Ten Per Cent Off for Three Years Is Not Thirty Per Cent Off
Use the amount formula for any number of years, halve the rate for half-yearly compounding, find whichever of P, r, n or A is missing, and handle population growth and machine depreciation.
Why is a machine losing 10% a year for three years not down by 30%?
A machine costs ₹50000 and loses 10% of its value each year. What is it worth after three years?
The tempting answer is to add the percentages: , so it has lost ₹15000 and is worth ₹35000.
Work it out properly and the machine is worth ₹36450 — one thousand four hundred and fifty rupees more than that.
The reason is that each year's 10% is taken on a smaller value than the year before. The first year loses ₹5000, from ₹50000. The second year loses only ₹4500, because it is 10% of ₹45000. The third loses ₹4050.
So percentages applied one after another never simply add up, and that is true whether the quantity is growing or shrinking. Compound interest, population growth and depreciation are all the same arithmetic, and all three are handled by one formula.
This page covers the ICSE Class 9 Mathematics chapter on compound interest using the formula: the amount formula itself, half-yearly compounding, finding any one of the quantities when the others are given, and problems on growth and depreciation.
The tempting answer is to add the percentages: , so it has lost ₹15000 and is worth ₹35000.
Work it out properly and the machine is worth ₹36450 — one thousand four hundred and fifty rupees more than that.
The reason is that each year's 10% is taken on a smaller value than the year before. The first year loses ₹5000, from ₹50000. The second year loses only ₹4500, because it is 10% of ₹45000. The third loses ₹4050.
So percentages applied one after another never simply add up, and that is true whether the quantity is growing or shrinking. Compound interest, population growth and depreciation are all the same arithmetic, and all three are handled by one formula.
This page covers the ICSE Class 9 Mathematics chapter on compound interest using the formula: the amount formula itself, half-yearly compounding, finding any one of the quantities when the others are given, and problems on growth and depreciation.
Formula
How does the compound interest formula work?
**Each year multiplies the amount by the same factor, so years multiply it by that factor times.**
If the rate is per cent per annum, then one year turns a principal into
The next year does the same thing to that amount, and so on. After years:
Worked example 1. Find the amount and the compound interest on ₹15000 at 12% per annum for 2 years.
Worked example 2, over three years. Find the amount and the compound interest on ₹20000 at 10% per annum for 3 years.
Check it year by year, as the previous chapter would have done. . The formula and the long method agree, as they must.
Two things to be careful about in the formula.
**First, counts the number of times interest is added, not necessarily the number of years — which is exactly what the next section is about.
Second, the interest is and never the formula's output on its own. The formula gives the amount**, which already contains the principal. Writing the value of as the answer to "find the compound interest" loses the mark, and it is the commonest slip in the chapter.
Powers worth knowing on sight, because they turn up constantly:
If the rate is per cent per annum, then one year turns a principal into
The next year does the same thing to that amount, and so on. After years:
Worked example 1. Find the amount and the compound interest on ₹15000 at 12% per annum for 2 years.
Worked example 2, over three years. Find the amount and the compound interest on ₹20000 at 10% per annum for 3 years.
Check it year by year, as the previous chapter would have done. . The formula and the long method agree, as they must.
Two things to be careful about in the formula.
**First, counts the number of times interest is added, not necessarily the number of years — which is exactly what the next section is about.
Second, the interest is and never the formula's output on its own. The formula gives the amount**, which already contains the principal. Writing the value of as the answer to "find the compound interest" loses the mark, and it is the commonest slip in the chapter.
Powers worth knowing on sight, because they turn up constantly:
What changes when interest is compounded half-yearly?
Halve the rate and double the number of periods — and the answer comes out slightly larger.
If a rate of per cent per annum is compounded half-yearly, then each half-year the amount grows by per cent, and in years there are such half-years. So
The number of times interest is added is called the number of conversion periods.
Worked example 1. Find the amount and the compound interest on ₹8000 at 10% per annum for 1 year, compounded half-yearly.
Rate per half-year ; number of periods .
Now compare with yearly compounding on the same money for the same time:
Half-yearly compounding earns ₹20 more. And that ₹20 is not a mystery — it can be located exactly.
After the first half-year the interest was . In the second half-year that ₹400 itself earned .
So the whole gain is the interest on the first half-year's interest — the same explanation as the compound-against-simple difference, now happening within a single year. The more often interest is added, the more often it starts earning, and that is the entire reason half-yearly beats yearly.
