Compound Interest

This is where the chapter starts: P, R and n (a whole number of years) are given, and you have to find the amount or the compound interest. Every later type is built on this one.

The basic method is the formula A = P(1 + R100)ⁿ. In the exam hall, though, use one of these two quicker routes.

Route 1: the fraction method. Write R as xy. Then P : A = yⁿ : (y + x)ⁿ. Divide P by yⁿ to get 1 unit and multiply by (y + x)ⁿ. At 10%, P : A is 100 : 121 for 2 years and 1000 : 1331 for 3 years.

Route 2: the effective percentage. Use the 2-year and 3-year tables from the foundation. At 10%, CI for 2 years is 21% of P and for 3 years 33.1% of P. At 5% for 3 years it is 15.7625% of P.

Radhika invests ₹40,000 at 20% for 2 years. Effective CI rate = 20 + 20 + 20 × 20100 = 44%, so CI = 44% of 40,000 = ₹17,600 and the amount is ₹57,600. With fractions: 25 : 36, 1 unit = 1600, A = 36 × 1600 = 57,600. Both routes give the same answer in a couple of lines.

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One more small idea that questions like to use: the amounts at the end of consecutive years are in the ratio 1 : (1 + R100). So at 10%, amount after the 1st year : amount after the 2nd year = 10 : 11, and the principal plays no part at all in such ratio questions.

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When the time is 4 or 5 years, keep applying the yearly factor one year at a time, or use the 2-year factor twice. Glance at the options before the last step; often only one of them has the right last digits.

Here the amount or the CI is given and you go backwards to the principal. You will also meet the term present value: the present value of a sum due after n years is the principal that would grow to that sum in n years. So "the present value of ₹X due 2 years hence" is the P whose amount after 2 years is X.

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When the amount is given: P = A(1 + R100)ⁿ. With fractions this is easy, because P : A = yⁿ : (y + x)ⁿ. The amount is (y + x)ⁿ units; find 1 unit and multiply by yⁿ.

When the CI is given: CI = P × (effective CI %), so P = CI × 100effective CI %. In units, CI = (y + x)ⁿ − yⁿ units.

Suppose Radhika wants ₹14,520 after 2 years at 10%. P : A = 100 : 121, so 121 units = 14,520, 1 unit = 120 and P = 12,000. If instead her CI for 2 years at 10% is ₹2,940, then 21 units = 2,940, 1 unit = 140 and P = 14,000.

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In the units method the amount always sits on the (y + x)ⁿ side and the principal on the yⁿ side. Write the ratio first, then put the given number on the correct side.

Now P and A are given and the rate or the time is missing. First find the growth ratio AP, reduce it to lowest terms, and see whose power it is.

Finding the rate: AP = (1 + R100)ⁿ, so take the n-th root. For example, AP = 121100 for 2 years ⇒ 1110 per year ⇒ R = 10%. If the CI is given instead of A, add it to P first.

Finding the time: the rate is known, so the yearly factor is known. Keep multiplying by it (or recognise the power) till you reach AP. At 10%, 13311000 means 3 years; at 8%, 729625 = (2725)² means 2 years.

Can Radhika's ₹7,000 become ₹9,261 in 3 years at a clean rate? Check: 92617000 = 13231000, which is not the ratio of two perfect cubes. But ₹40,000 becoming ₹46,305 gives 4630540000 = 92618000 = (2120)³, so R = 5%. The lesson: simplify AP first, then recognise squares and cubes. Learn squares up to 30 and cubes up to 20; most of these questions are only a matter of recognition.

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Some questions go one step further. After you find the rate, they ask for the SI or CI at that rate (or at a changed rate) on another sum. Do it in two clean steps: the rate first, then the value asked.

Population, the number of bacteria, the price of land: anything that grows every year by a fixed percentage of its current value grows exactly like compound interest. So the same formula works with new names:

Population after n years = present population × (1 + R100)ⁿ.

Depreciation is the opposite case. A machine, a car or a phone loses a fixed percentage of its current value every year, so the factor becomes (1 − R100):

Value after n years = present value × (1 − R100)ⁿ.

