Aptitude · Compound Interest · SI vs CI

The gap, and the formula that gives it in one line

Q: Give the two-year difference formula.

CI − SI = P(R/100)² . On ₹1,000 at 10% that is 1000 × 0.01 = ₹10. It is one year’s interest on one year’s interest, which is exactly where the whole gap comes from over two years.

Q: What is the three-year difference?

P(R/100)²(3 + R/100) . On ₹1,000 at 10% that is 10 × 3.1 = ₹31, which matches counting it year by year: ₹10 in the second year plus ₹21 in the third.

Examiners love the difference between simple and compound interest, because it can be handed to you as the only number in the question. For two years that difference has a clean closed form, and knowing it turns a long calculation into one multiplication.

Run both rules side by side and watch the gap open →

For two years, CI − SI = P(R/100)². That is the single most useful formula in the chapter.

01 The idea

Where the gap comes from, and how big it is

Take ₹1000 at 10% for three years. Simple interest adds ₹100 three times, giving ₹300. Compound interest adds 100, then 110, then 121, giving ₹331. The gap is ₹31, and it is not arbitrary.

Decompose it. In year two, compound interest earned ₹10 more — that is 10% of the ₹100 of interest sitting in the account. In year three it earned ₹21 more — 10% of the ₹210 of accumulated interest. Ten plus twenty-one is thirty-one. Every rupee of the gap is interest earned by interest.

For two years this becomes a formula worth memorising. The gap is one year’s interest on one year’s interest: P(R/100)². On ₹1000 at 10% that is 1000 × 0.01 = ₹10, which matches. Because it involves no time variable beyond the fixed two years, a question giving you the difference gives you the principal.

The gap widens with both time and rate, and faster than most people expect. On ₹2000 at 20% for four years, simple interest gives ₹1600 and compound gives ₹2147.20 — a gap of ₹547.20, more than a third of the simple interest itself.

Every rupee of the SI-CI gap is interest earned by interest. For two years that is exactly one year’s interest on one year’s interest.

The two-year differenceCI − SI = P(R/100)². Independent of anything but the principal and the rate, which is why it is so often the route into a question.

The three-year differenceP(R/100)²(3 + R/100). Less commonly needed but worth having; on ₹1000 at 10% it gives ₹31, matching the year-by-year count.

Ratio routeFor two years, CI/SI = (200 + R)/200. Handy when a question gives you the ratio of the two rather than their difference.

02 Worked example

₹1000 at 10% for 3 years, both rules side by side

The same sum as the previous lesson, extended a year so the gap is visible. Find the simple interest and the compound interest on ₹1000 at 10% per annum for 3 years, and account for the difference.

Year one — identicalBoth charge 10% of the original ₹1000. Nothing to compound yet.SI 100, CI 100 ⇒ both balances ₹1100

Year two — they separateSimple interest charges ₹1000 again; compound charges the ₹1100 now standing.SI +100 → 1200 CI +110 → 1210 gap ₹10

Year three — the gap widensCompound interest is now charging 10% of ₹1210.SI +100 → 1300 CI +121 → 1331 gap ₹31

Total the twoSimple interest is three equal payments; compound interest is three growing ones.SI = 100+100+100 = 300 CI = 100+110+121 = 331

Account for the ₹31The extra is precisely the interest that the interest earned, year by year.10% of 100 = 10 plus 10% of 210 = 21 ⇒ ₹31 ✓

That last step is the one worth internalising. The gap is not a formula to memorise but a quantity you can count: at the start of each year, take the accumulated interest and charge R% on it. Do that and you can compute the difference for any number of years without remembering the closed forms — which matters because the closed form for four years and beyond is rarely taught.

03 The method

The closed forms, and when each is worth using

Two formulas cover almost every question. The two-year one is the one examiners build questions around.

Two years: CI − SI = P(R/100)². Three years: CI − SI = P(R/100)²(3 + R/100), sometimes written PR²(300 + R)/10⁶.

The reverse question is the common one. Given that the two-year difference is ₹10 at 10%, the principal follows immediately: P = difference ÷ (R/100)² = 10 ÷ 0.01 = ₹1000. Also useful: for two years CI/SI = (200 + R)/200, so a ratio of 21 : 20 means R = 10%.

P and R	SI 2 yr	CI 2 yr	Difference
1,000 at 10%	200	210	`10`
5,000 at 10%	1,000	1,050	`50`
1,000 at 20%	400	440	`40`
2,000 at 20%	800	880	`80`
10,000 at 5%	1,000	1,025	`25`
1,000 at 10%, 3 yr	300	331	`31`
2,000 at 20%, 4 yr	1,600	2,147.20	`547.20`

05 Cheat sheet

The gap on one page

Two closed forms, two reverse routes, and the facts that let you check an answer instantly.

