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Compound Interest (Without Using Formula) — ICSE Class 9 Maths Important Questions

14 ICSE Class 9 Maths practice questions on Compound Interest (Without Using Formula), each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 16 October 2026

Last updated 25 October 2026

14
Questions
5
Topics
3
Key concepts
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With answers
Quick answer

ICSE Compound Interest (without formula) questions compute interest year by year: find Year 1 interest on the principal, add it to get the new principal, then repeat for later years. Common tasks are two- and three-year CI, difference between CI and SI, and growth/depreciation problems solved by successive yearly calculation.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Compound Interest (Without Using Formula)

Within the ICSE Class 9 Maths chapter Compound Interest (without using formula) you calculate interest year by year: the amount at the end of each year becomes the principal for the next year. You handle 2-3 year problems, the difference between compound and simple interest, and simple growth and depreciation.

Year-by-year compound interestAmount at the end of a periodDifference between CI and SIUniform growth (appreciation)Depreciation of value

Key concepts & formulas

Successive year method

Interest for a year =P×R×1100=\dfrac{P\times R\times 1}{100}=(P× R× 1)/100. Add it to PPP to get the amount, which becomes the principal for the next year; repeat for each year.

Compound interest

CI=Final Amount−Original Principal\text{CI}=\text{Final Amount}-\text{Original Principal}CI=Final Amount-Original Principal, after computing the amount year by year.

Growth and depreciation

Growth adds a percentage each period (new value === value +++ increase); depreciation subtracts it (new value === value −-- decrease), computed year by year.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The simple interest on Rs 2000\text{Rs }2000Rs 2000 for 111 year at 10%10\%10% per annum is:

  1. (a)

    Rs 100\text{Rs }100Rs 100

  2. (b)

    Rs 200\text{Rs }200Rs 200

  3. (c)

    Rs 220\text{Rs }220Rs 220

  4. (d)

    Rs 20\text{Rs }20Rs 20

Show model answer

Answer: (b) Rs 200\text{Rs }200Rs 200.

Interest=2000×10×1100=Rs 200.\text{Interest}=\dfrac{2000\times10\times1}{100}=\text{Rs }200.Interest=2000×10×1/100=Rs 200. For the first year, SI and CI are equal.

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Q2MCQEasy1 mark

In the year-by-year method, the amount at the end of the first year becomes the:

  1. (a)

    Interest for the second year

  2. (b)

    Principal for the second year

  3. (c)

    Rate for the second year

  4. (d)

    Compound interest

Show model answer

Answer: (b) Principal for the second year.

Under compounding, the previous year's amount (principal + interest) is treated as the principal for the next year.

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Q3MCQModerate1 mark

The compound interest on Rs 4000\text{Rs }4000Rs 4000 for 222 years, at 10%10\%10% per annum for the first year and 5%5\%5% per annum for the second year, is:

  1. (a)

    Rs 600\text{Rs }600Rs 600

  2. (b)

    Rs 620\text{Rs }620Rs 620

  3. (c)

    Rs 610\text{Rs }610Rs 610

  4. (d)

    Rs 640\text{Rs }640Rs 640

Show model answer

Answer: (b) Rs 620\text{Rs }620Rs 620.

Year 1 interest =4000×10100=400=\dfrac{4000\times10}{100}=400=4000×10/100=400, amount =4400=4400=4400. Year 2 interest =4400×5100=220=\dfrac{4400\times5}{100}=220=4400×5/100=220. CI =400+220=Rs 620.=400+220=\text{Rs }620.=400+220=Rs 620.

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Q4MCQHOTS1 mark

For 222 years at R%R\%R% per annum, the difference between CI and SI on a sum PPP equals:

  1. (a)

    P(R100)P\left(\dfrac{R}{100}\right)P(R/100)

  2. (b)

    P(R100)2P\left(\dfrac{R}{100}\right)^2P(R/100)²

  3. (c)

    2P(R100)2P\left(\dfrac{R}{100}\right)2P(R/100)

  4. (d)

    P(R100)3P\left(\dfrac{R}{100}\right)^3P(R/100)³

Show model answer

Answer: (b) P(R100)2P\left(\dfrac{R}{100}\right)^2P(R/100)².

