Interest (SI & CI)

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

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Finding P, R, T from amount data

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

The idea in one minute

Given amounts, work backwards to the inputs.

At CI, consecutive amounts divide: $\dfrac{A_{T+1}}{A_T} = 1 + \dfrac{R}{100}$ . At SI they subtract: the gap is one year's interest.

For P, divide a CI amount by the chip power, or subtract the accrued interest from an SI amount.

Amounts tell the story

Reverse questions hand you the amounts after whole years and ask for the principal or the rate. Everything lives in the gap between two amounts.

Rule: At CI divide consecutive amounts; at SI subtract them. Match the operation to the growth law.

CI: the ratio trick

\frac{A_{T+1}}{A_T} = 1 + \frac{R}{100}

Amounts Rs 4,840 (2 years) and Rs 5,324 (3 years): $\dfrac{5324}{4840} = 1.1$ , so the rate is 10%. The principal never entered. Walk back for it: $\dfrac{4840}{1.21} = 4000$ .

Another pair, smaller money: Rs 2,420 then Rs 2,662. The ratio $\dfrac{2662}{2420} = \dfrac{11}{10}$ again — 10%. The ratio never cares how large the sum is.

Watch: The ratio gives the chip 1.1; the rate is 10%. Reporting 1.1 as the rate is a real trap option.

SI: the constant gap

SI adds the same rupees every year, so consecutive amounts differ by exactly one year's interest:

A_{T+1} - A_T = \frac{P \times R}{100}

Amounts Rs 4,500 then Rs 5,000: yearly interest 500; P = 4500 − 500 = 4000; rate $= \dfrac{500 \times 100}{4000} = 12.5\%$ .

One more: Rs 6,000 then Rs 6,750 gives a gap of 750. If the question adds P = 4,500, the rate is $\dfrac{750 \times 100}{4500} = 16\dfrac{2}{3}\%$ .

Gaps wider than one year

Non-consecutive amounts: divide the rupee gap by the year gap first.

Rs 6,800 after 3 years and Rs 8,000 after 5 years: $\dfrac{8000 - 6800}{2} = 600$ per year. Then P = 6800 − 3 × 600 = 5000 and the rate is 12%.

Careful: 3 → 5 years is two years of interest. Dividing by 2 is the step students forget.

Two SIs on one sum

SI 1,200 for 2 years and 1,800 for 3 years: the Rs 600 difference is one year's interest, so $\dfrac{PR}{100} = 600$ .

That is one equation in two unknowns, so the question must supply P or R. With P = 10,000 the rate is 6%; with rate 6% the sum is 10,000. The two SIs alone never fix both.

The classic 2-year / 3-year pair at SI

"A sum amounts to A in 2 years and B in 3 years at SI." Yearly interest is B − A. Then P = A − 2(B − A) = 3A − 2B. The rate is $\dfrac{(B-A) \times 100}{P}$ .

Rs 5,200 after 3 years and Rs 5,600 after 4 years: yearly 400, P = 5200 − 3 × 400 = 4000, rate 10%.

Going back at CI

From a single amount, divide by the chip power: Rs 6,655 after 3 years at 10% → $\dfrac{6655}{1.331} = 5000$ .

And Rs 4,840 after 2 years at 10% is $\dfrac{4840}{1.21} = 4000$ — the same 4,000 the ratio walk gave above.

Note: Know the powers cold: 1.21, 1.331, 1.44, 1.5625, 1.728. They turn divisions into one step.

Question types you will see

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

Type 1very common2 practice Q

Principal from an amount at CI

How to spot it:

An amount after 2 or 3 whole years at a known CI rate is given; the original sum is asked.

P = \frac{A}{\left(1 + \frac{R}{100}\right)^T}

Method

Recall the chip powers: 1.21, 1.331, 1.44.
Divide the amount by the right power.
P must come out smaller than the amount.

Why it works:

Discounting undoes compounding exactly.

Try this

A sum amounts to Rs 6,655 in 3 years at 10% per annum compound interest. The sum is:

Show solution

$1.1^3 = 1.331$ .
$\dfrac{6655}{1.331} = 5000$ .

Answer

Rs 5,000

Type 2very common2 practice Q

Rate from consecutive CI amounts

How to spot it:

Amounts after T and T+1 years are given at CI; the rate is asked.

1 + \frac{R}{100} = \frac{A_{T+1}}{A_T}

Method

Divide the later amount by the earlier one.
Subtract 1 and read the rate.
For P, discount the given amount by the chips.

Why it works:

One year of growth is exactly one chip.

Try this

A sum amounts to Rs 4,840 in 2 years and to Rs 5,324 in 3 years at compound interest. The rate per annum is:

Show solution

$\dfrac{5324}{4840} = \dfrac{11}{10}$ .
$R = 10$ .

Answer

10%

Type 3very common2 practice Q

Rate from consecutive SI amounts

How to spot it:

Amounts after T and T+1 years at SI; the rate or the sum is asked.

