Age problems involve calculating present, past, or future ages of individuals based on given relationships and time periods. These problems test your ability to work with equations, ratios, and time calculations.
| Scenario | Problem Type | Example |
|---|---|---|
| Family Ages | Ratio comparison | Father is 3 times older than son |
| Past Ages | Historical calculation | 5 years ago, A was twice as old as B |
| Future Ages | Prediction | After 10 years, their combined age will be 80 |
| Age Difference | Constant gap | Age difference remains same over time |
| Benefit | Application |
|---|---|
| Algebraic Thinking | Converting words into mathematical equations |
| Ratio Understanding | Working with proportional relationships |
| Time Concept | Understanding how age changes with time |
| Logical Reasoning | Solving multi-step problems systematically |
| Exam Relevance | Common in all competitive examinations |
| Rule # | Principle | Example |
|---|---|---|
| Rule 1 | Age difference is constant | If A is 5 years older than B today, A will ALWAYS be 5 years older |
| Rule 2 | Ages change uniformly | After n years, everyone's age increases by n |
| Rule 3 | Ratio changes, difference doesn't | Ratio of ages changes over time, but difference stays same |
| Rule 4 | Present age is base | Always start from present age unless specified |
| Rule 5 | Sum/Difference method | Use sum and difference to find individual ages |
| Term | Meaning | Example | Notation |
|---|---|---|---|
| Present Age | Current age | Ram's present age is 20 | P or x |
| Past Age | Age before n years | 5 years ago | P - n |
| Future Age | Age after n years | After 10 years | P + n |
| Age Ratio | Proportion of ages | 3:2 ratio | a:b |
| Age Difference | Gap between ages | A is 10 years older | d |
| Sum of Ages | Combined age | Total age is 50 | S |
| Formula | Application | When to Use |
|---|---|---|
| Present Age = Past Age + Years | Finding present from past | 5 years ago age was 20 |
| Future Age = Present Age + Years | Finding future from present | After 10 years age will be? |
| Difference = Age₁ - Age₂ | Finding gap | How much older is A than B? |
| Sum = Age₁ + Age₂ | Combined ages | Total age of both |
| Ratio = Age₁ : Age₂ | Proportional relation | Ages are in ratio 3:2 |
Finding current age when total or difference is given.
Problem: The sum of ages of A and B is 50 years. The ratio of their ages is 3:2. Find their present ages.
Step-by-Step Solution:
| Step | Given Information | Calculation | Result |
|---|---|---|---|
| Step 1 | Sum of ages | A + B = 50 | - |
| Step 2 | Ratio of ages | A:B = 3:2 | - |
| Step 3 | Let ratio parts | A = 3x, B = 2x | - |
| Step 4 | Substitute in sum | 3x + 2x = 50 | 5x = 50 |
| Step 5 | Solve for x | x = 50 ÷ 5 | x = 10 |
| Step 6 | Find A's age | A = 3x = 3 × 10 | A = 30 years |
| Step 7 | Find B's age | B = 2x = 2 × 10 | B = 20 years |
Verification:
| Check | Calculation | Result |
|---|---|---|
| Sum check | 30 + 20 | 50 ✓ |
| Ratio check | 30:20 = 3:2 | 3:2 ✓ |
Answer: A = 30 years, B = 20 years ✅
Problem: The difference between ages of father and son is 30 years. The ratio of their ages is 5:2. Find their ages.
Step-by-Step Solution:
| Step | Information | Calculation | Result |
|---|---|---|---|
| Step 1 | Difference | F - S = 30 | - |
| Step 2 | Ratio | F:S = 5:2 | - |
| Step 3 | Let ratio parts | F = 5x, S = 2x | - |
| Step 4 | Substitute | 5x - 2x = 30 | 3x = 30 |
| Step 5 | Solve for x | x = 30 ÷ 3 | x = 10 |
| Step 6 | Father's age | F = 5 × 10 | F = 50 years |
| Step 7 | Son's age | S = 2 × 10 | S = 20 years |
Verification:
| Check | Calculation | Result |
|---|---|---|
| Difference | 50 - 20 | 30 ✓ |
| Ratio | 50:20 = 5:2 | 5:2 ✓ |
Answer: Father = 50 years, Son = 20 years ✅
Finding present age based on past age conditions.
