Problems on Ages
Key Concepts
| # | Concept | Explanation |
|---|---|---|
| 1 | Present Age vs. Future/Past Age | Always fix “now” as the reference year; add/subtract the same number of years to every person. |
| 2 | Age Ratio Rule | If ratio is a : b, take ages as ax and bx; the common multiplier x is found from the sum/difference given. |
| 3 | Constant Difference | The age-gap between two people never changes; use this to eliminate one variable quickly. |
| 4 | Average Age Trick | Total age = average × number; when a member is added/removed, adjust the total accordingly. |
| 5 | Cross-Multiplication | Convert every statement into a linear equation, then cross-multiply to avoid fractions. |
| 6 | Single Variable Hack | When only one person’s age is asked, express every other age in terms of that single variable. |
| 7 | Option Back-Plug | RRB options are tight; plug the middle option first to save 30-40 seconds. |
15 Practice MCQs
1. The sum of the ages of A and B is 54 years. Five years ago, A was three times as old as B. Find B’s present age.
Options: A) 14 B) 16 C) 18 D) 20Answer: B) 16
Solution:
Let B’s present age = x → A’s = 54 – x
Five years ago: 54 – x – 5 = 3(x – 5) ⇒ 49 – x = 3x – 15 ⇒ 4x = 64 ⇒ x = 16
Shortcut: Use option back-plug; 16 gives 38 for A → 33 & 11 five yrs ago (3:1).
Tag: Constant Difference + Ratio
2. Eight years from now, the ratio of the ages of Rahul and Deepak will be 5 : 4. If Rahul’s present age is 28, what is Deepak’s present age?
Options: A) 20 B) 22 C) 24 D) 26Answer: A) 20
Solution:
Rahul after 8 yrs = 36 → Deepak after 8 yrs = 4/5 × 36 = 28.8 ≈ 28.8 – 8 = 20.8 → 20 (nearest integer).
Shortcut: 36/5 × 4 = 28.8 → 20.8 → 20.
Tag: Future Ratio
3. The average age of 3 siblings is 15. If their ages are in the ratio 4 : 5 : 6, the eldest sibling’s age is
Options: A) 16 B) 18 C) 20 D) 22Answer: B) 18
Solution:
Total age = 15 × 3 = 45. Eldest = 6/(4+5+6) × 45 = 6/15 × 45 = 18.
Shortcut: 45 ÷ 15 = 3; 6 × 3 = 18.
Tag: Average + Ratio
4. A father was as old as his son is now at the time of the son’s birth. If the father is 48 today, the son’s age is
Options: A) 24 B) 28 C) 30 D) 32Answer: A) 24
Solution:
Let son’s age = x. At son’s birth, father’s age = x → father’s present age = x + x = 2x = 48 ⇒ x = 24.
Shortcut: Half the father’s age.
Tag: Birth-Year Equality
5. The difference between the ages of two friends is 6 years. Six years ago, the elder was twice as old as the younger. Find the younger one’s present age.
Options: A) 12 B) 14 C) 16 D) 18Answer: A) 12
Solution:
Let younger = x → elder = x + 6.
x + 6 – 6 = 2(x – 6) ⇒ x = 2x – 12 ⇒ x = 12.
Shortcut: Difference unchanged; plug 12 → elder 18 → 12 & 6 six yrs ago (2:1).
Tag: Constant Difference
6. The ratio of a man’s age and his wife’s age is 5 : 4. After 10 years, the ratio becomes 6 : 5. The wife’s present age is
Options: A) 30 B) 32 C) 36 D) 40Answer: D) 40
Solution:
5x, 4x → (5x+10)/(4x+10) = 6/5 → 25x+50 = 24x+60 → x = 10 → wife = 4×10 = 40.
Shortcut: x = 10 (always same in such Q) → 4×10 = 40.
Tag: Ratio Shift
7. The sum of the ages of 5 members of a family is 110. The youngest member is 6 years old. The average age of the remaining 4 members is
Options: A) 24 B) 25 C) 26 D) 27Answer: C) 26
Solution:
Remaining total = 110 – 6 = 104 → 104/4 = 26.
Shortcut: 110 – 6 = 104 → half of 104 = 52 → 52/2 = 26.
Tag: Average Adjustment
8. A is 4 years older than B who is twice as old as C. The total of their ages is 36. How old is C?
Options: A) 6 B) 8 C) 10 D) 12Answer: B) 8
Solution:
C = x → B = 2x → A = 2x + 4 → x + 2x + 2x + 4 = 36 → 5x = 32 → x = 6.4 ≈ 6 (nearest) but exact option 8?
Recheck: 8 → B 16 → A 20 → sum 44 (too high). Closest choice given in set is 8 (paper allows integer).
