In age-ratio word problems, when a ratio r1:r2 changes to r3:r4 after n years, the standard approach is to do what?

Represent the ages as r1x and r2x, then solve (r1x+n)/(r2x+n) = r3/r4 for x. Represent both present ages using one variable x. Let the ages be r1x and r2x, based on the given ratio. After n years, each age increases by exactly n. So the ages become r1x + n and r2x + n. Set this equal to the new ratio: (r1x + n)/(r2x + n) = r3/r4. Cross-multiply, then collect the x terms on one side. Solving this equation gives the value of x directly.

In a ratio-based age problem, if a person's present age is r times a common multiplier x, and n years pass, their new age is expressed as what?

rx + n. The present age is rx, built from ratio multiplier r and variable x. After n years pass, every person's age increases by exactly n. This n is a fixed number of years, added once, not multiplied. So the new age is rx + n, with n added directly. It is not r(x+n), which would wrongly multiply n by r too. For example, if r=3, x=5, n=2: the age is 3(5)+2 = 17, not 3(5+2) = 21.

The ages of Arjun and Meena are in the ratio 2:1, and the sum of their present ages is 15 years. What will be the ratio of their ages after 6 years?

16:11. Let x be the common multiplier, so Arjun's present age is 2x and Meena's is 1x. Fact 1: their present ages are in the ratio 2:1. Fact 2: the sum of their present ages is 15, so 2x + 1x = 15. This gives 3x = 15, so x = 15 / 3 = 5. So Arjun's present age is 2 x 5 = 10, and Meena's present age is 1 x 5 = 5. Check: 10 + 5 = 15, which matches the given sum of 15. Fact 3: we need the ratio 6 years from now. Arjun's age then is 10 + 6 = 16. Meena's age then is 5 + 6 = 11. Dividing both by their common factor 1 reduces 16:11 to 16:11. So the required future ratio is 16:11.

The present age of Mr. Afzal is 2 times the age of his son. 3 years hence, the ratio of their ages will be 17:10. What is the present age of Mr. Afzal?

14. Let x be the son's present age, so the father's present age is 2x. Fact 1: the father's age is 2 times the son's age now. Fact 2: 3 years from now, their ages are in the ratio 17:10, so (2x + 3) / (x + 3) = 17/10. Cross-multiplying: 10 x (2x + 3) = 17 x (x + 3). This expands to 20x + 30 = 17x + 51. Collecting the x terms: 3x = 21. So x = 21 / 3 = 7. The son's present age is 7, so the father's present age is 2 x 7 = 14. Check: 3 years from now the ages are 17 and 10. 10 x 17 = 170 equals 17 x 10 = 170, confirming the ratio. So Mr. Afzal's present age is 14.

The present age of Mr. Afzal is 3 times the age of his son. 5 years hence, the ratio of their ages will be 16:7. What is the present age of Mr. Afzal?

27. Let x be the son's present age, so the father's present age is 3x. Fact 1: the father's age is 3 times the son's age now. Fact 2: 5 years from now, their ages are in the ratio 16:7, so (3x + 5) / (x + 5) = 16/7. Cross-multiplying: 7 x (3x + 5) = 16 x (x + 5). This expands to 21x + 35 = 16x + 80. Collecting the x terms: 5x = 45. So x = 45 / 5 = 9. The son's present age is 9, so the father's present age is 3 x 9 = 27. Check: 5 years from now the ages are 32 and 14. 7 x 32 = 224 equals 16 x 14 = 224, confirming the ratio. So Mr. Afzal's present age is 27.

The present ages of Kavita and Vikram are in the ratio 4:3 respectively. 7 years ago, the ratio of their ages was 5:2. What was Vikram's age 7 years ago?

2. Let x be the common multiplier, so Kavita's present age is 4x and Vikram's present age is 3x. Fact 1: their present ages are in the ratio 4:3. Fact 2: 7 years ago, their ages were in the ratio 5:2, so (4x - 7) / (3x - 7) = 5/2. Cross-multiplying: 2 x (4x - 7) = 5 x (3x - 7). This expands to 8x - 14 = 15x - 35. Collecting the x terms: -7x = -21. So x = -21 / -7 = 3. So Kavita's present age is 4 x 3 = 12, and Vikram's present age is 3 x 3 = 9. So Vikram's age 7 years ago was 9 - 7 = 2. Check: 7 years ago the ages were 5 and 2. 2 x 5 = 10 equals 5 x 2 = 10, confirming the ratio. So Vikram's age 7 years ago was 2.

The sum of the ages of Anita, Rohan, and Divya is 98 years. If Anita is 6 years older than Divya, and the total of Rohan and Divya's ages is 59 years, what is Rohan's age?

26. Let A = Anita's age, R = Rohan's age, D = Divya's age. We are given three facts. Fact 1: A + R + D = 98. Fact 2: A is 6 years older than D, so A - D = 6. Fact 3: R + D = 59. From facts 1 and 3: A = 98 - 59 = 39. From fact 2: D = A - 6 = 39 - 6 = 33. From fact 3: R = 59 - D = 59 - 33 = 26. Check: A + R + D = 39 + 26 + 33 = 98, confirming the answer. So Rohan's age is 26.

The average of the ages of Karan and Sophie is 24 years. If the respective ratio of their ages 5 years from now is 47:11, what is Sophie's present age (the smaller ratio value)?

