Mathematics — Chapter 4

Problems on Ages — The Ratio Method and the Constant Difference

Why this chapter matters: Age questions are the most predictable block in one-paper arithmetic: every question is built from just three ingredients — a ratio, a difference, or a sum — plus the one fact that time moves equally for everyone. Our verified banks hold 68 topic-relevant items for this chapter: set 136’s 31 questions plus 37 tightly matched paper items. The set is the anchor — it contains every shape the papers ask — and several of its questions reappear almost verbatim in FPSC papers (the “24 years older, twice in two years” father–son question appears in three papers; the sum-60 father–son question in three more; the Kamal–Sarwar ratio question in two).

1. The three rules before any question

Rule 1 — The difference 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. Ages grow; gaps do not. Most age questions are solved by this one fact alone.

Rule 2 — Time is added to, or subtracted from, every person equally. “Six years ago” means subtract 6 from each person’s age; “after 4 years” means add 4 to each. A three-person total ten years ago is 30 less than today’s total (Q11); a three-person total three years ago is 9 less (paper item, FPSC 2018: 80 − 9 = 71).

Rule 3 — A ratio is a set of equal parts. If present ages are in the ratio 6 : 7, write them as 6x and 7x and find x from any other condition. The “unit-part” method of Section 3 is Rule 3 applied mechanically, and it solves more than half the set.

Two checks before you start any age question: which time is being asked (present, past, or future — Q22’s whole trap is this), and whose age is being asked (in the Kamal–Sarwar family, asking for Kamal instead of Sarwar changes the answer from 16 to 18).

2. One person, two times — the single-variable equation

When only one person’s age is unknown, let it be x and translate the sentence directly into an equation.

• Q1: Aslam’s age after 15 years will be 5 times his age 5 years back. x + 15 = 5(x − 5) → x + 15 = 5x − 25 → 4x = 40 → 10 years. Check: in 15 years he is 25; five years back he was 5; 25 = 5 × 5.
• Paper item (FPSC 2015, Huma’s version of the same question): age after 15 years = 5 times age 5 years back → the same equation, the same answer, 10 years. A set question repeated with a new name is still the same question.
• Q31 (the trick question): “Take my age three years hence, multiply by 3, subtract three times my age three years ago.” 3(x + 3) − 3(x − 3) = 3x + 9 − 3x + 9 = 18. The x cancels completely — the answer is 18 whatever the person’s age. If an equation collapses to a number with no x left, that number is the answer; do not hunt for a second equation.

The method: write “present = x”, then write each time-shifted age as x ± years, then translate “times”, “more than” and “ratio” into ×, + and :. Never solve in your head from the sentence — the equation is the solution.

3. The ratio method — one part = the key (the taxonomy’s named method)

This is the chapter’s core method, and it solves Q5–Q16. The pattern never varies: ratio → parts → one condition fixes the part → multiply back.

3.1 Present ratio + a present difference

• Q5: Present ratio P : Q = 6 : 7, and Q is 4 years older than P. One part = 4 years (7 − 6 = 1 part = 4). P = 24, Q = 28. After 4 years: 28 : 32 = 7 : 8. Note the ratio narrows as time passes — 6:7 becomes 7:8 — because the same 4 years are added to both.
• Paper item (FPSC 2010): ages differ by 12 years, ratio 3 : 7. Difference = 4 parts = 12 → 1 part = 3 → the ages are 9 and 21.
• Q6: Ratio 5 : 7; Q’s present age minus P’s age after 6 years is 2. 7x − (5x + 6) = 2 → 2x = 8 → x = 4. P = 20, Q = 28; total = 48 years. Read the condition exactly: it is Q now against P after 6 years, not both now.