The extra amount in CI is the interest on the first year's interest: interest of PR100\dfrac{PR}{100}PR/100 at R%R\%R% is PR100×R100=P(R100)2.\dfrac{PR}{100}\times\dfrac{R}{100}=P\left(\dfrac{R}{100}\right)^2.PR/100× R/100=P(R/100)².

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): For the first year, compound interest and simple interest on the same sum at the same rate are equal.

Reason (R): In the first year there is no interest earned on previous interest.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

Show model answer

Answer: (a) Both are true and R correctly explains A. Compounding differs from simple interest only when interest is earned on earlier interest, which first happens in the second year.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the amount on Rs 8000\text{Rs }8000Rs 8000 for 111 year at 12%12\%12% per annum compounded annually.

Show model answer

Interest for year 1 =8000×12×1100=Rs 960=\dfrac{8000\times12\times1}{100}=\text{Rs }960=8000×12×1/100=Rs 960.

Amount =8000+960=Rs 8960.=8000+960=\text{Rs }8960.=8000+960=Rs 8960.

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Q7Very ShortModerate2 marks

The population of a town is 600060006000 and increases by 5%5\%5% in one year. Find the population after one year.

Show model answer

Increase =6000×5100=300=\dfrac{6000\times5}{100}=300=6000×5/100=300.

Population after 111 year =6000+300=6300.=6000+300=6300.=6000+300=6300.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Calculate the compound interest on Rs 10000\text{Rs }10000Rs 10000 for 222 years at 8%8\%8% per annum, computing year by year.

Show model answer

Year 1: Interest =10000×8×1100=Rs 800=\dfrac{10000\times8\times1}{100}=\text{Rs }800=10000×8×1/100=Rs 800.

Amount after year 1 =10000+800=Rs 10800=10000+800=\text{Rs }10800=10000+800=Rs 10800 (principal for year 2).

Year 2: Interest =10800×8×1100=Rs 864=\dfrac{10800\times8\times1}{100}=\text{Rs }864=10800×8×1/100=Rs 864.

Amount after year 2 =10800+864=Rs 11664=10800+864=\text{Rs }11664=10800+864=Rs 11664.

∴ CI=11664−10000=Rs 1664.\therefore\ \text{CI}=11664-10000=\text{Rs }1664.∴ CI=11664-10000=Rs 1664.

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Q9Short AnswerModerate3 marks

Find the compound interest on Rs 6250\text{Rs }6250Rs 6250 for 333 years at 8%8\%8% per annum.

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Year 1: Interest =6250×8100=500=\dfrac{6250\times8}{100}=500=6250×8/100=500; amount =6250+500=6750=6250+500=6750=6250+500=6750.

Year 2: Interest =6750×8100=540=\dfrac{6750\times8}{100}=540=6750×8/100=540; amount =6750+540=7290=6750+540=7290=6750+540=7290.

Year 3: Interest =7290×8100=583.20=\dfrac{7290\times8}{100}=583.20=7290×8/100=583.20; amount =7290+583.20=7873.20=7290+583.20=7873.20=7290+583.20=7873.20.

∴ CI=7873.20−6250=Rs 1623.20.\therefore\ \text{CI}=7873.20-6250=\text{Rs }1623.20.∴ CI=7873.20-6250=Rs 1623.20.

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Q10Short AnswerModerate3 marks

The interest for the first year on Rs 15000\text{Rs }15000Rs 15000, compounded annually, is Rs 1500\text{Rs }1500Rs 1500. Find the rate of interest and the interest for the second year.

Show model answer

Rate: R=1500×10015000×1=10%R=\dfrac{1500\times100}{15000\times1}=10\%R=1500×100/15000×1=10% per annum.

Amount after year 1 =15000+1500=Rs 16500=15000+1500=\text{Rs }16500=15000+1500=Rs 16500 (principal for year 2).

