CSAT· Prelims
Problems on Ages for CSAT
Problems on ages for CSAT: the constant-difference rule, the k method for ratio ages, past and future ratios, and sum/average of ages, with worked examples and practice MCQs.
Problems on ages are word problems in which the ages of two or more people at different points in time (now, some years ago, some years hence) are linked by ratios, sums or differences, and you must find a present age. They matter for CSAT because Paper II is a qualifying paper needing a 33 percent score, and ages questions are among the most predictable marks on it: every one is a ratio or a linear equation in disguise, and the same three setups repeat year after year.
The golden rule: time shifts everyone equally
The golden rule of ages is that the difference between two people's ages never changes. If a father is 24 years older than his son today, he was 24 years older ten years ago and will be 24 years older ten years hence. So age differences are constant while ages themselves move together. Whenever a question gives a difference of ages, you can use it at any point in time.
Example 1. A man is 24 years older than his son. In two years, his age will be twice the age of his son. Find the father's present age. Let the son's present age be x; the father's is x + 24. In two years: (x + 24) + 2 = 2(x + 2), so x + 26 = 2x + 4, giving x = 22. The father's present age is 22 + 24 = 46 years.
Ratio-based ages: use the k method
When ages are given as a ratio such as 4:5, write the actual ages as 4k and 5k, where k is a common multiplier. Any extra condition (a future ratio, a difference, a sum) becomes one equation in k. This k method is the fastest route through most CSAT age questions.
Example 2. The ratio of the present ages of two brothers is 1:2, and five years back the ratio was 1:3. What will be the ratio of their ages after five years? Let the present ages be k and 2k. Five years back: (k - 5)/(2k - 5) = 1/3. Cross-multiplying: 3k - 15 = 2k - 5, so k = 10. Present ages are 10 and 20. After five years they will be 15 and 25, a ratio of 15:25 = 3:5.
Past and future ratios
A past ratio (for example, 'four years ago the ratio was 2:3') and a future ratio ('after four years it will be 5:7') both refer to the same present ages shifted by the given years. Write present ages as the unknown (x and y, or ak and bk), apply the shift, and equate the ratio. Two such conditions give two equations, which is exactly enough for two unknowns.
Example 3. Four years ago, the ratio of the ages of A and B was 2:3, and after four years it will become 5:7. Find their present ages. Let present ages be x and y. Four years ago: (x - 4)/(y - 4) = 2/3, so 3x - 12 = 2y - 8, giving 3x - 2y = 4. After four years: (x + 4)/(y + 4) = 5/7, so 7x + 28 = 5y + 20, giving 7x - 5y = -8. Solving: multiply the first by 5 (15x - 10y = 20) and the second by 2 (14x - 10y = -16); subtracting gives x = 36, and then 3(36) - 2y = 4 gives y = 52. Their present ages are 36 and 52 years.
Sum-of-ages and average-of-ages questions
Many questions hide behind a sum of ages: children born at fixed intervals, or a family's average age. If children are born at intervals of d years, their ages form an arithmetic sequence, and the youngest = average - (n-1)/2 x d. An average age question is a total question: multiply the average by the count to get the sum, then add or subtract the newcomer's age.
Example 4. The sum of the ages of 4 children born at intervals of 2 years each is 40 years. What is the age of the youngest child? The four ages are x, x+2, x+4, x+6 with sum 4x + 12 = 40, so 4x = 28 and x = 7. The youngest is 7 years old.
Example 5. The average age of A, B, C and D is 50 years. When E joins them, the average of all five becomes 49. How old is E? Old total = 4 x 50 = 200; new total = 5 x 49 = 245; so E = 245 - 200 = 45 years.
Common traps and speed tips
- Trap 1: applying the ratio to the wrong time. A ratio given 'six years ago' describes the ages six years ago, not the present ages. Shift first, then use the ratio.
- Trap 2: forgetting the difference is constant. If the elder is 30 years older, that gap holds in the past and the future. Use it to eliminate a variable.
- Trap 3: 'twice as old as' vs 'two years older'. 'Twice the age' is multiplication (2x); 'two years older' is addition (x + 2). Options are built on this confusion.
- Tip 1: always write the k form. Ages in ratio 4:5 are 4k and 5k. One equation gives k, and every asked age follows.
- Tip 2: check by substitution. Plug your ages back into the original conditions; a 10-second check catches sign errors.
