MathsICSE Class 10 · Banking

Rate of Interest on a Recurring Deposit: Method and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

To find the rate on a recurring deposit, first get the interest: I = maturity value − P × n. Then rearrange the interest formula: r = (I × 100 × 2 × 12)/(P × n(n + 1)), where P is the monthly instalment and n the number of months. The answer is a percentage per annum.

Subtract the total deposited from the maturity value to get interest, then divide by P×n(n+1)/24 and multiply by 100; never put MV into the formula.

Illustration drawn for QuestionBuddy.

Understand Rate of Interest on a Recurring Deposit

A typical ICSE banking question tells you the monthly instalment $P$ , the time and the maturity value, and asks for the rate of interest. It is the recurring deposit formula used backwards.

Step 1: find the interest. The maturity value is everything deposited plus interest, so

$I = M V - P \times n .$

The examiners reported, for 2025 Q2(ii), candidates who used the maturity value itself as the principal. The maturity value is never put into the interest formula.

Step 2: rearrange the formula. From $I = P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}$ ,

$r = \frac{I \times 100 \times 2 \times 12}{P \times n ( n + 1 )} = \frac{2400 I}{P n ( n + 1 )} .$

It is often easier to first compute $P \times \frac{n ( n + 1 )}{24}$ (the interest at 100%, in effect), and then

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

For example, ₹600 a month for 2 years gives $600 \times 25 = 15000$ . If the interest is ₹1,350, the rate is $\frac{1350}{15000} \times 100 = 9%$ .

Units. $n$ is in months (3 years = 36 months), and the rate comes out per cent per annum, because the 12 in the formula has already converted months to years.

Follow-up parts. Once the rate is known, the question may ask for the interest on a different instalment or time at the same rate. Since $I$ is proportional to $P$ , an extra ₹50 a month earns extra interest $50 \times \frac{n ( n + 1 )}{24} \times \frac{r}{100}$ ; you can find it directly without redoing the whole calculation. In 2025 Q2(ii), the examiners reported candidates who found the new interest but did not subtract to get the difference that was asked.

Sense check. Bank rates in ICSE questions are usually whole numbers or halves, such as 6%, 7.5% or 9%. An answer like 47% or 0.8% means a slip, usually years used for $n$ or the maturity value used as interest.

Rate of Interest on a Recurring Deposit: formulas to remember

Interest from the maturity value: $I = M V - P \times n$; Always the first step.
Rate of interest: $r = \frac{I \times 100 \times 2 \times 12}{P \times n ( n + 1 )}$; $r$ is per cent per annum; $n$ in months.
Rate using the 'interest at 1%': $r = \frac{I}{P \times \frac{n ( n + 1 )}{24} \times \frac{1}{100}}$; Divide the interest by the interest that 1% would give.
Recurring deposit interest: $I = P \times \frac{n ( n + 1 )}{2 \times 12} \times \frac{r}{100}$; The formula being rearranged.
Extra interest for an extra instalment amount: $Δ I = Δ P \times \frac{n ( n + 1 )}{24} \times \frac{r}{100}$; Same time and rate.

How to solve Rate of Interest on a Recurring Deposit questions

Two short steps, interest then rate, followed by any further parts.

1Write $P$ , $n$ (in months) and the maturity value.
2Total deposited $= P \times n$ ; interest $I = M V - P \times n$ .
3Compute $\frac{n ( n + 1 )}{24}$ and then $P \times \frac{n ( n + 1 )}{24}$ .
4Rate $r = \frac{I}{P \times \frac{n ( n + 1 )}{24}} \times 100$ per cent per annum.
5Sense check: is $r$ a sensible bank rate (roughly 4% to 12%)?
6For a follow-up with a different instalment, use the same $r$ and $n$ ; for a difference, subtract.

Rate of Interest on a Recurring Deposit in real life

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.

Rate from a passbook

Suppose your aunt's passbook shows that she paid ₹1,500 a month into a recurring deposit for 18 months and received ₹28,710 at maturity.

What rate of interest did the bank pay?

Try it first, then show the solutionSolution

Deposited $= 1500 \times 18 = 27000$ ; interest $= 28710 - 27000 = 1710$ .
$\frac{18 \times 19}{24} = 14.25$ ; $1500 \times 14.25 = 21375$ .
$r = \frac{1710}{21375} \times 100 = 8%$ .

Answer: 8% per annum.

Memory hook: Take out what you put in; what is left is interest, and interest tells you the rate.

Checking a claim about a rate

Imagine a friend says his recurring deposit of ₹1,000 a month for one year 'at 10%' paid him ₹12,650.

Is his claim about the rate correct?

Try it first, then show the solutionSolution

Interest $= 12650 - 12000 = 650$ .
$r = \frac{650}{1000 \times 6.5} \times 100 = 10%$ .

Answer: Yes, the rate was 10% per annum.

Memory hook: For one year, the interest at r% is $6.5 \times P \times \frac{r}{100}$ .