3.2 Present ratio + one person’s future age

• Q7: Ratio Ali : Danial = 4 : 3. After 6 years Ali will be 26, so Ali is 20 now → 1 part = 5 → Danial = 15 years.
• Q14 (ratio + product): Father : son = 7 : 3, product of ages = 756. 7x × 3x = 21x² = 756 → x² = 36 → x = 6. Father 42, son 18; after 6 years 48 : 24 = 2 : 1. A product condition still gives the part — just through a square.

3.3 Present ratio + a future ratio

• Q8: Present 5 : 6; seven years hence 6 : 7. The cross-multiplication shortcut: present parts 5, 6 and future parts 6, 7 differ by 1 part each way, so 1 part = 7 years (the gap). X = 5 × 7 = 35, Y = 42. Check: 42 : 49 = 6 : 7.
• Q9: Present 5 : 4; three years hence 11 : 9. Solve properly: (5x + 3)/(4x + 3) = 11/9 → 45x + 27 = 44x + 33 → x = 6. Samar 30, Anwar = 24. Check: 33 : 27 = 11 : 9. When the part-gaps are not both 1, cross-multiply — do not guess the part.
• Paper item (FPSC 2005, 2006): the same question as Q9 with new names (Anwar and Ali, 5 : 4 → 11 : 9) → Ali = 24. Another verbatim repeater.

3.4 Past ratio + future ratio

• Q10: Six years ago 6 : 5; four years hence 11 : 10. Let the past parts be 6x and 5x; present ages are 6x + 6 and 5x + 6; four years hence they are 6x + 10 and 5x + 10. (6x + 10)/(5x + 10) = 11/10 → 60x + 100 = 55x + 110 → x = 2. Six years ago Sarwar was 10, so Sarwar is 16 now; Kamal is 18. Check: 22 : 20 = 11 : 10.
• Paper items (FPSC 2009, 2010): the same question with Khalid and Sadaf. In the 2009 paper Sadaf is the second person (5 parts) → 16 years; in the 2010 paper the order is reversed and Sadaf is asked as the first person (6 parts) → 18 years. Same numbers, different person asked, different answer — this is why Section 1’s second check exists.
• Q4 (past ratio + future fraction): Ten years ago Ali : father = 1 : 5; after six years Ali will be three-sevenths of his father’s age. Let past ages be x and 5x; present: x + 10 and 5x + 10; in six years: x + 16 and 5x + 16. (x + 16) = (3/7)(5x + 16) → 7x + 112 = 15x + 48 → x = 8. Ali is 18, his father 50. Check: 24 = (3/7) × 56.
• Q15 (present ratio + future ratio + marriage ratio): Present man : wife = 4 : 3; after 4 years 9 : 7. (4x + 4)/(3x + 4) = 9/7 → 28x + 28 = 27x + 36 → x = 8. Man 32, wife 24. At marriage the ratio was 5 : 3: (32 − t)/(24 − t) = 5/3 → 96 − 3t = 120 − 5t → t = 12 years ago. Check: 20 : 12 = 5 : 3. Three ratios, one unknown each time — solve them in order.

3.5 Past multiple + present ratio

• Q16: Ten years ago A was half of B; present ratio 3 : 4. Present: 3x, 4x. Ten years ago: 3x − 10 = (4x − 10)/2 → 6x − 20 = 4x − 10 → x = 5. A = 15, B = 20; total = 35 years.

The ratio method in one line: parts first, one equation for the part, multiply back, and check the answer against every condition in the question — a part that satisfies only one condition is wrong.

4. The constant-difference method — father and son

Father–son questions are almost always difference questions in disguise: the difference is stated or implied, and a multiple at another time pins the ages down.

