Ratio, Proportion, Partnership & Ages

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high importance~2 Q in Tier 123 formulas⚡ 15 shortcuts5 subtopics

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Money ratios: income–expenditure, coins & mixed amounts

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⏱ 4 min read🧩 4 question types🎯 14 practice Q

The idea in one minute

Two classic money containers for ratios.

Income and expenditure: incomes in $a : b$ , expenditures in $p : q$ . Write incomes $ax, bx$ and expenditures $py, qy$ — two multipliers — then use each person's savings.

Coin bags: price one full set of coins in the ratio; divide the total value by the set value; multiply back for any count.

Two ratios, two multipliers

Incomes in $a : b$ and expenditures in $p : q$ . Write incomes $= ax$ , $bx$ and expenditures $= py$ , $qy$ .

Two different multipliers — mixing them is the standard error. Savings give one equation per person: $ax - py = s_1$ and $bx - qy = s_2$ .

Rule: Incomes take x, expenditures take y. Never one letter for both.

Equal savings — subtract

Each saves Rs 2,000 with incomes $9 : 7$ and expenditures $4 : 3$ :

9x - 4y = 2000, \qquad 7x - 3y = 2000

Subtract: $2x = y$ . Then $9x - 8x = 2000$ , so $x = 2000$ and A's income is 18,000.

Another: incomes $8 : 5$ , expenditures $5 : 3$ , each saves 1,200. Subtracting gives $3x = 2y$ , so $y = 1.5x$ . Then $8x - 7.5x = 0.5x = 1200$ , so $x = 2400$ and A's income is 19,200.

Equal savings cancel on subtraction — that is what makes these questions short.

Unequal savings — eliminate

When the savings differ, scale one equation so y matches, then subtract. Same two-line shape, one extra multiplication.

A complete walkthrough

Incomes $5 : 4$ , expenditures $3 : 2$ , each saves Rs 1,600. So $5x - 3y = 1600$ and $4x - 2y = 1600$ .

Equal savings, so subtract: $x - y = 0$ , meaning $x = y$ . Then $5x - 3x = 2x = 1600$ , so $x = 800$ .

Incomes 4,000 and 3,200; expenditures 2,400 and 1,600. Both savings check out at 1,600.

Coin bags

Coins in ratio $n_1 : n_2 : n_3$ of denominations $d_1, d_2, d_3$ . Price ONE set: $n_1d_1 + n_2d_2 + n_3d_3$ . Then sets = total value ÷ set value, and a coin's count = its term × sets.

Rs 1, 50-paise and 25-paise coins in $5 : 8 : 4$ worth Rs 210: a set is $500 + 400 + 100 = 1000$ paise $=$ Rs 10. So 21 sets, and the 50-paise coins number $8 \times 21 = 168$ .

Careful: Convert everything to paise (or everything to rupees) before pricing the set. Rs 1 = 100 paise.

The same engine counts total coins: Rs 5, Rs 2 and Rs 1 coins in $3 : 5 : 7$ worth Rs 512. A set is Rs 32, so 16 sets, and the total coins are $15 \times 16 = 240$ .

Wages in proportion to work

Wages follow work done. Days worked: shares $\propto$ days. Working together: shares $\propto$ work rates, $\dfrac{1}{\text{days}}$ .

Workers finishing alone in 6 and 9 days earn in ratio $\dfrac{1}{6} : \dfrac{1}{9} = 3 : 2$ . Of Rs 1,500, the faster worker gets 900.

Watch: Together, compare $\dfrac{1}{\text{days}}$ , never the days themselves.

Chained fractions

"A $= \dfrac{3}{4}$ of B and B $= \dfrac{4}{5}$ of C" gives $A : B : C = 3 : 4 : 5$ . Build the three-term ratio on a common base before touching any money.

Statement-defined shares

"A gets half of B; B gets two-thirds of C." Let the third hold 3: then B $=$ 2 and A $=$ 1. Ratio $1 : 2 : 3$ ; Rs 918 gives A $= \dfrac{918}{6} = 153$ .

Tip: Take the last-named person as the base unit; every statement becomes a whole-number multiple.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Income-expenditure-savings ratios

How to spot it:

Two ratios (income and expenditure) plus savings; find an income or an expenditure.

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

Method

Incomes = ax, bx; expenditures = py, qy (different letters).
Write both savings equations.
Equal savings: subtract. Unequal: eliminate y.
Build the asked quantity from x or y.