A_{T+1} - A_T = \frac{PR}{100}, \qquad P = A_T - (A_{T+1} - A_T)

Method

The difference is one year's interest — constant at SI.
Subtract it from the earlier amount for P.
Rate = difference ÷ P × 100.

Why it works:

SI growth is arithmetic, so equal gaps carry equal interest.

Try this

A sum amounts to Rs 4,500 in 1 year and Rs 5,000 in 2 years at simple interest. Find the rate per annum.

Show solution

Yearly interest $= 500$ ; P $= 4000$ .
Rate $= \dfrac{500 \times 100}{4000} = 12.5$ .

Answer

12.5%

Type 4common2 practice Q

Amounts spanning several years at SI

How to spot it:

'Amounts to A in m years and B in n years' with m and n not adjacent.

\text{yearly interest} = \frac{B - A}{n - m}

Method

Divide the amount gap by the year gap.
P = A − m × yearly interest.
Rate = yearly ÷ P × 100.

Why it works:

Linear growth means any two points fix the line.

Try this

A sum amounts to Rs 6,800 in 3 years and Rs 8,000 in 5 years at simple interest. Find the rate per annum.

Show solution

Yearly $= \dfrac{8000 - 6800}{2} = 600$ .
P $= 6800 - 1800 = 5000$ .
Rate $= \dfrac{600 \times 100}{5000} = 12$ .

Answer

12%

Type 5common3 practice Q

Two SIs on the same sum

How to spot it:

'SI for 2 years is x, for 3 years is y' — the difference pins the yearly interest.

\text{yearly interest} = \frac{y - x}{\Delta n}, \quad R = \frac{100 \times \text{yearly}}{P}

Method

Yearly interest = difference of the SIs ÷ year gap.
The yearly figure equals PR/100 — one equation.
Use the P or R the question supplies; recompute both SIs to check.

Why it works:

Same sum, same rate: every extra year adds the same rupees.

Try this

The simple interest on a sum of Rs 10,000 for 2 years is Rs 1,200 and for 3 years is Rs 1,800. The rate per annum is:

Show solution

Yearly $= 1800 - 1200 = 600$ .
$R = \dfrac{600 \times 100}{10000} = 6$ .

Answer

Formula sheet

Principal from amount

P = \frac{A}{\left(1 + \frac{r}{100}\right)^T}

CI: divide by the chip power.

Rate from consecutive amounts

1 + \frac{r}{100} = \frac{A_{t+1}}{A_t} \quad (\text{CI})

Divide at CI.

Yearly SI from amounts

SI_{\text{year}} = A_{t+1} - A_t \quad (\text{SI})

Subtract at SI.

Two-amount system

\frac{A_2}{A_1} = 1 + \frac{r}{100} \Rightarrow P = \frac{A_1}{1 + r/100}

Shortcuts that save time

⚡ Ratio first, then walk back

The ratio of consecutive CI amounts exposes the chip; divide once more for P.

Example

A sum amounts to Rs 1,331 in 3 years at 10% per annum CI. Find the sum.

Show solution

$P = 1331 \times \dfrac{10}{11} \times \dfrac{10}{11} \times \dfrac{10}{11}$ .
$= 1000$ .

Answer

Rs 1,000

⚡ Equal yearly steps at SI

Check the amounts differ by a constant — that constant is the yearly interest.

Example

A sum amounts to Rs 6,000 in 2 years and Rs 6,750 in 3 years at SI. Find the rate if P = Rs 4,500.

Show solution

Yearly SI $= 6750 - 6000 = 750$ .
Check: $4500 + 2 \times 750 = 6000$ .
$R = \dfrac{750 \times 100}{4500} = 16\dfrac{2}{3}\%$ .

Answer

16 2/3% p.a.

⚡ Two amounts, two unknowns

Dividing the amounts kills P and hands you the chip.

Example

A sum amounts to Rs 4,840 in 2 years and Rs 5,324 in 3 years at CI. Find the rate.

Show solution

$\dfrac{5324}{4840} = \dfrac{11}{10}$ .
$R = 10$ .

Answer

10% p.a.

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Dividing SI amounts to get the rate.

SI amounts grow by equal rupees; subtract them. Divide only at CI.

Mistake 02

Subtracting CI amounts to get the yearly interest.

CI growth is multiplicative; the ratio of consecutive amounts is the chip.

Mistake 03

Reporting the chip 1.1 as the rate.

Rate = (chip − 1) × 100 = 10%.

Mistake 04

Using the whole rupee gap when the years skip.

Divide the gap by the year gap first: 3 → 5 years is 2 years.

Mistake 05

Feeding SI amounts into a CI formula.

Check the growth law first: equal gaps mean SI.

Quick revision

Read this the night before the exam.

CI: $\dfrac{A_{T+1}}{A_T}$ is the chip; subtract 1 for the rate.
SI: $A_{T+1} - A_T$ is one year's interest, constant.
Year-skipping gaps: divide by the year gap.
SI amounts: P = amount $-$ years $\times$ yearly interest.
CI amounts: P = amount $\div$ 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 7 min · wrong answers go to your mistake notebook automatically.

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