• Q18: A man is 24 years older than his son; in two years his age will be twice his son’s. Difference = 24, always. In two years: son = s, man = s + 24 = 2s → s = 24, so the son is 22 now (man 46). Check: 24 and 48.
• Paper items (FPSC 2009, 2010, 2015): the same question in three papers → son 22 years every time. Learn it, don’t re-solve it.
• Q19: Eighteen years ago a father was three times his son; now he is twice. Difference is constant: let the son now be s, father 2s. Eighteen years ago: 2s − 18 = 3(s − 18) → 2s − 18 = 3s − 54 → s = 36. Son 36, father 72; sum = 108. Check: 18 years ago, 18 and 54.
• Q21: Ten years ago father was thrice son; ten years hence father will be twice son. Let present be F and S: F − 10 = 3(S − 10) and F + 10 = 2(S + 10). From the two: F = 3S − 20 and F = 2S + 10 → S = 30, F = 70; ratio = 7 : 3. Check: 60 and 20, then 80 and 40.
• Paper items (FPSC 2005, 2006): the same question → ratio 7 : 3 in both papers.
• Q27: Difference 10 years; fifteen years ago the elder was twice the younger. Let younger now be y: elder y + 10. Fifteen years ago: y + 10 − 15 = 2(y − 15) → y − 5 = 2y − 30 → y = 25. Elder = 35. Check: 10 and 20 fifteen years ago.
• Q20: A person’s age is two-fifths of his mother’s; after 8 years he will be one-half. Let person = 2x, mother = 5x. (2x + 8) = (5x + 8)/2 → 4x + 16 = 5x + 8 → x = 8. Person 16, mother = 40. Check: 24 = half of 48.
• Paper item (FPSC 2014): five years ago father’s age was thrice son’s; son is 30 now. Son five years ago was 25, father then 75, so father is 80 now. When one present age is given, work backwards through the past condition, then forwards to the present.
• Paper item (NTS 2012): a father is now three times his son; five years back he was four times. 3s − 5 = 4(s − 5) → s = 15.
• Paper item (FPSC 2009): Ali is now half his father’s age; after 20 years his father will be one and a half times Ali. F = 2A; 2A + 20 = 1.5(A + 20) → 0.5A = 10 → Ali = 20.

5. Sums, products and chains

5.1 Sums of ages

• Q23: Father + son = 60 now; six years ago father was five times son. Six years ago the sum was 60 − 12 = 48: son then 8, father 40 (5 × 8). Son now 14; after 6 years = 20. The step students skip is adjusting the sum for the time shift: a two-person sum changes by 2 × years, not by years.
• Paper items (FPSC 2009, 2010): the same question → son 20 years after 6 years.
• Q11: Three people total 93 now; ten years ago their ratio was 2 : 3 : 4. Ten years ago the total was 93 − 30 = 63 = 9 parts → 1 part = 7. Salman was 28 then, 38 now.
• Paper item (FPSC 2018): three people’s total now 80; three years ago 80 − 9 = 71 (Section 1, Rule 2).
• Paper item (FPSC 2010): A + B = 36, difference 6 → A = (36 + 6)/2 = 21, B = 15. Sum-and-difference: elder = (sum + difference)/2, younger = (sum − difference)/2.
• Q24 (the pure form): A + B is 12 more than B + C. (A + B) − (B + C) = A − C = 12 → C is 12 years younger than A. B cancels — you never need B’s age.
• Q25: A man’s age is three times the sum of his two sons’ ages; five years hence his age will be double the sum of his sons’ ages. Let sons’ sum = s, father = 3s. In five years: 3s + 5 = 2(s + 10) → s = 15. Father = 45. Check: 50 = double of 25. Note the sons’ sum grows by 10 (two sons × 5 years), not 5.
• Paper item (PPSC 2026): Moaz + father = 60; two years ago father was three times Moaz. (60 − M − 2) = 3(M − 2) → 58 − M = 3M − 6 → Moaz = 16.
• Paper item (FPSC 2008, 2015): a girl is 18 years younger than her mother; in 6 years their sum will be 54. Present sum = 54 − 12 = 42 → girl = (42 − 18)/2 = 12.

