ICSE Class 10 Maths Banking Solutions for Selina Chapter 2
ICSE Class 10 Maths Banking teaches recurring deposit accounts: a fixed amount is deposited every month, each deposit earns simple interest for a different number of months, and the maturity value is the total deposit plus interest. In Selina Concise Mathematics Class 10 ICSE Chapter 2 Banking (Recurring Deposit Accounts), most answers come from one formula, but marks are usually lost when students use the wrong number of months or treat the deposit as ordinary simple interest on P \times n.
Assumption for edition differences: some Selina/Concise editions label this chapter as Exercise 2, while others split it as Exercise 2(A), Exercise 2(B), Test Yourself and case-study questions. Match the question by the given values; the recurring-deposit method and final arithmetic shown below remain the same.
Recurring deposit formula reference
In a recurring deposit account, P is the monthly instalment, n is the number of months, r is the annual rate of interest, I is the interest, and M is the maturity value. The formula used in ICSE Class 10 Maths is:
I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}
The maturity value is:
M = Pn + I
| Quantity asked | Formula or step to use | Student note |
|---|---|---|
| Interest | \(I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\) | Use n in months, not years. |
| Total deposit | Pn | This is only the money deposited, not maturity value. |
| Maturity value | M = Pn + I | Add interest after finding the total deposit. |
| Monthly instalment | Put P=x, then solve M = nx + I | Keep x throughout the working. |
| Rate of interest | Use I = M - Pn, then solve for r | Do not use M directly as interest. |
Concept snapshot: why the formula uses \(n(n+1)\)
Think of the monthly deposits as standing in a queue. The first deposit waits for n months, the second waits for n-1 months, and the last deposit waits for 1 month. So the bank is effectively giving interest for \(n+(n-1)+\cdots+1\) month-units.
n+(n-1)+\cdots+1=\frac{n(n+1)}{2}
This is why the recurring deposit formula has \(\frac{n(n+1)}{2}\), and then divides by 12 because the rate is per annum.
Worked examples for Banking Recurring Deposit Accounts
These worked examples are included before the exercise answers because they show the three main question types: finding maturity value, finding monthly instalment, and finding rate of interest.
Worked Example 1: Find the maturity value
A student deposits \text{₹}500 per month for 18 months at 8\% per annum. Find the maturity value.
Step 1: Write the given values: P=\text{₹}500, n=18, r=8.
Step 2: Use the recurring deposit interest formula.
I = 500 \times \frac{18(18+1)}{2 \times 12} \times \frac{8}{100}
I = 500 \times \frac{18 \times 19}{24} \times 0.08 = 570
Step 3: Find the total deposit.
Pn = 500 \times 18 = 9000
Step 4: Add interest to the total deposit.
M = 9000 + 570 = 9570
Final answer: The maturity value is \text{₹}9570.
Worked Example 2: Find the monthly instalment
A recurring deposit account matures to \text{₹}26500 after 24 months at 10\% per annum. Find the monthly instalment.
Step 1: Let the monthly instalment be \text{₹}x. Then P=x, n=24, r=10.
Step 2: Find interest in terms of x.
I = x \times \frac{24(24+1)}{2 \times 12} \times \frac{10}{100}
I = x \times 25 \times \frac{10}{100} = 2.5x
Step 3: Use M=Pn+I.
26500 = 24x + 2.5x = 26.5x
x = \frac{26500}{26.5} = 1000
Final answer: The monthly instalment is \text{₹}1000.
Worked Example 3: Find the rate of interest
A depositor pays \text{₹}800 per month for 15 months and earns \text{₹}480 as interest. Find the rate of interest per annum.
Step 1: Write P=\text{₹}800, n=15, I=\text{₹}480, and let the rate be r\%.
Step 2: Substitute in the formula.
480 = 800 \times \frac{15(15+1)}{2 \times 12} \times \frac{r}{100}
480 = 800 \times 10 \times \frac{r}{100}
480 = 80r
Step 3: Solve for r.
r = \frac{480}{80} = 6
Final answer: The rate of interest is 6\% per annum.