Worked example 2. Find the amount on ₹6400 at 5% per annum for years, compounded half-yearly.
Rate per half-year ; number of periods .
A warning about the wording, because it is designed to catch you. "10% per annum compounded half-yearly" means 5% every six months — it does not mean 10% every six months, and it does not mean 20% per annum. The rate quoted is always per annum unless the question says otherwise, and your first line should be to write down the rate per period and the number of periods before touching the formula.
And note what happens with an odd half-year. For years there are conversion periods, not . Converting the time into periods correctly is the whole difficulty of these questions, and once is right the arithmetic is ordinary.
If a rate of per cent per annum is compounded half-yearly, then each half-year the amount grows by per cent, and in years there are such half-years. So
The number of times interest is added is called the number of conversion periods.
Worked example 1. Find the amount and the compound interest on ₹8000 at 10% per annum for 1 year, compounded half-yearly.
Rate per half-year ; number of periods .
Now compare with yearly compounding on the same money for the same time:
Half-yearly compounding earns ₹20 more. And that ₹20 is not a mystery — it can be located exactly.
After the first half-year the interest was . In the second half-year that ₹400 itself earned .
So the whole gain is the interest on the first half-year's interest — the same explanation as the compound-against-simple difference, now happening within a single year. The more often interest is added, the more often it starts earning, and that is the entire reason half-yearly beats yearly.
Worked example 2. Find the amount on ₹6400 at 5% per annum for years, compounded half-yearly.
Rate per half-year ; number of periods .
A warning about the wording, because it is designed to catch you. "10% per annum compounded half-yearly" means 5% every six months — it does not mean 10% every six months, and it does not mean 20% per annum. The rate quoted is always per annum unless the question says otherwise, and your first line should be to write down the rate per period and the number of periods before touching the formula.
And note what happens with an odd half-year. For years there are conversion periods, not . Converting the time into periods correctly is the whole difficulty of these questions, and once is right the arithmetic is ordinary.
How do you find P, r or n, and the difference between CI and SI?
Divide the amount by the principal to get the growth factor, then read whichever quantity is missing out of it.
Worked example 1 — find the rate. ₹12000 becomes ₹13230 in 2 years, compounded annually. Find the rate.
Check: , and .
Worked example 2 — find the time. In how many years will ₹1000 amount to ₹1331 at 10% per annum?
Worked example 3 — find the principal. A sum amounts to ₹9261 in 3 years at 5% per annum. Find the sum.
Check: .
Worked example 4 — difference between CI and SI over 2 years. Find it for ₹25000 at 4% per annum.
Using the shortcut for two years:
Verify it the long way. . And , so . The difference is . Agreed.
Worked example 5 — the same relation used backwards. The difference between the compound interest and the simple interest on a sum for 2 years at 5% per annum is ₹25. Find the sum.
Check: ; so ; difference . Agreed.
Now the point that ties all five together. In every one of them the first useful step was the growth factor , or the quantity that produces it.
- Rate unknown — take the th root of the factor and subtract
- Time unknown — see which power the factor is
- Principal unknown — divide the amount by the factor
So there is one equation in this chapter and four ways of being asked about it. Recognising the growth factor first turns every backwards problem into one line of arithmetic, and it is why writing down before doing anything else is a good habit.
Worked example 1 — find the rate. ₹12000 becomes ₹13230 in 2 years, compounded annually. Find the rate.
Check: , and .
Worked example 2 — find the time. In how many years will ₹1000 amount to ₹1331 at 10% per annum?
Worked example 3 — find the principal. A sum amounts to ₹9261 in 3 years at 5% per annum. Find the sum.
Check: .
Worked example 4 — difference between CI and SI over 2 years. Find it for ₹25000 at 4% per annum.
Using the shortcut for two years:
Verify it the long way. . And , so . The difference is . Agreed.
Worked example 5 — the same relation used backwards. The difference between the compound interest and the simple interest on a sum for 2 years at 5% per annum is ₹25. Find the sum.
Check: ; so ; difference . Agreed.
Now the point that ties all five together. In every one of them the first useful step was the growth factor , or the quantity that produces it.
- Rate unknown — take the th root of the factor and subtract
- Time unknown — see which power the factor is
- Principal unknown — divide the amount by the factor
So there is one equation in this chapter and four ways of being asked about it. Recognising the growth factor first turns every backwards problem into one line of arithmetic, and it is why writing down before doing anything else is a good habit.