5.2 Products and intervals

• Q14 (Section 3.2) is the product shape: ratio plus product gives the part through a square (21x² = 756 → x = 6).
• Paper item (PPSC 2025): product of Zahra’s and Ayesha’s ages is 195; Zahra is younger. 195 = 13 × 15 → Ayesha, the elder, is 15.
• Q26: Five children born at intervals of 3 years sum to 50. Youngest x: x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 5x + 30 = 50 → youngest = 4 years (ages 4, 7, 10, 13, 16).
• Paper item (FPSC 2016): the same shape, five children, sum 50 → youngest 4 years.
• Paper item (FPSC 2017): four children at 3-year intervals sum to 50: 4x + 18 = 50 → youngest = 8 years. Same method, one fewer child — the interval total changes, not the method.

5.3 Chains — three people, linked statements

• Q17: A is two years older than B, who is twice as old as C; total 27. C = c, B = 2c, A = 2c + 2. 5c + 2 = 27 → c = 5. B = 10 (A = 12). Check: 12 + 10 + 5 = 27.
• Paper items (FPSC 2016, 2017): the same chain → B = 10 in both papers.
• Q28: In 10 years A will be twice as old as B was 10 years ago; A is 9 years older than B. A = B + 9. (B + 9 + 10) = 2(B − 10) → B + 19 = 2B − 20 → B = 39 (A = 48). Check: in 10 years A is 58; B ten years ago was 29; 58 = 2 × 29.
• Q29 (a chain through a formula): Ali is 2 years younger than Moeen, who is 5 → Ali is 3. (Gulshan − 6)/18 = 3 → Gulshan = 60 years. Work the chain backwards from the known end.
• Q30 (a chain through birth years): Nadia’s father was 38 when she was born; her mother was 36 when her brother (four years younger than Nadia) was born. Father’s age = 38 + Nadia’s age; mother’s = 36 + (Nadia’s age − 4). Difference = 38 − 32 = 6 years — Nadia’s own age cancels, so it is never needed.
• Paper item (FPSC 2008; NTS 2015): a man is 5 years older than his wife, who is ten times her daughter; the daughter reaches 8 in 3 years (so she is 5 now). Wife 50, man = 55 years.
• Paper item (FPSC 2010): the same shape (daughter 4 years old three years back → 7 now; wife 35) → Asif = 40.
• Paper item (FPSC 2009, 2015): Lubna is 6 years older than Saima, Jaweria 5 years older than Lubna, total 41. Saima = s: s + (s + 6) + (s + 11) = 41 → Saima = 8.
• Paper item (NTS 2012): father’s age is twice the elder son’s; ten years hence father will be three times the younger son’s; sons differ by 15. Elder = e, younger = e − 15, father = 2e. 2e + 10 = 3(e − 15 + 10) → 2e + 10 = 3e − 15 → e = 25. Father = 50 years.
• Paper item (PPSC 2026): three years ago C was three times D; four years hence C will be twice D. C − 3 = 3(D − 3) and C + 4 = 2(D + 4) → C = 3D − 6 → 3D − 2 = 2D + 8 → D = 10, C = 24.
• Paper item (general-ability set): Rahan is 12, his mother three times his age (36); after 8 years his mother is 44.

5A. The OPC drill — average back-solving, large past multiples, and chain instances

The OPC book’s Age Problems block (pp. 70–76) mostly repeats this chapter’s set shapes — four of its ten examples are already taught here verbatim and are not re-taught: Ex7 (24 years older, twice in two years → son 22) is Section 4’s Q18; Ex3 (three times now, four times five years back → 15) is Section 4’s NTS 2012 paper item; Ex6 (A = B + 2, B = 2C, total 27 → B = 10) is Section 5.3’s Q17; Ex9 (two-fifths now, one-half after 8 years → mother 40) is Section 4’s Q20. What the block adds is one new shape and four new instances — all answers recomputed independently in this run:

Keyword translation (the block’s Rules 1–3, with one correction). “Ago / back / before” → subtract; “now” → present age; “after / hence / in” → add; “times” → multiply; “of” → multiply (as in “two-fifth of the age”). Present ages in a ratio a : b are written ax and bx. The block’s shorthand “to → divide” is not adopted: no bank stem supports reading “to” as division, and in ratio statements “a to b” is a ratio, not an operation.

