📑 Topics on this page (Jump to Topic):
📌 1. Key Principles
- If present age is x, after n years it is x + n and n years ago x − n.
- The difference between two people's ages never changes.
- If the ratio is a : b, take ages ak and bk.
- After t years the sum of two people's ages rises by 2t.
📘 Formulas:
- Ratio a:b, after t years c:d → k = t(d − c)/(ad − bc)
💡 Shortcut Trick: Option testing is usually the fastest method for age problems.
⚠️ Exam Traps:
- "n years ago the father was k times the son" — subtract n from both ages, not just one.
📌 2. Practice MCQs
📝 Practice MCQs (prepared by SF Prep — not PYQs)
Q1. The present ages of A and B are in the ratio 5 : 7. After 10 years the ratio will be 25 : 33. B's present age is: Practice
Correct answer: (B)
Solution: (5k + 10)/(7k + 10) = 25/33 → k = 8; B = 7k = 56 years.
Solution: (5k + 10)/(7k + 10) = 25/33 → k = 8; B = 7k = 56 years.
Q2. 4 years ago the father's age was 4 times his son's age. If the son is 6 years old now, the father's present age is: Practice
Correct answer: (C)
Solution: 4 years ago son = 2, father = 4 × 2 = 8; now father = 8 + 4 = 12 years.
Solution: 4 years ago son = 2, father = 4 × 2 = 8; now father = 8 + 4 = 12 years.
Q3. The sum of the present ages of two brothers is 20 years and their ratio is 1 : 3. After 6 years the younger brother's age will be: Practice
Correct answer: (D)
Solution: Younger = 20 × 1/4 = 5; after 6 years = 11 years.
Solution: Younger = 20 × 1/4 = 5; after 6 years = 11 years.
Q4. Two sisters differ in age by 4 years and the product of their ages is 192. The elder sister's age is: Practice
Correct answer: (A)
Solution: x(x + 4) = 192 → x = 12; elder = 16 years.
Solution: x(x + 4) = 192 → x = 12; elder = 16 years.
Q5. The father was 30 years old when his son was born. The son is now 8. The father's age 7 years from now will be: Practice
Correct answer: (B)
Solution: Father now = 30 + 8 = 38; after 7 years = 45 years.
Solution: Father now = 30 + 8 = 38; after 7 years = 45 years.
Q6. The present ages of A and B are in the ratio 5 : 7. After 6 years the ratio will be 31 : 41. B's present age is: Practice
Correct answer: (C)
Solution: (5k + 6)/(7k + 6) = 31/41 → k = 5; B = 7k = 35 years.
Solution: (5k + 6)/(7k + 6) = 31/41 → k = 5; B = 7k = 35 years.
Q7. 8 years ago the father's age was 4 times his son's age. If the son is 20 years old now, the father's present age is: Practice
Correct answer: (D)
Solution: 8 years ago son = 12, father = 4 × 12 = 48; now father = 48 + 8 = 56 years.
Solution: 8 years ago son = 12, father = 4 × 12 = 48; now father = 48 + 8 = 56 years.
Q8. The sum of the present ages of two brothers is 28 years and their ratio is 1 : 3. After 10 years the younger brother's age will be: Practice
Correct answer: (A)
Solution: Younger = 28 × 1/4 = 7; after 10 years = 17 years.
Solution: Younger = 28 × 1/4 = 7; after 10 years = 17 years.
Q9. Two sisters differ in age by 5 years and the product of their ages is 204. The elder sister's age is: Practice
Correct answer: (B)
Solution: x(x + 5) = 204 → x = 12; elder = 17 years.
Solution: x(x + 5) = 204 → x = 12; elder = 17 years.
Q10. The father was 35 years old when his son was born. The son is now 7. The father's age 6 years from now will be: Practice
Correct answer: (C)
Solution: Father now = 35 + 7 = 42; after 6 years = 48 years.
Solution: Father now = 35 + 7 = 42; after 6 years = 48 years.
Q11. The present ages of A and B are in the ratio 4 : 5. After 4 years the ratio will be 32 : 39. B's present age is: Practice
Correct answer: (D)
Solution: (4k + 4)/(5k + 4) = 32/39 → k = 7; B = 5k = 35 years.
Solution: (4k + 4)/(5k + 4) = 32/39 → k = 7; B = 5k = 35 years.
Q12. 5 years ago the father's age was 5 times his son's age. If the son is 10 years old now, the father's present age is: Practice
Correct answer: (A)
Solution: 5 years ago son = 5, father = 5 × 5 = 25; now father = 25 + 5 = 30 years.
Solution: 5 years ago son = 5, father = 5 × 5 = 25; now father = 25 + 5 = 30 years.
Q13. The sum of the present ages of two brothers is 78 years and their ratio is 4 : 9. After 8 years the younger brother's age will be: Practice
Correct answer: (B)
Solution: Younger = 78 × 4/13 = 24; after 8 years = 32 years.
Solution: Younger = 78 × 4/13 = 24; after 8 years = 32 years.
Q14. Two sisters differ in age by 3 years and the product of their ages is 270. The elder sister's age is: Practice
Correct answer: (C)
Solution: x(x + 3) = 270 → x = 15; elder = 18 years.
Solution: x(x + 3) = 270 → x = 15; elder = 18 years.
Q15. The present ages of A and B are in the ratio 5 : 7. After 5 years the ratio will be 25 : 33. B's present age is: Practice
Correct answer: (D)
Solution: (5k + 5)/(7k + 5) = 25/33 → k = 4; B = 7k = 28 years.
Solution: (5k + 5)/(7k + 5) = 25/33 → k = 4; B = 7k = 28 years.
Q16. The sum of the present ages of two brothers is 52 years and their ratio is 5 : 8. After 10 years the younger brother's age will be: Practice
Correct answer: (A)
Solution: Younger = 52 × 5/13 = 20; after 10 years = 30 years.
Solution: Younger = 52 × 5/13 = 20; after 10 years = 30 years.
Q17. A mother is 35 years old and her daughter is 8. After how many years will the mother be 4 times as old as the daughter? Practice
Correct answer: (B)
Solution: 35 + t = 4(8 + t) → t = (35 − 32)/3 = 1 years.
Solution: 35 + t = 4(8 + t) → t = (35 − 32)/3 = 1 years.
Q18. Two sisters differ in age by 5 years and the product of their ages is 50. The elder sister's age is: Practice
Correct answer: (C)
Solution: x(x + 5) = 50 → x = 5; elder = 10 years.
Solution: x(x + 5) = 50 → x = 5; elder = 10 years.
Q19. The father was 35 years old when his son was born. The son is now 9. The father's age 7 years from now will be: Practice
Correct answer: (D)
Solution: Father now = 35 + 9 = 44; after 7 years = 51 years.
Solution: Father now = 35 + 9 = 44; after 7 years = 51 years.
Q20. The present ages of A and B are in the ratio 3 : 5. After 10 years the ratio will be 31 : 45. B's present age is: Practice
Correct answer: (A)
Solution: (3k + 10)/(5k + 10) = 31/45 → k = 7; B = 5k = 35 years.
Solution: (3k + 10)/(5k + 10) = 31/45 → k = 7; B = 5k = 35 years.