Exercise 2 solutions: Banking Recurring Deposit Accounts
Question 1(a): Nature of interest in an R.D. account
\text{₹}P per month is deposited for n months in a recurring deposit account which pays interest at r\% per annum. The nature and time of interest calculated is: compound interest for n months, simple interest for n months, compound interest for one month, or simple interest for one month.
Step 1: The monthly deposits earn interest for n,n-1,n-2,\ldots,1 months respectively.
Step 2: The qualifying sum for one month is:
P(n+n-1+\cdots+1)=P \times \frac{n(n+1)}{2}
Step 3: The interest is calculated as simple interest for one month on this qualifying sum.
I = P \times \frac{n(n+1)}{2} \times \frac{1}{12} \times \frac{r}{100}
Final answer: The correct option is simple interest for one month.
Question 1(b): Interest on \text{₹}900 per month for 8 months
\text{₹}900 is deposited every month in a recurring deposit account at 10\% per annum. Find the interest earned in 8 months.
Step 1: Here P=\text{₹}900, n=8, r=10.
Step 2: Substitute in the formula.
I = 900 \times \frac{8(8+1)}{2 \times 12} \times \frac{10}{100}
I = 900 \times \frac{72}{24} \times \frac{1}{10} = 270
Final answer: The interest earned is \text{₹}270.
Question 1(c): Find the monthly instalment when interest is \text{₹}1404
A man gets \text{₹}1404 as interest at the end of one year. If the rate of interest is 12\% per annum in an R.D. account, find the monthly instalment.
Step 1: Let the monthly instalment be \text{₹}P. Here I=\text{₹}1404, n=12, r=12.
Step 2: Substitute in the formula.
1404 = P \times \frac{12(12+1)}{2 \times 12} \times \frac{12}{100}
1404 = P \times \frac{156}{24} \times \frac{12}{100} = 0.78P
Step 3: Solve for P.
P=\frac{1404}{0.78}=1800
Final answer: The monthly instalment is \text{₹}1800.
Question 1(d): Total money deposited by Manish
Manish opened an R.D. account and deposited \text{₹}1000 per month at 10\% per annum for 2 years. Find the total money deposited by him.
Step 1: Convert the time into months.
2 \text{ years} = 2 \times 12 = 24 \text{ months}
Step 2: Total deposit is monthly deposit multiplied by number of months.
Pn = 1000 \times 24 = 24000
Final answer: The total money deposited is \text{₹}24000.
Question 1(e): Find the rate of interest
\text{₹}800 per month is deposited in an R.D. account for 1\frac{1}{2} years. If the depositor gets \text{₹}2280 as interest at maturity, find the rate of interest.
Step 1: Convert time into months: 1\frac{1}{2} years =18 months. Here P=\text{₹}800, n=18, I=\text{₹}2280.
Step 2: Let the rate be r\%.
2280 = 800 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}
2280 = 800 \times \frac{342}{24} \times \frac{r}{100}
2280 = 114r
Step 3: Solve for r.
r=\frac{2280}{114}=20
Final answer: The rate of interest is 20\% per annum.
Question 2: Compare maturity amounts of A and B
Each of A and B opened a recurring deposit account in a bank. A deposited \text{₹}1200 per month for 3 years and B deposited \text{₹}1500 per month for 2\frac{1}{2} years. The bank pays 10\% per annum. Find who gets more on maturity and by how much.
Step 1: For A, P=\text{₹}1200, n=36, r=10.
I_A = 1200 \times \frac{36(36+1)}{2 \times 12} \times \frac{10}{100}
I_A = 1200 \times \frac{1332}{24} \times \frac{1}{10}=6660
M_A = 1200 \times 36 + 6660 = 49860
Step 2: For B, P=\text{₹}1500, n=30, r=10.
I_B = 1500 \times \frac{30(30+1)}{2 \times 12} \times \frac{10}{100}
I_B = 1500 \times \frac{930}{24} \times \frac{1}{10}=5812.50
M_B = 1500 \times 30 + 5812.50 = 50812.50
Step 3: Compare the maturity values.
50812.50 - 49860 = 952.50
Final answer: B gets more by \text{₹}952.50.