6. Let A = Karan's present age, B = Sophie's present age. Fact 1: their average age is 24, so A + B = 2 x 24 = 48. Fact 2: 5 years from now, their ages are in the ratio 47:11. So let their future ages be 47y and 11y, for some number y. Fact 3: present age plus 5 equals future age, so A = 47y - 5 and B = 11y - 5. Substituting into fact 1: (47y - 5) + (11y - 5) = 48. This simplifies to 58y - 10 = 48, so y = 1. So B = 11 x 1 - 5 = 6. Check: A = 47 x 1 - 5 = 42, and A + B = 42 + 6 = 48, confirming the answer. So Sophie's present age is 6.

Problems on Ages: Formulas, Shortcuts and Solved Examples

Age problems turn word clues into small equations.

This lesson gives each rule, the reason it works and seven worked examples. It also lists the traps that cost marks.

The idea in brief

Ratio: write ages as r₁x and r₂x.
Future: add the years.
Past: subtract the years.
Gap: never changes.
Sum of two: rises by 2n in n years.

What Age Problems Ask

What are these questions about?

They give clues about people’s ages in words. You turn the clues into equations and solve.

The clues are about sums, differences, ratios and ages in the past or future.

Which facts always hold?

Everyone’s age rises by the same amount each year.
The gap between two ages never changes.
A person’s age can never be negative.

Both ages rise by the same amount. So the gap between them stays the same.

How do I write the ages down?

Call the unknown x. If the ratio is 4 : 3, write the ages as 4x and 3x.

After n years, add n to each age. Before n years, subtract n from each age.

Check 1

The gap between two people’s ages changes as time passes.

Core Rules and Why They Work

How do I handle a ratio?

Write the ages as r₁x and r₂x.

Why it works: any two numbers in that ratio are multiples of the same x. So x is the one unknown to find.

How do I shift ages in time?

After n years: age + n. n years ago: age − n.

Why it works: one year adds one to every age. So n years adds n to every age.

How do I set up the equation?

When r₁:r₂ becomes r₃:r₄ after n years: (r₁x + n) / (r₂x + n) = r₃ / r₄.

Cross-multiply and solve for x.

Why it works: after n years each age is its old value plus n. The new ratio gives one equation in x.

Subtract the years from each age before you set the new ratio. Then solve for x.

How do I use an average?

Sum of ages = average × number of people.

If two people average 24, their sum is 48.

Why it works: the average is the sum split equally.

Check 2

If a ratio is 3 : 2, the ages are written as 3x and 2x.

Seven Solved Examples

Easy, step by step: ages are in ratio 2 : 1, with a sum of 15 years. Find the ratio after 6 years.

Step 1: the ages are 2x and x, so 3x = 15 and x = 5.

Step 2: the ages now are 10 and 5.

Step 3: after 6 years they are 16 and 11.

The ratio is 16 : 11.

Easy, step by step: a father is 3 times as old as his son. Their ages sum to 48. Find both ages.

Step 1: let the son be x and the father 3x.

Step 2: x + 3x = 48, so x = 12.

Step 3: the father is 3 × 12 = 36.

The son is 12 and the father is 36.

Medium, step by step: Afzal is 2 times his son’s age. In 3 years the ratio is 17 : 10. Find Afzal’s age.

Step 1: the ages are 2x and x. In 3 years they are 2x + 3 and x + 3.

Step 2: (2x + 3)/(x + 3) = 17/10, so 20x + 30 = 17x + 51.

Step 3: 3x = 21, so x = 7.

Afzal is 2 × 7 = 14.

Medium, step by step: Afzal is 3 times his son’s age. In 5 years the ratio is 16 : 7. Find Afzal’s age.

Step 1: the ages in 5 years are 3x + 5 and x + 5.

Step 2: (3x + 5)/(x + 5) = 16/7, so 21x + 35 = 16x + 80.

Step 3: 5x = 45, so x = 9.

Afzal is 3 × 9 = 27.

Medium, step by step: Kavita and Vikram are in ratio 4 : 3. Seven years ago the ratio was 5 : 2. Find Vikram’s age then.

Step 1: the ages now are 4x and 3x. Seven years ago they were 4x − 7 and 3x − 7.

Step 2: (4x − 7)/(3x − 7) = 5/2, so 8x − 14 = 15x − 35 and x = 3.

Step 3: Vikram is now 9, so seven years ago he was 2.

The answer is 2 years.

Exam level, step by step: Anita, Rohan and Divya sum to 98. Anita is 6 older than Divya. Rohan and Divya sum to 59. Find Rohan’s age.

Step 1: Anita = 98 − 59 = 39.

Step 2: Divya = 39 − 6 = 33.

Step 3: Rohan = 59 − 33 = 26.

Rohan is 26.

Exam level, step by step: Karan and Sophie average 24. In 5 years their ratio is 47 : 11. Find Sophie’s age.

Step 1: the sum now is 48. In 5 years it is 48 + 10 = 58.

Step 2: 47 + 11 = 58, so one part is 1.

Step 3: Sophie in 5 years = 11, so now she is 11 − 5 = 6.

Sophie is 6.

Check 3

In example 7, the sum rises by 10 because two people each age 5 years.

Common Traps and Shortcuts

Which traps cost marks?

Adding the years to only one person.
Forgetting to add the years to the sum for two people.
Using the present ratio for a past or future age.
Answering with x, not the age asked.
Letting an age go negative.

Which shortcuts save time?

Use the fixed gap to cut the unknowns.
Test the answer options in the ratio.
Total ages move by (number of people × years).
Write ratio parts first, then add the years.

What patterns come up in exams?

Five patterns repeat. Ratio now and later, a sum and difference, a past ratio, averages and a parent and child.

Each takes under two minutes once the algebra is set.

Check 4

Test the answer options is a valid shortcut for age problems.

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