How are population growth and depreciation the same problem?
Growth uses a plus sign in the bracket and depreciation uses a minus sign — and nothing else changes.
Worked example 1 — growth. The population of a town is 64000 and increases at 5% per annum. Find the population after 3 years.
Check year by year: .
Worked example 2 — depreciation. A machine costs ₹50000 and depreciates at 10% per annum. Find its value after 3 years and the total depreciation.
Check year by year: .
Now the two errors that this topic is built to expose.
Percentages do not add over years. Three years at 10% depreciation is not 30%. A flat 30% of ₹50000 would leave ₹35000, and the true value is ₹36450 — a difference of ₹1450, because each year's 10% is taken on a smaller amount than the last.
The same applies to growth in the other direction. Three years at 5% growth is not 15%. A flat 15% on 64000 gives , while the true figure is 74088 — this time 488 more, because each year's 5% is taken on a larger number.
So adding the percentages understates growth and overstates depreciation, and the two errors point in opposite directions for exactly the same reason.
A second warning about the depreciated value. The formula gives the value remaining, not the amount lost. If the question asks for the depreciation, you must subtract: . It is the same mistake as reporting the amount instead of the interest, and it costs the same mark.
What else this formula covers. Anything that changes by a fixed percentage per period:
- The population of a town, a state or a species
- The value of machinery, a vehicle or any asset that depreciates
- The production of a factory rising or falling by a fixed percentage
- The number of bacteria in a culture
And a mixed case worth expecting. A population may rise for two years and then fall for one, or a machine may depreciate at different rates in different years. Handle it exactly as the changing-rate problems of the previous chapter — one factor per period, multiplied together:
So there is really only one formula in these two chapters, and the sign in each bracket records whether that period was a gain or a loss.
Worked example 1 — growth. The population of a town is 64000 and increases at 5% per annum. Find the population after 3 years.
Check year by year: .
Worked example 2 — depreciation. A machine costs ₹50000 and depreciates at 10% per annum. Find its value after 3 years and the total depreciation.
Check year by year: .
Now the two errors that this topic is built to expose.
Percentages do not add over years. Three years at 10% depreciation is not 30%. A flat 30% of ₹50000 would leave ₹35000, and the true value is ₹36450 — a difference of ₹1450, because each year's 10% is taken on a smaller amount than the last.
The same applies to growth in the other direction. Three years at 5% growth is not 15%. A flat 15% on 64000 gives , while the true figure is 74088 — this time 488 more, because each year's 5% is taken on a larger number.
So adding the percentages understates growth and overstates depreciation, and the two errors point in opposite directions for exactly the same reason.
A second warning about the depreciated value. The formula gives the value remaining, not the amount lost. If the question asks for the depreciation, you must subtract: . It is the same mistake as reporting the amount instead of the interest, and it costs the same mark.
What else this formula covers. Anything that changes by a fixed percentage per period:
- The population of a town, a state or a species
- The value of machinery, a vehicle or any asset that depreciates
- The production of a factory rising or falling by a fixed percentage
- The number of bacteria in a culture
And a mixed case worth expecting. A population may rise for two years and then fall for one, or a machine may depreciate at different rates in different years. Handle it exactly as the changing-rate problems of the previous chapter — one factor per period, multiplied together:
So there is really only one formula in these two chapters, and the sign in each bracket records whether that period was a gain or a loss.
Exam tip
Exam tip: write the rate per period and the number of periods first
The formula gives the AMOUNT. , always. Reporting as the interest is the commonest lost mark in the chapter.
Before substituting, write down two things: the rate per conversion period and the number of periods. Every half-yearly question is decided in that one line.
Half-yearly means halve the rate and double the periods: .
"10% per annum compounded half-yearly" means 5% every six months — not 10% every six months.
**For years half-yearly there are 3 periods**, not . Convert the time to periods carefully.
Explain why half-yearly earns more — the first half-year's interest starts earning in the second half-year. On ₹8000 at 10% that is , which is exactly the gain.
**For a backwards problem, compute first.** Then take the th root for the rate, match a power for the time, or divide for the principal.
Learn these powers: , , , , , , .
** is for TWO years only**, and it works backwards to find just as well as forwards.
Depreciation uses a MINUS sign in the bracket: .
