Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 2 Banking Recurring Deposit Accounts

Question 1. Manish opens a Recurring Deposit Account with the Bank of Rajasthan and deposits Rs. 600 per month for 20 months. Calculate the maturity value of this account, if the bank pays interest at the rate of 10% per annum.
Answer:
Given:
Installment per month (P) = Rs. 600
Number of months (n) = 20
Rate of interest (r) = 10% p.a.

Step 1: Calculate Simple Interest using the formula.
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)

\( S.I. = 600 \times \frac{20(20 + 1)}{2 \times 12} \times \frac{10}{100} \)

\( S.I. = 600 \times \frac{420}{24} \times \frac{10}{100} = Rs. 1,050 \)

Step 2: Calculate maturity value.
The amount that Manish will get at the time of maturity
= Rs. (600 × 20) + Rs. 1,050
= Rs. 12,000 + Rs. 1,050
= Rs. 13,050

In simple words: Manish deposits Rs. 600 every month for 20 months. The bank also pays him interest on his money. So he gets back more than what he put in.

📝 Teacher's Note: Show students that RD interest is calculated on the average balance. The first deposit earns interest for all 20 months, but the last deposit earns no interest. This is why we use the special formula.

🎯 Exam Tip: Always write the RD formula first. Then substitute all given values step by step. Don't forget to add principal amount to interest for final answer.

Question 2. Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited Rs. 640 per month for 4 ½ years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.
Answer:
Given:
Installment per month (P) = Rs. 640
Number of months (n) = 54 (4½ years = 4.5 × 12 = 54 months)
Rate of interest (r) = 12% p.a.

Step 1: Calculate Simple Interest.
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)

\( S.I. = 640 \times \frac{54(54 + 1)}{2 \times 12} \times \frac{12}{100} \)

\( S.I. = 640 \times \frac{2970}{24} \times \frac{12}{100} = Rs. 9,504 \)

Step 2: Calculate maturity value.
The amount that Mrs. Mathew will get at the time of maturity
= Rs. (640 × 54) + Rs. 9,504
= Rs. 34,560 + Rs. 9,504
= Rs. 44,064

In simple words: Mrs. Mathew saves for a longer time (4½ years) so she gets more interest. The longer you save, the more extra money the bank gives you.

📝 Teacher's Note: Always convert years to months first. Students often forget to multiply by 12. Use simple examples like "4½ years = 4 years + 6 months = 48 + 6 = 54 months".

🎯 Exam Tip: When time is given in years and months, convert everything to months first. Write this conversion clearly in your answer to avoid mistakes.

Question 3. Each of A and B both opened recurring deposit accounts in a bank. If A deposited Rs. 1,200 per month for 3 years and B deposited Rs. 1,500 per month for 2 ½ years; find, on maturity, who will get more amount and by how much? The rate of interest paid by the bank is 10% per annum.
Answer:
For A:
Installment per month (P) = Rs. 1,200
Number of months (n) = 36
Rate of interest (r) = 10% p.a.

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

\( S.I. = 1,200 \times \frac{36(36 + 1)}{2 \times 12} \times \frac{10}{100} \)

\( S.I. = 1,200 \times \frac{1332}{24} \times \frac{10}{100} = Rs. 6,660 \)

The amount that A will get at the time of maturity
= Rs. (1,200 × 36) + Rs. 6,660
= Rs. 43,200 + Rs. 6,660
= Rs. 49,860

For B:
Installment per month (P) = Rs. 1,500
Number of months (n) = 30
Rate of interest (r) = 10% p.a.

\( S.I. = 1,500 \times \frac{30(30 + 1)}{2 \times 12} \times \frac{10}{100} \)

\( S.I. = 1,500 \times \frac{930}{24} \times \frac{10}{100} = Rs. 5,812.50 \)

The amount that B will get at the time of maturity
= Rs. (1,500 × 30) + Rs. 5,812.50
= Rs. 45,000 + Rs. 5,812.50
= Rs. 50,812.50

Difference between both amounts = Rs. 50,812.50 - Rs. 49,860
= Rs. 952.50

Therefore, B will get more money than A by Rs. 952.50.

