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Compound Interest (Using Formula) — ICSE Class 9 Maths Important Questions

13 ICSE Class 9 Maths practice questions on Compound Interest (Using Formula), each with a full model answer, covering 6 topics from the chapter in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 18 October 2026

Last updated 25 October 2026

13
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6
Topics
3
Key concepts
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Quick answer

ICSE Compound Interest (using formula) questions apply A=P(1+r100)nA=P\left(1+\dfrac{r}{100}\right)^{n}A=P(1+r/100)^n to find amount and interest, handle half-yearly compounding (halve the rate, double the periods), and solve growth and depreciation problems. Practise finding rate, time, or principal by substituting into the formula.

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

About Compound Interest (Using Formula)

Inside the ICSE Class 9 Maths chapter Compound Interest (using formula) you use A=P(1+r100)nA=P\left(1+\dfrac{r}{100}\right)^{n}A=P(1+r/100)^n to find the amount and compound interest quickly, adapt it for half-yearly compounding, and apply the same growth formula to population increase and machine depreciation.

The compound interest formulaFinding amount and CIHalf-yearly compoundingGrowth and appreciationDepreciation of valueCI as repeated simple interest; CI minus SI

Key concepts & formulas

CI formula

A=P(1+r100)nA=P\left(1+\dfrac{r}{100}\right)^{n}A=P(1+r/100)^n where AAA is the amount, PPP the principal, rrr the annual rate and nnn the number of years; CI=A−P\text{CI}=A-PCI=A-P.

Half-yearly compounding

For half-yearly compounding use rate r2%\dfrac{r}{2}\%r/2% per half-year and 2n2n2n periods: A=P(1+r/2100)2nA=P\left(1+\dfrac{r/2}{100}\right)^{2n}A=P(1+r/2/100)^2n.

Growth and depreciation

Appreciation: A=P(1+r100)nA=P\left(1+\dfrac{r}{100}\right)^{n}A=P(1+r/100)^n; depreciation: A=P(1−r100)nA=P\left(1-\dfrac{r}{100}\right)^{n}A=P(1-r/100)^n.

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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)

Q1MCQModerate1 mark

The difference between the compound interest and the simple interest on Rs 5000\text{Rs }5000Rs 5000 for 222 years at 10%10\%10% per annum (compounded annually) is:

  1. (a)

    Rs 0\text{Rs }0Rs 0

  2. (b)

    Rs 50\text{Rs }50Rs 50

  3. (c)

    Rs 100\text{Rs }100Rs 100

  4. (d)

    Rs 500\text{Rs }500Rs 500

Show model answer

Answer: (b) Rs 50\text{Rs }50Rs 50. SI =5000×10×2100=Rs 1000=\dfrac{5000\times10\times2}{100}=\text{Rs }1000=5000×10×2/100=Rs 1000. CI =5000(1.1)2−5000=6050−5000=Rs 1050=5000(1.1)^2-5000=6050-5000=\text{Rs }1050=5000(1.1)²-5000=6050-5000=Rs 1050. Difference =Rs 50=\text{Rs }50=Rs 50 (the interest on the first year's interest, 10%10\%10% of 500500500).

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Q2MCQEasy1 mark

For half-yearly compounding at 10%10\%10% per annum for 111 year, the rate and number of periods used are:

  1. (a)

    10%10\%10%, 111

  2. (b)

    5%5\%5%, 222

  3. (c)

    20%20\%20%, 222

  4. (d)

    5%5\%5%, 111

Show model answer

Answer: (b) 5%5\%5%, 222.

Halve the rate (10%→5%10\%\to5\%10%→5% per half-year) and double the periods (111 year →2\to2→2 half-years).

