Ratio & Proportion · Topic 5 of 6

🎂 Ages

1 exam question type, fully solved

An age problem is just a ratio with a clock attached: the same two people grow older together, so once you fix the ratio you only have to add or subtract the years.

How to crack any age-ratio problem

When two ages are in the ratio , never treat 5 and 7 as the real ages — write them as and with the same multiplier . That single unknown carries the whole problem.

• years ago means subtract from each person: and . years hence (later) means add : and . Time passes for everybody, so the change goes on both ages.

• Set the second moment equal to its given ratio, e.g. , and cross-multiply: .

• Open the brackets, gather the -terms on one side and the plain numbers on the other, and solve the simple linear equation for .

• Finally the actual age — so A , B , and their sum .

• Speed trick: the age difference never changes with time, so a fixed gap (like “30 years older”) pins down at once.

📊Ratio ScalerEnter A and B to explore the ratio — add C for compound ratio

C (optional)

Simplified Ratio

3 : 5

3rd Proportional (x where A:B = B:x)

x = 8.3333

Mean Proportional (√A×B)

√15≈ 3.873

Proportion Checker — Is a:b = c:d ?

:=:

The two ages are and , so their gap is . If that gap is 6 years then , and the younger one is years old.

Q1A is twice as old as B. Three years ago, A was 3 times as old as B. How old is A now?

Q2A's age is 3 times B's age. 10 years ago, A was 5 times B's age. Find the sum of their present ages.

Q3The present ages of A and B are in the ratio 3:2. After 4 years the ratio becomes 7:5. Find A’s present age.

📚

Real exam questions — Ages

1 question types · 4 solved examples from real SSC papers

An age problem is just a ratio with a clock attached: the two people get older together, so once you write both ages with one shared multiplier, you only add or subtract the years and solve a single equation.

How to solve this type

Write both ages with the SAME multiplier — if the ratio is

5:7

take the ages as

5x

and

7x

. For "

k

years ago" subtract

k

from each age, for "

k

years hence" add

k

to each (time passes for everyone, so change BOTH). Set the second moment equal to its given ratio, cross-multiply, and solve the linear equation for

x

. The real age

=

multiplier

\times x

. Speed trick: the age difference

7x-5x=2x

never changes, so any fixed gap (e.g. "30 years older") pins down

x

at once.

The present ages of A and B are in the ratio 5:3. After 6 years the ratio becomes 7:5. Find A's present age.

A15B9C12D18

A father is 30 years older than his son. 6 years ago the father's age was 3 times the son's age. Find their present ages.

AF=42, S=12BF=51, S=21CF=45, S=15DF=48, S=18

The ratio of the ages of A and B 4 years ago was 4:5. Eight years from now it will be 11:13. Find the sum of their present ages.

A96B80C72D76

10 years ago the ages of A and B were in the ratio 1:2. Their present ages are in the ratio 3:5. Find A's present age.

A25B20C30D15