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Banking (Recurring Deposit Accounts) — ICSE Class 10 Maths Important Questions

13 ICSE Class 10 Maths practice questions on Banking (Recurring Deposit Accounts), 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

13
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3
Key concepts
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Banking (Recurring Deposit Accounts) — ICSE Class 10 Maths Important Questions

RD Maturity Value, No Confusion

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Quick answer

Regularly-tested ICSE Banking questions give a monthly deposit PPP, a period of nnn months and a rate rrr, and ask for the interest I=P×n(n+1)2×12×r100I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}I=P×n(n+1)/2×12×r/100 and the maturity value MV=Pn+I\text{MV}=Pn+IMV=Pn+I. Reverse problems that ask you to find the rate, the monthly deposit or the number of months from a given interest or maturity value appear regularly.

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

About Banking (Recurring Deposit Accounts)

In the ICSE Class 10 Maths chapter Banking you work with a Recurring Deposit (RD) account, where a fixed sum is deposited every month. You compute the interest using I=P×n(n+1)2×12×r100I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}I=P×n(n+1)/2×12×r/100 and the maturity value as total deposit plus interest, and you also reverse the formula to find the rate, the monthly instalment or the number of months.

Monthly deposit and total sum depositedInterest formula for RDMaturity value of an RDFinding the rate of interestFinding the monthly deposit or number of months

Key concepts & formulas

Total sum deposited

If PPP is the monthly deposit and the money is kept for nnn months, the total sum deposited is P×nP\times nP× n.

Interest on an RD

I=P×n(n+1)2×12×r100I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}I=P×n(n+1)/2×12×r/100, where the factor n(n+1)2\dfrac{n(n+1)}{2}n(n+1)/2 counts the equivalent months for which the instalments earn interest.

Maturity value

MV=total deposit+interest=Pn+I.\text{MV}=\text{total deposit}+\text{interest}=Pn+I.MV=total deposit+interest=Pn+I. Reverse problems substitute known values and solve for rrr, PPP or nnn.

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Important questions with answers

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Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

Mr. Sharma deposits ₹500\text{₹}500₹500 per month in a recurring deposit account for 222 years. The total money he deposits is:

  1. (a)

    ₹12000\text{₹}12000₹12000

  2. (b)

    ₹6000\text{₹}6000₹6000

  3. (c)

    ₹10000\text{₹}10000₹10000

  4. (d)

    ₹1000\text{₹}1000₹1000

Show model answer

Answer: (a) ₹12000\text{₹}12000₹12000.

222 years =24=24=24 months, so total deposit =500×24=₹12000.=500\times24=\text{₹}12000.=500×24=₹12000.

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Q2MCQEasy1 mark

For a recurring deposit kept for n=12n=12n=12 months, the value of n(n+1)2\dfrac{n(n+1)}{2}n(n+1)/2 (the equivalent number of months for interest) is:

  1. (a)

    787878

  2. (b)

    156156156

  3. (c)

    666666

  4. (d)

    909090

Show model answer

Answer: (a) 787878.

n(n+1)2=12×132=1562=78.\dfrac{n(n+1)}{2}=\dfrac{12\times13}{2}=\dfrac{156}{2}=78.n(n+1)/2=12×13/2=156/2=78.

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Q3MCQModerate1 mark

The interest on an RD of ₹400\text{₹}400₹400 per month for 121212 months at 8%8\%8\% per annum is:

  1. (a)

    ₹208\text{₹}208₹208

  2. (b)

    ₹260\text{₹}260₹260

  3. (c)

    ₹416\text{₹}416₹416

  4. (d)

    ₹200\text{₹}200₹200

Show model answer

Answer: (a) ₹208\text{₹}208₹208.

I=400×12×132×12×8100=400×6.5×0.08=₹208.I=400\times\dfrac{12\times13}{2\times12}\times\dfrac{8}{100}=400\times6.5\times0.08=\text{₹}208.I=400×12×13/2×12×8/100=400×6.5×0.08=₹208.

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Q4MCQHOTS1 mark

The maturity value of an RD of ₹1000\text{₹}1000₹1000 per month for 111 year at 10%10\%10\% per annum is:

  1. (a)

    ₹12650\text{₹}12650₹12650

  2. (b)

    ₹13200\text{₹}13200₹13200

  3. (c)

    ₹12600\text{₹}12600₹12600

  4. (d)

    ₹13000\text{₹}13000₹13000

Show model answer

Answer: (a) ₹12650\text{₹}12650₹12650.