Problem: 5 years ago, the ratio of ages of A and B was 2:3. The sum of their present ages is 50. Find their present ages.
Step-by-Step Solution:
| Step | Analysis | Calculation | Result |
|---|---|---|---|
| Step 1 | 5 years ago ages | (A-5):(B-5) = 2:3 | - |
| Step 2 | Let past ages | A-5 = 2x, B-5 = 3x | - |
| Step 3 | Present ages | A = 2x+5, B = 3x+5 | - |
| Step 4 | Sum given | A + B = 50 | - |
| Step 5 | Substitute | (2x+5) + (3x+5) = 50 | - |
| Step 6 | Simplify | 5x + 10 = 50 | 5x = 40 |
| Step 7 | Solve for x | x = 40 ÷ 5 | x = 8 |
| Step 8 | A's present age | 2(8) + 5 | A = 21 years |
| Step 9 | B's present age | 3(8) + 5 | B = 29 years |
Verification Table:
| Check | Calculation | Result |
|---|---|---|
| 5 years ago A | 21 - 5 = 16 | - |
| 5 years ago B | 29 - 5 = 24 | - |
| Past ratio | 16:24 = 2:3 | ✓ |
| Present sum | 21 + 29 = 50 | ✓ |
Answer: A = 21 years, B = 29 years ✅
Problem: 10 years ago, a father was 3 times as old as his son. The father's present age is 50 years. Find the son's present age.
Step-by-Step Solution:
| Step | Information | Calculation | Result |
|---|---|---|---|
| Step 1 | Father's present age | F = 50 | Given |
| Step 2 | Father's age 10 years ago | F - 10 = 50 - 10 | 40 years |
| Step 3 | Son's age 10 years ago | Let S - 10 = x | - |
| Step 4 | Condition given | F - 10 = 3(S - 10) | - |
| Step 5 | Substitute | 40 = 3(S - 10) | - |
| Step 6 | Solve | 40 = 3S - 30 | - |
| Step 7 | Rearrange | 3S = 70 | - |
| Step 8 | Find S | S = 70 ÷ 3 | S = 23.33 years |
Wait! Age should be whole number. Let's recalculate:
| Correct Step | Calculation | Result |
|---|---|---|
| 10 years ago father | 40 | - |
| 10 years ago son | 40 ÷ 3 = 13.33 | - |
| Present son's age | 13.33 + 10 | 23.33 years |
Note: If problem expects whole numbers, verify the question or round appropriately.
Answer: Son's present age ≈ 23 years ✅
Finding present age based on future age conditions.
Problem: After 8 years, the ratio of ages of P and Q will be 5:4. The sum of their present ages is 63 years. Find their present ages.
Step-by-Step Solution:
| Step | Analysis | Calculation | Result |
|---|---|---|---|
| Step 1 | After 8 years ratio | (P+8):(Q+8) = 5:4 | - |
| Step 2 | Let future ages | P+8 = 5x, Q+8 = 4x | - |
| Step 3 | Present ages | P = 5x-8, Q = 4x-8 | - |
| Step 4 | Sum given | P + Q = 63 | - |
| Step 5 | Substitute | (5x-8) + (4x-8) = 63 | - |
| Step 6 | Simplify | 9x - 16 = 63 | - |
| Step 7 | Solve | 9x = 79 | x = 79÷9 = 8.78 |
| Step 8 | P's age | 5(8.78) - 8 | P = 35.9 ≈ 36 |
| Step 9 | Q's age | 4(8.78) - 8 | Q = 27.1 ≈ 27 |
Let me recalculate with exact values:
| Correct Approach | Calculation | Result |
|---|---|---|
| Let P = present age of P | P + Q = 63 | Q = 63 - P |
| After 8 years | (P+8):(Q+8) = 5:4 | - |
| Cross multiply | 4(P+8) = 5(Q+8) | - |
| Expand | 4P + 32 = 5Q + 40 | - |
| Substitute Q = 63-P | 4P + 32 = 5(63-P) + 40 | - |
| Simplify | 4P + 32 = 315 - 5P + 40 | - |
| Combine | 9P = 323 | - |
| Solve | P = 323 ÷ 9 | P = 35.89 years |
| Find Q | Q = 63 - 35.89 | Q = 27.11 years |
For whole number answer, let's verify if question has correct values.
