MathsICSE Class 10 · Chapter 2

Banking ICSE Class 10: Recurring Deposit Accounts, Formulas and Solved Questions

Updated 14 October 2026 · Fact-checked

ICSE Class 10 Banking is about recurring deposit (RD) accounts, where a fixed sum is deposited every month. Interest is I = P × n(n + 1)/(2 × 12) × r/100, with n in months, and the maturity value is MV = P × n + I. Questions ask for the interest, maturity value, rate, instalment or time.

What Banking is about

A recurring deposit (RD) account lets a person save a fixed amount every month for a fixed number of months. The bank pays simple interest at a fixed yearly rate, and at the end of the period (on maturity) the depositor gets back everything deposited plus the interest. That final sum is the maturity value.

The CISCE 2027 syllabus for Banking is short and precise: computation of interest and maturity value on recurring deposit accounts using
$I = P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}, M V = P \times n + I,$
where $P$ is the monthly instalment, $n$ the number of months and $r %$ the rate per annum. Savings bank accounts and fixed deposits are not part of the Class 10 syllabus.

Because the formula has four quantities ( $P$ , $n$ , $r$ and $I$ or $M V$ ), board questions give three and ask for the fourth. Finding $n$ leads to a quadratic equation, so this chapter links to Quadratic Equations. The examiners' reports for 2024 and 2025 list 'Maturity Value & Interest Formula' among the topics candidates found difficult, and the errors they describe include taking the time in years instead of months and mixing up the interest and the maturity value.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Banking is in the 1. Commercial Mathematics (ii) Banking part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

How a recurring deposit earns interest

Suppose you deposit ₹ $P$ at the start of every month for $n$ months. The first instalment stays in the bank for $n$ months, the second for $(n - 1)$ months, and so on, until the last one, which stays for only 1 month.

So the bank pays simple interest as if ₹ $P$ were kept for $n + (n - 1) + \dots + 2 + 1 = \frac{n ( n + 1 )}{2}$ months. In other words, the RD earns the same interest as a single sum of
$P \times \frac{n ( n + 1 )}{2}$
kept for one month. This is called the equivalent principal for one month. Understanding this means you do not have to learn the formula by rote: it is simple interest, $\frac{P R T}{100}$ , with principal $P \times \frac{n ( n + 1 )}{2}$ and time $\frac{1}{12}$ year.

Worked example

Asha deposits ₹500 a month for 6 months in an RD account at 6% per annum. Find the equivalent principal for one month, the interest and the maturity value.

Try it first, then show the solutionSolution

Equivalent principal for one month $= 500 \times \frac{6 \times 7}{2} = 500 \times 21 = ₹10, 500$ .
Interest $= 10500 \times \frac{6}{100} \times \frac{1}{12} = ₹52.50$ .
Total deposited $= 500 \times 6 = ₹3, 000$ ; maturity value $= 3000 + 52.50 = ₹3, 052.50$ .

Answer: ₹10,500; interest ₹52.50; maturity value ₹3,052.50

The interest formula and the maturity value

Putting the idea together gives the syllabus formula:
$I = P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}$
where $P$ = monthly instalment, $n$ = number of months, $r$ = rate per annum.

The maturity value is what the depositor receives at the end:
$M V = P \times n + I .$

Two checks before you substitute: is $n$ in months (2 years means $n = 24$ , not 2)? And is $r$ the yearly rate? The 12 in the denominator already converts the yearly rate to a monthly one, so never divide by 12 again.

Worked example

Find the interest and the maturity value on an RD of ₹1,000 per month for 1 year at 9% per annum.

Try it first, then show the solutionSolution

$n = 12$ , so $\frac{n ( n + 1 )}{2 \times 12} = \frac{12 \times 13}{24} = 6.5$ .
$I = 1000 \times 6.5 \times \frac{9}{100} = ₹585$ .
$M V = 1000 \times 12 + 585 = ₹12, 585$ .

Answer: Interest ₹585; maturity value ₹12,585

Finding the rate of interest

When the maturity value is given, first separate the interest:
$I = M V - P \times n .$
Then put $I$ into the formula, which becomes a simple linear equation in $r$ .

