Ratio, Proportion, Partnership & Ages
🔒 Log in to trackRatio technique with its three biggest exam containers: proportion (mean/third/fourth proportional), partnership (capital × time profit splits) and ages (present vs shifted ratios). Tier 1 reliably has 2 questions from this cluster; most are 30–45 second marks once the multiplier habit is built.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
5 exam-level questions worked step by step.
63 questions — untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (63 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
Divide an amount in a given ratio
very commonA sum (inheritance, bonus, scholarship) split among 2-3 people in ratio m : n : p; find a share, or a share given the total/difference.
How to solve: One part = total ÷ sum of terms; multiply by the asked term. If one share or the difference is given, recover the multiplier x first.
Example: A sum of ₹2,500 is divided among A, B and C in the ratio 2 : 3 : 5. What is the largest share?
Total parts = 10 → 1 part = 250 → largest (C) = 5 × 250 = ₹1,250.
Combine two ratios into A : B : C
very common'A : B = …, B : C = …, find A : B : C' or a quantity-based version ('twice of A = thrice of B').
How to solve: LCM bridge: scale both ratios so the shared term is equal. 'kA = lB = mC' → A : B : C = (LCM/k) : (LCM/l) : (LCM/m).
Example: If A : B = 2 : 5 and B : C = 6 : 7, find A : B : C.
Bridge B to 30: A : B = 12 : 30, B : C = 30 : 35 → A : B : C = 12 : 30 : 35.
Mean / third / fourth proportional
commonDirect definition question: 'mean proportional between a and b', 'third proportional to a and b', 'fourth proportional to a, b, c'.
How to solve: √(ab), b²/a, bc/a respectively — write the proportion first (a : b :: b : ? repeats the middle; a : b :: c : ? does not). Pure formula recall.
Example: Find the third proportional to 12 and 18.
12 : 18 :: 18 : x → x = 18²/12 = 27.
Partnership profit division
very common2-3 partners, capitals (sometimes changed mid-year or joined late), profit at year end; find a share or the total.
How to solve: Ratio of capital × months (sum capital × months per stretch for mid-year changes), then fraction × profit. Working partner: their cut comes off the profit first.
Example: A invests ₹30,000 for 12 months and B invests ₹40,000 for 6 months. Out of a profit of ₹28,800, find A's share.
Capital-months 360000 : 240000 = 3 : 2 → A = 3/5 × 28800 = ₹17,280.
Ages: present and shifted ratios
very common'Ratio of present ages is a : b; after/before n years it is p : q' or 'father is k times the son'.
How to solve: Write ages ax, bx; shift both by n (add for hence, subtract for ago); equate to p : q; cross-multiply once. The constant age-difference is a free checker.
Example: The present ages of two persons are in the ratio 4 : 5. After 16 years the ratio will be 6 : 7. Find the sum of their present ages.
7(4x + 16) = 6(5x + 16) → x = 8 → ages 32 and 40 → sum 72 years.
Income–expenditure–savings ratios
commonTwo ratios (income, expenditure) plus equal or unequal savings; find an income or an expenditure.
How to solve: Incomes ax, bx and expenditures py, qy (different multipliers); two savings equations; subtract or eliminate y.
Example: Incomes of A and B are in ratio 8 : 5 and expenditures in 5 : 3. If each saves ₹1,200, find A's income.
8x − 5y = 1200, 5x − 3y = 1200 → y = 1.5x → 0.5x = 1200 → A = 8x = ₹19,200.
Coin denomination ratios
occasional50p/25p/10p (or ₹1/50p) coins in a ratio with total value; find a coin count or the total number of coins.
How to solve: Price one full set (Σ count × value, one unit only), divide the total value by the set value to get sets, multiply back for counts.
Example: A bag has 50-paise, 25-paise and 10-paise coins in the ratio 7 : 6 : 5 amounting to ₹550. Find the number of 25-paise coins.
Set value = 550 paise = ₹5.50 → 100 sets → 25p coins = 6 × 100 = 600.
Componendo–dividendo
occasional(x + y)/(x − y) given as a ratio; find x : y or x : (x + y), sometimes with coefficients on x and y.
How to solve: Jump formula: x/y = (p + q)/(p − q). With coefficients, C&D first gives mx : ny, then divide off the coefficients. Check the direction (sum over difference).
Example: If (x + y)/(x − y) = 5/3, find x : y.
x : y = (5 + 3) : (5 − 3) = 8 : 2 = 4 : 1.