Tricks to remember Rate of Interest on a Recurring Deposit

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

2400 on top
$r = \frac{2400 I}{P n ( n + 1 )}$ : 2 × 12 × 100 = 2400 goes on top with the interest.
Use it when: Writing the rearranged formula quickly.
Out minus in
Interest = what comes out (MV) − what went in ( $P \times n$ ).
Use it when: The first line of every rate question.
Sensible-rate alarm
If your rate is above 15% or below 2%, stop and look for months-versus-years or MV-versus-I slips.
Use it when: Catching errors before moving on.

Rate of Interest on a Recurring Deposit: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Rate for a one-year deposit · 1 mark · MCQBoard level

A recurring deposit of ₹1,000 per month for 1 year earns ₹520 as interest. The rate of interest per annum is:
(a) 6%
(b) 7%
(c) 8%
(d) 9%

Try it first, then show the solutionSolution

$P \times \frac{12 \times 13}{24} = 1000 \times 6.5 = 6500$ .
$r = \frac{520}{6500} \times 100 = 8%$ .

Answer: (c) 8%

Example 2: Rate from the maturity value · 3 marks · Short answerBoard level

Sameer deposits ₹600 per month in a recurring deposit for 2 years and receives ₹15,750 at maturity. Find the rate of interest.

Try it first, then show the solutionSolution

Deposited $= 600 \times 24 = 14400$ ; $I = 15750 - 14400 = 1350$ .
$\frac{24 \times 25}{24} = 25$ ; $600 \times 25 = 15000$ .
$r = \frac{1350}{15000} \times 100 = 9%$ .

Answer: 9% per annum

Example 3: A three-year deposit · 3 marks · Short answerHarder board level

A recurring deposit of ₹750 per month for 3 years matures to ₹30,330. Find the rate of interest.

Try it first, then show the solutionSolution

$n = 36$ ; deposited $= 750 \times 36 = 27000$ ; $I = 30330 - 27000 = 3330$ .
$\frac{36 \times 37}{24} = 55.5$ ; $750 \times 55.5 = 41625$ .
$r = \frac{3330}{41625} \times 100 = 8%$ .

Answer: 8% per annum

Example 4: Rate, then the effect of a bigger instalment · 4 marks · Short answerHarder board level

Neha deposits ₹400 per month for 18 months in a recurring deposit and receives ₹7,884 at maturity. Find (a) the rate of interest, (b) how much more interest she would have earned if she had deposited ₹500 per month instead.

Try it first, then show the solutionSolution

Deposited $= 400 \times 18 = 7200$ ; $I = 7884 - 7200 = 684$ .
$\frac{18 \times 19}{24} = 14.25$ ; $400 \times 14.25 = 5700$ .
(a) $r = \frac{684}{5700} \times 100 = 12%$ .
(b) Extra instalment ₹100: extra interest $= 100 \times 14.25 \times \frac{12}{100} = 171$ .

Answer: (a) 12% per annum (b) ₹171 more

Example 5: Using the rate from one deposit for another · 4 marks · Short answerHOTS / competency

At a bank, a recurring deposit of ₹800 per month for 1 year earns ₹390 as interest. At the same rate, find the maturity value of a recurring deposit of ₹1,200 per month for 2 years.

Try it first, then show the solutionSolution

$r = \frac{390}{800 \times 6.5} \times 100 = \frac{390}{5200} \times 100 = 7.5%$ .
New deposit: $I = 1200 \times 25 \times \frac{7.5}{100} = 2250$ .
$M V = 1200 \times 24 + 2250 = 28800 + 2250 = 31050$ .

Answer: Rate 7.5% per annum; maturity value ₹31,050.

Example 6: Case study: choosing a bank · 4 marks · Case-basedHOTS / competency

Mr Das wants to deposit ₹2,000 a month for 2 years. Bank A says such a deposit would mature to ₹52,500. Bank B offers 8.5% per annum.
(a) Find Bank A's rate of interest.
(b) Find the maturity value at Bank B.
(c) Which bank should he choose, and how much more will he get?

Try it first, then show the solutionSolution

Deposited $= 2000 \times 24 = 48000$ ; Bank A interest $= 52500 - 48000 = 4500$ .
(a) $r = \frac{4500}{2000 \times 25} \times 100 = 9%$ .
(b) Bank B: $I = 2000 \times 25 \times \frac{8.5}{100} = 4250$ ; $M V = 48000 + 4250 = 52250$ .
(c) Bank A gives $52500 - 52250 = 250$ more.

Answer: (a) 9% (b) ₹52,250 (c) Bank A, by ₹250.