Year 2 interest =16500×10×1100=Rs 1650.=\dfrac{16500\times10\times1}{100}=\text{Rs }1650.=16500×10×1/100=Rs 1650.

So the rate is 10%10\%10% per annum and the second-year interest is Rs 1650\text{Rs }1650Rs 1650.

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Q11Short AnswerHOTS3 marks

The difference between the compound interest and the simple interest on a sum for 222 years at 10%10\%10% per annum is Rs 50\text{Rs }50Rs 50. Find the sum, using year-by-year working.

Show model answer

Let the sum be Rs P\text{Rs }PRs P.

Simple interest for 222 years =P×10×2100=P5=\dfrac{P\times10\times2}{100}=\dfrac{P}{5}=P×10×2/100=P/5.

Compound interest: Year 1 interest =P×10100=P10=\dfrac{P\times10}{100}=\dfrac{P}{10}=P×10/100=P/10; amount =P+P10=11P10=P+\dfrac{P}{10}=\dfrac{11P}{10}=P+P/10=11P/10.

Year 2 interest =110×11P10=11P100=\dfrac{1}{10}\times\dfrac{11P}{10}=\dfrac{11P}{100}=1/10× 11P/10=11P/100. So CI=P10+11P100=10P+11P100=21P100\text{CI}=\dfrac{P}{10}+\dfrac{11P}{100}=\dfrac{10P+11P}{100}=\dfrac{21P}{100}CI=P/10+11P/100=(10P+11P)/100=21P/100.

Difference =CI−SI=21P100−P5=21P−20P100=P100=\text{CI}-\text{SI}=\dfrac{21P}{100}-\dfrac{P}{5}=\dfrac{21P-20P}{100}=\dfrac{P}{100}=CI-SI=21P/100-P/5=(21P-20P)/100=P/100.

Given P100=50⇒P=Rs 5000.\dfrac{P}{100}=50\Rightarrow P=\text{Rs }5000.P/100=50⇒ P=Rs 5000.

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Long answer questions (5 marks)

Q12Long AnswerModerate5 marks

A machine is purchased for Rs 80000\text{Rs }80000Rs 80000. Its value depreciates by 10%10\%10% in the first year and by 8%8\%8% in the second year. Find its value at the end of 222 years and the total depreciation.

Show model answer

Year 1 (depreciation 10%10\%10%):

Decrease =80000×10100=Rs 8000=\dfrac{80000\times10}{100}=\text{Rs }8000=80000×10/100=Rs 8000.

Value at end of year 1 =80000−8000=Rs 72000=80000-8000=\text{Rs }72000=80000-8000=Rs 72000.

Year 2 (depreciation 8%8\%8%):

Decrease =72000×8100=Rs 5760=\dfrac{72000\times8}{100}=\text{Rs }5760=72000×8/100=Rs 5760.

Value at end of year 2 =72000−5760=Rs 66240=72000-5760=\text{Rs }66240=72000-5760=Rs 66240.

Total depreciation =80000−66240=Rs 13760.=80000-66240=\text{Rs }13760.=80000-66240=Rs 13760.

So the machine is worth Rs 66240\text{Rs }66240Rs 66240 after 222 years, having depreciated by Rs 13760\text{Rs }13760Rs 13760.

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Q13Long AnswerHOTS5 marks

A sum of money amounts to Rs 9680\text{Rs }9680Rs 9680 in 222 years and to Rs 10648\text{Rs }10648Rs 10648 in 333 years, compounded annually. Find the sum and the rate of interest, using the year-by-year idea.

Show model answer

The interest for the third year is the difference of the two amounts:

Third-year interest =10648−9680=Rs 968=10648-9680=\text{Rs }968=10648-9680=Rs 968.

This Rs 968\text{Rs }968Rs 968 is the interest for one year on the amount after 222 years, i.e. on Rs 9680\text{Rs }9680Rs 9680.