Question 3: Find Ashish’s monthly deposit
Ashish deposits a certain sum every month for 12 months. The bank pays 11\% per annum, and the maturity value is \text{₹}12715. Find the monthly deposit.
Step 1: Let the monthly deposit be \text{₹}x. Here n=12, r=11, M=\text{₹}12715.
Step 2: Express interest in terms of x.
I = x \times \frac{12(12+1)}{2 \times 12} \times \frac{11}{100}
I = x \times \frac{156}{24} \times \frac{11}{100}=\frac{143x}{200}
Step 3: Use maturity value.
12715 = 12x + \frac{143x}{200}
12715 = \frac{2543x}{200}
x=\frac{12715 \times 200}{2543}=1000
Final answer: Ashish deposited \text{₹}1000 per month.
Question 4: Find monthly instalment from maturity value
A man has an R.D. account for 3\frac{1}{2} years. The rate of interest is 12\% per annum and the maturity value is \text{₹}10206. Find the monthly instalment.
Step 1: Let the monthly instalment be \text{₹}x. Convert 3\frac{1}{2} years into months: 3\frac{1}{2}\times 12=42.
Step 2: Express interest in terms of x.
I = x \times \frac{42(42+1)}{2 \times 12} \times \frac{12}{100}
I = x \times \frac{42 \times 43}{24} \times \frac{12}{100}=\frac{903x}{100}
Step 3: Use M=Pn+I.
10206 = 42x+\frac{903x}{100}=\frac{5103x}{100}
x=\frac{10206 \times 100}{5103}=200
Final answer: The monthly instalment is \text{₹}200.
Question 5: Amit’s maturity value
Amit deposited \text{₹}150 per month for 8 months under a recurring deposit scheme. Find the maturity value if the rate is 8\% per annum.
Step 1: Here P=\text{₹}150, n=8, r=8.
Step 2: Find the interest.
I = 150 \times \frac{8(8+1)}{2 \times 12} \times \frac{8}{100}
I = 150 \times \frac{72}{24} \times 0.08=36
Step 3: Add the total deposit and interest.
M = 150 \times 8 + 36 = 1236
Final answer: The maturity value is \text{₹}1236.
Question 6: Find the time of Mr. Gulati’s R.D. account
Mr. Gulati has a recurring deposit account of \text{₹}300 per month. If the rate is 12\% per annum and the maturity value is \text{₹}8100, find the time in years.
Step 1: Let the time be n months.
Step 2: Write the maturity equation.
I = 300 \times \frac{n(n+1)}{2 \times 12} \times \frac{12}{100}
I = \frac{3n(n+1)}{2}
8100 = 300n+\frac{3n(n+1)}{2}
Step 3: Clear the denominator and simplify.
16200 = 600n+3n(n+1)
3n^2+603n-16200=0
n^2+201n-5400=0
Step 4: Factorise.
n^2+225n-24n-5400=0
n(n+225)-24(n+225)=0
(n-24)(n+225)=0
Step 5: Reject the negative value because time cannot be negative.
n=24 \text{ months}=2 \text{ years}
Final answer: The time is 2 years.
Question 7: Mr. Gupta’s interest, rate and extra interest
Mr. Gupta deposited \text{₹}2500 per month for 2 years. At maturity he got \text{₹}67500. Find the interest earned, the rate of interest per annum, and the extra interest if he deposits \text{₹}100 more per month at the same rate and for the same time.
Step 1: Find the total deposit.
Pn = 2500 \times 24 = 60000
Step 2: Find the interest earned.
I = 67500 - 60000 = 7500
Step 3: Let the rate be r\%. Use the interest formula.
7500 = 2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100}
7500 = 2500 \times 25 \times \frac{r}{100}=625r
r=\frac{7500}{625}=12
Step 4: If the monthly deposit becomes \text{₹}2600, compute the new interest.
I_{\text{new}} = 2600 \times \frac{24(24+1)}{2 \times 12} \times \frac{12}{100}
I_{\text{new}} = 2600 \times 25 \times 0.12 = 7800
Step 5: Find the extra interest.