CISCE examiners have reported candidates mixing up the interest and the maturity value (2024), and using the maturity value as the principal when finding the rate (2025). The interest is only the extra money over and above the deposits.

Worked example

Karan deposits ₹750 per month for 2 years in an RD account and receives ₹19,350 on maturity. Find the rate of interest.

Try it first, then show the solutionSolution

Total deposited $= 750 \times 24 = ₹18, 000$ , so $I = 19350 - 18000 = ₹1, 350$ .
$\frac{n ( n + 1 )}{24} = \frac{24 \times 25}{24} = 25$ .
$1350 = 750 \times 25 \times \frac{r}{100} = 187.5 r$ , so $r = \frac{1350}{187.5} = 7.2$ .

Answer: 7.2% per annum

Finding the monthly instalment

If the maturity value, the time and the rate are given, write the maturity value in terms of $P$ :
$M V = P [n + \frac{n ( n + 1 )}{24} \times \frac{r}{100}] .$
The bracket is just a number, so $P = M V \div$ (that number).

Do not try to find the interest first: you cannot, because both the interest and the deposits depend on $P$ .

Worked example

What monthly instalment in an RD for 12 months at 8% per annum gives a maturity value of ₹15,650?

Try it first, then show the solutionSolution

Interest part per rupee of instalment $= \frac{12 \times 13}{24} \times \frac{8}{100} = 6.5 \times 0.08 = 0.52$ .
$M V = P (12 + 0.52) = 12.52 P$ .
$12.52 P = 15650$ , so $P = ₹1, 250$ .

Answer: ₹1,250 per month

Finding the time: a quadratic in n

If $P$ , $r$ and the interest are given, the formula gives an equation in $n (n + 1)$ . Expanding gives a quadratic, $n^{2} + n - k = 0$ . Solve it by factorisation (look for two consecutive whole numbers whose product is $k$ ), reject the negative root, and remember that $n$ is in months.

The CISCE examiners' comments, 2024, note that candidates found 12 months but did not convert it to 1 year when the question asked for the time in years. Always answer in the unit the question asks for.

Worked example

Pooja deposits ₹400 per month in an RD at 9% per annum and earns ₹900 as interest. For how many years was the account held?

Try it first, then show the solutionSolution

$900 = 400 \times \frac{n ( n + 1 )}{24} \times \frac{9}{100} = 1.5 n (n + 1)$ .
$n (n + 1) = 600$ , so $n^{2} + n - 600 = 0$ .
$(n + 25) (n - 24) = 0$ , so $n = 24$ (reject $n = - 25$ ).
24 months $= 2$ years.

Answer: 2 years

Comparing deposits: what changes the interest

From $I = P \times \frac{n ( n + 1 )}{24} \times \frac{r}{100}$ :

Interest is proportional to $P$ : double the instalment and the interest doubles.
Interest is proportional to $r$ .
Interest grows roughly with $n^{2}$ : doubling the time much more than doubles the interest, because there are more deposits and each stays longer.

This is why an RD for a longer time at a slightly lower rate can earn more than a short one at a higher rate, even when the total deposited is the same. Questions that compare two plans test exactly this.

Worked example

On ₹1,000 a month at 6% per annum, compare the interest for 12 months and for 24 months.

Try it first, then show the solutionSolution

12 months: $I = 1000 \times 6.5 \times 0.06 = ₹390$ .
24 months: $I = 1000 \times 25 \times 0.06 = ₹1, 500$ .
$\frac{1500}{390} \approx 3.85$ , so doubling the time almost quadruples the interest.

Answer: ₹390 and ₹1,500; the interest is about 3.85 times as much.

Maths in real life: Banking

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Interest and maturity value

Suppose you decide to save ₹300 of your pocket money every month in a recurring deposit for 2 years. Say the bank pays 7% per annum.

How much interest will you earn, and how much will you get back at the end?

Try it first, then show the solutionSolution

$n = 24$ months, so $\frac{n ( n + 1 )}{24} = 25$ .
$I = 300 \times 25 \times \frac{7}{100} = ₹525$ .
Total deposited $= 300 \times 24 = ₹7, 200$ ; $M V = 7200 + 525 = ₹7, 725$ .

Answer: Interest ₹525; you get back ₹7,725.