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For the past (the value n years ago), divide instead of multiplying: value n years ago = present value(1 − R100)ⁿ for depreciation, and present population(1 + R100)ⁿ for growth.

Radhika's scooter is worth ₹50,000 and loses 20% of its value every year. 20% = 15, so each year it is worth 45 of the year before. After 2 years: 50,000 × 45 × 45 = ₹32,000. Two years ago it was worth 50,000 × 54 × 54 = ₹78,125.

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A 10% fall for 3 years is not a 30% fall. The fall is 1 − (0.9)³ = 27.1%, because each year the 10% is taken on a smaller value.

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Births, deaths and migration. Sometimes the growth is given in two pieces: a birth rate and a death rate. Only the difference matters, so the net growth rate = birth rate − death rate, and then the usual formula runs. With a birth rate of 5% and a death rate of 3%, a town of 40,000 grows at 2% a year and has 40,000 × 1.02 × 1.02 = 41,616 people after 2 years.

People leaving or arriving work differently. A fixed number of migrants (say 2,000 people leave at the end of a year) is added or subtracted at that point of the timeline, and the next year's growth starts on the changed figure. Do the years one by one; never apply (1 + R100)ⁿ across a year in which people moved. A migration given as a percentage is just another year factor: a 10% outflow multiplies by 0.9.

Sometimes the rate is not the same every year: 8% in the first year, 10% in the second, and so on. The logic does not change. Each year the money is multiplied by that year's own factor:

A = P × (1 + R₁100) × (1 + R₂100) × (1 + R₃100) …

This is plain successive percentage change. For two years the effective rate is R₁ + R₂ + R₁R₂100; 8% and 10%, for example, give 8 + 10 + 0.8 = 18.8%.

Radhika invests ₹25,000 at 4% in the first year and 20% in the second. Factor = 2625 × 65, so A = 25,000 × 2625 × 65 = ₹31,200 and CI = ₹6,200. With percentages: 4 + 20 + 80100 = 24.8%, and 24.8% of 25,000 = 6,200.

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The question can also be reversed: the amount is given and the principal is asked. Then divide by the product of the factors.

One more idea you may need is the equivalent annual rate. If the growth is different each year, the equivalent rate r is the single rate that gives the same final result: (1 + r100)ⁿ = product of all the yearly factors. The order of the rates does not matter, because the order of multiplication does not change a product.

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What if the time is 2 years 6 months but the interest is compounded only once a year? The bank adds interest only at the end of each full year. For the leftover part of a year it gives simple interest on the amount reached so far.

So for time = n years + a fraction f of a year:

A = P × (1 + R100)ⁿ × (1 + f × R100)

Radhika deposits ₹50,000 at 20% for 1 year 3 months, compounded yearly. After 1 year: 50,000 × 1.2 = 60,000. For the next 3 months (14 year) the rate is 20 × 14 = 5%, given as SI on 60,000, which is 3,000. Amount = ₹63,000, CI = ₹13,000.

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Seen as one factor, this is 1.2 × 1.05 = 1.26. In percentage language, 20% followed by 5% gives 20 + 5 + 1 = 26%.

The same factor works backwards too. If the CI for 112 years at 10% (compounded yearly) is given, the factor is 1.1 × 1.05 = 1.155, so CI = 15.5% of P.

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Under yearly compounding, the fractional part never gets its own compounding. Six months at 10% p.a. simply add 5% as simple interest on the amount.

So far interest was added once a year. Banks often add it every six months (half-yearly), every three months (quarterly) or after any fixed period. The rule fits in one line: change both the rate and the time to the period.

If interest is compounded k times a year:

• rate per period = Rk
• number of periods = n × k

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Half-yearly (k = 2): the rate is halved and the periods are doubled. Quarterly (k = 4): the rate becomes one-fourth and the periods four times. For "compounded every 8 months", one period is 8 months, so the rate per period is R × 812 and the number of periods is total months8.

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Radhika deposits ₹40,000 at 10% per annum for 1 year, compounded half-yearly. Rate per half-year = 5%, number of half-years = 2. A = 40,000 × (1.05)² = 44,100, so CI = ₹4,100. With yearly compounding she would have got only ₹4,000. The extra ₹100 is the interest on the first half-year's interest.