7800 - 7500 = 300
Final answer: Interest earned =\text{₹}7500, rate =12\% per annum, and extra interest =\text{₹}300.
Question 8: Mohan’s instalment, maturity value and reduced interest
Mohan has an R.D. account for 2 years at 6\% per annum. He gets \text{₹}1200 as interest at maturity. Find the monthly instalment, the maturity value, and the reduction in interest if the instalment is decreased by 20\%.
Step 1: Let the monthly instalment be \text{₹}x. Here n=24, r=6, I=\text{₹}1200.
1200 = x \times \frac{24(24+1)}{2 \times 12} \times \frac{6}{100}
1200 = x \times 25 \times 0.06 = 1.5x
x=\frac{1200}{1.5}=800
Step 2: Find the maturity value.
M = 800 \times 24 + 1200 = 20400
Step 3: Reduce the instalment by 20\%.
20\% \text{ of } 800 = \frac{20}{100}\times 800=160
\text{New instalment}=800-160=640
Step 4: Find the new interest.
I_{\text{new}} = 640 \times \frac{24(24+1)}{2 \times 12} \times \frac{6}{100}
I_{\text{new}} = 640 \times 25 \times 0.06 = 960
Step 5: Find the reduction in interest.
1200 - 960 = 240
Final answer: Monthly instalment =\text{₹}800, maturity value =\text{₹}20400, and reduced interest =\text{₹}240.
Test Yourself and case-study practice
The following revision problems use the same Selina Banking Recurring Deposit Accounts method. They are written to practise the question types that commonly appear after the main exercise: MCQ reasoning, finding the least instalment, and case-study interpretation.
Revision Question 1: Find the least monthly deposit
Mr. Bajaj needs \text{₹}30000 after 2 years. What least monthly deposit, in a multiple of \text{₹}5, should he make in an R.D. account at 8\% per annum?
Step 1: Let the monthly deposit be \text{₹}x. Here n=24, r=8, M=\text{₹}30000.
Step 2: Find interest in terms of x.
I = x \times \frac{24(24+1)}{2 \times 12} \times \frac{8}{100}
I = x \times 25 \times 0.08 = 2x
Step 3: Use maturity value.
30000 = 24x+2x=26x
x=\frac{30000}{26}=1153.846\ldots
Step 4: The deposit must be a multiple of \text{₹}5 and should be enough to reach the target. The next multiple of \text{₹}5 after \text{₹}1153.846\ldots is \text{₹}1155.
Final answer: The least monthly deposit is \text{₹}1155.
Revision Question 2: Rate of interest from maturity value
Shahrukh deposited \text{₹}800 per month for 1\frac{1}{2} years and received \text{₹}15084 at maturity. Find the rate of interest per annum.
Step 1: Convert time into months: 1\frac{1}{2} years =18 months. Here P=\text{₹}800, n=18, M=\text{₹}15084.
Step 2: Find the total deposit and interest.
Pn = 800 \times 18 = 14400
I = 15084 - 14400 = 684
Step 3: Let the rate be r\%.
684 = 800 \times \frac{18(18+1)}{2 \times 12} \times \frac{r}{100}
684 = 800 \times \frac{342}{24} \times \frac{r}{100}=114r
r=\frac{684}{114}=6
Final answer: The rate of interest is 6\% per annum.
Case-study question: Revised bank rate
Rina deposits \text{₹}1000 per month for 2 years. The bank originally gives 8\% per annum, but later compares it with 6\% per annum for the same period. Find the qualifying principal for one month, the maturity value at 8\%, and the reduction in interest if the rate is 6\%.
Step 1: Here P=\text{₹}1000, n=24. The qualifying principal for one month is:
P \times \frac{n(n+1)}{2}=1000 \times \frac{24 \times 25}{2}=300000
Step 2: Find interest at 8\% per annum.
I_{8\%}=300000 \times \frac{1}{12} \times \frac{8}{100}=2000
Step 3: Find maturity value at 8\%.
M=1000 \times 24 + 2000 = 26000
Step 4: Find interest at 6\%.
