Problems on Ages Explained with Formulas and Examples - Aptitude Questions & Answers

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Problems on Ages Explained with Formulas and Examples is one of the most important topics in Quantitative Aptitude. In this lesson, you will learn concepts, formulas, shortcuts, solved examples, and aptitude questions with answers. This topic is useful for exams like SSC, Bank, CAT, TCS, and other competitive exams.

Ages: Complete Guide with Formulas, Concepts & Examples

What are Age Problems?

Age problems are a type of word problem in which the ages of people are compared either in the present, past, or future. These problems typically involve linear equations and ratios, and require understanding of how ages change over time.

Example: If Arvind is 10 years old now, in 5 years he will be 15, and 3 years ago he was 7.

Fundamental Concepts of Age Problems

Core Principles

1. Age Difference is Constant

The difference between two people\'s ages remains constant throughout their lives.

If A is x years older than B, then:
Age difference = A - B = x (constant)

Example: If Ram is 5 years older than Shyam, this difference will always be 5 years.

2. Ages Change Equally Over Time

All ages increase at the same rate (1 year per year).

After n years: Everyone\'s age increases by n
Before n years: Everyone\'s age decreases by n

3. Ratio Representation

When ages are in ratio a:b, we represent them as ax and bx.

If ratio of A:B = a:b, then:
A\'s age = ax, B\'s age = bx (where x is the common multiplier)

4. Age at Different Times

Present age = a
After n years = a + n
Before n years = a - n

Types of Age Problems

Type 1: Ratio of Ages Changing Over Time

Example Problem: Ratio of present ages of Amisha and Nandana is 13:11. Four years later, the ratio becomes 15:13. Find Amisha\'s present age.

Let present ages: Amisha = 13x, Nandana = 11x
After 4 years: Amisha = 13x + 4, Nandana = 11x + 4
Given ratio: (13x + 4)/(11x + 4) = 15/13
Cross multiply: 13(13x + 4) = 15(11x + 4)
169x + 52 = 165x + 60
169x - 165x = 60 - 52
4x = 8 → x = 2
Amisha\'s present age = 13x = 13 × 2 = 26 years

Amisha\'s present age = 26 years

Example 2: Ratio of ages of A and B is 3:4. After 5 years, ratio becomes 4:5. Find their present ages.

Let present ages: A = 3x, B = 4x
After 5 years: A = 3x + 5, B = 4x + 5
Given: (3x + 5)/(4x + 5) = 4/5
5(3x + 5) = 4(4x + 5)
15x + 25 = 16x + 20
x = 5
A\'s age = 3 × 5 = 15 years, B\'s age = 4 × 5 = 20 years

Type 2: Product of Ages Given

Example Problem: Ratio of father\'s age to son\'s age is 9:5. Product of their ages is 2205. Find ratio of their ages after 5 years.

Let ages: Father = 9x, Son = 5x
Product: 9x × 5x = 2205
45x² = 2205
x² = 2205/45 = 49
x = 7 (age cannot be negative)
Father\'s age = 9 × 7 = 63 years
Son\'s age = 5 × 7 = 35 years
After 5 years: Father = 68, Son = 40
Ratio = 68:40 = 17:10

After 5 years, ratio = 17:10

Type 3: Sum of Ages Given

Example Problem: Sum of present ages of father and son is 60 years. 5 years ago, father was 4 times as old as son. Find their present ages.

Let present ages: Father = F, Son = S
F + S = 60 ...(1)
5 years ago: Father = F - 5, Son = S - 5
Given: (F - 5) = 4(S - 5)
F - 5 = 4S - 20
F - 4S = -15 ...(2)
From (1): F = 60 - S
Substitute in (2): (60 - S) - 4S = -15
60 - 5S = -15
5S = 75 → S = 15
F = 60 - 15 = 45

Father = 45 years, Son = 15 years

Type 4: Age Difference Problems

Example Problem: A is 5 years older than B. After 10 years, A will be twice as old as B. Find their present ages.

