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

13 hand-picked ICSE Class 9 Maths important questions for Compound Interest (Without Using Formula), each with a full model answer — the formats and topics most likely to appear in your board exam.

13
Questions
6
Question types
32
Total marks
₹0
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Quick answer

High-yield 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): CISCE (Council for the Indian School Certificate Examinations)

About Compound Interest (Without Using Formula)

In 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.

Question typeCountMarks
MCQ41
Assertion–Reason11
Very Short22
Short Answer33
Long Answer25
Case-based14

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 5000\text{Rs }5000Rs 5000 for 222 years at 10%10\%10\% per annum is:

  1. (a)

    Rs 1000\text{Rs }1000Rs 1000

  2. (b)

    Rs 1050\text{Rs }1050Rs 1050

  3. (c)

    Rs 1100\text{Rs }1100Rs 1100

  4. (d)

    Rs 500\text{Rs }500Rs 500

Show model answer

Answer: (b) Rs 1050\text{Rs }1050Rs 1050.

Year 1 interest =5000×10100=500=\dfrac{5000\times10}{100}=500=5000×10/100=500, amount =5500=5500=5500. Year 2 interest =5500×10100=550=\dfrac{5500\times10}{100}=550=5500×10/100=550. CI =500+550=Rs 1050.=500+550=\text{Rs }1050.=500+550=Rs 1050.

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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.

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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.

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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.

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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 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)

Q11Long 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.

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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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Q12Long 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. Working backwards at 10%10\%10\%:

Amount after year 1 =96801.1=Rs 8800=\dfrac{9680}{1.1}=\text{Rs }8800=9680/1.1=Rs 8800.

Sum P=88001.1=Rs 8000P=\dfrac{8800}{1.1}=\text{Rs }8000P=8800/1.1=Rs 8000.

Check: Year 1: 8000→88008000\to880080008800; Year 2: 8800→96808800\to968088009680; Year 3: 9680→106489680\to10648968010648. 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)

Q13Case-basedModerate4 marks

Rohan deposits Rs 20000\text{Rs }20000Rs 20000 in a bank that pays 10%10\%10\% per annum compounded annually. He wants to track his money year by year.

(i) Find the interest for the first year.

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

(iii) Find the interest for the second year.

(iv) Find the compound interest for 222 years.

Show model answer

(i) First-year interest =20000×10×1100=Rs 2000=\dfrac{20000\times10\times1}{100}=\text{Rs }2000=20000×10×1/100=Rs 2000.

(ii) Amount at end of year 1 =20000+2000=Rs 22000=20000+2000=\text{Rs }22000=20000+2000=Rs 22000.

(iii) Second-year interest =22000×10×1100=Rs 2200=\dfrac{22000\times10\times1}{100}=\text{Rs }2200=22000×10×1/100=Rs 2200.

(iv) Amount at end of year 2 =22000+2200=Rs 24200=22000+2200=\text{Rs }24200=22000+2200=Rs 24200; CI =24200−20000=Rs 4200.=24200-20000=\text{Rs }4200.=24200-20000=Rs 4200.

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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

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    A full mix — multiple-choice questions, assertion–reason questions, very short answer questions, short answer questions, long answer questions, case-based questions — so every format in the ICSE paper is covered.

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