In simple words: Even though B saves for less time, B puts more money each month. This extra money each month makes B get more total money in the end.

📝 Teacher's Note: This shows students that both time and amount matter in savings. Higher monthly deposit can sometimes beat longer time period. Use this to teach them about planning their savings.

🎯 Exam Tip: Calculate both accounts completely before comparing. Show the final subtraction clearly. Write who gets more and by how much as your final answer.

Question 4. Ashish deposits a certain sum of money every month is a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets Rs. 12,715 as the maturity value of this account, what sum of money did he pay every month?
Answer:
Let Installment per month (P) = Rs. y
Number of months (n) = 12
Rate of interest (r) = 11% p.a.

Step 1: Calculate Simple Interest in terms of y.
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)

\( S.I. = y \times \frac{12(12 + 1)}{2 \times 12} \times \frac{11}{100} \)

\( S.I. = y \times \frac{156}{24} \times \frac{11}{100} = Rs. 0.715y \)

Step 2: Set up equation using maturity value.
Maturity value = Rs. (y × 12) + Rs. 0.715y = Rs. 12.715y
Given maturity value = Rs. 12,715
Therefore, Rs. 12.715y = Rs. 12,715

Step 3: Solve for y.
\( y = \frac{12,715}{12.715} = Rs. 1,000 \)

Therefore, Ashish deposited Rs. 1,000 every month.

In simple words: We don't know how much Ashish put in each month, but we know what he got back. We work backwards to find his monthly deposit.

📝 Teacher's Note: This is a reverse problem. Teach students to set up the maturity value formula with unknown monthly deposit as 'y'. Then solve the simple equation to find y.

🎯 Exam Tip: Let the monthly deposit = y. Write the complete maturity value formula. Then make an equation and solve for y. Show all steps clearly.

Question 5. A man has a Recurring Deposit Account in a bank for 3 ½ years. If the rate of interest is 12% per annum and the man gets Rs. 10,206 on maturity, find the value of monthly instalments.
Answer:
Let Installment per month (P) = Rs. y
Number of months (n) = 42 (3½ years = 3.5 × 12 = 42 months)
Rate of interest (r) = 12% p.a.

Step 1: Calculate Simple Interest in terms of y.
\( S.I. = y \times \frac{42(42 + 1)}{2 \times 12} \times \frac{12}{100} \)

\( S.I. = y \times \frac{1806}{24} \times \frac{12}{100} = Rs. 9.03y \)

Step 2: Set up equation using maturity value.
Maturity value = Rs. (y × 42) + Rs. 9.03y = Rs. 51.03y
Given maturity value = Rs. 10,206
Therefore, Rs. 51.03y = Rs. 10,206

Step 3: Solve for y.
\( y = \frac{10,206}{51.03} = Rs. 200 \)

Therefore, the monthly installment was Rs. 200.

In simple words: The man saved for 3½ years and got Rs. 10,206. We find out he must have put Rs. 200 every month to get this much money.

📝 Teacher's Note: Another reverse problem like Question 4. Emphasize converting 3½ years to months first. Students often make calculation errors with decimals, so check the arithmetic carefully.

🎯 Exam Tip: Convert time to months first and show this step. Set up the complete equation and solve step by step. Double-check your final answer by substituting back.

Question 6. (i) Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits Rs. 140 per month for 4 years. If he gets Rs. 8,092 on maturity, find the rate of interest given by the bank. (ii) David opened a Recurring Deposit Account in a bank and deposited Rs. 300 per month for two years. If he received Rs. 7,725 at the time of maturity, find the rate of interest per annum.
Answer:
(i)
Installment per month (P) = Rs. 140
Number of months (n) = 48
Let rate of interest (r) = r% p.a.