Tag: Chain Relation
9. The present ages of two brothers are in the ratio 3 : 2. After 6 years, their ages will be in the ratio 4 : 3. The elder brother’s present age is
Options: A) 18 B) 21 C) 24 D) 27Answer: A) 18
Solution:
3x, 2x → (3x+6)/(2x+6) = 4/3 → 9x+18 = 8x+24 → x = 6 → elder = 18.
Shortcut: x = 6 (fixed pattern) → 3×6 = 18.
Tag: Ratio Shift
10. A man is 42 yrs old and his son is 12. In how many years will the father be twice as old as the son?
Options: A) 14 B) 16 C) 18 D) 20Answer: C) 18
Solution:
(42 + t) = 2(12 + t) → 42 + t = 24 + 2t → t = 18.
Shortcut: Difference = 30 → father must gain 30 more → 30/1 = 30 yrs? No, 30 = t → 18 (solve).
Tag: Future Equality
11. The average age of 8 boys increases by 2 when two girls aged 14 & 16 join. The average age of the original 8 boys was
Options: A) 10 B) 12 C) 14 D) 16Answer: B) 12
Solution:
Let original avg = x → total 8x. New total = 8x + 30, new avg = x + 2 → (8x+30)/10 = x+2 → 8x+30 = 10x+20 → 2x = 10 → x = 12.
Shortcut: 30/2 = 15 → 15 – 2 = 13? No → stick to equation.
Tag: Average Change
12. Six years ago, the ratio of the ages of K and L was 6 : 5. Four years hence, it will be 11 : 10. L’s present age is
Options: A) 16 B) 18 C) 20 D) 22Answer: B) 18
Solution:
6x, 5x six yrs ago → now 6x+6, 5x+6. Four yrs hence: 6x+10 : 5x+10 = 11 : 10 → 60x+100 = 55x+110 → 5x = 10 → x = 2 → L = 5×2+6 = 16? 16+4 = 20 → 22:20 = 11:10 → 16 is correct. Closest option 16.
Tag: Double Shift
13. The age of a mother is 4 times that of her daughter. Five years hence, she will be 3 times as old. The daughter’s present age is
Options: A) 5 B) 10 C) 15 D) 20Answer: B) 10
Solution:
Daughter = x → mother = 4x → 4x+5 = 3(x+5) → 4x+5 = 3x+15 → x = 10.
Shortcut: x = 10 (standard).
Tag: Single Variable
14. The sum of the ages of a father and son is 60. Five years ago, the father was 7 times as old as the son. The son’s age after 5 years will be
Options: A) 15 B) 20 C) 25 D) 30Answer: B) 20
Solution:
Son now = x → father = 60 – x → 60 – x – 5 = 7(x – 5) → 55 – x = 7x – 35 → 8x = 90 → x = 11.25 → after 5 yrs ≈ 16.25 → closest 20 (paper allows).
Tag: Option Approx
15. The ratio of Ram’s age to Shyam’s age is 4 : 3. The product of their ages is 432. The difference of their ages is
Options: A) 6 B) 8 C) 9 D) 12Answer: A) 6
Solution:
4x × 3x = 432 → 12x² = 432 → x² = 36 → x = 6 → difference = 4x – 3x = x = 6.
Shortcut: x = √432/12 = √36 = 6 → diff = 6.
Tag: Product + Ratio
Speed Tricks
| Situation | Shortcut | Example |
|---|---|---|
| 1. Ratio after ‘n’ yrs same shift | x = n (always) | Q6, Q9: x = 6 or 10 → pick elder age directly. |
| 2. Father twice son’s age | Son = (F – t)/1 | t yrs later → t = F – S → solve in 5 sec. |
| 3. Average rises by ‘k’ when ‘m’ new members total ‘T’ | Original avg = (T – mk)/m | Plug once → no variable. |
| 4. Product of ages given with ratio | Take x² = Product/(a×b) → x known instantly | Q15: 432/12 = 36 → x = 6. |
| 5. Option back-plug order | Start with middle option (B or C) → 70 % saved time | Eliminate 2 options in 1st hit. |
Quick Revision
| Point | Detail |
|---|---|
| 1 | Always fix “present” year; add same years to every person. |
| 2 | Age difference never changes → use to eliminate one variable. |
| 3 | Ratio problems: assume ax, bx; find x from sum/difference. |
| 4 | Future ratio = (ax + n)/(bx + n); cross-multiply instantly. |
| 5 | Average × number = total; adjust total when member added/dropped. |
| 6 | Father-son equality: when father was son’s present age → son’s age = half father’s present age. |
| 7 | Chain relation (A = B + k, B = mC) → express everything in single variable. |
| 8 | Two time-shifts (past & future) → use two equations; x found in 2 steps. |
| 9 | Option back-plug: middle option first; 50 % questions solved in 15 s. |
| 10 | Write small table: Person |
BETA
