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

13 hand-picked ICSE Class 9 Maths important questions for Compound Interest (Using Formula), each with a full model answer — the formats and topics most likely to appear in your board exam.

13
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6
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32
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Quick answer

High-yield 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. Finding rate, time, or principal by substituting into the formula appears frequently.

Syllabus reference (2026–27): CISCE (Council for the Indian School Certificate Examinations)

About Compound Interest (Using Formula)

In 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 value

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.

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

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

The compound interest formula for the amount is:

  1. (a)

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

  2. (b)

    A=P+PRT100A=P+\dfrac{PRT}{100}A=P+PRT/100

  3. (c)

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

  4. (d)

    A=Pr100nA=P\dfrac{r}{100}nA=Pr/100n

Show model answer

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

This gives the amount after nnn years at rate r%r\%r\% per annum compounded annually.

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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\to22 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\dfrac1212

  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

Show model answer

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

Find the amount on Rs 12000\text{Rs }12000Rs 12000 for 222 years at 5%5\%5\% per annum compounded annually.

Show model answer

A=P(1+r100)n=12000(1+5100)2A=P\left(1+\dfrac{r}{100}\right)^{n}=12000\left(1+\dfrac{5}{100}\right)^2A=P(1+r/100)^n=12000(1+5/100)^2.

A=12000×(1.05)2=12000×1.1025=Rs 13230.A=12000\times(1.05)^2=12000\times1.1025=\text{Rs }13230.A=12000×(1.05)^2=12000×1.1025=Rs 13230.

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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=11/10/100=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=19360A_2=P(1+r/100)^2=19360 ...(1)

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

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    Yes — they are aligned to the CISCE latest syllabus syllabus for ICSE Class 9 Maths, so nothing here is outside the current course.
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  • What types of questions are covered for Compound Interest (Using Formula)?
    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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