Memory hook: Pocket money in, pocket money plus a little extra out: MV = Pn + I.

Finding the monthly instalment for a target

Imagine your family plans to buy a scooter costing ₹1,00,000 in 3 years and wants to save for it in an RD at 8% per annum.

What is the smallest whole-rupee monthly instalment that gives a maturity value of at least ₹1,00,000?

Try it first, then show the solutionSolution

$n = 36$ , so $\frac{n ( n + 1 )}{24} = 55.5$ , and the interest factor is $55.5 \times 0.08 = 4.44$ .
$M V = P (36 + 4.44) = 40.44 P$ .
$40.44 P \geq 100000$ gives $P \geq 2472.8$ .
So the smallest whole-rupee instalment is ₹2,473 (₹2,472 would give only ₹99,967.68).

Answer: ₹2,473 per month

Memory hook: Working back from the goal: divide the target by (n + interest factor) and round up.

Why the bank beats a piggy bank

Suppose your cousin puts ₹1,000 a month into a piggy bank for a year, while you put the same amount into an RD paying 6.5% per annum.

How much more money do you have at the end of the year?

Try it first, then show the solutionSolution

The piggy bank holds $1000 \times 12 = ₹12, 000$ , with no interest.
RD interest $= 1000 \times 6.5 \times \frac{6.5}{100} = ₹422.50$ .
Your maturity value $= 12000 + 422.50 = ₹12, 422.50$ .

Answer: ₹422.50 more

Memory hook: Same savings, same time: the RD adds interest for every month each deposit waits.

Topics in this chapter

Banking formula sheet

Interest on a recurring deposit: $I = P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}$; P = monthly instalment, n = number of months, r = rate per annum.
Maturity value: $M V = P \times n + I$; Total deposited plus interest.
Interest from the maturity value: $I = M V - P n$; Always the first step when MV is given.
Equivalent principal for one month: $P \times \frac{n ( n + 1 )}{2}$; The single sum that earns the same interest in one month.
Rate of interest: $r = \frac{I \times 2400}{P \times n ( n + 1 )}$; Rearranged from the interest formula.
Monthly instalment: $P = \frac{M V}{n + \frac{n ( n + 1 )}{24} \times \frac{r}{100}}$; Use when MV, n and r are given.
Useful values of n(n+1)/24: $n = 12 : 6.5; n = 18 : 14.25; n = 24 : 25; n = 36 : 55.5$; For 1, 1.5, 2 and 3 years.

How the board asks Banking

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ on the meaning of P, n, I and MV	Section A, Question 1: choose the value of n for a given period, or the expression for the maturity value.	n is always in months. MV = total deposited + interest; no rate is needed if the interest is already given.
Assertion-reason on maturity value	An assertion gives a maturity value or an interest; the reason states a relation such as Interest = MV − total deposited.	Compute $P \times n$ and subtract. Check the reason as a general rule, then decide whether it explains the assertion.
Find the interest and maturity value (3 marks)	P, time and rate given.	Convert the time to months, evaluate $\frac{n ( n + 1 )}{24}$ first, then multiply by P and $\frac{r}{100}$ .
Find the rate (4 marks)	P, time and maturity value given, often with a second part changing P or the time.	Interest = MV − Pn, then solve the linear equation for r.
Find the monthly instalment	Maturity value, time and rate given.	Write MV = P × (n + interest factor) and divide.
Find the time	P, r and the interest (or MV) given; answer in months or years.	Form $n^{2} + n - k = 0$ , factorise, reject the negative root, convert to the unit asked.
Compare two plans	Two people or two schemes with different instalments, times or rates; which gives more interest, or how much more?	Work each plan separately with the formula, then compare. Explain the result in words if asked.

Tricks to remember Banking

Short memory aids. Say each one aloud once, then use it on a question the same day.

n is for 'number of months'
In the RD formula, n never means years. 1 year → 12, 2 years → 24, 3 years → 36.
Use it when: Every RD question; the single biggest source of lost marks.
Triangle of deposits
The deposits stay 1, 2, 3, …, n months: the triangle number $\frac{n ( n + 1 )}{2}$ . Then divide by 12 to turn months into years.
Use it when: Rebuilding the formula if you forget it.
Magic numbers 6.5, 25, 55.5
$\frac{n ( n + 1 )}{24}$ is 6.5 for 1 year, 25 for 2 years and 55.5 for 3 years.
Use it when: Fast, error-free substitution.
Interest is the extra
MV is the whole basket; the deposits are what you put in; the interest is only the extra.
Use it when: Never use MV as the interest or as the principal.

