Ratio, Proportion, Partnership & Ages

high importance~2 Q in Tier 122 formulas⚡ 15 shortcuts5 subtopics

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Money ratios: income–expenditure, coins & mixed amounts

Two classic containers for ratios:

Income–expenditure: incomes and expenditures are given as two ratios, and each person's savings (income − expenditure) is known. Set incomes $= ax, bx$ and expenditures $= py, qy$ and solve the two saving equations.
Coin bags: numbers of coins of denominations are in a given ratio. Value per 'set' = Σ(count × denomination); total value = (value per set) × k.

The two-variable (x, y) income-expenditure setup is the standard CGL hard question — keep the arithmetic in fractions and it stays small.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 14 practice Q

Detailed notes

Income–expenditure–savings (two ratios)

Incomes in ratio $a : b$ and expenditures in ratio $p : q$ → write incomes $= ax, bx$ and expenditures $= py, qy$ . Two different multipliers — that is the trap. Savings give one linear equation per person: $ax - py = s_1, \qquad bx - qy = s_2.$ Subtract the equations (equal savings make the right sides cancel) or solve by elimination; then build any income, expenditure or saving. Keep the arithmetic in fractions and it stays two lines. Always sanity-check both savings come out positive.

Coin and denomination bags

Coins of denominations $d_1, d_2, d_3$ in ratio $n_1 : n_2 : n_3$ : price ONE full set, $\text{set value} = n_1d_1 + n_2d_2 + n_3d_3,$ then number of sets = total value ÷ set value. Count of a coin = its ratio term × sets. Convert everything to paise (or everything to rupees) first — ₹1 = 100 paise.

Wages in proportion to work

Fair division follows work done. Days worked → shares ∝ days. Working together → shares ∝ work rates ( $\frac{1}{\text{days}}$ ), not the days themselves: workers finishing alone in 6 and 9 days earn in ratio $\frac{1}{6} : \frac{1}{9} = 3 : 2$ .

Chained ratios on money

"A = 3/4 of B, B = 4/5 of C" → convert to a common base: $A : B : C = 3 : 4 : 5$ . Then any total or share is plain part-division. Whenever two linked fractions appear, build the three-term ratio before touching the money.

A complete income–expenditure walkthrough

"Incomes 5 : 4, expenditures 3 : 2, each saves ₹1,600."

Incomes $5x, 4x$ ; expenditures $3y, 2y$ — note the two different letters.
$5x - 3y = 1600$ and $4x - 2y = 1600$ .
Equal savings → subtract the equations: $(5x - 4x) - (3y - 2y) = 0$ , i.e. $x - y = 0$ , so $x = y$ .
Substitute: $5x - 3x = 1600$ → $x = 800$ . Incomes ₹4,000 and ₹3,200; expenditures ₹2,400 and ₹1,600 — both savings check out at ₹1,600 ✓. With equal savings, the subtraction step is what makes the question short; with unequal savings, eliminate y by cross-multiplying the two equations.

Money-ratio hygiene

Convert paise/rupees and hours/minutes before forming any ratio.
Answer what is asked: coin COUNT vs coin VALUE, income vs expenditure — the options always carry both.
Check plausibility: savings must come out positive for everyone; a negative y means the equations were set up with swapped multipliers.

Quick revision

Incomes and expenditures take different multipliers x and y.
Subtract savings equations to eliminate.
Coin bags: price one set, divide, multiply.
Wages together ∝ 1/days each.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Income–expenditure–savings ratiosvery common3 practice Q

How to spot it:

Two income ratios, two expenditure ratios, savings given — find an income or expenditure.

ax - py = s_1, \quad bx - qy = s_2

Incomes = ax, bx; expenditures = py, qy (different letters!).
Write both savings equations.
Equal savings → subtract the equations to kill the constant; else eliminate y by cross-multiplying.

Why: two ratios plus two savings give exactly two linear equations in x and y.

Example: The incomes of A and B are in the ratio 9 : 7 and their expenditures in the ratio 4 : 3. If each saves ₹2,000, A's income is:

$9x - 4y = 2000$ , $7x - 3y = 2000$ → subtract: $2x = y$ . Then $9x - 8x = 2000$ → x = 2000 → A $= ₹18{,}000$ .