Total deposit =1000×12=₹12000=1000\times12=\text{₹}12000=1000×12=₹12000; I=1000×12×1324×10100=1000×6.5×0.10=₹650.I=1000\times\dfrac{12\times13}{24}\times\dfrac{10}{100}=1000\times6.5\times0.10=\text{₹}650.I=1000×12×13/24×10/100=1000×6.5×0.10=₹650.

MV=12000+650=₹12650.\text{MV}=12000+650=\text{₹}12650.MV=12000+650=₹12650.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): The maturity value of a recurring deposit equals the total money deposited plus the interest earned.

Reason (R): In an RD, interest is calculated on the whole sum deposited for the entire period of the account.

  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: (c) A is true — MV=Pn+I\text{MV}=Pn+IMV=Pn+I. R is false: each instalment earns interest for a different number of months, so interest uses the equivalent-months factor n(n+1)2\dfrac{n(n+1)}{2}n(n+1)/2, not the whole sum for the whole period.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Find the interest on a recurring deposit of ₹600\text{₹}600₹600 per month for 202020 months at 9%9\%9\% per annum.

Show model answer

I=P×n(n+1)2×12×r100=600×20×2124×9100.I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}=600\times\dfrac{20\times21}{24}\times\dfrac{9}{100}.I=P×n(n+1)/2×12×r/100=600×20×21/24×9/100.

=600×17.5×0.09=₹945.=600\times17.5\times0.09=\text{₹}945.=600×17.5×0.09=₹945.

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Q7Very ShortModerate2 marks

A person deposits ₹250\text{₹}250₹250 per month for 222 years in an RD at 8%8\%8\% per annum. Find the maturity value.

Show model answer

Total deposit =250×24=₹6000.=250\times24=\text{₹}6000.=250×24=₹6000.

I=250×24×2524×8100=250×25×0.08=₹500.I=250\times\dfrac{24\times25}{24}\times\dfrac{8}{100}=250\times25\times0.08=\text{₹}500.I=250×24×25/24×8/100=250×25×0.08=₹500.

MV=6000+500=₹6500.\text{MV}=6000+500=\text{₹}6500.MV=6000+500=₹6500.

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Short answer questions (3 marks)

Q8Short AnswerEasy3 marks

Mrs. Rao deposits ₹800\text{₹}800₹800 per month in a recurring deposit account for 333 years at 7%7\%7\% per annum. Find (i) the interest earned and (ii) the maturity value.

Show model answer

Here P=800P=800P=800, n=36n=36n=36, r=7r=7r=7.

(i) I=800×36×372×12×7100=800×55.5×0.07=₹3108.I=800\times\dfrac{36\times37}{2\times12}\times\dfrac{7}{100}=800\times55.5\times0.07=\text{₹}3108.I=800×36×37/2×12×7/100=800×55.5×0.07=₹3108.

(ii) Total deposit =800×36=₹28800.=800\times36=\text{₹}28800.=800×36=₹28800.

MV=28800+3108=₹31908.\text{MV}=28800+3108=\text{₹}31908.MV=28800+3108=₹31908.

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Q9Short AnswerModerate3 marks

Mr. Gupta deposits ₹700\text{₹}700₹700 per month in an RD for 222 years and receives ₹1050\text{₹}1050₹1050 as interest at maturity. Find the rate of interest per annum.

Show model answer

Here P=700P=700P=700, n=24n=24n=24.

I=P×n(n+1)2×12×r100=700×24×2524×r100=700×25×r100=175r.I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}=700\times\dfrac{24\times25}{24}\times\dfrac{r}{100}=700\times25\times\dfrac{r}{100}=175r.I=P×n(n+1)/2×12×r/100=700×24×25/24×r/100=700×25×r/100=175r.

So 175r=1050⇒r=1050175=6.175r=1050\Rightarrow r=\dfrac{1050}{175}=6.175r=1050 r=1050/175=6.

Rate =6%=6\%=6\% per annum.

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Q10Short AnswerHOTS3 marks

David deposits ₹150\text{₹}150₹150 per month in a recurring deposit account at 8%8\%8\% per annum and earns ₹300\text{₹}300₹300 as interest at maturity. Find the number of months for which he deposited and the maturity value.

Show model answer

I=P×n(n+1)2×12×r100=150×n(n+1)24×8100.I=P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}=150\times\dfrac{n(n+1)}{24}\times\dfrac{8}{100}.I=P×n(n+1)/2×12×r/100=150×n(n+1)/24×8/100.

=150×824×100 n(n+1)=0.5 n(n+1).=\dfrac{150\times8}{24\times100}\,n(n+1)=0.5\,n(n+1).=150×8/24×100\,n(n+1)=0.5\,n(n+1).