Average back-solving (new shape). Five years ago, the average of P and Q was 15 years; today the average of P, Q and R is 20 years. How old will R be after 10 years? First convert the old average back to a present sum: five years ago P + Q = 2 × 15 = 30, so today P + Q = 30 + (2 × 5) = 40. Today P + Q + R = 3 × 20 = 60, so R = 60 − 40 = 20 now, and 30 after 10 years. The step students skip is the first one: an old average must be walked forward to the present (heads × years, Section 1 Rule 2) before it can be combined with a present average.

Large past multiple. A father is twice as old as his son; 20 years ago the father was 12 times the son. F = 2S; F − 20 = 12(S − 20) → 2S − 20 = 12S − 240 → 10S = 220 → S = 22, father = 44. The multiple (12×) looks extreme, but the equation is Section 4’s shape unchanged — a large multiple only makes the past ages small (son 2, father 24 twenty years ago; check: 24 = 12 × 2).

Past multiple + present ratio (new instance). A is twice as old as B; three years ago A was three times B. A = 2B; A − 3 = 3(B − 3) → 2B − 3 = 3B − 9 → B = 6, A = 12. Check: three years ago, 9 and 3 — three times, as stated.

Mother chain (new instance). A mother is twice as old as A’s brother. A is 6 years younger than his brother but 4 years older than his sister. If A’s sister is 15, how old is the mother? Sister 15 → A = 15 + 4 = 19 → brother = 19 + 6 = 25 → mother = 2 × 25 = 50. Work the chain from the one known age, one link at a time — never from the asked end backwards.

Single person, two times (new bank row). After 32 years from now, a boy will be five times as old as he was 8 years back. B + 32 = 5(B − 8) → B + 32 = 5B − 40 → 4B = 72 → B = 18. This is Section 2’s shape (compare Q1’s x + 15 = 5(x − 5) → 10), and its stem is a genuine bank item — FPSC Assistant Director FIA paper 2000 — that the original count script missed because it required the word “age” in the question text and this stem says “how old”. It is the one addition that changes this chapter’s bank count: 67 → 68 (see the practice block). No 🔁/⭐/🆕 mark is claimed for it — a single 2000 paper row is below both 🔁 limbs.

6. Master table — the age methods at a glance

7. Rapid-fire one-liners

• One person, two times: x + 15 = 5(x − 5) → Aslam 10; the 3(x + 3) − 3(x − 3) riddle → 18 always.
• Father–son difference: 24 older, twice in 2 years → son 22; 18 years ago thrice, now twice → son 36, father 72, sum 108.
• Ratio + difference: 6 : 7 with 4 years’ gap → 24 and 28 → after 4 years 7 : 8; 3 : 7 with 12 years’ gap → 9 and 21.
• Ratio now + later: 5 : 6 → 6 : 7 in 7 years → X 35; 5 : 4 → 11 : 9 in 3 years → Anwar 24.
• Past + future ratios: 6 : 5 six years ago → 11 : 10 four years hence → Sarwar 16 (Kamal 18); 1 : 5 ten years ago → Ali 18, father 50.
• Three ratios: man 32, wife 24 → married 12 years ago (20 : 12 = 5 : 3).
• Past multiple + present ratio: 10 years ago half, now 3 : 4 → 15 and 20, total 35.
• Ratio + product: 7 : 3, product 756 → 42 and 18 → after 6 years 2 : 1.
• Sums: father + son 60, six years ago 5× → son 14 now, 20 in 6 years; three people 93, ten years ago 2 : 3 : 4 → Salman 38; A + B exceeds B + C by 12 → C 12 younger than A.
• Sons’ sum: father 3× sum now, 2× sum in 5 years → father 45.
• Intervals: five children, 3-year gaps, sum 50 → youngest 4; four children, sum 50 → youngest 8.
• Chains: A = B + 2, B = 2C, total 27 → B 10; A = B + 9, A + 10 = 2(B − 10) → B 39; Gulshan: (G − 6)/18 = 3 → 60; parents’ gap from birth years → 6.
• Paper repeaters: “24 older, twice in 2 years” → 22 (three papers); sum-60 father–son → son 20 in 6 years (three papers); Kamal–Sarwar ratio → 16/18 (two papers, by who is asked).
• OPC drill: average back-solve → R 30 after 10 years; twice now, 12× twenty years ago → father 44; twice now, thrice 3 years ago → A 12; sister 15 → mother 50; single person B + 32 = 5(B − 8) → 18 (FPSC 2000).