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After this conversion, everything you learnt earlier (fractions, effective percentages, units) works exactly as before. For example, 20% p.a. compounded half-yearly for 112 years = 10% for 3 periods = 33.1%.

If deposits are made on different dates (say 1st January and 1st July), treat each deposit separately: each one grows only for the periods it actually stays in the bank, and then you add them.

Compounded monthly (fixed deposits, recurring loans). The same rule, with 12 periods in a year: rate per month = R12% and the number of periods = the number of months. At 9% per annum compounded monthly, the monthly rate is 0.75%. Radhika's ₹10,000 for 2 months grows to 10,000 × (1.0075)² = ₹10,150.56, while simple interest would give ₹10,150. Over a few months the gap is only paise, so questions usually ask for the exact amount or for the effective annual rate: (1 + R1200)¹² − 1. At 6% p.a. compounded monthly, the effective rate is about 6.17%.

Once you can convert the rate and time for half-yearly or quarterly compounding, three more kinds of questions open up.

1. Reverse questions. The amount or CI is given and P, R or the time is asked. Convert to periods first, then go backwards exactly as in Types 2 and 3. If 20% p.a. compounded half-yearly turns ₹X into ₹Y, then YX is a power of 1.1, and that power is the number of half-years.

2. Effective annual rate. "R% per annum compounded half-yearly" actually gives a little more than R% in a year. The true one-year growth is called the effective annual rate:
effective rate = (1 + R200)² − 1 = R + R²400 (in %).
At 10% compounded half-yearly: 5% twice = 10.25%. So 10% half-yearly is the same as 10.25% compounded yearly.

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3. Yearly vs half-yearly. For 1 year, yearly compounding gives R% and half-yearly gives R + R²400%. The extra, (R2)²% of P, is just the interest on the first half-year's interest.

Radhika lends ₹50,000 for 1 year at 12%. Yearly: interest = 6,000. Half-yearly: 6% twice = 12.36%, interest = 6,180. Difference = 0.36% of 50,000 = ₹180, and 0.36 is simply 6²100.

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The nominal rate is what is written; the effective rate is what you actually earn. The more often interest is compounded, the bigger the effective rate.

"Find the interest for the 3rd year" does not mean the CI for 3 years. It means only the interest earned during the third year, which is the interest on the amount at the end of the 2nd year.

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Interest of the n-th year = A_n−1 × R100 = P(1 + R100)ⁿ⁻¹ × R100.

The most useful fact of this type: the interest of each year is (1 + R100) times the interest of the year before, so the yearly interests form a GP. Hence

• the interest of the 1st year = SI of one year = PR100;
• if the interests of two consecutive years are I₁ and I₂, the rate R = I₂ − I₁I₁ × 100;
• the interest of a later year = interest of an earlier year × (1 + R100)^gap.

Radhika's deposit earns ₹1,500 interest in the 2nd year and ₹1,800 in the 3rd. Rate = 3001500 × 100 = 20%. Interest of the 1st year = 15001.2 = 1250, which is 20% of the principal, so P = ₹6,250. And the 4th year's interest = 1800 × 1.2 = ₹2,160.

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Notice where the extra 300 came from: it is 20% of 1500. The extra interest of a year over the year before is the interest on the previous year's interest.

A familiar SSC pattern: "A sum amounts to ₹X in 2 years and to ₹Y in 3 years. Find the rate (or the sum)."

Note that the amount after 3 years is the amount after 2 years grown for one more year. So the 2-year amount acts as the principal for the 3rd year:
Y = X × (1 + R100) ⇒ R = Y − XX × 100.

If the gap is more than one year, use the power: A_t+kA_t = (1 + R100)^k.

Once the rate is known, go back to the principal: P = A_t(1 + R100)^t. There is a shorter way through the GP property. The principal, the amount after 1 year, after 2 years and so on all form a GP, so if A_t and A_2t are given, P, A_t, A_2t are in GP and P = A_t²A_2t.

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Radhika's sum becomes ₹7,200 in 2 years and ₹8,640 in 3 years. R = 14407200 × 100 = 20%. P = 72001.44 = ₹5,000. And the amount after 4 years = 8640 × 1.2 = ₹10,368.