Why it works:

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

Try this

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

Show solution

$9x - 4y = 2000$ and $7x - 3y = 2000$ .
Subtract: $2x = y$ . Then $9x - 8x = 2000$ .
$x = 2000$ ; A $= 9 \times 2000 = 18000$ .

Answer

Rs 18,000

Type 2common3 practice Q

Coin denominations in a ratio

How to spot it:

Bags of Rs 1 / 50p / 25p (or Rs 5 / Rs 2 / Rs 1) coins in a ratio with the total value given.

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

Method

Convert all values to ONE unit (paise or rupees).
Price one full set of coins in the given ratio.
Sets = total value ÷ set value; a coin's count = its term × sets.

Why it works:

The ratio fixes how many identical sets the bag holds — the set is the repeating block.

Try this

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

Show solution

Set $= 500 + 400 + 100 = 1000$ paise $=$ Rs 10.
$210 \div 10 = 21$ sets.
50-paise coins $= 8 \times 21 = 168$ .

Answer

168

Type 3common4 practice Q

Money shared in proportion to work

How to spot it:

Wages divided by days worked, or by work rates when the workers finish alone in different times.

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

Method

Convert the story into one common ratio (days, or 1/days together).
Add the parts; one part = money ÷ parts.
Multiply by the asked worker's parts.

Why it works:

Fair division always reduces to one three-term ratio on a common base.

Try this

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

Show solution

Rates $\dfrac{1}{6} : \dfrac{1}{9} = 3 : 2$ .
Share $= \dfrac{3}{5} \times 1500 = 900$ .

Answer

Rs 900

Type 4common2 practice Q

Condition-driven division

How to spot it:

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

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

Method

Let the LAST-named person hold the base unit.
Translate each statement into a multiple of that unit.
Read the ratio, add parts, divide the total.

Why it works:

Chain statements fix the mutual multiples; the last-named base untangles them in one pass.

Try this

Rs 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:

Show solution

Third $= 3$ , second $= 2$ , first $= 1$ .
Ratio $1 : 2 : 3$ .
First $= \dfrac{918}{6} = 153$ .

Answer

Rs 153

Formula sheet

Savings equations

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

Two multipliers: x for incomes, y for expenditures.

Coin value

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

Price one set, then scale.

Wages together

\text{shares} \propto \frac{1}{\text{days alone}}

Difference of shares

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

Shortcuts that save time

⚡ Subtract the savings equations

Write both savings equations and subtract — equal savings kill the constant and one variable.

Example

Incomes of A and B are in the ratio 8 : 5 and expenditures in 5 : 3. If each saves Rs 1,200, find A's income.

Show solution

$8x - 5y = 1200$ , $5x - 3y = 1200$ ; subtract: $3x = 2y$ .
$y = 1.5x$ ; $0.5x = 1200$ , so $x = 2400$ .
A's income $= 8 \times 2400 = 19200$ .

Answer

Rs 19,200

⚡ Value per set for coins

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

Example

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

Show solution

Set $= 350 + 150 + 50 = 550$ paise $=$ Rs 5.50.
$550 \div 5.50 = 100$ sets.
25-paise coins $= 6 \times 100 = 600$ .

Answer

600

⚡ Base the chain on the last person

For 'A gets half of B, B gets two-thirds of C', give the last-named person the unit and read the ratio off.

Example

Rs 918 is divided so that the first gets half of the second and the second gets two-thirds of the third. Find the first share.

Show solution

Third $= 3$ , second $= 2$ , first $= 1$ .
Ratio $1 : 2 : 3$ ; first $= \dfrac{918}{6} = 153$ .

Answer

Rs 153

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using one multiplier for both income and expenditure ratios.

Incomes take x, expenditures take y — two letters, two equations.

Mistake 02

Mixing paise and rupees in coin problems.

Convert everything to one unit first; Rs 1 = 100 paise.

Mistake 03

Answering the value of the coins when the count is asked (or the reverse).

Read whether the question wants a count or a value; both sit in the options.

Mistake 04

Comparing days instead of rates when workers work together.

Together, shares follow 1/days.

Mistake 05

Accepting a negative expenditure multiplier.

A negative y means the equations were set up with swapped multipliers.

Quick revision

Read this the night before the exam.

Incomes $ax$ , $bx$ ; expenditures $py$ , $qy$ — two multipliers.
Equal savings: subtract the two equations.
Unequal savings: eliminate y by scaling.
Coin bags: price one set, divide, multiply back.
Wages together $\propto$ 1/days.
Statement shares: base on the last-named person.

Practice: 14 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

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

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