Answer: P ≈ 36 years, Q ≈ 27 years ✅
Problem: After 5 years, a father will be twice as old as his son. The father's present age is 35 years. Find the son's present age.
Step-by-Step Solution:
| Step | Information | Calculation | Result |
|---|---|---|---|
| Step 1 | Father's present age | F = 35 | Given |
| Step 2 | Father after 5 years | F + 5 = 40 | - |
| Step 3 | Condition | F + 5 = 2(S + 5) | - |
| Step 4 | Substitute | 40 = 2(S + 5) | - |
| Step 5 | Divide by 2 | 20 = S + 5 | - |
| Step 6 | Solve for S | S = 20 - 5 | S = 15 years |
Verification:
| Time | Father | Son | Relation |
|---|---|---|---|
| Present | 35 | 15 | - |
| After 5 years | 40 | 20 | 40 = 2 × 20 ✓ |
Answer: Son's present age = 15 years ✅
Problems involving both past and future conditions.
Problem: 4 years ago, A was 3 times as old as B. After 4 years, A will be twice as old as B. Find their present ages.
Step-by-Step Solution:
| Step | Condition | Equation | Result |
|---|---|---|---|
| Step 1 | 4 years ago | A - 4 = 3(B - 4) | Equation 1 |
| Step 2 | Expand Eq. 1 | A - 4 = 3B - 12 | - |
| Step 3 | Simplify Eq. 1 | A = 3B - 8 | ...(i) |
| Step 4 | After 4 years | A + 4 = 2(B + 4) | Equation 2 |
| Step 5 | Expand Eq. 2 | A + 4 = 2B + 8 | - |
| Step 6 | Simplify Eq. 2 | A = 2B + 4 | ...(ii) |
| Step 7 | Equate (i) and (ii) | 3B - 8 = 2B + 4 | - |
| Step 8 | Solve for B | B = 12 | B = 12 years |
| Step 9 | Find A from (ii) | A = 2(12) + 4 | A = 28 years |
Verification Table:
| Time Period | A's Age | B's Age | Condition | Check |
|---|---|---|---|---|
| 4 years ago | 24 | 8 | A = 3B | 24 = 3×8 ✓ |
| Present | 28 | 12 | - | - |
| After 4 years | 32 | 16 | A = 2B | 32 = 2×16 ✓ |
Answer: A = 28 years, B = 12 years ✅
Problems involving average ages of groups.
Problem: The average age of a family of 5 members is 24 years. If the age of the youngest member is 8 years, what was the average age of the family at the birth of the youngest member?
Step-by-Step Solution:
| Step | Information | Calculation | Result |
|---|---|---|---|
| Step 1 | Present average | 24 years | Given |
| Step 2 | Number of members | 5 | Given |
| Step 3 | Total present age | 5 × 24 | 120 years |
| Step 4 | Youngest's age | 8 years | Given |
| Step 5 | 8 years ago members | 5 - 1 = 4 | (Youngest not born) |
| Step 6 | Total age 8 years ago | 120 - (5 × 8) | 120 - 40 = 80 |
| Step 7 | Average 8 years ago | 80 ÷ 4 | 20 years |
Detailed Breakdown:
| Period | Members | Total Age | Average |
|---|---|---|---|
| Present | 5 | 120 | 24 |
| 8 years ago | 4 | 80 | 20 |
Answer: Average age at birth of youngest = 20 years ✅
Problem: The average age of 10 students is 20 years. If a student aged 20 is replaced by a new student, the average becomes 21 years. Find the age of the new student.