Setup | Method |
|---|---|
Ratio of present ages | write as ak, bk; use the extra condition for k |
Ratio in the past/future | shift present ages by the years, then equate ratio |
Difference of ages | constant at all times; use at whichever time is convenient |
Sum with fixed intervals | arithmetic sequence; youngest = average - ((n-1)/2) x interval |
Average age changes | new total - old total = newcomer's age |
The ratio of the ages of two persons is 4:7, and the elder is 30 years older than the younger. The sum of their ages is:
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Answer: (B) Ages are 4k and 7k; the difference 3k = 30 gives k = 10. The ages are 40 and 70, whose sum is 110 years.
The present ages of Sameer and Anand are in the ratio 5:4. Three years hence, the ratio will become 11:9. Anand's present age is:
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Answer: (A) Let the ages be 5k and 4k. Then (5k + 3)/(4k + 3) = 11/9; cross-multiplying: 45k + 27 = 44k + 33, so k = 6. Anand's present age is 4 x 6 = 24 years.
A mother is four times as old as her daughter. Twenty years hence, she will be twice as old as her daughter. The sum of their present ages is:
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Answer: (B) Let the daughter's age be x; the mother's is 4x. In 20 years: 4x + 20 = 2(x + 20), so 4x + 20 = 2x + 40, giving 2x = 20 and x = 10. The ages are 10 and 40, summing to 50 years.
The sum of the ages of 5 children born at intervals of 3 years each is 50 years. The age of the youngest child is:
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Answer: (A) The ages are x, x+3, x+6, x+9, x+12 with sum 5x + 30 = 50, so 5x = 20 and x = 4. The youngest is 4 years old.
In a family, the average age of the father and mother is 35 years. The average age of the father, mother and their only son is 27 years. The son's age is:
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Answer: (C) Father + mother = 2 x 35 = 70. All three = 3 x 27 = 81. The son's age = 81 - 70 = 11 years.
Maria is younger than John by 6 years. If their ages are in the ratio 5:7, Maria's age is:
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Answer: (B) Ages are 5k and 7k; the difference 2k = 6 gives k = 3. Maria's age is 5 x 3 = 15 years.
Ajay is 40 years old and Sumit is 60 years old. How many years ago was the ratio of their ages 3:5?
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Answer: (B) Let it be x years ago: (40 - x)/(60 - x) = 3/5. Cross-multiplying: 200 - 5x = 180 - 3x, so 2x = 20 and x = 10 years ago.
Answer key
- Q1 - (b). Ages are 4k and 7k; the difference 3k = 30 gives k = 10. The ages are 40 and 70, whose sum is 110 years.
- Q2 - (a). Let the ages be 5k and 4k. Then (5k + 3)/(4k + 3) = 11/9; cross-multiplying: 45k + 27 = 44k + 33, so k = 6. Anand's present age is 4 x 6 = 24 years.
- Q3 - (b). Let the daughter's age be x; the mother's is 4x. In 20 years: 4x + 20 = 2(x + 20), so 4x + 20 = 2x + 40, giving 2x = 20 and x = 10. The ages are 10 and 40, summing to 50 years.
- Q4 - (a). The ages are x, x+3, x+6, x+9, x+12 with sum 5x + 30 = 50, so 5x = 20 and x = 4. The youngest is 4 years old.
- Q5 - (c). Father + mother = 2 x 35 = 70. All three = 3 x 27 = 81. The son's age = 81 - 70 = 11 years.
- Q6 - (b). Ages are 5k and 7k; the difference 2k = 6 gives k = 3. Maria's age is 5 x 3 = 15 years.
- Q7 - (b). Let it be x years ago: (40 - x)/(60 - x) = 3/5. Cross-multiplying: 200 - 5x = 180 - 3x, so 2x = 20 and x = 10 years ago.
Why does the age difference never change?
Because time passes equally for everyone. If A is 10 years older than B today, then after 5 years A has aged 5 years and B has also aged 5 years, so the gap is still 10. This lets you carry a known difference into any past or future equation.
When should I use x and y instead of the k method?
Use the k method when the question gives a ratio (4:5 becomes 4k, 5k). Use x and y when the question gives no ratio but two independent conditions, such as a past ratio and a future ratio; the two equations then solve directly for the two present ages.
How do interval-birth questions work?
Children born at fixed intervals have ages in an arithmetic sequence. With 5 children at 3-year intervals and a total of 50, write the ages as x, x+3, x+6, x+9, x+12; their sum is 5x + 30 = 50, so x = 4. The youngest is 4.
Can an age question have a fractional answer?
Occasionally, but UPSC's CSAT age questions are designed to give whole-number ages. If your equation yields a fraction like 13.5 years, recheck the setup: you have probably applied a ratio to the wrong point in time.