The depreciation formula gives the value LEFT. Subtract from the original cost if the question asks how much value was lost.
Never add percentages across years. Three years at 10% is not 30% — and the error goes one way for growth and the other for depreciation.
And for a mixed problem, write one bracket per period and multiply them, with a plus for a gain and a minus for a loss.
Before substituting, write down two things: the rate per conversion period and the number of periods. Every half-yearly question is decided in that one line.
Half-yearly means halve the rate and double the periods: .
"10% per annum compounded half-yearly" means 5% every six months — not 10% every six months.
**For years half-yearly there are 3 periods**, not . Convert the time to periods carefully.
Explain why half-yearly earns more — the first half-year's interest starts earning in the second half-year. On ₹8000 at 10% that is , which is exactly the gain.
**For a backwards problem, compute first.** Then take the th root for the rate, match a power for the time, or divide for the principal.
Learn these powers: , , , , , , .
** is for TWO years only**, and it works backwards to find just as well as forwards.
Depreciation uses a MINUS sign in the bracket: .
The depreciation formula gives the value LEFT. Subtract from the original cost if the question asks how much value was lost.
Never add percentages across years. Three years at 10% is not 30% — and the error goes one way for growth and the other for depreciation.
And for a mixed problem, write one bracket per period and multiply them, with a plus for a gain and a minus for a loss.
Did you know
Why adding percentages understates growth and overstates loss
Two of the errors in this chapter look like the same mistake and point in opposite directions, which is worth understanding once rather than memorising twice.
Add the percentages for growth and you get too little. Three years at 5% growth, treated as 15%, gives 73600 where the truth is 74088.
Add the percentages for depreciation and you get too much loss. Three years at 10% depreciation, treated as 30%, gives ₹35000 where the truth is ₹36450.
The reason is the same in both cases: each year's percentage is applied to whatever the quantity has become, not to what it started as.
In growth, the quantity has got bigger, so each successive percentage is a percentage of more — and the total gain exceeds the naive sum.
In depreciation, the quantity has got smaller, so each successive percentage is a percentage of less — and the total loss falls short of the naive sum.
One consequence is worth noticing because it surprises people. A 10% rise followed by a 10% fall does not bring you back to where you started.
Take ₹1000. A 10% rise gives ₹1100. A 10% fall on ₹1100 takes off ₹110, leaving ₹990. You are ₹10 down.
And the order makes no difference — a 10% fall first gives ₹900, then a 10% rise gives ₹990 again. The factors are either way, and is less than .
So a percentage increase and the same percentage decrease never cancel. The reason is that they are percentages of different bases — the rise is 10% of the smaller number and the fall is 10% of the larger one.
Which is why a shopkeeper who raises a price by 20% and then offers "20% off" is not back at the original price — the customer pays of it, and that arithmetic is the entire content of this chapter applied to two periods instead of three.
Add the percentages for growth and you get too little. Three years at 5% growth, treated as 15%, gives 73600 where the truth is 74088.
Add the percentages for depreciation and you get too much loss. Three years at 10% depreciation, treated as 30%, gives ₹35000 where the truth is ₹36450.
The reason is the same in both cases: each year's percentage is applied to whatever the quantity has become, not to what it started as.
In growth, the quantity has got bigger, so each successive percentage is a percentage of more — and the total gain exceeds the naive sum.
In depreciation, the quantity has got smaller, so each successive percentage is a percentage of less — and the total loss falls short of the naive sum.
One consequence is worth noticing because it surprises people. A 10% rise followed by a 10% fall does not bring you back to where you started.
Take ₹1000. A 10% rise gives ₹1100. A 10% fall on ₹1100 takes off ₹110, leaving ₹990. You are ₹10 down.
And the order makes no difference — a 10% fall first gives ₹900, then a 10% rise gives ₹990 again. The factors are either way, and is less than .
So a percentage increase and the same percentage decrease never cancel. The reason is that they are percentages of different bases — the rise is 10% of the smaller number and the fall is 10% of the larger one.
Which is why a shopkeeper who raises a price by 20% and then offers "20% off" is not back at the original price — the customer pays of it, and that arithmetic is the entire content of this chapter applied to two periods instead of three.
Exam relevance
Why does JEE Main keep returning to exponential growth?
Because a quantity multiplied by a fixed factor each period is a geometric progression, and that idea runs from Class 11 sequences right through to Class 12 differential equations.