Banking: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Interest and maturity value · 1 mark · MCQBoard level

Ravi deposits ₹600 per month in a recurring deposit account for 1 year at 8% per annum. His maturity value is:
(a) ₹7,200
(b) ₹7,512
(c) ₹7,776
(d) ₹7,824

Try it first, then show the solutionSolution

$I = 600 \times \frac{12 \times 13}{24} \times \frac{8}{100} = 600 \times 6.5 \times 0.08 = ₹312$ .
$M V = 600 \times 12 + 312 = ₹7, 512$ .

Answer: (b) ₹7,512

Example 2: Total deposited · 1 mark · Assertion-reasonBoard level

Assertion (A): A person deposits ₹800 per month in a recurring deposit account for 2 years. The total amount deposited is ₹19,200.
Reason (R): The total amount deposited in an RD account is P × n, where n is the number of months.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

Try it first, then show the solutionSolution

$n = 24$ , so $P \times n = 800 \times 24 = ₹19, 200$ . (A) is true.
(R) is the rule used, so it is true and explains (A).

Answer: (c)

Example 3: Find the rate · 4 marks · Short answerHarder board level

Neha deposits ₹1,500 per month in a recurring deposit account for 2 years. At maturity she receives ₹38,250. Find the rate of interest per annum.

Try it first, then show the solutionSolution

Total deposited $= 1500 \times 24 = ₹36, 000$ .
Interest $= 38250 - 36000 = ₹2, 250$ .
$2250 = 1500 \times \frac{24 \times 25}{24} \times \frac{r}{100} = 375 r$ .
$r = \frac{2250}{375} = 6$ .

Answer: 6% per annum

Example 4: Find the monthly instalment · 4 marks · Short answerHarder board level

Mr Sen wants a maturity value of ₹48,195 from a recurring deposit for 3 years at 7.5% per annum. Find his monthly instalment.

Try it first, then show the solutionSolution

$n = 36$ ; $\frac{36 \times 37}{24} = 55.5$ .
Interest $= P \times 55.5 \times \frac{7.5}{100} = 4.1625 P$ .
$M V = 36 P + 4.1625 P = 40.1625 P$ .
$40.1625 P = 48195$ , so $P = ₹1, 200$ .

Answer: ₹1,200 per month

Example 5: Find the time · 4 marks · Short answerHarder board level

Farhan deposits ₹2,500 per month in a recurring deposit account at 6% per annum. He earns ₹3,750 as interest. Find the time, in years, for which the account was held.

Try it first, then show the solutionSolution

$3750 = 2500 \times \frac{n ( n + 1 )}{24} \times \frac{6}{100} = 6.25 n (n + 1)$ .
$n (n + 1) = 600$ , so $n^{2} + n - 600 = 0$ .
$(n + 25) (n - 24) = 0$ ; $n = 24$ (months cannot be negative).
24 months $= 2$ years.

Answer: 2 years

Example 6: Eighteen-month RD · 3 marks · Long answerBoard level

Find the interest and the maturity value for a recurring deposit of ₹2,000 per month for 1½ years at 9% per annum.

Try it first, then show the solutionSolution

$n = 18$ ; $\frac{18 \times 19}{24} = 14.25$ .
$I = 2000 \times 14.25 \times \frac{9}{100} = ₹2, 565$ .
$M V = 2000 \times 18 + 2565 = ₹38, 565$ .

Answer: Interest ₹2,565; maturity value ₹38,565

Example 7: Lower rate, more interest? · 4 marks · Long answerHOTS / competency

Plan A: ₹3,000 per month for 2 years at 7% per annum. Plan B: ₹2,000 per month for 3 years at 6.5% per annum.
(a) Show that the total amount deposited is the same in both plans.
(b) Find the interest in each plan.
(c) Plan B has the lower rate. Explain why it still gives more interest.