Step-by-Step Solution:
| Step | Analysis | Calculation | Result |
|---|---|---|---|
| Step 1 | Original total age | 10 × 20 | 200 years |
| Step 2 | Student removed | 20 years | - |
| Step 3 | New average | 21 years | Given |
| Step 4 | New total age | 10 × 21 | 210 years |
| Step 5 | Increase in total | 210 - 200 | 10 years |
| Step 6 | New student's age | 20 + 10 | 30 years |
Formula Method:
| Component | Value |
|---|---|
| Increase in total = New total - Old total | 210 - 200 = 10 |
| New student = Removed student + Increase | 20 + 10 = 30 |
Answer: Age of new student = 30 years ✅
Using the constant difference principle.
Problem: The age of a man is 4 times that of his son. 5 years ago, the man was 9 times as old as his son. Find their present ages.
Step-by-Step Solution:
| Step | Condition | Equation | Result |
|---|---|---|---|
| Step 1 | Present condition | M = 4S | ...(i) |
| Step 2 | 5 years ago | M - 5 = 9(S - 5) | ...(ii) |
| Step 3 | Expand equation (ii) | M - 5 = 9S - 45 | - |
| Step 4 | Simplify | M = 9S - 40 | ...(iii) |
| Step 5 | Equate (i) and (iii) | 4S = 9S - 40 | - |
| Step 6 | Solve | 5S = 40 | - |
| Step 7 | Find S | S = 8 | S = 8 years |
| Step 8 | Find M from (i) | M = 4 × 8 | M = 32 years |
Verification:
| Time | Man | Son | Ratio/Multiple | Check |
|---|---|---|---|---|
| Present | 32 | 8 | 4 times | 32 = 4×8 ✓ |
| 5 years ago | 27 | 3 | 9 times | 27 = 9×3 ✓ |
Answer: Man = 32 years, Son = 8 years ✅
| Step | Action | Details | Example |
|---|---|---|---|
| Step 1 | Read Carefully | Identify all given information | "Sum is 50, ratio is 3:2" |
| Step 2 | Identify Type | Past, present, future, or combination? | "5 years ago" = Past problem |
| Step 3 | Define Variables | Let present ages be x, y, etc. | Let son's age = S, father's = F |
| Step 4 | Form Equations | Convert statements to mathematical equations | F = 3S (father 3 times older) |
| Step 5 | Solve Systematically | Use substitution or elimination | Substitute and solve |
| Step 6 | Calculate Answer | Find the required values | S = 12, F = 36 |
| Step 7 | Verify | Check answer against all conditions | 36 = 3×12 ✓ |
| Q# | Question | Options |
|---|---|---|
| Q1 | The sum of ages of A and B is 40. A is twice as old as B. Find A's age. | A) 20 yearsbrB) 26.67 yearsbrC) 30 yearsbrD) 25 years |
Solution:
| Step | Equation | Calculation |
|---|---|---|
| Sum | A + B = 40 | ...(i) |
| Relation | A = 2B | ...(ii) |
| Substitute | 2B + B = 40 | 3B = 40 |
| Solve | B = 40 ÷ 3 | B = 13.33 |
| Find A | A = 2 × 13.33 | A = 26.67 years |
Answer: B) 26.67 years ✅
| Q# | Question | Options |
|---|---|---|
| Q2 | Father is 30 years older than son. After 10 years, father will be twice as old as son. Find son's present age. | A) 10 yearsbrB) 15 yearsbrC) 20 yearsbrD) 25 years |
Solution:
| Step | Equation | Calculation |
|---|---|---|
| Difference | F = S + 30 | ...(i) |
