Interest (SI & CI)

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CI vs SI: differences & doubling

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⏱ 4 min read🧩 5 question types🎯 14 practice Q

The idea in one minute

The gap between CI and SI is the interest on earlier interest.

For 2 years: $CI - SI = P\left(\dfrac{R}{100}\right)^2$ . For 3 years: multiply that by $\left(3 + \dfrac{R}{100}\right)$ .

At CI a sum doubles on a fixed schedule: 2 times in T years means $2^k$ times in kT years.

Why CI and SI differ

For the first year, CI and SI on the same money at the same rate are equal. From year two they part ways: CI charges interest on the interest already earned; SI never does.

The whole gap between them is exactly this interest-on-interest.

Rule: Two years: $CI - SI = P\left(\dfrac{R}{100}\right)^2$ . Three years: multiply by $\left(3 + \dfrac{R}{100}\right)$ .

Two years in one line

Rs 12,500 at 12%: $12500 \times \dfrac{12}{100} \times \dfrac{12}{100} = 180$ .

Run it backwards: a difference of Rs 50 at 10% means $P = \dfrac{50 \times 10000}{100} = 5000$ .

At 10% the difference is only 1% of P. That gives a quick size check on every answer.

Watch: CI is bigger than SI for 2 or more years, but the two are equal for a single year. Say which one is larger before computing.

Three years

CI - SI = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right)

Rs 10,000 at 10%: $10000 \times 0.01 \times 3.1 = 310$ .

At small rates the bracket sits near 3, so the three-year gap runs about three times the two-year gap.

Both figures handed to you

When a question gives both SI and CI for two years, the difference isolates the rate:

R = \frac{200 \times \text{difference}}{SI}

SI 800 and CI 820: $R = \dfrac{200 \times 20}{800} = 5\%$ . Then $P = \dfrac{800 \times 100}{2 \times 5} = 8000$ .

Another pair: SI 800, CI 832. Difference 32 $= P\left(\dfrac{R}{100}\right)^2$ and $PR = 40000$ from the SI. Divide: $R = 8\%$ , P = 5,000.

Tip: Two numbers in, two numbers out — rate first, then the principal.

Doubling chains at CI

CI multiplies the money by the same factor every T years. So doubles in T years means 4 times in 2T, 8 times in 3T, 16 times in 4T.

A sum doubling in 6 years becomes 8 times in 18 years. Sixteen times is $2^4$ , so it needs $4 \times 6 = 24$ years on the same clock.

Careful: The chain belongs to CI only. At simple interest, doubling in T years means tripling in 2T — not quadrupling.

From a multiplier to a rate

"Becomes 1.44 times in 2 years at CI" gives chip $= \sqrt{1.44} = 1.2$ , so 20%. Know the squares: 1.21 → 10%, 1.44 → 20%, 2.25 → 50%. And the cubes: 1.331 → 10%, 1.728 → 20%.

One pair, two laws

Rs 8,000 at 5% for 2 years: SI $= 800$ , CI $= 8000 \times 1.05^2 - 8000 = 820$ . The Rs 20 gap is exactly $8000 \times 0.05^2$ .

The same pair of amounts read as SI growth would give yearly interest Rs 484, P = 4,356, rate $11\dfrac{1}{9}\%$ — a different law gives a different rate. Always check which interest the question means.

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

CI minus SI for 2 years

How to spot it:

Same sum, same rate, exactly 2 years — the difference is asked, or the sum from the difference.

CI - SI = P\left(\frac{R}{100}\right)^2

Method

Apply the formula directly for the difference.
Reverse it: P = difference × 10000 ÷ R².
Check size: at 10% the difference is 1% of P.

Why it works:

Only the first year's interest gets re-earned, and only once.

Try this

The difference between the compound and simple interest on a sum for 2 years at 10% per annum is Rs 50. Find the sum.

Show solution

$P = \dfrac{50 \times 10000}{100}$ .
$= 5000$ . Check: $5000 \times 0.01 = 50$ .

Answer

Rs 5,000

Type 2very common2 practice Q

CI minus SI for 3 years

How to spot it:

'For 3 years' with the difference or a sum asked.

CI - SI = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right)

Method

Compute (R/100)² and multiply by (3 + R/100).
Multiply by P, or divide the difference by the factor.
At 10% the factor is 3.1.

Why it works:

Two interest-on-interest effects stack by year three.

Try this

The difference between CI and SI on Rs 10,000 for 3 years at 10% per annum is:

Show solution

$10000 \times (0.1)^2 \times 3.1$ .
$= 310$ .

Answer

Rs 310

Type 3common2 practice Q

CI and SI both given

How to spot it:

Both the SI and the CI for the same sum are stated; the rate or the sum is asked.