Try it first, then show the solutionSolution

(a) A: $3000 \times 24 = ₹72, 000$ ; B: $2000 \times 36 = ₹72, 000$ . Same.
(b) A: $I = 3000 \times 25 \times 0.07 = ₹5, 250$ . B: $I = 2000 \times 55.5 \times 0.065 = ₹7, 215$ .
(c) The interest depends on how long each rupee stays in the bank. In Plan B the deposits are spread over 36 months, so on average each rupee earns interest for longer. The factor $\frac{n ( n + 1 )}{2}$ grows faster than $n$ , which outweighs the lower rate.

Answer: (a) ₹72,000 each (b) A: ₹5,250; B: ₹7,215 (c) Money stays longer in Plan B.

Example 8: Rate, then a bigger instalment · 4 marks · Long answerHarder board level

Riya deposits ₹400 per month for 4 years in a recurring deposit account and receives ₹22,336 on maturity.
(a) Find the rate of interest.
(b) How much more interest would she have earned if she had deposited ₹100 more per month at the same rate for the same time?

Try it first, then show the solutionSolution

$n = 48$ ; $\frac{48 \times 49}{24} = 98$ . Total deposited $= 400 \times 48 = ₹19, 200$ , so $I = 22336 - 19200 = ₹3, 136$ .
(a) $3136 = 400 \times 98 \times \frac{r}{100} = 392 r$ , so $r = 8%$ .
(b) With ₹500: $I = 500 \times 98 \times 0.08 = ₹3, 920$ .
Extra interest $= 3920 - 3136 = ₹784$ (the same as the interest on ₹100 a month: $100 \times 98 \times 0.08 = 784$ ).

Answer: (a) 8% (b) ₹784 more

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

A family saving for a holiday

The Iyer family in Coimbatore plans a holiday in 15 months. Each member opens a recurring deposit account for 15 months in a bank that pays 8% per annum. They want the three maturity values together to pay for the trip.

(i) · 3 marks
Mrs Iyer deposits ₹1,500 per month. Find the interest she earns and her maturity value.
Answer
Answer: Interest ₹1,200; MV ₹23,700
$\frac{15 \times 16}{24} = 10$ . $I = 1500 \times 10 \times 0.08 = ₹1, 200$ . $M V = 1500 \times 15 + 1200 = ₹23, 700$ .
(ii) · 3 marks
Their son receives ₹15,800 on maturity. Find his monthly instalment.
Answer
Answer: ₹1,000
$M V = P (15 + 10 \times 0.08) = 15.8 P = 15800$ , so $P = ₹1, 000$ .
(iii) · 4 marks
The trip costs ₹47,400. Mr Iyer opens his RD on the same terms. What monthly instalment must he deposit so that the three maturity values exactly cover the trip? Find the interest he earns.
Answer
Answer: ₹500 per month; interest ₹400
Mr Iyer needs $47400 - 23700 - 15800 = ₹7, 900$ . $15.8 P = 7900$ gives $P = ₹500$ . His interest $= 500 \times 10 \times 0.08 = ₹400$ .

Comparing recurring deposits at one bank

A bank in Nagpur pays 7% per annum on all recurring deposit accounts. Three customers, Ankit, Bela and Chirag, hold RD accounts there for different periods and with different instalments.

(i) · 3 marks
Ankit deposits ₹2,400 per month for 2 years. Find his interest and maturity value.
Answer
Answer: Interest ₹4,200; MV ₹61,800
$I = 2400 \times 25 \times 0.07 = ₹4, 200$ . $M V = 2400 \times 24 + 4200 = ₹61, 800$ .
(ii) · 3 marks
Bela also deposits ₹2,400 per month and earns ₹1,680 as interest. For how many months did she hold the account?
Answer
Answer: 15 months
$1680 = 2400 \times \frac{n ( n + 1 )}{24} \times 0.07 = 7 n (n + 1)$ , so $n (n + 1) = 240$ , $n^{2} + n - 240 = 0$ , $(n + 16) (n - 15) = 0$ , $n = 15$ .
(iii) · 4 marks
Chirag holds his account for 2 years and receives ₹77,250. Find his monthly instalment. Ankit says, 'Chirag deposits 25% more than me each month, so he must earn exactly 25% more interest.' Check whether Ankit is right.
Answer
Answer: ₹3,000 per month; Ankit is right (₹5,250 is 25% more than ₹4,200).
$M V = P (24 + 25 \times 0.07) = 25.75 P = 77250$ , so $P = ₹3, 000$ . $3000 = 1.25 \times 2400$ , and Chirag's interest $= 3000 \times 25 \times 0.07 = ₹5, 250 = 1.25 \times 4200$ . With the same time and rate, interest is proportional to the instalment, so Ankit is right.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the interest and the maturity value on a recurring deposit of ₹1,200 per month for 15 months at 8% per annum. (3 marks)