This is the foundation for Class 11 Mathematics Sequences and Series and Binomial Theorem, and Class 12 Differential Equations and Application of Derivatives, examined in JEE Main. The amounts are a geometric progression with common ratio , and the formula on this page is its th term. **Questions on the th term and the sum of a GP are standard, and instalment or annuity problems are compound interest applied to a series** of deposits summed with .
The half-yearly idea generalises to continuous compounding. Class 11 and 12 show that compounding times a year gives , and that as grows this tends to — which is where the number comes from. **The observation on this page that more frequent compounding earns more is exactly the limit that produces , and it is examined as a limit.
Logarithms answer the time question properly.** Matching a power works only for the convenient numbers examiners choose. Class 11 gives , and questions asking in how many years a sum doubles or a population triples are standard JEE Main items — with an answer that is not a whole number.
The binomial theorem explains the CI-minus-SI shortcut. Expanding , the first two terms give precisely the simple interest and everything after them is the compound excess. For that excess is , which is the formula — now derived instead of remembered, and extendable to three years. **Approximation questions using for small appear in JEE Main.
Growth and decay become a differential equation.** Class 12 solves to get , and population growth, radioactive decay, cooling and the depreciation of an asset are its standard applications. The minus sign that distinguishes depreciation from growth on this page becomes a negative there, and half-life problems are the depreciation formula in disguise.
The successive-percentage point is examined directly. The observation that a 20% rise followed by a 20% fall leaves you at of the original appears in Class 11 as a question on successive percentage change, and the general result — a rise of followed by a fall of gives a net change of — is worth deriving once.
What the questions look like. For board work, expect find the amount and the compound interest using the formula, compute with half-yearly compounding, **find , or from given data, find the difference between CI and SI for 2 years, and solve a growth or depreciation problem. The rate per period and the number of periods must be stated. For JEE Main, expect GP terms and sums, annuities, logarithms for the time, binomial approximation and exponential growth by differential equation.
How board and competitive emphasis differ. A board paper rewards the stated periods, the substituted formula and at the end. A competitive paper assumes all of it and asks for a sum of deposits, a doubling time, or the solution of a growth equation.
The single trap that costs the most marks.** Answering "find the compound interest" with the value of . The formula produces the amount, which already includes the principal — the interest is . The same error appears in depreciation questions, where the formula gives the value remaining and the question often asks for the value lost. The defence is to underline what the question actually asked for before you start, because in this chapter the formula's output and the required answer are different quantities more often than they are the same.
This is the foundation for Class 11 Mathematics Sequences and Series and Binomial Theorem, and Class 12 Differential Equations and Application of Derivatives, examined in JEE Main. The amounts are a geometric progression with common ratio , and the formula on this page is its th term. **Questions on the th term and the sum of a GP are standard, and instalment or annuity problems are compound interest applied to a series** of deposits summed with .
The half-yearly idea generalises to continuous compounding. Class 11 and 12 show that compounding times a year gives , and that as grows this tends to — which is where the number comes from. **The observation on this page that more frequent compounding earns more is exactly the limit that produces , and it is examined as a limit.
Logarithms answer the time question properly.** Matching a power works only for the convenient numbers examiners choose. Class 11 gives , and questions asking in how many years a sum doubles or a population triples are standard JEE Main items — with an answer that is not a whole number.
The binomial theorem explains the CI-minus-SI shortcut. Expanding , the first two terms give precisely the simple interest and everything after them is the compound excess. For that excess is , which is the formula — now derived instead of remembered, and extendable to three years. **Approximation questions using for small appear in JEE Main.
Growth and decay become a differential equation.** Class 12 solves to get , and population growth, radioactive decay, cooling and the depreciation of an asset are its standard applications. The minus sign that distinguishes depreciation from growth on this page becomes a negative there, and half-life problems are the depreciation formula in disguise.
The successive-percentage point is examined directly. The observation that a 20% rise followed by a 20% fall leaves you at of the original appears in Class 11 as a question on successive percentage change, and the general result — a rise of followed by a fall of gives a net change of — is worth deriving once.
What the questions look like. For board work, expect find the amount and the compound interest using the formula, compute with half-yearly compounding, **find , or from given data, find the difference between CI and SI for 2 years, and solve a growth or depreciation problem. The rate per period and the number of periods must be stated. For JEE Main, expect GP terms and sums, annuities, logarithms for the time, binomial approximation and exponential growth by differential equation.
