← Reasoning

Problems Based on Ages

Reasoning · Verbal Reasoning

8 Part2 worked examples11 practice questions

Age means the time that has passed since birth. In the exam the questions arrive as a small story: “The father is four times as old as his son; after five years…”. Your job is to translate the story into the language of maths and then solve it.

This chapter stands not on memorised tricks but on three simple ideas: (1) everyone grows by the same amount every year, so the gap between two ages never changes; (2) a ratio can be thought of in units; (3) whatever is unknown can be called x and an equation built. These three solve every type of question.

If you are weak in maths, do not worry — we start with the easiest ideas and give the reason at every step.

How to use this chapter

  1. Read — the idea first, then the method.
  2. Study the solved example — the “Example” boxes show the full solution. At every step ask: why was this done?
  3. Try it yourself — solve the “Try it yourself first” questions on paper, then choose an option. If stuck, take the hint; open the solution last.
  4. Know the level — each question is marked Basic, Standard or Advanced.
NoteThe questions you get right are remembered in this browser. Check your progress in the panel on the right (the “Contents” button on a phone).
1

Foundation — the maths of age

How ages change with time

Counting forward and back, why the gap stays fixed, and three ways to turn words into maths.

Age with time — three firm rules

Add for later, subtract for earlier, the gap is fixed

Counting n years ahead or back, the gap never changes, the ratio does, and how a group's total changes.

Every age problem has its roots in three small facts. Once you understand them, all the later methods will make sense on their own.

Rule 1 — add for “after”, subtract for “ago”

If someone is 25 today, after 6 years the person will be 25 + 6 = 31, and 6 years ago was 25 − 6 = 19. This is so simple that we often take it lightly — yet most mistakes happen right here: adding for “ago” or subtracting for “after”.

Rule 2 — the gap between two ages never changes

The father is 32 and the son is 6. The father is 26 years older. After one year both are one year older, so the gap is still 26. Look at the table — the gap row is the same in every column.

Now4 years later10 years later5 years ago
Father32364227
Son610161
Gap26262626
Ratio16 : 318 : 521 : 827 : 1

The gap never changes, but the ratio keeps changing (see the last row). This is the biggest secret of age problems: when the ratio changes, hold on to the gap.

Rule 3 — the total of a group

One person's age rises by n in n years. For m people the total rises by m × n. If four people total 100 today, after 5 years the total is 100 + 4 × 5 = 120.

FactDoes it change?Example (father 32, son 6)
Each ageYes, by the same amountAfter 10 years 42 and 16
Gap of two peopleNo26 → 26
Ratio of two peopleYes16 : 3 → 21 : 8
Sum of two peopleYes, by 2 × n38 → 58
RememberAt birth the age is 0. So “the mother was 25 when the child was born” means the gap between their ages is 25 years.
Example 1 · Basic
Nitin is 30 years older than Chetan. How old will Nitin be when Chetan is 23?
Solution, step by step
  1. 1The gap of 30 years always stays the same.
  2. 2So Nitin = 23 + 30 = 53 years.
Answer 53 years
The gap is fixed, so no equation is needed here.
BasicTry it yourself first
The mother is 36 years old now. What will the mother's age be after 11 years?
BasicTry it yourself first
Bindu is 41 and Vikas is 10 now. What will be the sum of their ages after 12 years?
BasicTry it yourself first
The father is 44 and the son is 13. What was the sum of their ages 3 years ago?
BasicTry it yourself first
Bindu is 31 years older than Rajesh. How old will Bindu be when Rajesh is 17?
BasicTry it yourself first
The present ages of 4 people (Bindu, Mahesh, Seema, Indu) add up to 70 years. What will be the sum of their ages after 5 years?
BasicTry it yourself first
The mother was 21 when her daughter was born. The daughter is 6 today. How old is the mother today?

From words to equations — three methods

Unit method, x method and table method

A dictionary from words to maths and how to pick the right one of the three methods.

Age questions arrive in words. The first job is to translate. The table below is your dictionary.

The question saysIn mathsExample
Ratio p : qp units and q units (or px and qx)3 : 2 → 3x, 2x
A is k times BB = x, A = kxthree times → x, 3x
After n years+ n to every agex + 5
n years ago− n from every agex − 5
A is d years older than BA = B + dA = B + 7
Sum SA + B = Sx + 3x = 48
Average a, m peopleTotal = a × m4 × 30 = 120

Method 1 — the unit method (for ratio questions)

The ratio 5 : 3 means one person has 5 equal pieces of age and the other has 3. Call each piece 1 unit. Whatever amount is given (a sum or a difference), match it to the sum or difference of units and find the value of 1 unit.

Steps of the unit method
Ratio → unitsSum/difference → units1 unit = amount ÷ unitsunit × own ratio number

Method 2 — the x method (equation)

When the conditions say “n years ago / after”, call the unknown age x, write every other age in terms of x, turn the condition into an equation and solve it. At the end re-read the question to see what x stood for and what is asked — sometimes x is not the age but a part of it.

