Interest (SI & CI)

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

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Subtopic 4 of 5·← CI vs SI: differences & doubling·Finding P, R, T from amount data →

Equal annual instalments

A loan at interest is cleared by equal instalments at the end of each year. Two clean methods:

Tabular (works for both SI and CI): grow the outstanding debt by one year's interest, subtract the instalment, repeat. The last row must reach zero.

Present value (CI): with instalment x and rate r, $P = \dfrac{x}{1+\frac{r}{100}} + \dfrac{x}{(1+\frac{r}{100})^2} + \cdots + \dfrac{x}{(1+\frac{r}{100})^n}$ .

Simple-interest shortcut: $P = nx - \frac{x\,r}{100}\cdot\frac{n(n-1)}{2}$ (each instalment saved earns interest for the years it was NOT outstanding).

CGL numbers are chosen so the tabular method runs on small fractions — do it year by year.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 13 practice Q

Detailed notes

The idea: instalments repay principal plus interest

A loan of P is cleared by n equal annual instalments of ₹x paid at the end of each year. Each instalment must cover (a) its share of the principal and (b) interest for the time the money stayed with the borrower.

Present value — the one rule

The instalment paid after k years is worth only $\frac{x}{(1+r)^k}$ today, where $r = \frac{R}{100}$ . The loan equals the present values of all instalments: $P = \frac{x}{1+r} + \frac{x}{(1+r)^2} + \cdots + \frac{x}{(1+r)^n}$ Take the common denominator and solve for x. At 10% for 2 years: $P = x\left(\frac{10}{11} + \frac{100}{121}\right) = \frac{210x}{121}$ so P = 2,520 → $x = \frac{2520 \times 121}{210} = ₹1{,}452$ . The tabular check: year 1 debt $= 2520 \times 1.1 - 1452 = 1320$ ; year 2: $1320 \times 1.1 - 1452 = 0$ . Zero after the last instalment confirms the answer.

Ready denominators

Rate	2 instalments	3 instalments
10%	$\frac{210}{121}x$	$\frac{3310}{1331}x$
20%	$\frac{55}{36}x$	$\frac{455}{216}x$
25%	$\frac{36}{25}x$	$\frac{244}{125}x$
Each entry: P = factor × x, so $x = P \div$ factor. Every fraction comes from chips: $\frac{10}{11} + \frac{100}{121}$ etc.

Simple-interest instalments

Here interest accrues on the outstanding principal, but the standard exam shortcut (which is exact for the usual 'interest on whole period deducted' reading) is: $P = nx - \frac{r}{100}\,x\,\frac{n(n-1)}{2}$ The second term knocks off the interest not payable because later instalments return their money earlier. ₹1,200 at 10% in 5 annual instalments: $1200 = 5x - 0.1x \times 10 = 4x$ → x = ₹300. A tabular SI walk (add interest on the balance, subtract the instalment) must end at zero; use it to verify small cases.

Hire purchase / down payment

Two shapes:

Balance carried at a rate, cleared by equal instalments — apply the present-value rule to the balance, not the cash price. Cash price 2,500, down 400, balance 2,100 at 10% in 2 instalments: $\frac{210x}{121} = 2100$ → x = ₹1{,}210.
Flat interest on the whole balance: instalment $= \frac{\text{balance} + \text{balance} \times R \times T/100}{\text{number of instalments}}$ . Balance 6,000 at 12% for 1 year in 12 monthly pieces: $\frac{6000 + 720}{12} = ₹560$ .

Present worth of a future payment

The same rule with one term: ₹1,815 due 2 years from now at 10% CI is worth $\frac{1815}{1.21} = ₹1{,}500$ today. Discounting is the reverse of compounding.

Quick revision

$P = \sum \frac{x}{(1+r)^k}$ — one equation, one unknown x.
10%, 2 years: $P = \frac{210x}{121}$ ; 20%: $\frac{55x}{36}$ ; 25%: $\frac{36x}{25}$ .
Verify by the zero-balance table.
SI shortcut: $P = nx - \frac{rx}{100} \cdot \frac{n(n-1)}{2}$ .
H.P. flat rate: (balance + flat interest) ÷ instalments.
Present worth = future value ÷ chip.