∴ R=968×1009680×1=10%\therefore\ R=\dfrac{968\times100}{9680\times1}=10\%∴ R=968×100/9680×1=10% per annum.

Finding the sum PPP: After 222 years the amount is 968096809680. At 10%10\%10%, each year's amount is 110%110\%110% of the year before, so work backwards:

Amount after year 1 =9680×100110=Rs 8800=\dfrac{9680\times100}{110}=\text{Rs }8800=9680×100/110=Rs 8800.

Sum P=8800×100110=Rs 8000P=\dfrac{8800\times100}{110}=\text{Rs }8000P=8800×100/110=Rs 8000.

Check: Year 1: 8000→88008000\to88008000→8800; Year 2: 8800→96808800\to96808800→9680; Year 3: 9680→106489680\to106489680→10648. Correct.

Hence the sum is Rs 8000\text{Rs }8000Rs 8000 and the rate is 10%10\%10% per annum.

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Case-based questions (4 marks)

Q14Case-basedModerate4 marks

The population of a town is 200002000020000. It rises by 10%10\%10% in the first year, and then, after many people move away, it falls by 10%10\%10% in the second year.

(i) Find the increase in the first year.

(ii) Find the population at the end of the first year.

(iii) Find the decrease in the second year.

(iv) Find the population at the end of the second year. Is it the same as at the start? Explain.

Show model answer

(i) Increase =20000×10100=2000=\dfrac{20000\times10}{100}=2000=20000×10/100=2000.

(ii) Population at the end of year 1 =20000+2000=22000=20000+2000=22000=20000+2000=22000.

(iii) Decrease =22000×10100=2200=\dfrac{22000\times10}{100}=2200=22000×10/100=2200.

(iv) Population at the end of year 2 =22000−2200=19800=22000-2200=19800=22000-2200=19800. It is 200200200 less than at the start, because the 10%10\%10% fall is taken on the larger number 220002200022000, not on 200002000020000.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • How do you calculate compound interest without the formula?
    Work year by year: find one year's interest on the principal, add it to get the amount, and use that amount as the next year's principal. For ₹100001000010000 at 8%8\%8%: year 1 gives ₹800800800 (amount ₹108001080010800), year 2 gives ₹864864864 (amount ₹116641166411664), so the CI is ₹166416641664.
  • What is the difference between CI and SI for two years?
    It is the interest earned in year 2 on year 1's interest: P(R100)2P\left(\dfrac{R}{100}\right)^2P(R/100)². For the first year, CI and SI are always equal, which is the point of this page's assertion–reason question.
  • How do you find the rate and principal from two successive amounts?
    The difference between the amounts is one year's interest on the earlier amount. If a sum becomes ₹968096809680 after 2 years and ₹106481064810648 after 3, the interest ₹968968968 is 10%10\%10% of ₹968096809680, so R=10%R=10\%R=10%, and the principal is 96801.21=\dfrac{9680}{1.21}=9680/1.21= ₹800080008000.
  • How do you calculate depreciation year by year?
    Subtract each year's percentage from the value at the start of that year. A ₹800008000080000 machine losing 10%10\%10% then 8%8\%8% is worth ₹720007200072000 after one year and ₹662406624066240 after two. Population that rises one year and falls the next is handled the same way.
  • Which ICSE Class 9 chapter teaches the compound interest formula?
    Compound Interest (Using Formula), Chapter 3, follows this chapter and introduces A=P(1+R100)nA=P\left(1+\dfrac{R}{100}\right)^nA=P(1+R/100)^n. Learning the year-by-year method here first makes that formula easy to understand, and the formula chapter has its own important-questions page on this site.
  • Why does ICSE teach compound interest without a formula first?
    Because many questions change the rate each year, ask for one year's interest, or work backwards from interest to rate, and the step-by-step method handles all of them. It also shows why the formula works.
  • Can I download compound interest without formula questions?
    Yes. Enter your email in the free PDF box on this page and all 14 year-by-year compound interest questions open with worked answers as a printable worksheet, ready to save as a PDF.

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