Let B\'s present age = x
Then A\'s present age = x + 5
After 10 years: A = (x + 5) + 10 = x + 15
After 10 years: B = x + 10
Given: x + 15 = 2(x + 10)
x + 15 = 2x + 20
x = -5? This means our assumption about \"older\" might be wrong
Correct approach: Let\'s re-read: A is 5 years older than B
So A = B + 5
After 10 years: (B + 5 + 10) = 2(B + 10)
B + 15 = 2B + 20
-B = 5 → B = -5 (impossible)
This means the problem might have different interpretation or needs checking

Let\'s try a working example: A is 10 years older than B. After 5 years, A will be twice as old as B.

Let B = x, then A = x + 10
After 5 years: A = x + 15, B = x + 5
x + 15 = 2(x + 5)
x + 15 = 2x + 10
x = 5
B = 5 years, A = 15 years

Type 5: Multiple People Age Problems

Example Problem: The average age of 3 people is 30 years. Their ages are in ratio 2:3:5. Find the age of the youngest.

Let ages be 2x, 3x, 5x
Sum of ages = 2x + 3x + 5x = 10x
Average = 10x/3 = 30
10x = 90 → x = 9
Youngest = 2x = 18 years

Example 2: A father is 4 times as old as his son. After 10 years, he will be twice as old as his son. After another 10 years, what will be the ratio?

Let son\'s age = x, father\'s age = 4x
After 10 years: 4x + 10 = 2(x + 10)
4x + 10 = 2x + 20
2x = 10 → x = 5
Present: Son = 5, Father = 20
After another 10 years (total 20 years): Son = 25, Father = 40
Ratio = 40:25 = 8:5

Type 6: \"n times\" Age Problems

Example Problem: A father is 3 times as old as his son. After 12 years, he will be twice as old as his son. Find their present ages.

Let son\'s age = x, father\'s age = 3x
After 12 years: 3x + 12 = 2(x + 12)
3x + 12 = 2x + 24
x = 12
Son = 12 years, Father = 36 years

Example 2: A is twice as old as B. 10 years ago, A was 4 times as old as B. Find their present ages.

Let B\'s age = x, then A\'s age = 2x
10 years ago: A = 2x - 10, B = x - 10
Given: 2x - 10 = 4(x - 10)
2x - 10 = 4x - 40
2x = 30 → x = 15
B = 15 years, A = 30 years

Type 7: \"Years Hence/Ago\" Problems

Example Problem: 10 years ago, A was half of B\'s age. If the ratio of their present ages is 3:4, what will be the sum of their ages after 5 years?

Let present ages: A = 3x, B = 4x
10 years ago: A = 3x - 10, B = 4x - 10
Given: 3x - 10 = ½(4x - 10)
2(3x - 10) = 4x - 10
6x - 20 = 4x - 10
2x = 10 → x = 5
Present: A = 15, B = 20
After 5 years: A = 20, B = 25
Sum = 20 + 25 = 45 years

Advanced Age Problem Strategies

Strategy 1: Using Age Difference Constant

Example: The ratio of ages of A and B is 5:3. After 6 years, the ratio becomes 7:5. Find their ages.

Let present ages: A = 5x, B = 3x
Age difference = 5x - 3x = 2x (constant)
After 6 years: A = 5x + 6, B = 3x + 6
Ratio (5x+6):(3x+6) = 7:5
5(5x+6) = 7(3x+6)
25x + 30 = 21x + 42
4x = 12 → x = 3
A = 15 years, B = 9 years

Alternative using constant difference: Difference in ratio terms = 7-5=2 and 5-3=2. Constant difference corresponds to 2x.

Strategy 2: Backward Calculation

Example: The sum of ages of father and son is 50 years. 5 years ago, father was 7 times as old as son. Find current ages.

5 years ago, sum of ages = 50 - (5+5) = 40 years
Let son\'s age 5 years ago = x
Father\'s age 5 years ago = 7x
x + 7x = 40 → 8x = 40 → x = 5
5 years ago: Son = 5, Father = 35
Present: Son = 10, Father = 40

Strategy 3: Using Equations with Multiple Variables

Example: A\'s age is twice B\'s age. 8 years later, A\'s age will be 4 years more than B\'s age. Find their ages.