8. Most-asked facts (from our verified past-paper banks)

• Set 136 is the topic’s anchor: 31 questions — ratio-method items 12 (Q4–Q16 except Q11), father–son difference items 7 (Q2, Q3, Q18–Q21, Q27), sum/product/chain items 10 (Q11, Q17, Q22–Q26, Q28–Q30), single-person items 2 (Q1, Q31). The ratio family alone is over a third of the set — master Section 3 first.
• The paper items (37) are led by multiple-of-age items (13) and ratio-of-ages items (8) — and most of them are set questions with new names. The papers test the set’s shapes more than new shapes: if you can solve set 136 cold, the paper items are recognition, not new work. (37, not 36: the OPC drill added the FPSC 2000 single-person item, whose stem says “how old” rather than “age” and so escaped the original count script.)
• The verbatim repeaters: Q18’s father–son (three papers), Q23’s sum-60 father–son (three papers), Q9’s 5 : 4 → 11 : 9 ratio (two papers), Q21’s thrice-then-twice father–son (two papers), Q10’s Kamal–Sarwar ratio (two papers, answer by who is asked), Q17’s three-person chain (two papers). A question that appears in two or three papers is the highest-yield item in the chapter.
• The three-person total shift (Q11’s 93 → 63; FPSC 2018’s 80 → 71) is the papers’ favourite small trap: the total changes by n × years, and forgetting the n is the error.
• The newest papers (PPSC 2025–2026) still ask the same shapes: product 195 → 15, Moaz + father sum 60 → 16, C/D three-years-ago/ four-years-hence → 24. The topic has not changed in twenty years — only the names have.

9. Traps & confusions

• Present vs then — read what is asked. Q22 asks the father’s present age: 40. The 36 in the options is his age four years ago — the right number for the wrong question. Always write “now” next to the answer you circle.
• Who is asked in a ratio pair. The Kamal–Sarwar question gives 16 for Sarwar and 18 for Kamal; the Khalid–Sadaf papers give 16 or 18 for Sadaf depending on which side of the ratio she sits. Solve both ages, then pick the one named.
• Ratio narrows with time. 6 : 7 becomes 7 : 8 after 4 years; it never stays 6 : 7. If your future ratio equals your present ratio, you added the years to only one person.
• Sum shifts by n × years. Three people ten years ago: subtract 30, not 10 (Q11). Two people’s sum in six years: add 12, not 6 (Q23). The multiplier is the head-count.
• Difference never changes — use it. In Q18 and Q19 the difference is the whole solution: set the two ages at the stated time in terms of one unknown and let the constant difference do the work.
• “Times more” vs “times as old”. “Three times as old” means × 3. In this chapter’s banks the papers use “times as old” consistently; if a question ever says “three times more”, treat it with suspicion and check the options — the two readings differ by one whole age.
• The riddle that needs no solving. Q31’s expression collapses to 18 for any age. If your equation’s variable cancels out, stop — the constant is the answer, not an error.
• Age at birth chains (Q30). The parents’ age difference comes from their ages at two different children’s births; the child’s current age cancels. Do not invent it.
• Old averages must be walked forward first. The OPC average question fails if the five-years-ago average (15) is combined directly with today’s average (20): convert it to today’s sum (P + Q = 40) first, then R = 60 − 40 = 20.
• “Of” multiplies; “to” does not divide. “Two-fifth of the age” is × 2/5; the OPC shorthand “to → divide” is not a rule and is not used in this chapter — in “ratio a to b”, “to” is the ratio itself.
• Do not re-solve a duplicate. Four OPC age examples are already this chapter’s Q18, Q17, Q20 and the NTS 2012 item — recognising a known question is faster than solving it again, and solving it twice risks two different answers.