| After 10 years | F + 10 = 2(S + 10) | ...(ii) |
| Expand | F + 10 = 2S + 20 | - |
| Simplify | F = 2S + 10 | ...(iii) |
| Equate (i) & (iii) | S + 30 = 2S + 10 | - |
| Solve | S = 20 | S = 20 years |
Answer: C) 20 years ✅
| Q# | Question | Options |
|---|---|---|
| Q3 | 5 years ago, the ratio of ages of P and Q was 3:4. After 5 years, the ratio will be 4:5. Find P's present age. | A) 25 yearsbrB) 30 yearsbrC) 35 yearsbrD) 40 years |
Solution:
| Step | Condition | Equation |
|---|---|---|
| 5 years ago | (P-5):(Q-5) = 3:4 | 4(P-5) = 3(Q-5) |
| Expand | 4P - 20 = 3Q - 15 | 4P = 3Q + 5 ...(i) |
| After 5 years | (P+5):(Q+5) = 4:5 | 5(P+5) = 4(Q+5) |
| Expand | 5P + 25 = 4Q + 20 | 5P = 4Q - 5 ...(ii) |
| From (i) | Q = (4P-5)/3 | Substitute in (ii) |
| Solve | 5P = 4[(4P-5)/3] - 5 | - |
| Simplify | 15P = 16P - 20 - 15 | P = 35 |
Answer: C) 35 years ✅
| Q# | Question | Options |
|---|---|---|
| Q4 | The average age of 4 children is 12 years. If the age of the mother is included, the average becomes 16 years. Find the mother's age. | A) 32 yearsbrB) 36 yearsbrC) 40 yearsbrD) 44 years |
Solution:
| Step | Calculation | Result |
|---|---|---|
| Total age of 4 children | 4 × 12 | 48 years |
| Total with mother | 5 × 16 | 80 years |
| Mother's age | 80 - 48 | 32 years |
Answer: A) 32 years ✅
| Q# | Question | Options |
|---|---|---|
| Q5 | 10 years ago, father was 12 times as old as son. 10 years from now, he will be twice as old as son. Find their present ages. | A) F=34, S=8brB) F=44, S=12brC) F=46, S=14brD) F=50, S=15 |
Solution:
| Step | Condition | Equation |
|---|---|---|
| 10 years ago | F - 10 = 12(S - 10) | F - 10 = 12S - 120 |
| Simplify | F = 12S - 110 | ...(i) |
| After 10 years | F + 10 = 2(S + 10) | F + 10 = 2S + 20 |
| Simplify | F = 2S + 10 | ...(ii) |
| Equate (i) & (ii) | 12S - 110 = 2S + 10 | 10S = 120 |
| Solve | S = 12 | S = 12 years |
| Find F | F = 2(12) + 10 | F = 34 years |
Wait, let me verify:
Recalculating with F=34:
Answer: A) F=34 years, S=8 years
Let me recalculate more carefully:
| Recalculation | Working |
|---|---|
| From S = 12 | F = 2(12) + 10 = 34 |
| Check 10 yrs ago | F-10=24, S-10=2 → 24=12×2 ✓ |
| Check after 10 yrs | F+10=44, S+10=22 → 44=2×22 ✓ |
Correct Answer: F=34, S=12 (Not in options exactly, closest is A)
| Q# | Question | Options |
|---|---|---|
| Q6 | A man's age is 3 times his son's age. 12 years ago, he was 7 times as old as his son. Find the man's present age. | A) 36 yearsbrB) 42 yearsbrC) 48 yearsbrD) 54 years |
Solution:
| Step | Equation | Calculation |
|---|---|---|
| Present | M = 3S | ...(i) |
| 12 years ago | M - 12 = 7(S - 12) | ...(ii) |
| Expand | M - 12 = 7S - 84 | - |
| Simplify | M = 7S - 72 | ...(iii) |
| Equate (i) & (iii) | 3S = 7S - 72 | 4S = 72 |
| Solve | S = 18 | S = 18 years |
| Find M | M = 3 × 18 | M = 54 years |
Verification:
Answer: D) 54 years ✅
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