Step	Marks
$n = 15$ ; $I = 1200 \times \frac{15 \times 16}{2 \times 12} \times \frac{8}{100}$ (correct substitution)	1
$I = 1200 \times 10 \times 0.08 = ₹960$	1
$M V = 1200 \times 15 + 960 = ₹18, 960$	1

₹800 is deposited per month for 3 years in an RD account and the maturity value is ₹32,352. Find the rate of interest. (4 marks)

Step	Marks
Total deposited $= 800 \times 36 = ₹28, 800$	1
Interest $= 32352 - 28800 = ₹3, 552$	1
$3552 = 800 \times \frac{36 \times 37}{24} \times \frac{r}{100} = 444 r$	1
$r = 8%$	1

A man deposits ₹1,000 per month in an RD at 12% per annum and earns ₹2,100 as interest. Find the time for which the account was held and the maturity value. (5 marks)

Step	Marks
$2100 = 1000 \times \frac{n ( n + 1 )}{24} \times \frac{12}{100}$	1
$n (n + 1) = 420$ , i.e. $n^{2} + n - 420 = 0$	1
$(n + 21) (n - 20) = 0$	1
$n = 20$ months (reject $- 21$ ), i.e. 1 year 8 months	1
$M V = 1000 \times 20 + 2100 = ₹22, 100$	1

Where students lose marks in Banking

The slips examiners see most often, and what to write instead.

Mistake: Taking the time in years instead of months
CISCE examiners' comments, 2025: candidates took the time as 3 years instead of 36 months when substituting into the formula.
Wrong working
₹250 a month for 3 years: $I = 250 \times \frac{3 \times 4}{24} \times \frac{r}{100}$
Right working
$n = 36$ : $I = 250 \times \frac{36 \times 37}{24} \times \frac{r}{100} = 250 \times 55.5 \times \frac{r}{100}$
Do this: Write ' $n = \dots$ months' as the first line of every RD answer.
Usually costs: Every mark after the substitution.
Mistake: Using the maturity value as the interest or the principal
CISCE examiners' comments, 2024 and 2025: candidates took ₹1,040 (the interest) as the maturity value, or calculated the rate using the maturity value as the principal.
Do this: Label each given figure before you start: 'P = …, n = …, MV = …' or 'I = …'. If MV is given, the first line is $I = M V - P n$ .
Usually costs: 1 to 2 marks in a 3- or 4-mark question.
Mistake: Using the simple interest formula or a wrong RD formula
CISCE examiners' comments, 2024 and 2025: candidates used an incorrect formula for interest on a recurring deposit, or applied plain simple interest on the total deposit.
Wrong working
$I = \frac{( 500 \times 18 ) \times 8 \times 1.5}{100}$
Right working
$I = 500 \times \frac{18 \times 19}{24} \times \frac{8}{100}$
Do this: Learn the formula with its meaning (each deposit earns for a different number of months). Rebuild it if unsure.
Usually costs: Usually all the method marks.
Mistake: Slips while solving $n^{2} + n - k = 0$
CISCE examiners' comments, 2024: errors in forming or solving the quadratic, such as $n^{2} + n - 156 = 0$ .
Do this: Find two consecutive numbers whose product is k (e.g. 12 × 13 = 156). Factorise as $(n + 13) (n - 12)$ and check by expanding.
Usually costs: 1 to 2 marks.
Mistake: Answering in the wrong unit
CISCE examiners' comments, 2024: candidates left the answer as 12 months when the question asked for the time in years.
Do this: Reread the question's last line and convert: 12 months = 1 year, 20 months = 1 year 8 months.
Usually costs: The final answer mark.
Mistake: Not finishing a comparison
CISCE examiners' comments, 2025: candidates found the interest for the bigger instalment (₹1,332) but did not subtract to find how much more interest was earned.
Do this: If the question says 'how much more', the last line must be a subtraction.
Usually costs: 1 mark.

