Interest (SI & CI)

high importance~1 Q in Tier 122 formulas⚡ 15 shortcuts5 subtopics

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Equal annual instalments

⏱ 4 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A loan is cleared by equal yearly payments; each payment covers part of the principal plus interest on the money still unpaid.

Value every payment in today's money: $P = \dfrac{x}{c} + \dfrac{x}{c^2} + \cdots + \dfrac{x}{c^n}$ with $c = 1 + \dfrac{R}{100}$ .

Check any answer by growing the debt year by year until it hits exactly zero.

What an instalment does

A loan of P is cleared by n equal yearly payments of Rs x, each paid at the end of a year. Each payment must cover its share of the principal plus interest for the years that share stayed unpaid.

Rule: The loan equals the value today of all the instalments. Money paid later is worth less today.

Present value — the working rule

An instalment x paid after k years is worth only its discounted value today. With chip $c = 1 + \dfrac{R}{100}$ :

P = \frac{x}{c} + \frac{x}{c^2} + \cdots + \frac{x}{c^n}

At 10% for 2 years: $P = x\left(\dfrac{10}{11} + \dfrac{100}{121}\right) = \dfrac{210x}{121}$ .

So a loan of Rs 2,520 needs $x = 2520 \times \dfrac{121}{210} = 1452$ .

The table that checks everything

Grow the debt one year, subtract the instalment, repeat:

Year	Debt × 1.1 − instalment
1	$2520 \times 1.1 - 1452 = 1320$
2	$1320 \times 1.1 - 1452 = 0$

Tip: The final row must land exactly on zero. A miss of even Rs 1 means the instalment is wrong.

Ready-made loan factors

Rate	2 instalments	3 instalments
10%	P = 210x/121	P = 3310x/1331
20%	P = 55x/36	P = 455x/216
25%	P = 36x/25	P = 244x/125

Every entry is just the discounted fractions added. So Rs 3,310 at 10% over 3 years needs $x = 3310 \times \dfrac{1331}{3310} = 1331$ — and indeed $1210 + 1100 + 1000 = 3310$ .

Instalments at simple interest

The exam shortcut subtracts the interest the early payments save:

P = nx - \frac{r}{100} \cdot x \cdot \frac{n(n-1)}{2}

Rs 1,200 at 10% in 5 yearly payments: $1200 = 5x - 0.1x \times 10 = 4x$ , so x = 300. Two payments at 5%: $1950 = 2x - 0.05x$ , so x = 1,000.

Careful: The deduction uses n(n−1)/2, one per pair of years the later instalments save. Writing n(n+1)/2 is the standard slip.

Hire purchase and down payments

Cash price minus down payment gives the balance; only the balance carries interest.

Flat interest shape: add balance × R × T/100 and split evenly. Balance 6,000 at 12% for 1 year in 12 monthly parts: $\dfrac{6000 + 720}{12} = 560$ per month.

Compounding shape: apply the present-value rule to the balance. Rs 2,100 at 10% in 2 yearly payments: $\dfrac{210x}{121} = 2100$ , so x = 1,210.

Present worth of one future payment

The same rule with a single term. Rs 1,815 due after 2 years at 10% CI is worth $\dfrac{1815}{1.21} = 1500$ today.

Note: Discounting is compounding run backwards: divide by the chip once per year of delay.

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

Two equal annual instalments at CI

How to spot it:

'A loan is returned in 2 equal annual instalments at R% CI' — find the instalment, or the loan from it.

P = \frac{x}{1+r} + \frac{x}{(1+r)^2}

Method

Set the chips: 10% → 10/11 and 100/121; 20% → 5/6 and 25/36; 25% → 4/5 and 16/25.
Add the two fractions; P = x times their sum.
Divide P by the sum to get x; verify with the debt table.

Why it works:

Each instalment is discounted for the years it is delayed.

Try this

A borrows Rs 2,520 from B at 10% CI and returns it in 2 equal annual instalments. Find each instalment.

Show solution

$P = x\left(\dfrac{10}{11} + \dfrac{100}{121}\right) = \dfrac{210x}{121}$ .
$x = 2520 \times \dfrac{121}{210} = 1452$ .
Check: $2520 \times 1.1 - 1452 = 1320$ ; $1320 \times 1.1 - 1452 = 0$ .

Answer

Rs 1,452

Type 2common3 practice Q

Three equal annual instalments at CI

How to spot it:

'Cleared in 3 equal annual instalments' — same rule, one more term.

P = x\left(\frac{1}{1+r} + \frac{1}{(1+r)^2} + \frac{1}{(1+r)^3}\right)

Method

Build the three discounted fractions.
At 10% they add to 3310x/1331.
Divide P by the factor; check the three present values.

Why it works:

One extra year of delay means one more discount factor.

Try this

A loan of Rs 3,310 at 10% CI is repaid in 3 equal annual instalments. Each instalment is:

Show solution

$x = 3310 \times \dfrac{1331}{3310} = 1331$ .
Present values: $1210 + 1100 + 1000 = 3310$ .