Case	Formula	On ₹1,000 at 10%
Difference, 2 years	`P(R/100)²`	1000 × 0.01 = 10
Difference, 3 years	`P(R/100)²(3+R/100)`	10 × 3.1 = 31
Principal from the 2-yr gap	`P = diff ÷ (R/100)²`	10 / 0.01 = 1,000
Rate from the CI : SI ratio	`CI/SI = (200+R)/200`	21:20 → R = 10%
Year 1	`SI = CI always`	both 100
Which is larger	`CI > SI for T > 1`	331 > 300
Gap for 4+ years	`count it year by year`	no standard closed form taught

The two-year gap is quadratic in the rateDoubling the rate quadruples the gap, since it is proportional to R squared. At 10% on ₹1,000 it is ₹10; at 20% it is ₹40.

The gap is interest on interestYou can always compute it by charging R% on the accumulated interest at the start of each year. That works for any number of years, closed form or not.

A given difference gives the principalFor two years the difference determines P once R is known, which is why the reverse question is so commonly set.

06 Where & why

Where this shows up

The difference between the two interests is one of the most reliably set items in the whole quantitative syllabus.

Bank PO · SSC CGL

“The difference between CI and SI is ₹X”

The standard reverse question. Two years and a rate give the principal in one division; students who compute both interests from scratch lose a minute.

TCS Digital · Infosys

Three-year differences

Set less often but with the same structure. The closed form P(R/100)²(3 + R/100) handles it, or count the interest-on-interest directly.

Ratio phrasings

“CI and SI are in the ratio 21 : 20”

For two years the ratio is (200 + R)/200, so the rate falls out immediately with no principal needed at all.

Interviews

“Why is compound interest larger?”

A clean answer names the mechanism — the base grows, so later periods charge more — rather than just quoting that a power beats a product.

If a question mentions both interests in one sentence, reach for the difference formula before computing either of them. That instinct is worth more marks in this chapter than any amount of arithmetic speed.

07 Interview questions

What gets asked

Ten, and the reverse question in the middle is the one you will actually meet.

Which is larger, simple or compound interest?

Compound, for any period beyond the first. They are exactly equal for one period, because there is no accumulated interest to compound. Beyond that, compound interest charges a growing base while simple interest keeps charging the original principal.

Give the two-year difference formula.

CI − SI = P(R/100)². On ₹1,000 at 10% that is 1000 × 0.01 = ₹10. It is one year’s interest on one year’s interest, which is exactly where the whole gap comes from over two years.

The difference between CI and SI on a sum for 2 years at 10% is ₹50. Find the sum.

₹5,000. Rearranging the formula, P = difference ÷ (R/100)² = 50 ÷ 0.01 = ₹5,000. This is the standard reverse question and it is a single division once you have the formula.

What is the three-year difference?

P(R/100)²(3 + R/100). On ₹1,000 at 10% that is 10 × 3.1 = ₹31, which matches counting it year by year: ₹10 in the second year plus ₹21 in the third.

Explain the ₹31 without using a formula.

In year two, compound interest charges 10% on the ₹100 of interest already there, earning ₹10 extra. In year three it charges 10% on the ₹210 of accumulated interest, earning ₹21 extra. Ten plus twenty-one is thirty-one. Every rupee of the gap is interest on interest.

CI and SI on a sum for two years are in the ratio 21 : 20. Find the rate.

10%. For two years CI/SI = (200 + R)/200, so (200 + R)/200 = 21/20 gives 200 + R = 210 and R = 10%. No principal is needed, which is what makes ratio phrasings quick.

How does the gap change if you double the rate?

It roughly quadruples, because the two-year difference is proportional to R squared. On ₹1,000 the gap is ₹10 at 10% and ₹40 at 20%. That quadratic dependence is worth knowing as a sanity check on any answer.

Is there a closed form for the four-year difference?

Not one that is standard or worth memorising. For four years and beyond, compute both interests directly or accumulate the interest-on-interest year by year. The two- and three-year forms are the only ones examiners expect.

₹2,000 at 20% for 4 years — how big is the gap?

Hard

(a) Starting from CI = P[(1 + r)² − 1] and SI = 2Pr, where r = R/100, derive the two-year difference formula. (b) Do the same for three years and confirm you get P·r²(3 + r). (c) Using the interest-on-interest argument rather than algebra, work out the four-year difference for ₹1,000 at 10% and check it against a direct calculation of both interests.

Follow-up

Part (c) is the one that matters. The algebra gets ugly at four years, but the interest-on-interest count does not — you simply charge 10% on the accumulated interest at the start of each year and add up. That method has no upper limit, which the closed forms do.

Show the hint

For (c), the accumulated interest at the start of years 2, 3 and 4 is 100, 210 and 331 respectively under compounding.