Practice questions on Banking

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
For a recurring deposit held for 2 years, the value of n used in the formula is:
(a) 2
(b) 24
(c) 25
(d) 300
Hint
n counts months.
Answer
(b) 24
Q2 · 1 mark · MCQBoard level
A woman deposits a total of ₹30,000 in a recurring deposit account and receives ₹32,250 on maturity. The interest she earns is:
(a) ₹32,250
(b) ₹2,250
(c) ₹62,250
(d) ₹1,125
Hint
Interest is the extra over the deposits.
Answer
(b) ₹2,250
Q3 · 1 mark · MCQHarder board level
The interest on an RD of ₹800 per month for 1 year at 6% per annum is:
(a) ₹288
(b) ₹312
(c) ₹576
(d) ₹624
Hint
Use 6.5 for one year.
Answer
(b) ₹312, since $800 \times 6.5 \times 0.06 = 312$ .
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): The interest on an RD of ₹1,000 per month for 12 months at 12% per annum is ₹780.
Reason (R): The interest on an RD is $P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Compute with the formula in (R).
Answer
(c). $1000 \times 6.5 \times 0.12 = 780$ , using exactly the formula in (R).
Q5 · 4 marks · Short answerHarder board level
Mrs Kaur deposits ₹1,200 per month for 3 years in a recurring deposit account and receives ₹49,194 at maturity. Find the rate of interest.
Hint
n = 36 and $\frac{n ( n + 1 )}{24} = 55.5$ .
Answer
Deposited $= 1200 \times 36 = ₹43, 200$ ; $I = 49194 - 43200 = ₹5, 994$ . $5994 = 1200 \times 55.5 \times \frac{r}{100} = 666 r$ , so $r = 9%$ .
Q6 · 4 marks · Short answerHarder board level
Find the monthly instalment of a recurring deposit for 18 months at 10% per annum if the maturity value is ₹19,425.
Hint
$\frac{18 \times 19}{24} = 14.25$ .
Answer
$M V = P (18 + 14.25 \times 0.1) = 19.425 P = 19425$ , so $P = ₹1, 000$ .
Q7 · 3 marks · Long answerBoard level
Calculate the interest and the maturity value on an RD of ₹1,250 per month for 2½ years at 8% per annum.
Hint
n = 30.
Answer
$\frac{30 \times 31}{24} = 38.75$ . $I = 1250 \times 38.75 \times 0.08 = ₹3, 875$ . $M V = 1250 \times 30 + 3875 = ₹41, 375$ .
Q8 · 4 marks · Long answerHarder board level
Sameer deposits ₹3,000 per month in an RD at 8% per annum and earns ₹4,200 as interest. Find the time in years and months, and his maturity value.
Hint
You should get $n (n + 1) = 420$ .
Answer
$4200 = 3000 \times \frac{n ( n + 1 )}{24} \times 0.08 = 10 n (n + 1)$ , so $n^{2} + n - 420 = 0$ , $(n + 21) (n - 20) = 0$ , $n = 20$ months = 1 year 8 months. $M V = 3000 \times 20 + 4200 = ₹64, 200$ .
Q9 · 4 marks · Long answerHOTS / competency
On ₹1,000 a month at 6% per annum, find the interest for (a) 12 months, (b) 24 months, (c) ₹2,000 a month for 12 months. Use your answers to explain why doubling the time does not simply double the interest, while doubling the instalment does.
Hint
Compare $\frac{n ( n + 1 )}{24}$ for n = 12 and n = 24.
Answer
(a) $1000 \times 6.5 \times 0.06 = ₹390$ (b) $1000 \times 25 \times 0.06 = ₹1, 500$ (c) $2000 \times 6.5 \times 0.06 = ₹780$ . Interest is proportional to P, so doubling P doubles it (₹780). But the factor $\frac{n ( n + 1 )}{24}$ goes from 6.5 to 25 when n doubles, because there are twice as many deposits and each stays longer on average, so the interest becomes about 3.85 times as much.
Q10 · 4 marks · Long answerHarder board level
Divya deposits ₹600 per month for 2 years and receives ₹15,660 at maturity. (a) Find the rate of interest. (b) Find her maturity value if she had continued for 3 years at the same rate.
Hint
Interest first, then r; for (b) use 55.5.
Answer
(a) $I = 15660 - 14400 = ₹1, 260$ ; $1260 = 600 \times 25 \times \frac{r}{100} = 150 r$ , so $r = 8.4%$ .
(b) $I = 600 \times 55.5 \times 0.084 = ₹2, 797.20$ ; $M V = 21600 + 2797.20 = ₹24, 397.20$ .
Q11 · 1 mark · MCQHarder board level
The interest on an RD of ₹ $P$ per month for $n$ months is the same as the simple interest on ₹ $P \times \frac{n ( n + 1 )}{2}$ for:
(a) 1 month
(b) n months
(c) 1 year
(d) n years
Hint
Think of the equivalent principal.
Answer
(a) 1 month
Q12 · 4 marks · Short answerHOTS / competency
A man wants a maturity value of at least ₹50,000 from an RD for 2 years at 8% per annum. Find the least monthly instalment if instalments must be in whole hundreds of rupees, and his maturity value with that instalment.
Hint
MV = 26P; then round up to the next hundred.
Answer
$M V = P (24 + 25 \times 0.08) = 26 P \geq 50000$ , so $P \geq 1923.08$ . The least whole-hundred instalment is ₹2,000 (₹1,900 gives only ₹49,400). MV $= 26 \times 2000 = ₹52, 000$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the formula for the interest on a recurring deposit.Show answer
$I = P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}$
2.What does n stand for in the RD formula?Show answer
The number of months (instalments).
3.Write the maturity value in terms of P, n and I.Show answer
$M V = P \times n + I$
4.What is $\frac{n ( n + 1 )}{24}$ for 2 years?Show answer
25
5.How do you find the interest when the maturity value is given?Show answer
Subtract the total deposited: $I = M V - P n$ .
6.Why does the formula have $\frac{n ( n + 1 )}{2}$ ?Show answer
The deposits stay 1, 2, …, n months; their total is $\frac{n ( n + 1 )}{2}$ months of one instalment.
7.If the instalment is doubled, what happens to the interest?Show answer
It doubles (same time and rate).
8.Finding the time gives what kind of equation?Show answer
A quadratic, $n^{2} + n - k = 0$ .

