Problems on ages and numbers

Placement essentials (aptitude, reasoning, verbal, communication, interviews, Excel, workplace) (All) · Quantitative aptitude · free study notes for exams, placements and interviews

In short

Problems on ages and numbers are word problems in which an unknown quantity - somebody's age, or a number and its digits - is described in words and must be converted into one or two simple equations. They test translation skill rather than heavy computation: nearly every question reduces to a linear equation, a ratio with a common multiplier, or a place-value expression like 10a + b. They appear in the quantitative section of almost every Indian recruitment test (TCS NQT, Infosys, Wipro NLTH, Capgemini, Accenture, Cognizant, AMCAT, eLitmus), in bank and SSC exams, in CAT/CMAT, in the General Aptitude section of GATE, and in the first-year engineering mathematics and business-mathematics papers of most universities. Because the arithmetic is light, these are marks you are expected to score in under a minute - the whole skill is a clean setup and a quick verification.

Explained

Setting up an age problem: the three time frames

Fix the present as the variable. If a person's present age is x years, then n years hence the age is x + n and n years ago it was x − n.
Every sentence is a statement about ONE instant of time. Translate each sentence separately, then combine.
"n years ago" must be subtracted from both people's ages - the commonest error is ageing one person and not the other.
The difference between two ages never changes: if A is 8 years older than B today, A was 8 years older 10 years ago and will be 8 years older 20 years hence. Only the *ratio* changes.
A ratio of two ages always moves towards 1:1 as time passes, and away from 1:1 as you go back. Use this as an instant plausibility check on a question or an answer.
Choose the variable that makes the fewest fractions: for "A is thrice B", let B = x and A = 3x, not A = x and B = x/3.
Standard phrase-to-equation translation:

Statement in words	Equation
A is twice as old as B	A = 2B
A is older than B by 6 years	A = B + 6
A's age is 6 years less than B's	A = B − 6
Five years ago A was thrice B	A − 5 = 3(B − 5)
Four years hence A will be twice B	A + 4 = 2(B + 4)
Sum of present ages is 60	A + B = 60
A's age 10 years hence is thrice his age 10 years ago	A + 10 = 3(A − 10)
A is as much older than B as he is younger than C	A − B = C − A, i.e. 2A = B + C
Present ages are in the ratio 4 : 5	A = 4k, B = 5k
B is 40% older than A	B = 1.4A

Ratio-based age problems

Never write a ratio as a fraction and cross-multiply at once. Introduce a common multiplier k: ages 4 : 5 become 4k and 5k, so one unknown covers both.
With three people in ratio 3 : 4 : 5 and a given sum, 12k = sum - a one-step solve.
Two ratios at two different instants give one equation in k. From ratio a : b now becoming c : d after t years: k = t(c − d)/(ad − bc); if the second ratio c : d was t years *ago*, k = t(d − c)/(ad − bc).
A ratio alone can never fix an age - a second piece of information (sum, difference, or a ratio at another time) is always needed. If a question gives only one ratio, the answer must be "cannot be determined", and that is a real option in bank exams.
Watch out for ratios of *differences*: "the ratio of the differences between their ages 5 years hence and 5 years ago" is always 1 : 1, because differences are constant.

Group, family and average-age problems

Total age of a group = average × number of members. Work with totals, convert back to an average only at the end.
Going back t years, the total age of a group of n people falls by n × t - but only for members who were already born.
When a new member (a baby, a new employee) joins, the total rises by that member's age and the count by 1; the average change tells you the member's age.
Replacement problems: if a member aged p is replaced by one aged q, the total changes by (q − p), so the average changes by (q − p)/n.
"Average age of a family remained the same after x years" means the fall caused by a new birth exactly cancels the rise from everyone ageing.
Captain/team type: average age of 11 players, excluding the captain and one other, the remaining 9 average is given - again pure totals.