Set 0.5 n(n+1)=300⇒n(n+1)=600.0.5\,n(n+1)=300\Rightarrow n(n+1)=600.0.5\,n(n+1)=300 n(n+1)=600. Since 24×25=60024\times25=60024×25=600, n=24n=24n=24 months.

Total deposit =150×24=₹3600=150\times24=\text{₹}3600=150×24=₹3600, so MV=3600+300=₹3900.\text{MV}=3600+300=\text{₹}3900.MV=3600+300=₹3900.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Mr. Mehta opens a recurring deposit account and deposits ₹2500\text{₹}2500₹2500 per month for 444 years at 6%6\%6\% per annum. Find:

(i) the total money deposited,

(ii) the interest earned,

(iii) the maturity value of the account.

Show model answer

Here P=2500P=2500P=2500, n=48n=48n=48, r=6r=6r=6.

(i) Total deposit =2500×48=₹120000.=2500\times48=\text{₹}120000.=2500×48=₹120000.

(ii) I=2500×48×492×12×6100=2500×98×0.06=₹14700.I=2500\times\dfrac{48\times49}{2\times12}\times\dfrac{6}{100}=2500\times98\times0.06=\text{₹}14700.I=2500×48×49/2×12×6/100=2500×98×0.06=₹14700.

(iii) MV=120000+14700=₹134700.\text{MV}=120000+14700=\text{₹}134700.MV=120000+14700=₹134700.

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Q12Long AnswerHOTS5 marks

A recurring deposit account matures in 222 years at a rate of 10%10\%10\% per annum and gives a maturity value of ₹13250\text{₹}13250₹13250. Find the monthly deposit.

Show model answer

Let the monthly deposit be ₹P\text{₹}P₹P, with n=24n=24n=24, r=10r=10r=10.

Interest I=P×24×252×12×10100=P×25×0.10=2.5P.I=P\times\dfrac{24\times25}{2\times12}\times\dfrac{10}{100}=P\times25\times0.10=2.5P.I=P×24×25/2×12×10/100=P×25×0.10=2.5P.

Total deposit =24P=24P=24P, so MV=24P+2.5P=26.5P.\text{MV}=24P+2.5P=26.5P.MV=24P+2.5P=26.5P.

26.5P=13250⇒P=1325026.5=500.26.5P=13250\Rightarrow P=\dfrac{13250}{26.5}=500.26.5P=13250 P=13250/26.5=500.

Monthly deposit =₹500.=\text{₹}500.=₹500.

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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

Sonia opens a recurring deposit account in a bank. She deposits ₹1200\text{₹}1200₹1200 per month for 303030 months, and the bank pays interest at 8%8\%8\% per annum.

(i) Find the total sum she deposits.

(ii) Find the equivalent number of months, n(n+1)2\dfrac{n(n+1)}{2}n(n+1)/2.

(iii) Find the interest she earns.

(iv) Find the maturity value of her account.

Show model answer

Here P=1200P=1200P=1200, n=30n=30n=30, r=8r=8r=8.

(i) Total deposit =1200×30=₹36000.=1200\times30=\text{₹}36000.=1200×30=₹36000.

(ii) n(n+1)2=30×312=465\dfrac{n(n+1)}{2}=\dfrac{30\times31}{2}=465n(n+1)/2=30×31/2=465 months.

(iii) I=1200×46512×8100=1200×38.75×0.08=₹3720.I=1200\times\dfrac{465}{12}\times\dfrac{8}{100}=1200\times38.75\times0.08=\text{₹}3720.I=1200×465/12×8/100=1200×38.75×0.08=₹3720.

(iv) MV=36000+3720=₹39720.\text{MV}=36000+3720=\text{₹}39720.MV=36000+3720=₹39720.

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All ICSE Class 10 Maths Chapters

GST (Goods and Services Tax)Banking (Recurring Deposit Accounts)Shares and DividendsLinear Inequations (In One Variable)Quadratic EquationsSolving (Simple) Problems Based on Quadratic EquationsRatio and ProportionRemainder and Factor TheoremsMatricesArithmetic ProgressionGeometric ProgressionReflectionSection and Mid-Point FormulaEquation of a LineSimilarity (With Applications to Maps and Models)Loci (Locus and Its Constructions)CirclesTangents and Intersecting ChordsConstructions (Circles)Cylinder, Cone and Sphere (Surface Area and Volume)Trigonometrical IdentitiesHeights and DistancesGraphical Representation (Histograms and Ogives)Measures of Central Tendency (Mean, Median, Quartiles and Mode)Probability

Frequently asked questions

  • Do these Banking (Recurring Deposit Accounts) questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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