10. Memory tricks

• “Difference never changes” — say it before every father–son question. It turns Q18 into one line: in two years, son = s, man = s + 24 = 2s.
• “Parts, then multiply back” — the ratio mantra. Ratio → parts → one condition → part value → both ages. Every Q5–Q16 solution is those four steps in that order.
• “Shift the sum by heads × years” — three people, ten years back: 3 × 10 = 30 off the total. Heads × years, every time.
• “Sum-and-difference halves” — elder = (sum + difference)/2, younger = (sum − difference)/2. 36 and 6 → 21 and 15, no equations needed.
• “Ask who, ask when” — before circling, check the name (Kamal or Sarwar?) and the time (present or four years ago?). Those two checks catch the chapter’s two most common errors.
• “Narrowing ratios” — as time passes, any age ratio creeps towards 1 : 1 (6 : 7 → 7 : 8). If your computed future ratio widened, you subtracted where you should have added.

11. Visuals (site build specs)

• DIAGRAM — The constant difference: two horizontal age bars (father and son) at “10 years ago”, “now” and “10 years hence”, with the gap bracketed as the same 24 years in all three — placed in Section 4 beside Q18.
• FLOWCHART — Which method? “Ratio given?” → unit-part method (Section 3); “Difference given?” → constant-difference equation (Section 4); “Sum given?” → adjust by heads × years, then ratio/multiple (Section 5) — placed in Section 1 as the chapter’s decision graphic, with Q8, Q18 and Q23 as the three leaves.
• TABLE — Master table as a reference card: the Section 6 table rendered colour-banded, with the ratio rows and the constant-difference row highlighted — placed in Section 6 as the chapter’s revision graphic.
• DIAGRAM — Average back-solving: two average bars (P + Q five years ago at 15, walked forward to 40 today) feeding the three-person total of 60, with R = 20 now and 30 after 10 years — placed in Section 5A beside the average drill.

12. Practice this topic

• Practice set: 68 topic-relevant items in the verified banks serve this topic — set 136’s 31 problems-on-ages questions and 37 tightly matched paper items (multiple-of-age 13, ratio-of-ages 8, sum-of-ages 8, age-difference 5, past/future-age 2, plus the FPSC Assistant Director FIA 2000 single-person item added by the OPC drill in Section 5A). The site serves them as this chapter’s practice set; work set 136 first for every shape, then the papers’ verbatim repeaters (Q18, Q23, Q9, Q21, Q10, Q17 families) for speed. Average-age questions, stone-age GK items, and vocabulary items that merely contain the letters “age” are excluded from this count and served by their own chapters.
• Book mode: ← Previous: Trains, Time & Distance · Next: Partnership & Chain Rule →

Goldmine Notes · Mathematics · Chapter 4 · Sources: FPSC Edition 32, Arithmetic set 136 — Problems on Ages (31 questions, all answers rechecked independently); verified PPSC/FPSC past-paper banks (36 tightly matched paper items, script-counted — counting script hidden_files/goldmine-notes-count-ma04.py); ratio-part, constant-difference and sum-adjustment methods as standard arithmetic. Standard answers only.