Number problems: place value and digit reversal

A two-digit number with tens digit a and units digit b is 10a + b, not ab. Its reverse is 10b + a. Constraints: a is 1 to 9, b is 0 to 9, both integers.
Key identities: (10a + b) + (10b + a) = 11(a + b) and (10a + b) − (10b + a) = 9(a − b).
Consequences used directly by examiners: the difference of a two-digit number and its reverse is always a multiple of 9 (and equals 9 × difference of digits); their sum is always a multiple of 11; if the sum is a perfect square, a + b = 11.
Three digits: N = 100a + 10b + c, reverse = 100c + 10b + a, so N − reverse = 99(a − c) (the middle digit cancels) and N + reverse = 101(a + c) + 20b.
"A number is k times the sum of its digits" → 10a + b = k(a + b). "The number exceeds the sum of its digits by 27" → 10a + b − (a + b) = 27 → 9a = 27 → a = 3, which fixes only the tens digit.
"On interchanging the digits the number increases by 36" → 9(b − a) = 36 → b − a = 4.
Digits must come out as single non-negative integers; a fractional or two-digit "digit" means the setup is wrong.

Sum, difference, product and consecutive-number problems

Two numbers from sum S and difference D: x = (S + D)/2, y = (S − D)/2. S and D must have the same parity for integer answers.
Link the three symmetric quantities: (x + y)² = (x − y)² + 4xy, and x² + y² = [(x + y)² + (x − y)²]/2. These let you find x − y without finding x and y.
Consecutive integers: take them as n, n + 1, n + 2, ... or, better, centre them - three consecutive integers as (n − 1), n, (n + 1) so the sum is 3n.
For an odd count of consecutive terms, sum = count × middle term: sum of 3 = 3 × middle, sum of 5 = 5 × middle. Sum of k consecutive integers from n = kn + k(k − 1)/2.
Consecutive even or odd numbers differ by 2: n, n + 2, n + 4; their sum of three = 3(n + 2) = 3 × middle. Sum of k such terms from n = kn + k(k − 1).
Difference of squares of two consecutive integers equals their sum: (n + 1)² − n² = 2n + 1.
Fraction problems: let the fraction be x/y, write one equation for the stated relation between numerator and denominator and one for the value after the change; "denominator exceeds numerator by 3" → y = x + 3.
Products lead to quadratics: n(n + 2) = 168 → n² + 2n − 168 = 0 → n = 12. Reject the negative root unless the question allows negative numbers; reject a non-integer root for ages and digits.

Speed methods, back-solving and sanity checks

Options first. In a multiple-choice test, substituting the four options into the sentence is usually faster than solving, especially when the question asks for only one person's age.
The ages of the *other* person are often implied by the options: if "son's age" options are 8, 10, 12, 14, the father's age follows instantly from "thrice as old".
Two useful one-line shortcuts, derived from the standard setup (A = n × B now):
- A will be m times B after t years → B = t(m − 1)/(n − m).
- A was m times B t years ago → B = t(m − 1)/(m − n).
Divisibility screening: if ages are in ratio 4 : 7 and the sum is given, the sum must be a multiple of 11 - this kills wrong options at a glance.
Always substitute the answer back into the original words, not into your own equation; that catches a mistranslation, which is the real risk.
Expect whole numbers. Ages, digits and "two numbers" in these questions are almost always positive integers; a fractional answer is a warning, though a few questions legitimately give half-years.
Time budget in a placement test: 40 to 60 seconds per question. If the setup needs more than two variables, you have probably over-modelled it.

How it is examined in India

Placement tests (TCS NQT, Infosys, Wipro NLTH, Capgemini, Cognizant, Accenture, AMCAT, eLitmus): typically one or two questions on ages and one on number/digit problems in the quantitative section; marks are equal to the hardest questions, so they are high-value.
Banking (IBPS PO/Clerk, SBI, RBI Assistant) and insurance: ages appear in the "miscellaneous word problems" set, sometimes with a "cannot be determined" option to test whether you notice missing data.
SSC CGL/CHSL, RRB NTPC, state PSC, CDS/NDA: direct single-step ages and digit questions in the arithmetic section.
CAT/XAT/CMAT and GATE General Aptitude: the same ideas appear dressed as algebra or as a short logical puzzle; GATE's General Aptitude section carries 15 marks out of 100 and includes such numerical word problems.
University papers: these are the standard worked examples of "linear equations in one and two variables" in business mathematics and first-year engineering mathematics.
Exam patterns, section timings and cut-offs change from year to year - confirm the current pattern from the official notification rather than from an old question bank.