How board and competitive emphasis differ. A board paper rewards the stated periods, the substituted formula and at the end. A competitive paper assumes all of it and asks for a sum of deposits, a doubling time, or the solution of a growth equation.
The single trap that costs the most marks.** Answering "find the compound interest" with the value of . The formula produces the amount, which already includes the principal — the interest is . The same error appears in depreciation questions, where the formula gives the value remaining and the question often asks for the value lost. The defence is to underline what the question actually asked for before you start, because in this chapter the formula's output and the required answer are different quantities more often than they are the same.
Key takeaways
The compound interest formula, half-yearly compounding and depreciation: quick revision
- ** and . The formula gives the amount, never the interest.
- ₹15000 at 12% for 2 years**: , so and .
- ₹20000 at 10% for 3 years: , so and . Year by year: .
- Half-yearly: halve the rate and double the periods — , where is the number of conversion periods.
- ₹8000 at 10% for 1 year half-yearly: for periods, , so and — against ₹8800 and ₹800 yearly.
- The extra ₹20 is the interest on the first half-year's interest: . More frequent compounding earns more because the interest starts earning sooner.
- **₹6400 at 5% for years half-yearly**: for 3 periods, , so and .
- "10% per annum compounded half-yearly" means 5% every six months. The quoted rate is per annum unless stated otherwise.
- **Find **: ₹12000 to ₹13230 in 2 years. , , so .
- **Find **: ₹1000 to ₹1331 at 10%. , so .
- **Find **: amount ₹9261 at 5% for 3 years. .
- **Always compute the growth factor first** — then take the th root for the rate, match a power for the time, or divide for the principal.
- ** for 2 years.** ₹25000 at 4% gives , confirmed by and . Backwards: a difference of ₹25 at 5% gives .
- Growth: . A population of 64000 at 5% for 3 years becomes .
- Depreciation: . A machine of ₹50000 at 10% for 3 years is worth , so the total depreciation is ₹13550.
- Percentages never add across years. 30% flat would leave ₹35000 against the true ₹36450; 15% flat would give 73600 against the true 74088. Adding understates growth and overstates loss.
- The depreciation formula gives the value LEFT — subtract to find the value lost.
- A rise and an equal fall do not cancel: , so ₹1000 becomes ₹990 either way round.
- For mixed rates, write one bracket per period and multiply, with a plus for a gain and a minus for a loss.
Take any sum, apply a 20% rise and then a 20% fall, and see how far short of the original you land — then explain it in one sentence about the two different bases.
- ₹15000 at 12% for 2 years**: , so and .
- ₹20000 at 10% for 3 years: , so and . Year by year: .
- Half-yearly: halve the rate and double the periods — , where is the number of conversion periods.
- ₹8000 at 10% for 1 year half-yearly: for periods, , so and — against ₹8800 and ₹800 yearly.
- The extra ₹20 is the interest on the first half-year's interest: . More frequent compounding earns more because the interest starts earning sooner.
- **₹6400 at 5% for years half-yearly**: for 3 periods, , so and .
- "10% per annum compounded half-yearly" means 5% every six months. The quoted rate is per annum unless stated otherwise.
- **Find **: ₹12000 to ₹13230 in 2 years. , , so .
- **Find **: ₹1000 to ₹1331 at 10%. , so .
- **Find **: amount ₹9261 at 5% for 3 years. .
- **Always compute the growth factor first** — then take the th root for the rate, match a power for the time, or divide for the principal.
- ** for 2 years.** ₹25000 at 4% gives , confirmed by and . Backwards: a difference of ₹25 at 5% gives .
- Growth: . A population of 64000 at 5% for 3 years becomes .
- Depreciation: . A machine of ₹50000 at 10% for 3 years is worth , so the total depreciation is ₹13550.
- Percentages never add across years. 30% flat would leave ₹35000 against the true ₹36450; 15% flat would give 73600 against the true 74088. Adding understates growth and overstates loss.
- The depreciation formula gives the value LEFT — subtract to find the value lost.
- A rise and an equal fall do not cancel: , so ₹1000 becomes ₹990 either way round.
- For mixed rates, write one bracket per period and multiply, with a plus for a gain and a minus for a loss.
Take any sum, apply a 20% rise and then a 20% fall, and see how far short of the original you land — then explain it in one sentence about the two different bases.