Let B\'s age = x, A\'s age = 2x
After 8 years: A = 2x + 8, B = x + 8
Given: (2x + 8) = (x + 8) + 4
2x + 8 = x + 12
x = 4
B = 4 years, A = 8 years

Ages: Frequently Asked Questions

Why is age difference constant in these problems?

Age difference remains constant because both people age at the same rate (1 year per year). If A is 5 years older than B today, A will always be 5 years older than B.

How to set up equations for age problems?

1. Define variables for unknown ages. 2. Express all conditions in terms of these variables. 3. Account for time changes (past/future). 4. Set up equations based on given ratios, sums, or products.

What\'s the most common mistake in age problems?

Forgetting that when we go back or forward in time, all ages change. If going back 5 years, subtract 5 from everyone\'s age, not just one person.

How to handle problems with \"n times as old\"?

If A is n times as old as B, then A = n×B. Remember this relationship changes over time unless n=1.

Can ages be negative in solutions?

No, ages cannot be negative. If you get a negative age, recheck your equation setup or interpretation of the problem.

Practice Problems

Problem 1

A\'s age is 3 times B\'s age. After 10 years, A\'s age will be twice B\'s age. Find their present ages.

Solution: Let B=x, A=3x. After 10 years: 3x+10=2(x+10) → 3x+10=2x+20 → x=10. B=10 years, A=30 years.

Problem 2

The sum of ages of mother and daughter is 50 years. 5 years ago, mother was 7 times as old as daughter. Find mother\'s present age.

Solution: 5 years ago, sum=50-10=40. Let daughter=x, mother=7x. x+7x=40 → x=5. 5 years ago: daughter=5, mother=35. Present: daughter=10, mother=40.

Problem 3

Ratio of ages of A and B is 4:3. After 6 years, ratio becomes 11:9. Find B\'s present age.

Solution: Let A=4x, B=3x. (4x+6)/(3x+6)=11/9 → 9(4x+6)=11(3x+6) → 36x+54=33x+66 → 3x=12 → x=4. B=3×4=12 years.

Problem 4

A father said to his son, \"I was as old as you are now when you were born.\" If father is 40 years now, find son\'s age.

Solution: Age difference = x (constant). When son was born (0 years), father was x years. Now son=x, father=40. So 40-x=x → 2x=40 → x=20. Son=20 years.

Problem 5

The average age of 5 children is 8 years. If the age of father is included, average becomes 15 years. Find father\'s age.

Solution: Sum of 5 children=5×8=40. With father: 6 people, average=15, sum=90. Father\'s age=90-40=50 years.

Important Formulas and Relationships

Basic Relationships:
• Present age = a
• After n years = a + n
• Before n years = a - n
• Age difference = constant

Ratio Problems:
If ratio = a:b, then ages = ax and bx
After n years: (ax+n):(bx+n) = new ratio

\"n times\" Problems:
If A = n×B now, and after m years A = k×B, then:
nB + m = k(B + m)

Sum/Difference Problems:
A + B = S (sum)
A - B = D (difference, constant)
A = (S + D)/2, B = (S - D)/2

Frequently Asked Questions

What is Problems on Ages Explained with Formulas and Examples?

Problems on Ages Explained with Formulas and Examples is an important aptitude topic used in competitive exams that tests your logical reasoning and problem-solving abilities.

Is Problems on Ages Explained with Formulas and Examples important for competitive exams?

Yes, Problems on Ages Explained with Formulas and Examples is frequently asked in SSC, Bank, CAT, TCS, and other placement exams. It's essential to master this topic for better scores.

How to prepare Problems on Ages Explained with Formulas and Examples easily?

Practice solved examples, learn formulas and shortcuts, and attempt practice questions regularly to master Problems on Ages Explained with Formulas and Examples.

What are the important formulas in Problems on Ages Explained with Formulas and Examples?

Key formulas vary by topic, but generally include basic concepts, shortcuts, and standard problem-solving approaches specific to Problems on Ages Explained with Formulas and Examples.

How many questions come from Problems on Ages Explained with Formulas and Examples?

Typically 5-10 questions come from Problems on Ages Explained with Formulas and Examples in most competitive exams, making it a high-scoring section.

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