Key points

Present-age variable - set x as the present age; past is x − n, future is x + n, and apply n to everyone in the sentence.
Constant difference - the gap between two ages never changes with time; only their ratio does.
Ratio needs an anchor - a single ratio cannot fix ages; a sum, a difference or a second-instant ratio is required.
Common multiplier k - write ratio a : b as ak, bk to reduce two unknowns to one.
Ratio drift - an age ratio always moves towards 1 : 1 with time; use it to sanity-check answers.
Two-ratio formula - a : b now and c : d after t years gives k = t(c − d)/(ad − bc).
n-times/m-times shortcut - if A = nB now and A = m·B after t years, then B = t(m − 1)/(n − m).
Group totals - total age = average × members; going back t years reduces the total by (members × t).
Place value - a two-digit number is 10a + b, a three-digit number 100a + 10b + c; never write it as ab.
Reversal identities - sum with reverse = 11(a + b); difference = 9(a − b); for three digits the difference = 99(a − c).
Multiple-of-9 rule - any change produced by interchanging two digits is a multiple of 9.
Sum and difference - x = (S + D)/2, y = (S − D)/2; S and D must share parity for integers.
Symmetric identity - (x + y)² = (x − y)² + 4xy connects sum, difference and product without solving for x and y.
Odd count of consecutives - sum = count × middle term; centring the terms removes most of the algebra.
Products give quadratics - expect two roots and reject the one that is negative or non-integral.
Integrality check - ages and digits must be whole numbers in the given range (digits 0-9, tens digit at least 1).
Back-substitution - verify against the original sentence; it catches mistranslation, the main source of error.
Option elimination - substituting the four choices is often faster than solving, and divisibility screening removes wrong options instantly.

Formulas and examples

Age after n years = x + n; age n years ago = x − n (x = present age in years)
Difference of ages is constant: A − B = (A + n) − (B + n) for every n
Ratio method: ages in ratio a : b → ak and bk, where k > 0 is the common multiplier
Ratio a : b now, c : d after t years → k = t(c − d)/(ad − bc); ratio c : d t years ago → k = t(d − c)/(ad − bc)
A = nB now and A = m·B after t years → B = t(m − 1)/(n − m), A = nB
A = nB now and A was m·B t years ago → B = t(m − 1)/(m − n)
Total age of a group = average age × number of members; total t years ago = present total − (members × t)
Two-digit number N = 10a + b (a = tens digit 1-9, b = units digit 0-9); reverse = 10b + a
N + reverse = 11(a + b); N − reverse = 9(a − b); three-digit: N − reverse = 99(a − c)
Sum S and difference D of two numbers: x = (S + D)/2, y = (S − D)/2
(x + y)² = (x − y)² + 4xy; x² + y² = [(x + y)² + (x − y)²]/2; (n + 1)² − n² = 2n + 1
Sum of k consecutive integers from n = kn + k(k − 1)/2; of k consecutive even/odd numbers from n = kn + k(k − 1)
For an odd number of consecutive terms: sum = (number of terms) × (middle term)

Example 1 - father and son, two time frames. A father is three times as old as his son. After 12 years he will be twice as old as the son. Find their present ages.

Let the son be x, father 3x (choose the smaller as x to avoid fractions).
After 12 years: 3x + 12 = 2(x + 12) → 3x + 12 = 2x + 24 → x = 12.
Son = 12 years, father = 36 years.
Shortcut check with B = t(m − 1)/(n − m), n = 3, m = 2, t = 12: B = 12 × 1/1 = 12. ✔
Verify in words: after 12 years, 48 and 24 - exactly twice. ✔

Example 2 - ratio at two instants. The present ages of A and B are in the ratio 5 : 7. Four years hence the ratio will be 3 : 4. Find the present ages and the ratio four years ago.