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Q3MCQModerate1 mark

The amount on Rs 5000\text{Rs }5000Rs 5000 at 10%10\%10% per annum for 222 years compounded annually is:

  1. (a)

    Rs 6000\text{Rs }6000Rs 6000

  2. (b)

    Rs 6050\text{Rs }6050Rs 6050

  3. (c)

    Rs 5500\text{Rs }5500Rs 5500

  4. (d)

    Rs 6100\text{Rs }6100Rs 6100

Show model answer

Answer: (b) Rs 6050\text{Rs }6050Rs 6050.

A=5000(1+10100)2=5000(1.1)2=5000×1.21=Rs 6050.A=5000\left(1+\dfrac{10}{100}\right)^2=5000(1.1)^2=5000\times1.21=\text{Rs }6050.A=5000(1+10/100)²=5000(1.1)²=5000×1.21=Rs 6050.

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Q4MCQHOTS1 mark

A sum doubles itself in a certain time under CI. The value of (1+r100)n\left(1+\dfrac{r}{100}\right)^{n}(1+r/100)^n for this is:

  1. (a)

    111

  2. (b)

    222

  3. (c)

    12\dfrac121/2

  4. (d)

    444

Show model answer

Answer: (b) 222.

If the amount is twice the principal, A=2PA=2PA=2P, so P(1+r100)n=2P⇒(1+r100)n=2.P\left(1+\dfrac{r}{100}\right)^n=2P\Rightarrow\left(1+\dfrac{r}{100}\right)^n=2.P(1+r/100)^n=2P⇒ (1+r/100)^n=2.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): When interest is compounded half-yearly, the interest is more than when compounded annually at the same annual rate for the same time.

Reason (R): More frequent compounding means interest is added to the principal more often.

  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

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Answer: (a) Both are true and R correctly explains A. Because half-yearly compounding adds interest twice a year, the principal grows sooner, giving a larger amount.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Without using the formula, find the compound interest on Rs 10000\text{Rs }10000Rs 10000 for 222 years at 10%10\%10% per annum, compounded annually.

Show model answer

Year 1: interest =10%=10\%=10% of 10000=Rs 100010000=\text{Rs }100010000=Rs 1000; amount =Rs 11000=\text{Rs }11000=Rs 11000.

Year 2: interest =10%=10\%=10% of 11000=Rs 110011000=\text{Rs }110011000=Rs 1100; amount =Rs 12100=\text{Rs }12100=Rs 12100.

CI =1000+1100=Rs 2100=1000+1100=\text{Rs }2100=1000+1100=Rs 2100. (Compound interest is simple interest computed each year on a growing principal.)

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Q7Very ShortModerate2 marks

Find the compound interest on Rs 8000\text{Rs }8000Rs 8000 for 111 year at 10%10\%10% per annum compounded half-yearly.

Show model answer

Rate per half-year =5%=5\%=5%, periods =2=2=2.

A=8000(1+5100)2=8000×(1.05)2=8000×1.1025=Rs 8820A=8000\left(1+\dfrac{5}{100}\right)^2=8000\times(1.05)^2=8000\times1.1025=\text{Rs }8820A=8000(1+5/100)²=8000×(1.05)²=8000×1.1025=Rs 8820.

CI =8820−8000=Rs 820.=8820-8000=\text{Rs }820.=8820-8000=Rs 820.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Find the compound interest on Rs 15625\text{Rs }15625Rs 15625 for 333 years at 12%12\%12% per annum compounded annually.

Show model answer

A=15625(1+12100)3=15625×(1.12)3A=15625\left(1+\dfrac{12}{100}\right)^3=15625\times(1.12)^3A=15625(1+12/100)³=15625×(1.12)³.

(1.12)3=1.404928(1.12)^3=1.404928(1.12)³=1.404928.

A=15625×1.404928=Rs 21952A=15625\times1.404928=\text{Rs }21952A=15625×1.404928=Rs 21952.

∴ CI=21952−15625=Rs 6327.\therefore\ \text{CI}=21952-15625=\text{Rs }6327.∴ CI=21952-15625=Rs 6327.