Where students lose marks in Rate of Interest on a Recurring Deposit

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

Mistake: Putting the maturity value in place of the interest
Only the interest part of the maturity value comes from the rate.
Wrong working
$r = \frac{2400 \times 15750}{600 \times 24 \times 25}$ .
Right working
$I = 15750 - 14400 = 1350$ ; $r = \frac{2400 \times 1350}{600 \times 600} = 9$ .
Do this: Write $I = M V - P n$ before any rate working.
Usually costs: The examiners named this error for 2025 Q2(ii).
Mistake: Taking n in years
n counts monthly instalments.
Wrong working
3 years: $n (n + 1) = 3 \times 4 = 12$ .
Right working
$n = 36$ : $n (n + 1) = 36 \times 37 = 1332$ .
Do this: Convert to months on the first line.
Usually costs: All the marks; reported for 2025 Q2(ii).
Mistake: Finding the new interest but not the difference
The follow-up often asks 'how much more'.
Wrong working
Interest with ₹500 a month $= 855$ . (Stops.)
Right working
Extra interest $= 855 - 684 = 171$ .
Do this: Re-read the last line of the question before writing the answer.
Usually costs: The final mark.

Exam tips

Write three lines: deposited, interest, rate. Each earns a mark in a 3- or 4-mark question.
In 2025 Q2(ii) (₹250 a month for 3 years, maturity ₹10,110), the expected points were interest ₹1,110 and rate 8%; the examiners reported candidates who took the time as 3 instead of 36 months and who used the maturity value as the principal.
Keep fractions until the end; $\frac{684}{5700}$ simplifies exactly to 0.12.
State the rate as '…% per annum'.

Practice questions on Rate of Interest on a Recurring Deposit

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, the rate of interest r% per annum is given by:
(a) $\frac{2400 I}{P n ( n + 1 )}$
(b) $\frac{1200 I}{P n ( n + 1 )}$
(c) $\frac{100 I}{P n}$
(d) $\frac{2400 P}{I n ( n + 1 )}$
Hint
Rearrange $I = P \times \frac{n ( n + 1 )}{24} \times \frac{r}{100}$ .
Answer
(a)
Q2 · 3 marks · Short answerBoard level
A recurring deposit of ₹500 per month for 1 year earns ₹325 interest. Find the rate.
Hint
$500 \times 6.5 = 3250$ .
Answer
$r = \frac{325}{3250} \times 100 = 10%$ per annum.
Q3 · 3 marks · Short answerHarder board level
Pooja deposits ₹1,200 per month for 2 years and gets ₹31,200 at maturity. Find the rate of interest.
Hint
Interest = 31,200 − 28,800.
Answer
$I = 2400$ ; $r = \frac{2400}{1200 \times 25} \times 100 = 8%$ per annum.
Q4 · 3 marks · Short answerHarder board level
A recurring deposit of ₹250 per month for 4 years matures to ₹13,960. Find the rate of interest.
Hint
$\frac{48 \times 49}{24} = 98$ .
Answer
$I = 13960 - 12000 = 1960$ ; $r = \frac{1960}{250 \times 98} \times 100 = 8%$ per annum.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If ₹100 per month for 12 months earns ₹65 interest in a recurring deposit, the rate is 10% per annum.
Reason (R): For a recurring deposit, $r = \frac{2400 I}{P n ( n + 1 )}$ .
(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
$\frac{2400 \times 65}{100 \times 12 \times 13}$ .
Answer
(c). $\frac{2400 \times 65}{100 \times 156} = 10$ .
Q6 · 4 marks · Short answerHarder board level
A recurring deposit of ₹1,500 per month for 3 years earns ₹6,660 interest. (a) Find the rate. (b) At what rate would the same deposit earn ₹8,325?
Hint
$1500 \times 55.5 = 83250$ .
Answer
(a) $\frac{6660}{83250} \times 100 = 8%$ . (b) $\frac{8325}{83250} \times 100 = 10%$ .
Q7 · 4 marks · Short answerHOTS / competency
In a 2-year recurring deposit, the interest earned equals exactly two monthly instalments. Find the rate of interest.
Hint
Write $I = 2 P$ and use $\frac{n ( n + 1 )}{24} = 25$ .
Answer
$2 P = P \times 25 \times \frac{r}{100} \Rightarrow r = 8%$ per annum, whatever the instalment.

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.MV ₹6,260, ₹500 a month for 12 months. Interest?Show answer
₹260.
2.Interest ₹260 on ₹500 a month for 12 months. Rate?Show answer
$\frac{260}{3250} \times 100 = 8%$ .
3.Write r in terms of I, P and n.Show answer
$r = \frac{2400 I}{P n ( n + 1 )}$ .
4.₹1,000 a month for 2 years earns ₹2,000. Rate?Show answer
$\frac{2000}{25000} \times 100 = 8%$ .
5.What is the first step in any rate question when the maturity value is given?Show answer
Interest = maturity value − (instalment × number of months).

Rate of Interest on a Recurring Deposit: frequently asked questions

How do I find the rate of interest in a recurring deposit?

Find the interest as maturity value minus total deposits, then use r = 2400 × I ÷ (P × n(n + 1)), with n in months. The answer is the rate per cent per annum.

Why do we use 2400 in the rate formula?

It is 2 × 12 × 100: the 2 from the sum n(n + 1)/2, the 12 that turns months into years, and the 100 for per cent.

Can I use the maturity value directly to find the rate?

No. Only the interest depends on the rate, so subtract the total deposited from the maturity value first.