# Problems on Ages – Aptitude Questions and Answers

Problems on ages are an important topic in the quantitative aptitude section of many competitive exams.

Age aptitude questions involve using Algebraic Equations to determine the age of an individual based on the given data. While they may seem confusing at first, practice and a good understanding of the underlying concepts can help candidates become proficient in solving age-related problems.

By mastering problems of age, candidates can develop valuable problem-solving and analytical skills that can be applied in various aspects, making them well-prepared for a range of quantitative aptitude questions.

To learn more about the Quantitative Aptitude syllabus and sample questions related to age, candidates can refer to the linked article.

**Practice Quiz:**

## Age Problem Formulas

Here are some formulas and tricks to solve age problems that can make it simple for candidates to solve problems with ease:

- An individual’s age after n years will be (x+n) years old, and their age before n years will be (x-n) years if their present age is x.
- The age of one person can be considered as px and the age of the other person as qx, if the age is expressed as a ratio of p:q.
- If a person is currently x years old, they will be (x × n) years old in n years.
- If a person is currently x years old, then 1/n of their age will be (x/n) years.

By mastering these tricks and formulas, you can easily solve various age-related problems and improve your quantitative aptitude skills.

## Sample Problems on Ages

### Q1: A’s age after 15 years would be equal to 5 times his age 5 years ago. Find his age 3 years hence.

** Solution **:

Let A’s present age be ‘n’ years.

According to the question,

n + 15 = 5 (n – 5)

=> n + 15 = 5 n – 25

=> 4n = 40

=> n = 10

=> A’s present age = 10 years

Therefore, A’s age 3 years hence = 10 + 3 = 13 years

### Q2: The product of the ages of A and B is 240. If twice the age of B is more than A’s age by 4 years, what was B’s age 2 years ago?

** Solution **:

Let A’s present age be x years. Then, B’s present age = 240 / x years

So, according to question

2 (240 / x ) – x = 4

=> 480 – x2 = 4 x

=> x2 + 4 x – 480 = 0

=> (x + 24) (x – 20) = 0

=> x = 20

=> B’s present age = 240 / 20 = 12 years

Thus, B’s age 2 years ago = 12 – 2 = 10 years

### Q3: The present age of a mother is 3 years more than three times the age of her daughter. Three years hence, the mother’s age will be 10 years more than twice the age of the daughter. Find the present age of the mother.

** Solution **:

Let the daughter’s present age be ‘n’ years.

=> Mother’s present age = (3n + 3) years

So, according to the question

(3n + 3 + 3) = 2 (n + 3) + 10

=> 3n + 6 = 2n + 16

=> n = 10

Hence, mother’s present age = (3n + 3) = ((3 x 10) + 3) years = 33 years

### Q4: The ratio of present ages of A and B is 6 : 7. Five years hence, this ratio would become 7 : 8. Find the present age of A and B.

** Solution **:

Let the common ratio be ‘n’.

=> A’s present age = 6 n years

=> B’s present age = 7 n years

So, according to the question

(6 n + 5) / (7 n + 5) = 7 / 8

=> 48 n + 40 = 49 n + 35

=> n = 5

Thus, A’s present age = 6 n = 30 years and B’s present age = 7 n = 35 years

Test your knowledge of Ages in Quantitative Aptitude with the quiz linked below, containing numerous practice questions to help you master the topic:-

## Also Practice:

H.C.F and L.C.M Aptitude Questions

Work and Wages Aptitude Questions

Pipes and Cistern Aptitude Questions

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