Let A = 5k, B = 7k. Then (5k + 4)/(7k + 4) = 3/4 → 4(5k + 4) = 3(7k + 4) → 20k + 16 = 21k + 12 → k = 4.
A = 20 years, B = 28 years. Formula check: k = t(c − d)/(ad − bc) = 4(3 − 4)/(5×4 − 7×3) = (−4)/(−1) = 4. ✔
Four years ago: 16 and 24 → 2 : 3. Note the ratio was further from 1 : 1 in the past, as expected.

Example 3 - family average and a new birth. The average age of a family of 6 members is 22 years. The youngest member is 7 years old. What was the average age of the family when the youngest was born?

Present total = 22 × 6 = 132 years.
Seven years ago the youngest did not exist, so remove his 7 years: 132 − 7 = 125 is the present total of the other 5.
Each of those 5 was 7 years younger: total then = 125 − (5 × 7) = 90 years.
Average then = 90/5 = 18 years.
Common trap: dividing 90 by 6. Only 5 members existed.

Example 4 - two-digit number with reversal. The sum of the digits of a two-digit number is 12, and the number obtained by reversing the digits is 18 less than the original number. Find the number and check what its sum with the reverse gives.

Let the number be 10a + b. Given a + b = 12 and (10a + b) − (10b + a) = 18 → 9(a − b) = 18 → a − b = 2.
Solving: a = (12 + 2)/2 = 7, b = (12 − 2)/2 = 5 → number = 75.
Verify: reverse 57, and 75 − 57 = 18. ✔
Sum with reverse = 75 + 57 = 132 = 11 × 12 = 11(a + b). ✔

Example 5 - sum, product and consecutive numbers. (a) The sum of two numbers is 25 and their product is 144. Find the difference and the numbers.

(x − y)² = (x + y)² − 4xy = 625 − 576 = 49 → x − y = 7.
x = (25 + 7)/2 = 16, y = (25 − 7)/2 = 9; check 16 × 9 = 144. ✔

(b) The product of two consecutive even numbers is 168.

n(n + 2) = 168 → n² + 2n − 168 = 0 → (n + 14)(n − 12) = 0 → n = 12 (reject −14 for natural numbers) → 12 and 14.

Sum = 3 × middle → middle = 19 → 17, 19, 21.

Example 6 - the reversed-age puzzle (interview favourite). A man's age is the reverse of his wife's age. One-eleventh of the sum of their ages equals the difference between their ages. Find both ages.

Man = 10a + b, wife = 10b + a with a > b. Sum = 11(a + b), difference = 9(a − b).
(1/11) × 11(a + b) = 9(a − b) → a + b = 9a − 9b → 10b = 8a → 4a = 5b → a = 5, b = 4.
Man = 54, wife = 45. Check: sum 99, 99/11 = 9; difference 54 − 45 = 9. ✔

Problem type	Standard setup	Typical equation
"n times now, m times later"	son = x, father = nx	nx + t = m(x + t)
Ratio at two instants	ak, bk	(ak + t)/(bk + t) = c/d
Age difference given	x and x + d	(x + d) − t = m(x − t)
Group average	total = average × n	total − n·t for t years ago
Two-digit number	10a + b	relation in a + b and a − b
Fraction	x/(x + d)	(x + p)/(x + d + p) = r/s
Consecutive terms	centre at n	3n, 5n, ... for odd counts
Sum and product	S, P	x, y are roots of z² − Sz + P = 0

Where it is used

Campus recruitment screening: the quantitative round of IT services hiring (TCS NQT, Infosys, Wipro, HCL, Capgemini, Cognizant, LTIMindtree) and of the AMCAT and eLitmus aggregator tests, where sectional cut-offs decide who reaches the technical interview.
Banking, insurance and government recruitment: IBPS and SBI PO/Clerk, RBI, LIC AAO, SSC CGL/CHSL, RRB NTPC, CDS and NDA arithmetic sections.
Management entrance and B-school prep: CAT, XAT, CMAT, MAT and NMAT quantitative ability; also the aptitude papers of many state CET examinations.
University coursework: linear equations in business mathematics (B.Com, BBA) and engineering mathematics foundations, and in AICTE-model "soft skills and aptitude" or employability-skills courses that engineering colleges run in the pre-final year.
Teaching, training and content roles: aptitude trainers at T.I.M.E., Career Launcher, Unacademy and in-house college training cells; question-setters for ed-tech item banks.
On the job: the same place-value and date arithmetic shows up in validating ages against dates of birth in HR and payroll systems (EPF/ESI eligibility, retirement-date computation), in cheque and account-number check logic, and in writing test cases for age-eligibility rules in banking and insurance software.