Method 3 — the table method

When there are two people and two times (now and n years later), draw a small table with people as rows and times as columns. Nothing gets left out.

NowAfter 6 years
Father4x4x + 6
Sonxx + 6

Now write the condition (say “after 6 years the father will be three times the son”) from the right-hand column: 4x + 6 = 3(x + 6). Solve: 4x + 6 = 3x + 18, x = 12.

SituationBest method
Ratio + sum or differenceUnit method
Ratio now and ratio laterTable + x method
“After / before how many years”The gap method (in the “when” part later)
AveragesTotal = average × number
A riddle-like sentenceA direct equation
Example 2 · Standard
Lokesh's age is now 6 times Zoya's age. After 8 years, Lokesh's age will be four times Zoya's age. What is the present age of Lokesh?
Solution, step by step
  1. 1Let Zoya be x now; then Lokesh is 6x.
  2. 2After 8 years: Lokesh = 6x + 8, Zoya = x + 8. Condition: 6x + 8 = 4(x + 8).
  3. 36x + 8 = 4x + 32, so 2x = 24, x = 12.
  4. 4Zoya = 12 years, Lokesh = 6 × 12 = 72 years.
Answer 72 years
Where two points in time appear, a table is the safest.
StandardTry it yourself first
Uma's age is now 5 times Rajesh's age. After 3 years, Uma's age will be four times Rajesh's age. What is the present age of Uma?
BasicTry it yourself first
The ages of Deepa and Indu add up to 28 years and are in the ratio 4 : 3. What is the present age of Indu?
BasicTry it yourself first
Neha is 8 years older than Anil, and their ages are in the ratio 3 : 2. What is the present age of Anil?
BasicTry it yourself first
Pooja's age is three times Lokesh's age. Together they are 36 years old. What is the present age of Lokesh?
StandardTry it yourself first
A man was asked his age. He said, “4 years ago my age was half of what it will be 4 years from now.” What is his present age?
2

Ratio problems

The most frequently asked questions

Ratio with a sum or difference, three persons, and a ratio that changes after or before n years.

Sum or difference with a ratio

Straight use of the unit method

Finding both ages from a sum or difference and a ratio (or “k times”, or “a part equals a part”).

🔒

Three persons and a chain of gaps

Three-way ratio, average, and “A is d older than B”

A three-way ratio, the total from an average, and a chain of gaps starting from the youngest.

🔒

A ratio that changes after or before n years

The most popular age question

Finding present ages from the present ratio and the ratio after or before n years; or the other age when one age is given.

🔒
3

Conditions “n years ago / after”

A times-condition, when it happens, and a group total

Ages being k times n years ago or after, “after / before how many years” questions, and how a group total shifts.

Age k times, n years ago or after

One equation, one unknown

Solving a condition like “n years ago the father was k times the son” together with a sum or a present times-condition.

🔒

After / before how many years?

The gap method: the fastest route

When the gap is fixed and you are asked when an age is/was k times, the answer comes straight from gap ÷ (k − 1).

🔒

The total age of a group

Each year the total rises by m

How the total of m members' ages changes with time, and in how many years a given total is reached.

🔒
4

Average age

Total = average × number

Total from average, members joining or leaving, the teacher, a baby's birth and replacing a player.

From average to total and back

When a member joins or leaves

Getting the total from an average and a head-count; the effect on the average and the age when one member joins or leaves.

🔒

The teacher, the baby and a replacement

Three classic average questions

The average rising when a teacher is included, staying the same when a baby is born, and a player being replaced.

🔒
5

Riddles, births and digits

The types that come in different clothes

Age riddles, the age at a child's birth and birth years, and ages whose digits are reversed.

Age riddles

Make the sentence a direct equation

Sentences like “my age n years ago…” where the same person's age is compared at two times.

🔒

Age at birth and birth years

The link between birth year and gap

The parent's age at a child's birth, and age in a year = that year − birth year.

🔒

Digits reversed give the other age

10a + b and 10b + a

A two-digit age whose digits reversed give another age; the sum is a multiple of 11 and the difference a multiple of 9.

🔒
6

Strategy guide

Choosing the method and the common traps

How to pick the right method at a glance, the common slips and ways to check.

Spot it and choose the method

Method from the look of the question

Decide from what is given whether the unit, equation, gap or average method applies.

🔒

Common traps

The mistakes that keep happening

The commonest ways of losing marks in age questions.

🔒
7

Mixed practice

All types mixed — first spot the type

Age questions of every type in three levels.

Basic practice

One method, one or two steps

Direct counting, ratio with a sum or difference, times-conditions and simple averages.

🔒

Standard practice

Equations, “when”, and the classic averages

A changing ratio, times-conditions, “after how many years”, chains of gaps, averages and birth years.

🔒

Advanced practice

Two conditions, reversed questions, riddles

“Ago” ratios, riddles, the baby average, reversed digits and longer equations.

🔒
8

Revision — cards

Flip each card and test yourself

Think of the answer yourself first, then flip.

🔒