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The difference of two consecutive amounts is the interest of that one year, and that interest is R% of the earlier amount, not of the principal.

"A sum doubles in 4 years at CI. In how many years will it become 8 times?" This kind of question is pure indices.

If a sum becomes k times in t years, then in every block of t years it again becomes k times (the rate stays the same). So:

• in 2t years it becomes k² times,
• in 3t years it becomes k³ times, and so on.

After 0, t, 2t, 3t … years the sum is P, kP, k²P, k³P …. To find the time for N times, write N as a power of k: N = k^m ⇒ time = m × t.

Radhika's money triples in 6 years. 27 times = 3³ ⇒ 3 × 6 = 18 years. 9 times = 3² ⇒ 12 years. And in 3 years (half of 6) it becomes √3 times.

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Compare this with SI. If a sum doubles under SI (interest = 1 × P) in 4 years, becoming 8 times (interest = 7P) needs 7 × 4 = 28 years. Under CI it needs only 12 years. SI adds, CI multiplies, and that one line explains the whole difference.

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The same idea gives the rate too: if a sum becomes k times in n years, then 1 + R100 = k¹ⁿ. For example, 4 times in 2 years ⇒ factor 2 per year ⇒ R = 100%.

Go back to the foundation for this type. In the first year CI = SI. In the second year CI earns one extra thing, the interest on the first year's interest. So:

CI − SI (2 years) = interest on PR100 for 1 year = PR100 × R100

CI − SI = P(R100)²

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That one formula handles three kinds of questions:

• P and R given ⇒ difference = P × R²10000.
• Difference and R given ⇒ P = difference × (100R)².
• Difference and P given ⇒ (R100)² = differenceP, so R = 100√dP.

Radhika lends ₹12,500 for 2 years at 8%. Difference = 12,500 × 6410000 = ₹80. Now reverse it: if the difference at 8% is ₹80, P = 80 × (1008)² = 80 × 156.25 = ₹12,500.

The percentage method is just as useful. For 2 years the difference is (R²100)% of P: 0.25% at 5%, 1% at 10%, 2.25% at 15% and 4% at 20%. So at 10%, "difference ₹50" means 1% of P = 50, and P = ₹5,000 in one line.

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For 3 years the extra part of CI over SI has more pieces. Use the year-wise picture from the foundation, with a = SI of one year:

• 2nd year extra = interest on a (1 piece)
• 3rd year extra = interest on 2a (the SI of the first two years) + interest on the 2nd year's extra

So CI − SI (3 years) = 3 × P(R100)² + P(R100)³, that is

CI − SI (3 years) = P(R100)²(3 + R100)

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At 10% this is P × 1100 × 3.1 = 3.1% of P; at 5%, 0.7625% of P; at 20%, 12.8% of P. These values are already in your foundation table.

The second family of questions links CI and SI of the same sum.

• If SI for 2 years and CI for 2 years are both given: SI of one year = SI2, and the extra (CI − SI) is the interest on one year's SI. So R = CI − SISI2 × 100, and then P = SI × 100R × 2.
• If CI for some years is given and SI is asked (or the reverse), find P from the CI percentage and then the SI, or simply use the ratio of the two percentages. At 20% for 3 years, CI : SI = 72.8 : 60.

Radhika's sum gives an SI of ₹900 and a CI of ₹954 in 2 years. One year's SI = 450, and the extra 54 is the interest on 450 ⇒ R = 54450 × 100 = 12%. P = 450 × 10012 = ₹3,750.

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For 4 years or for half-yearly cases, don't hunt for a new formula: find CI and SI separately and subtract.

Here the question has more than one stage. Money is withdrawn or added in between, the amount of one scheme is put into another, a loan taken at SI is lent out at CI, or one sum is split between an SI scheme and a CI scheme. None of this needs a new formula. It needs discipline: follow the money stage by stage.

The method:

• Draw a small timeline: what is in the account at the start of each stage?
• Grow it for that stage (SI or CI, whichever the stage says).
• Apply the change (withdrawal, addition, transfer) at the exact moment it happens.
• Move to the next stage with the new balance.