Interview questions

Q: A father is 30 years older than his son and in 12 years he will be twice as old as the son. Find their ages? A: Son x, father x + 30; x + 42 = 2(x + 12) = 2x + 24 → x = 18, so the son is 18 and the father 48.
Q: Why can a question that gives only the ratio of two ages never be solved? A: A ratio fixes only the proportion; ages 20 and 28 and ages 40 and 56 both give 5 : 7, so a second independent fact - a sum, a difference, or the ratio at another time - is essential.
Q: The ratio of two ages is 5 : 7 today. Can it ever be 7 : 5? A: No. The older person stays older and the ratio moves monotonically towards 1 : 1 with time, so it can approach but never cross equality.
Q: What stays constant as two people age, and how do you use it? A: The difference of their ages. It lets you write both ages with one variable (x and x + d) and immediately rejects any question whose data imply a changing difference.
Q: Why is the difference between a two-digit number and its reverse always divisible by 9? A: (10a + b) − (10b + a) = 9(a − b), which is 9 times an integer.
Q: The sum of a two-digit number and the number formed by reversing its digits is 121. What are the possible numbers? A: 11(a + b) = 121 → a + b = 11, so 29, 38, 47, 56, 65, 74, 83, 92 all work - the data fix only the digit sum.
Q: The sum of a father's and son's ages is 45. Five years ago the product of their ages was four times the father's age then. Find the ages? A: (F − 5)(S − 5) = 4(F − 5) → S − 5 = 4 → S = 9 and F = 36; check 31 × 4 = 124 = 4 × 31.
Q: The average age of 5 employees is 30. A sixth joins and the average becomes 29. How old is he? A: New total = 29 × 6 = 174, old total = 150, so he is 24 years old.
Q: Sum of two numbers is 20 and the sum of their squares is 232. Find the numbers without solving a quadratic by formula? A: 2xy = 400 − 232 = 168 → xy = 84; (x − y)² = 400 − 336 = 64 → x − y = 8, giving 14 and 6.
Q: Three consecutive integers sum to 72 - can that be? A: Yes. Three consecutive integers sum to 3 × the middle term, so the sum must be a multiple of 3; 72 is a multiple of 3, so the numbers are 23, 24, 25. (The screen is the divisibility test: 71 or 73 would be impossible.)
Q: How do you spot a wrongly framed age question in an exam? A: The answer comes out negative, fractional where it should be whole, or implies a child older than the parent; also check that the stated ratios move the right way in time.
Q: In a timed test, when do you back-solve from options instead of forming equations? A: When the question asks for a single person's age or a single number, the options are few and small, and the sentence can be checked in one line - typically faster than setting up and solving.

Common mistakes

Ageing only one person: writing A − 5 = 3B instead of A − 5 = 3(B − 5) for "five years ago".
Writing a two-digit number as the product of its digits (ab) instead of 10a + b, or forgetting that the tens digit cannot be 0.
Cross-multiplying a ratio straight away instead of introducing the common multiplier k, which creates two unknowns from one.
Assuming the age *ratio* is constant over time, or that a ratio alone is enough data to find ages.
In family-average problems, dividing the past total by the present number of members, or forgetting that an unborn member contributes no years.
Keeping both roots of a quadratic - reporting a negative age, a non-integral digit or a digit greater than 9.
Verifying the answer against your own equation rather than the original wording, so a mistranslated sentence goes undetected.
Over-modelling with three or four variables and running out of time when one variable and the constant age difference would have done.