Answer

Rs 1,331

Type 3common2 practice Q

Instalments with simple interest

How to spot it:

'Repaid in n equal annual instalments at R% simple interest.'

P = nx - \frac{r}{100}\,x\,\frac{n(n-1)}{2}

Method

Compute the bracket n(n−1)/2.
Write P = x times (n − r·bracket/100) and divide.
Cross-check small cases with a balance table.

Why it works:

Instalments paid early save part of the interest.

Try this

A man borrows Rs 1,200 at 10% simple interest and pays it in 5 equal annual instalments. Each instalment is:

Show solution

$1200 = 5x - \dfrac{10x}{100} \times 10 = 5x - x$ .
$1200 = 4x$ , so $x = 300$ .

Answer

Rs 300

Type 4common2 practice Q

Hire purchase with a down payment

How to spot it:

'Cash price Rs C, Rs D paid down, the balance with interest in equal instalments.'

\text{instalment} = \frac{\text{balance}(1 + RT/100)}{n} \ \text{(flat)} \quad \text{or PV on the balance}

Method

Balance = cash price − down payment; work only with the balance.
Flat interest: add balance × R × T/100 and split evenly.
At a compounding rate, use present value on the balance.

Why it works:

The down payment is settled today, so it never earns interest.

Try this

A TV costs Rs 7,200 cash. A buyer pays Rs 1,200 down and the balance with 12% per annum interest for 1 year, in 12 equal monthly instalments (interest on the full balance). Find the monthly instalment.

Show solution

Balance $= 6000$ ; interest $= 6000 \times \dfrac{12}{100} = 720$ .
$\dfrac{6000 + 720}{12} = 560$ .

Answer

Rs 560

Type 5occasional2 practice Q

Present worth of a future payment

How to spot it:

'What is the present value of Rs A due in T years at R%?'

\text{PW} = \frac{A}{(1 + r)^T}

Method

Divide the future amount by the chip, T times.
For a stream of payments, do it term by term and add.
Recognise it as the instalment formula with a single term.

Why it works:

Discounting is compounding in reverse.

Try this

Find the present worth of Rs 1,815 due 2 years hence at 10% per annum compound interest.

Show solution

$\dfrac{1815}{1.1 \times 1.1} = \dfrac{1815}{1.21}$ .
$= 1500$ .

Answer

Rs 1,500

Formula sheet

Present value (CI)

P = \sum_{k=1}^{n} \frac{x}{\left(1 + \frac{r}{100}\right)^k}

One term per instalment.

Simple interest instalment

P = nx - \frac{x\,r}{100} \cdot \frac{n(n-1)}{2}

Interest the early payments save.

Tabular step

\text{debt}_{k+1} = \text{debt}_k\left(1 + \frac{r}{100}\right) - x

Must end at zero.

Shortcuts that save time

⚡ Grow, subtract, repeat

Two or three rows of the debt table finish any small question.

Example

A man borrows Rs 1,050 at 10% per annum compound interest and repays it in two equal annual instalments. Find each instalment.

Show solution

Year 1: $1050 \times 1.1 = 1155$ ; debt $= 1155 - x$ .
Year 2: $(1155 - x) \times 1.1 = x$ .
$1270.5 = 2.1x$ , so $x = 605$ .

Answer

Rs 605

⚡ Present-value one-liner

Discount each instalment by the chip power for its year and add.

Example

A loan of Rs 2,520 at 10% CI is repaid in 2 equal annual instalments. Find each instalment.

Show solution

$2520 = x\left(\dfrac{10}{11} + \dfrac{100}{121}\right) = \dfrac{210x}{121}$ .
$x = 2520 \times \dfrac{121}{210} = 1452$ .

Answer

Rs 1,452

⚡ SI instalment shortcut

Use $P = nx - \dfrac{xr}{100} \cdot \dfrac{n(n-1)}{2}$ straight off for simple-interest loans.

Example

Rs 1,950 is repaid in two equal annual instalments at 5% per annum simple interest. Find each instalment.

Show solution

$1950 = 2x - \dfrac{5x}{100} \times 1$ .
$1950 = 1.95x$ , so $x = 1000$ .

Answer

Rs 1,000

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Charging a full year's interest on money already repaid.

Interest accrues on the outstanding balance only.

Mistake 02

Discounting every instalment by the same chip power.

The instalment after k years divides by the k-th power of the chip.

Mistake 03

Taking n(n+1)/2 in the SI shortcut.

The bracket is n(n−1)/2.

Mistake 04

Stopping the table one row early.

The debt must read exactly zero after the last instalment.

Mistake 05

Applying interest to the full cash price in hire purchase.

Only the balance after the down payment carries interest.

Quick revision

Read this the night before the exam.

Loan $=$ sum of instalments discounted to today.
10%, 2 years: $P = \dfrac{210x}{121}$ ; 20%: $\dfrac{55x}{36}$ .
Always verify with the zero-balance table.
SI shortcut: $P = nx - \dfrac{xr}{100} \cdot \dfrac{n(n-1)}{2}$ .
Hire purchase: interest on the balance only.
Present worth $=$ future value $\div$ chip power.

Practice: 13 questions

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

Topic test · 13 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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