 3 times as old as B” means A = 3B. Read carefully!
• Constant Difference: The age difference between two people never changes. This is a very useful property for cross-verifying your answers.
• Present Age Base: Always try to assume the “present age” as ‘x’ unless specified otherwise. This prevents confusion with + and – signs.
• Option Checking: Before solving complex equations, glance at the options. Often, if a ratio is given as 3:4, the answer must be a multiple of 3 or 4, which helps eliminate wrong choices immediately.

5. Practice Questions

1. A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of the son is?
2. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
3. A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, then how old is B?
4. Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
5. Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?