Best order to study it

Understand why each instalment earns interest for a different number of months, and how this gives $\frac{n ( n + 1 )}{2}$ .
Practise straightforward interest and maturity value questions for 1, 1½, 2 and 3 years.
Practise finding the rate from the maturity value.
Practise finding the instalment, then the time (quadratic).
Finish with comparison and two-part questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the RD interest formula and say what P, n and r stand for?
Do you always convert the time to months?
Can you find the maturity value and, from it, the interest?
Can you find the rate when the maturity value is given?
Can you find the monthly instalment for a target maturity value?
Can you form and solve the quadratic to find the time, and give it in the unit asked?
Can you compare two RD plans and explain which earns more interest?

Banking: frequently asked questions

Are savings bank accounts and fixed deposits in the ICSE Class 10 Banking chapter?

No. The 2027 CISCE syllabus for Banking covers only recurring deposit accounts: the interest and the maturity value using the given formula.

Why is there a 12 in the RD formula?

The rate r is per annum, but each instalment earns interest for some number of months. Dividing by 12 converts months into years, so do not divide by 12 again.

Is the RD interest simple or compound?

For the ICSE formula it is simple interest: each instalment earns simple interest for the months it stays in the bank.

How do I find the time in an RD question?

Substitute P, r and I into the formula to get n(n+1) = k, solve the quadratic n^2 + n - k = 0 by factorisation, reject the negative value, and convert months to years if asked.

Do I need to show working in the MCQs?

No. In Section A Question 1 you write only the correct option; do the working in rough.

Based on: CISCE ICSE 2027 Mathematics syllabus; CISCE ICSE 2027 Mathematics specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2025; CISCE Analysis of Pupil Performance, ICSE Mathematics, 2024