Type 2: Coin denominations in a ratiocommon3 practice Q

How to spot it:

Bags of ₹1/50p/25p (or ₹5/₹2/₹1) coins in a ratio with the total value given; find a coin count or the total.

\text{set value} = \sum n_i d_i, \quad \text{sets} = \frac{\text{total value}}{\text{set value}}

Convert all values to ONE unit (paise or rupees).
Price one full set of coins in the given ratio.
Divide the total value by the set value → sets; multiply back for counts.

Why: the ratio fixes how many full sets the bag contains — the set is the repeating block.

Example: A bag contains ₹1, 50-paise and 25-paise coins in the ratio 5 : 8 : 4, amounting to ₹210. The number of 50-paise coins is:

Set $= 500 + 400 + 100 = 1000$ paise = ₹10 → 21 sets → 50p coins $= 8 \times 21 = 168$ .

Type 3: Money shared in proportion to work or linked ratioscommon4 practice Q

How to spot it:

Wages divided in proportion to work/days; or shares linked by 'A = 3/4 of B, B = 4/5 of C'.

\text{wage} \propto \text{work done}, \quad \text{together} \propto \frac{1}{\text{days alone}}

Convert the story into one common ratio (days, or 1/days when working together).
Add the parts; one part = money ÷ parts.
For chains like A = (3/4)B = (3/4)(4/5)C, build A : B : C first.

Why: fair division always reduces to a three-term ratio on a common base.

Example: Two workers, working alone, can finish a job in 6 days and 9 days. They work together and earn ₹1,500. The first worker's share is:

Rates: $\frac{1}{6} : \frac{1}{9} = 3 : 2$ → share $= \frac{3}{5} \times 1500 = ₹900$ .

Type 4: Condition-driven division ('half of', 'twice of')common2 practice Q

How to spot it:

'A gets twice as much as B', 'the first gets half of the second' — shares defined by statements, not by a stated ratio.

\text{translate each statement into a multiple}, \ \text{then read off } A : B : C

Let the LAST person hold x (or 1 part).
Translate each statement into multiples of that part.
Read the ratio, add parts, divide the total.

Why: chain statements fix the mutual multiples; taking the last person as the base untangles them in one pass.

Example: ₹918 is divided among three friends such that the first gets half of the second and the second gets two-thirds of the third. The first friend's share is:

Let the third = 3 → second = 2 → first = 1 → ratio $1 : 2 : 3$ → first $= \frac{918}{6} = ₹153$ .

Formulas

Savings equations

ax - py = s_1,\quad bx - qy = s_2

Coin value

\text{total value} = k \sum_i (n_i \times d_i)

Difference of shares

\text{given excess} = (b - a)x

Shortcut tricks

⚡ Solve for x, y in two lines

Write both saving equations, subtract them to kill one variable.

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 and 5x − 3y = 1200 ⇒ subtract: 3x − 2y = 0 ⇒ y = 1.5x ⇒ 0.5x = 1200 ⇒ x = 2400 ⇒ A's income = 8 × 2400 = ₹19,200.

⚡ Value per set for coins

Take one full set of coins in the given ratio, price it, and scale.

Example: A bag has 50-paise, 25-paise and 10-paise coins in the ratio 7 : 6 : 5 amounting to ₹550 in total. Find the number of 25-paise coins.

Set value = 350 + 150 + 50 = 550 paise ⇒ 550 ÷ 5.50 = 100 sets ⇒ 25p coins = 6 × 100 = 600.

⚡ Sum-difference shortcuts

Sum of shares = (a + b)x, difference = (b − a)x — jump straight to x from whichever is given.

Example: Two numbers in ratio 5 : 7 have difference 12. Find the smaller.

2x = 12 ⇒ x = 6 ⇒ smaller = 30.

Where students lose marks

Using the expenditure ratio with the income multiplier x (they are different variables x and y).
Mixing paise and rupees in coin problems (1 rupee = 100 paise).
In coin questions, giving the value of the coins instead of their count (or vice versa).
Forgetting to check that savings come out positive for both persons.

Practice sets — 14 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.