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Q9Short AnswerModerate3 marks

At what rate per cent per annum will Rs 6400\text{Rs }6400Rs 6400 amount to Rs 7744\text{Rs }7744Rs 7744 in 222 years, compounded annually?

Show model answer

A=P(1+r100)nA=P\left(1+\dfrac{r}{100}\right)^{n}A=P(1+r/100)^n

7744=6400(1+r100)2⇒(1+r100)2=77446400=1211007744=6400\left(1+\dfrac{r}{100}\right)^2\Rightarrow\left(1+\dfrac{r}{100}\right)^2=\dfrac{7744}{6400}=\dfrac{121}{100}7744=6400(1+r/100)²⇒ (1+r/100)²=7744/6400=121/100.

Taking square roots: 1+r100=1110⇒r100=1101+\dfrac{r}{100}=\dfrac{11}{10}\Rightarrow\dfrac{r}{100}=\dfrac{1}{10}1+r/100=11/10⇒ r/100=1/10.

∴ r=10%\therefore\ r=10\%∴ r=10% per annum.

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Q10Short AnswerHOTS3 marks

In how many years will Rs 5000\text{Rs }5000Rs 5000 amount to Rs 5832\text{Rs }5832Rs 5832 at 8%8\%8% per annum compounded annually?

Show model answer

A=P(1+r100)nA=P\left(1+\dfrac{r}{100}\right)^{n}A=P(1+r/100)^n

5832=5000(1+8100)n⇒(1.08)n=58325000=1.16645832=5000\left(1+\dfrac{8}{100}\right)^{n}\Rightarrow(1.08)^n=\dfrac{5832}{5000}=1.16645832=5000(1+8/100)^n⇒(1.08)^n=5832/5000=1.1664.

Now (1.08)2=1.1664(1.08)^2=1.1664(1.08)²=1.1664, so n=2n=2n=2.

∴\therefore∴ The required time is 222 years.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

The population of a city was 500005000050000. It increased by 8%8\%8% in the first year and 10%10\%10% in the second year, but decreased by 5%5\%5% in the third year. Find the population at the end of 333 years.

Show model answer

Apply the growth/decline factor year by year.

End of year 1 (+8%+8\%+8%): 50000×(1+8100)=50000×1.08=5400050000\times\left(1+\dfrac{8}{100}\right)=50000\times1.08=5400050000× (1+8/100)=50000×1.08=54000.

End of year 2 (+10%+10\%+10%): 54000×1.10=5940054000\times1.10=5940054000×1.10=59400.

End of year 3 (−5%-5\%-5%): 59400×(1−5100)=59400×0.95=5643059400\times\left(1-\dfrac{5}{100}\right)=59400\times0.95=5643059400× (1-5/100)=59400×0.95=56430.

∴\therefore∴ The population at the end of 333 years is 56430.56430.56430.

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Q12Long AnswerHOTS5 marks

A sum of money is invested at compound interest. It amounts to Rs 21296\text{Rs }21296Rs 21296 in 333 years and to Rs 19360\text{Rs }19360Rs 19360 in 222 years. Using the formula, find the rate of interest and the sum.

Show model answer

Let the sum be PPP and rate r%r\%r% per annum. Then:

A2=P(1+r100)2=19360A_2=P\left(1+\dfrac{r}{100}\right)^2=19360A₂=P(1+r/100)²=19360 ...(1)

A3=P(1+r100)3=21296A_3=P\left(1+\dfrac{r}{100}\right)^3=21296A₃=P(1+r/100)³=21296 ...(2)

Dividing (2) by (1):

(1+r100)=2129619360=1.1\left(1+\dfrac{r}{100}\right)=\dfrac{21296}{19360}=1.1(1+r/100)=21296/19360=1.1.

∴ r100=0.1⇒r=10%\therefore\ \dfrac{r}{100}=0.1\Rightarrow r=10\%∴ r/100=0.1⇒ r=10% per annum.