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: Two equal annual instalments (CI)very common3 practice Q

How to spot it:

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

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

Set chips: 10% → $\frac{10}{11}, \frac{100}{121}$ ; 20% → $\frac{5}{6}, \frac{25}{36}$ ; 25% → $\frac{4}{5}, \frac{16}{25}$ .
Add the two fractions; P = x × their sum.
$x = P \times \frac{\text{denominator}}{\text{numerator}}$ . Verify with the zero-balance table.

Why: each instalment is discounted for the years it is delayed.

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

$x = 2520 \times \frac{121}{210} = ₹1{,}452$ . Table: 2520×1.1 − 1452 = 1320; 1320×1.1 − 1452 = 0.

Type 2: Three equal annual instalments (CI)common3 practice Q

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)

Build the three discounted fractions.
10%: $\frac{13310 + 12100 + 11000}{14641}x = \frac{36410x}{14641} = \frac{3310x}{1331}$ .
$x = P \times \frac{1331}{3310}$ at 10% (for P = 3,310 → x = 1,331).

Why: one extra year means one extra discount factor.

Example: A loan of ₹3,310 at 10% CI is to be repaid in 3 equal annual instalments. The value of each instalment is:

$x = 3310 \times \frac{1331}{3310} = ₹1{,}331$ . (1210 + 1100 + 1000 = 3310.)

Type 3: Instalments with simple interestcommon2 practice Q

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}

Compute the deduction term $\frac{n(n-1)}{2}$ .
Write $P = x\left[n - \frac{r}{100}\cdot\frac{n(n-1)}{2}\right]$ and divide.
Cross-check small cases with a balance table.

Why: instalments paid earlier save part of the interest.

Example: A man borrows ₹1,200 at 10% simple interest and agrees to pay it in 5 equal annual instalments. Each instalment is:

$1200 = 5x - 0.1x(10) = 4x$ → x = ₹300.

Type 4: Hire purchase with a down paymentcommon2 practice Q

How to spot it:

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

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

Balance = cash price − down payment; work only with the balance.
Flat interest: add balance × R × T/100 and split into instalments.
Rate on balance: use the present-value rule with P = balance.

Why: the down payment never earns interest for the seller.

Example: A TV costs ₹7,200 cash. A buyer pays ₹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.

Balance 6,000; interest $= 6000 \times \frac{12}{100} = 720$ ; total 6,720 → ₹560 per month.

Type 5: Present worth of a future paymentoccasional2 practice Q

How to spot it:

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

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

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

Why: discounting is compounding in reverse.

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

$\frac{1815}{1.1 \times 1.1} = ₹1{,}500$ .

Formulas

Present value (CI)

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

Simple interest instalment

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

Tabular step

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

Shortcut tricks

⚡ Tabular year-by-year

Grow, subtract, repeat. Two or three rows finish the question.

Example: A man borrows ₹1,050 at 10% p.a. compound interest and repays it in two equal annual instalments. Find each instalment.

Year 1: 1050 → 1155; pay x → 1155 − x. Year 2: (1155 − x) × 1.1 = x ⇒ 1270.5 = 2.1x ⇒ x = 605.

⚡ Present-value fraction add

Discount each instalment by $\left(1+\frac{r}{100}\right)^{-k}$ and add.

Example: $P = \frac{x}{1.1} + \frac{x}{1.21}$ ⇒ x in one line for 2-year cases.

For P = 1050, r = 10%: $\frac{1.1x + x}{1.21} = 1050$ ⇒ $x = 605$ .

⚡ SI instalment formula

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

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

$1950 = 2x - \frac{x \times 5}{100} \times 1 = 1.95x$ ⇒ $x = 1000$ , i.e. ₹1,000.

Where students lose marks

Adding a full year's interest on money already repaid (use the outstanding balance, not the original P).
Counting n instalments but compounding only n − 1 times (or the reverse).
In the SI formula, taking $\frac{n(n+1)}{2}$ instead of $\frac{n(n-1)}{2}$ .
Forgetting the final row must hit exactly zero — always verify the tabular closure.

Practice sets — 13 questions

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

Topic test · 13 questions

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