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For "one sum in two schemes", write the interest or amount of each scheme as a percentage of its own part (SI at 13% for 4 years = 52%; CI at 12% for 2 years = 25.44%) and form one equation.

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Radhika deposits ₹30,000 at 10% CI and withdraws ₹3,000 after one year. Balance after year 1 = 33,000 − 3,000 = 30,000; after year 2 = 30,000 × 1.1 = ₹33,000. Because she withdrew exactly the first year's interest, the second year started again from ₹30,000.

A withdrawal or an addition changes the principal only from that moment onwards. Money added in the middle grows only for the time that is left.

"Divide ₹X between A and B so that A's amount after 8 years equals B's amount after 10 years." The idea behind this is neat.

Let A's part be a and B's part b, at the same rate R. Equal amounts mean
a(1 + R100)⁸ = b(1 + R100)¹⁰ ⇒ ab = (1 + R100)².

B's money stays 2 years longer, so B needs a smaller part. The part invested for the shorter time must be bigger, by the factor of the extra years. Only the difference of the times matters, not the times themselves.

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So a : b = (1 + R100)^gap : 1. At 10% with a gap of 2 years, a : b = 121 : 100. Now divide the total in this ratio.

Radhika divides ₹4,100 between her two nephews, aged 16 and 17, so that each gets the same amount at 18, at 5% CI. The younger one's money stays 2 years and the elder one's 1 year, a gap of 1 year. Elder : younger = 1.05 : 1 = 21 : 20. The elder gets 2141 × 4100 = ₹2,100 and the younger ₹2,000.

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The same logic works when one part is at SI and the other at CI for the same time: equal amounts ⇒ the parts are in the inverse ratio of their amount factors.

Instalment questions are set along with interest, so we learn both kinds here: first under SI, then under CI.

Equal annual instalments that clear a debt due after n years. Suppose a debt of ₹D is due at the end of n years at R% SI, and you pay it in n equal instalments of ₹x at the end of each year. The lender is fair: every instalment paid early earns SI from the day it is paid till the due date.

• The 1st instalment earns interest for (n − 1) years,
• the 2nd for (n − 2) years, …,
• and the last instalment earns nothing.

So D = nx + xR100[(n − 1) + (n − 2) + … + 1] = nx + xR100 × n(n − 1)2.

Radhika must clear ₹6,600 due in 3 years at 10% SI with 3 equal annual instalments. D = x(3 + 0.1 × 3) = 3.3x ⇒ x = ₹2,000. Check: 2000 × 1.2 + 2000 × 1.1 + 2000 = 2400 + 2200 + 2000 = 6600 ✓.

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Cash price vs instalment price. "₹X cash, or ₹D down payment and ₹Y after m months." The amount borrowed is (X − D), and the interest paid is Y − (X − D). Then
rate = interestborrowed × 12m × 100.

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Monthly instalments work like the annual ones, except that the rate per month is R12 and each instalment earns interest for the months that remain.

Under CI, the cleanest way to think about instalments is the present value idea of Type 2. A loan of ₹P taken today and repaid by equal yearly instalments x is fair when

P = present value of all the instalments = x(1 + R100) + x(1 + R100)² + … + x(1 + R100)ⁿ.

Fraction method. Let the rate be ab, so the yearly factor is a + bb.

• 2 instalments: P = x(ba + b + b²(a + b)²). At 10%: P = x(1011 + 100121) = x × 210121.
• 3 instalments at 10%: P = x × 1210 + 1100 + 10001331 = x × 33101331.

Radhika borrows ₹8,200 at 5% CI and repays it in 2 equal yearly instalments. 5% = 120, so P = x(2021 + 400441) = x × 820441 and x = 8200 × 441820 = ₹4,410. Check: 8200 × 1.05 = 8610; 8610 − 4410 = 4200; 4200 × 1.05 = 4410 ✓.

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Part payments. When the borrower pays some money at the end of a year, just follow the balance: balance at the end of the year = previous balance × (1 + R100) − payment. Carry on year by year; the last payment clears whatever is left. It is the same "follow the money" method as in Type 14.

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In a shop question, the instalments are only for the balance left after the cash down payment. Subtract the down payment first.