Substitute in (1): P(1.1)2=19360⇒P×1.21=19360⇒P=193601.21=Rs 16000P(1.1)^2=19360\Rightarrow P\times1.21=19360\Rightarrow P=\dfrac{19360}{1.21}=\text{Rs }16000P(1.1)²=19360⇒ P×1.21=19360⇒ P=19360/1.21=Rs 16000.

Hence the sum is Rs 16000\text{Rs }16000Rs 16000 and the rate is 10%10\%10% per annum.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

A car is bought for Rs 600000\text{Rs }600000Rs 600000. Its value depreciates at 10%10\%10% per annum. The owner uses the formula A=P(1−r100)nA=P\left(1-\dfrac{r}{100}\right)^{n}A=P(1-r/100)^n.

(i) Write the depreciation factor for one year.

(ii) Find the value of the car after 111 year.

(iii) Find the value of the car after 222 years.

(iv) Find the total depreciation over 222 years.

Show model answer

(i) Depreciation factor =1−10100=0.9=1-\dfrac{10}{100}=0.9=1-10/100=0.9.

(ii) Value after 111 year =600000×0.9=Rs 540000=600000\times0.9=\text{Rs }540000=600000×0.9=Rs 540000.

(iii) Value after 222 years =600000×(0.9)2=600000×0.81=Rs 486000=600000\times(0.9)^2=600000\times0.81=\text{Rs }486000=600000×(0.9)²=600000×0.81=Rs 486000.

(iv) Total depreciation =600000−486000=Rs 114000.=600000-486000=\text{Rs }114000.=600000-486000=Rs 114000.

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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

  • What is the compound interest formula for ICSE Class 9?
    A=P(1+r100)nA=P\left(1+\dfrac{r}{100}\right)^nA=P(1+r/100)^n, and compound interest is A−PA-PA-P. ₹500050005000 at 10%10\%10% for 222 years gives A=5000×1.21=A=5000\times1.21=A=5000×1.21= ₹605060506050, so the interest is ₹105010501050. Here nnn counts the number of times interest is added, so it changes with half-yearly compounding.
  • How do you calculate compound interest compounded half-yearly?
    Halve the yearly rate and double the number of periods. ₹800080008000 for 111 year at 10%10\%10% compounded half-yearly uses 5%5\%5% for 222 periods: 8000×1.052=8000\times1.05^2=8000×1.05²= ₹882088208820, so the interest is ₹820820820, more than the ₹800800800 earned with yearly compounding.
  • How do you find the rate or time from the CI formula?
    Divide the amount by the principal and recognise the power. ₹640064006400 becoming ₹774477447744 in 222 years gives 1.21=1.121.21=1.1^21.21=1.1², so the rate is 10%10\%10%. ₹500050005000 becoming ₹583258325832 at 8%8\%8% gives 1.1664=1.0821.1664=1.08^21.1664=1.08², so the time is 222 years.
  • How is depreciation calculated with the formula?
    Use A=P(1−r100)nA=P\left(1-\dfrac{r}{100}\right)^nA=P(1-r/100)^n. A car bought for ₹600000600000600000 that loses 10%10\%10% a year is worth 600000×0.92=600000\times0.9^2=600000×0.9²= ₹486000486000486000 after two years. This page's case study follows such a car.
  • Should I learn compound interest without the formula first?
    Yes. Compound Interest (Without Using Formula), Chapter 2, builds the year-by-year method, and this Chapter 3 turns it into the formula. Both have important-questions pages on Classmate AI; Expansions (Chapter 4) comes next.
  • When should I use the formula and when the year-by-year method?
    Use the formula when the rate stays the same each year. If the rate changes, as with population growing 8%8\%8% then 10%10\%10%, multiply the factors year by year: 50000×1.08×1.10=5940050000\times1.08\times1.10=5940050000×1.08×1.10=59400.
  • Can I download Compound Interest using formula ICSE questions?
    Yes. Enter your email in the free PDF box on this page and all 13 compound interest formula questions open with step-by-step answers as a printable worksheet.

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