I_{6\%}=300000 \times \frac{1}{12} \times \frac{6}{100}=1500
Step 5: Find the reduction in interest.
2000-1500=500
Final answer: Qualifying principal =\text{₹}300000, maturity value at 8\% =\text{₹}26000, and reduction in interest =\text{₹}500.
Quick answer index
| Section | Question | Answer / final result |
|---|---|---|
| Exercise 2 | 1(a) | Simple interest for one month. |
| Exercise 2 | 1(b) | \text{₹}270 |
| Exercise 2 | 1(c) | \text{₹}1800 |
| Exercise 2 | 1(d) | \text{₹}24000 |
| Exercise 2 | 1(e) | 20\% per annum |
| Exercise 2 | 2 | B gets more by \text{₹}952.50, using the printed rate 10\% per annum. |
| Exercise 2 | 3 | \text{₹}1000 per month |
| Exercise 2 | 4 | \text{₹}200 per month |
| Exercise 2 | 5 | \text{₹}1236 |
| Exercise 2 | 6 | 2 years |
| Exercise 2 | 7 | Interest =\text{₹}7500, rate =12\%, extra interest =\text{₹}300 |
| Exercise 2 | 8 | Instalment =\text{₹}800, maturity value =\text{₹}20400, less interest =\text{₹}240 |
| Revision | Least monthly deposit | \text{₹}1155 |
| Revision | Rate from maturity | 6\% per annum |
| Case study | Revised rate | Reduction in interest =\text{₹}500 |
Examiner’s mindset for recurring deposit problems
In ICSE Class 10 Maths Banking answers, credit is usually tied to the method: identifying P, n, r, applying the correct R.D. formula, and then writing the maturity value or final comparison clearly. A correct formula with n in years instead of months often leads to the wrong answer, even if the arithmetic after that is neat.
For comparison questions, show both maturity values before subtracting. For rate or instalment questions, write I=M-Pn wherever maturity value is given. This one line prevents a common error: treating maturity value as interest.
Common mistakes students make
- Using years instead of months: For 2\frac{1}{2} years, use n=30, not n=2.5, in the R.D. formula.
- Forgetting the total deposit: Maturity value is M=Pn+I. Interest alone is not the final maturity amount.
- Using ordinary simple interest on Pn: The first deposit earns interest for more months than the last deposit, so use \(P \times \frac{n(n+1)}{2}\) as the qualifying sum.
- Rounding too early: In questions asking for the least instalment in a multiple of \text{₹}5, first solve the exact value, then round upward to the next multiple of 5.
- Missing the printed rate: If a question says 10\%, solve with 10\%. Do not copy a rate from a different example.
Related ICSE Class 10 Maths resources
For chapter-wise practice, use the ICSE Class 10 solutions page and the Selina Concise Mathematics Class 10 solutions index. Students revising the full paper can also use the ICSE Class 10 Maths study page.
For board-level syllabus and examination references, check the official CISCE website. For simple-interest basics that support commercial mathematics, students may also consult NCERT learning resources where the concept overlaps with school arithmetic.
Frequently Asked Questions
What is the recurring deposit formula in ICSE Class 10 Maths?
The recurring deposit formula in ICSE Class 10 Maths is \(I=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\), where P is the monthly instalment, n is the number of months, r is the annual rate, and I is the interest.
Why do we use \(n(n+1)\) in Banking Recurring Deposit Accounts?
We use \(n(n+1)\) because the monthly deposits earn interest for n,n-1,n-2,\ldots,1 months. The sum of these month-units is \(\frac{n(n+1)}{2}\).
How do I find maturity value in Selina Chapter 2 Banking?
First find interest using \(I=P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}\). Then add the total deposit: M=Pn+I. Do not write the interest as the maturity value.
What should I do if my Selina edition has different question numbers?
Match the question by the given values: monthly instalment P, number of months n, rate r, interest I, and maturity value M. Edition numbering may vary, but the recurring deposit method is the same.
Is the R.D. account interest compound interest or simple interest?
For the ICSE Class 10 Maths Banking formula, the interest is treated as simple interest for one month on the qualifying sum \(P \times \frac{n(n+1)}{2}\).