R = \frac{200 \times (CI - SI)}{SI} \quad \text{(2 years)}

Method

Difference = interest-on-interest; SI = 2 × first-year interest.
Divide: R/100 = difference × 2 ÷ SI.
Then P = 100 × SI ÷ (2R).

Why it works:

Both figures share the same P and the same first-year interest.

Try this

The simple interest on a sum for 2 years is Rs 800 and the compound interest is Rs 820. Find the rate per cent.

Show solution

Difference $= 20$ .
$R = \dfrac{200 \times 20}{800} = 5$ ; P $= \dfrac{800 \times 100}{10} = 8000$ .

Answer

5% (P = Rs 8,000)

Type 4common2 practice Q

Doubling chain at CI

How to spot it:

'A sum doubles in T years at CI — when is it 4, 8, 16 times?'

\times 2 \text{ in } T \ \Rightarrow\ \times 2^k \text{ in } kT

Method

Write the target as a power of 2: 8 = 2³, 16 = 2⁴.
Multiply T by that power.
At SI the same question needs T × (n − 1) instead.

Why it works:

CI multiplies by a constant factor per period.

Try this

A sum doubles itself in 6 years at compound interest. In how many years will it become 8 times itself?

Show solution

$8 = 2^3$ .
$3 \times 6 = 18$ years.

Answer

18 years

Type 5common2 practice Q

Multiplier in 2 years — root the chip

How to spot it:

'Becomes 2.25 times / 1.44 times itself in 2 years at CI' — find the rate.

R = 100\left(\sqrt{m} - 1\right)

Method

Take the square root of the multiplier for 2 years, cube root for 3.
Subtract 1 and convert to a per cent.
Known pairs: 1.21 → 10%, 1.44 → 20%, 2.25 → 50%.

Why it works:

The multiplier is the chip raised to the number of years.

Try this

A sum becomes 2.25 times itself in 2 years at compound interest. The rate per annum is:

Show solution

$\sqrt{2.25} = 1.5$ .
$R = 50$ .

Answer

50%

Formula sheet

2-year difference

CI - SI = P\left(\frac{R}{100}\right)^2

3-year difference

CI - SI = P\left(\frac{R}{100}\right)^2\left(3 + \frac{R}{100}\right)

CI doubling chain

2\times \text{ in } T \Rightarrow 2^k\times \text{ in } kT

SI multiple pace

n\times \text{ in } T \Rightarrow n'\times \text{ in } T',\ (n'-1) = (n-1)\frac{T'}{T}

Linear, not powers.

Rate from both figures

R = \frac{200 \times (CI_2 - SI_2)}{SI_2}

Two-year case.

Shortcuts that save time

⚡ The rate-fraction squared gap

Two equations — the SI and the difference — hand you the rate and the principal.

Example

The simple interest on a sum for 2 years is Rs 800 and the compound interest is Rs 832. Find the rate.

Show solution

Difference $32 = P\left(\dfrac{R}{100}\right)^2$ and $PR = 40000$ .
$R = \dfrac{32 \times 10000}{40000} = 8$ .

Answer

8% (P = Rs 5,000)

⚡ Count the doublings

Every T years at CI the money multiplies by the same factor, so count powers.

Example

A sum doubles in 8 years at CI. In how many years will it become 8 times?

Show solution

$8 = 2^3$ .
$3 \times 8 = 24$ years.

Answer

24 years

⚡ First-year interest lent again

For 2 years, think of the difference as the first year's interest earning R% once more.

Example

The CI−SI difference on a sum for 2 years at 10% is Rs 50. Find the sum.

Show solution

$P \times 0.1^2 = 50$ .
$P = 5000$ .

Answer

Rs 5,000

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $P\left(\dfrac{R}{100}\right)^2$ for 3 years.

Multiply by $\left(3 + \dfrac{R}{100}\right)$ for the third year.

Mistake 02

Running the CI doubling chain on an SI question.

At SI, doubles in T means triples in 2T, not 4 times.

Mistake 03

Differencing two amounts without stripping the principal.

Differences live on interest: take SI = A − P first.

Mistake 04

Expecting a CI−SI gap within a single year.

Year one is identical; the gap starts with year two.

Quick revision

Read this the night before the exam.

2 years: $CI - SI = P\left(\dfrac{R}{100}\right)^2$ .
3 years: multiply by $\left(3 + \dfrac{R}{100}\right)$ .
Both given: $R = \dfrac{200 \times \text{diff}}{SI}$ , then P.
CI chain: $\times 2$ in T $\Rightarrow$ $\times 2^k$ in kT.
Multiplier m in 2 years: rate $= 100(\sqrt{m} - 1)$ .

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 6 min · wrong answers go to your mistake notebook automatically.

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