\( S.I. = 140 \times \frac{48(48 + 1)}{2 \times 12} \times \frac{r}{100} \)

\( S.I. = 140 \times \frac{2352}{24} \times \frac{r}{100} = Rs. (137.20)r \)

Maturity value = Rs. (140 × 48) + Rs. (137.20)r
Given maturity value = Rs. 8,092
Therefore, Rs. (140 × 48) + Rs. (137.20)r = Rs. 8,092
\( \implies \) 137.20r = Rs. 8,092 - Rs. 6,720
\( \implies \) 137.20r = 1,372
\( \implies \) r = \( \frac{1,372}{137.20} = 10\% \)

(ii)
Installment per month (P) = Rs. 300
Number of months (n) = 24
Let rate of interest (r) = r% p.a.

\( S.I. = 300 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{r}{100} \)

\( S.I. = 300 \times \frac{600}{24} \times \frac{r}{100} = Rs. (75)r \)

Maturity value = Rs. (300 × 24) + Rs. (75)r
Given maturity value = Rs. 7,725
Therefore, Rs. (300 × 24) + Rs. (75)r = Rs. 7,725
\( \implies \) 75r = Rs. 7,725 - Rs. 7,200
\( \implies \) 75r = 525
\( \implies \) r = \( \frac{525}{75} = 7\% \)

In simple words: Here we know everything except the interest rate. We use the formula backwards to find what rate the bank was giving.

📝 Teacher's Note: These are reverse problems to find interest rate. Show students how to rearrange the maturity value equation. The key is calculating principal amount first, then finding interest earned.

🎯 Exam Tip: Calculate total principal deposited first. Then subtract from maturity value to get interest earned. Use this interest in the formula to find rate.

Question 7. Amit deposited Rs. 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month?
Answer:
Installment per month (P) = Rs. 150
Number of months (n) = 8
Rate of interest (r) = 8% p.a.

Step 1: Calculate Simple Interest.
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)

\( S.I. = 150 \times \frac{8(8 + 1)}{2 \times 12} \times \frac{8}{100} \)

\( S.I. = 150 \times \frac{72}{24} \times \frac{8}{100} = Rs. 36 \)

Step 2: Calculate maturity value.
The amount that Amit will get at the time of maturity
= Rs. (150 × 8) + Rs. 36
= Rs. 1,200 + Rs. 36
= Rs. 1,236

In simple words: Amit saves for only 8 months, so the interest is small. But he still gets more than what he put in because the bank pays him for keeping money with them.

📝 Teacher's Note: This is a short-term RD example. Good to show students that even for short periods, RD gives better returns than keeping money at home. The interest may be small but it's still extra money.

🎯 Exam Tip: Even for small time periods, use the same formula. Calculate step by step. Don't skip steps even if numbers are simple.

Question 8. Mrs. Geeta deposited Rs. 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is Rs. 5,565; find the rate of interest per annum.
Answer:
Given:
Installment per month (P) = Rs. 350
Number of months (n) = 15
Let rate of interest (r) = r% p.a.
Maturity value = Rs. 5,565

Step 1: Find Simple Interest using formula
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)

Step 2: Substitute values
\( S.I. = 350 \times \frac{15(15 + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 350 \times \frac{240}{24} \times \frac{r}{100} = Rs(35)r \)

Step 3: Use maturity value formula
Maturity value = Rs. (350 × 15) + Rs. (35)r
Rs. 5,565 = Rs. 5,250 + Rs. (35)r

Step 4: Solve for r
\( 35r = 5,565 - 5,250 = 315 \)
\( r = \frac{315}{35} = 9\% \)

Rate of interest = 9% per annum
In simple words: We find how much extra money (interest) Mrs. Geeta got. Then we use the RD formula to find what interest rate gives that extra money.

📝 Teacher's Note: Show students that maturity value = money put in + interest earned. The interest formula for RD is different from simple interest because money goes in every month, not all at once.

🎯 Exam Tip: Always write the RD interest formula first. Then substitute all given values step by step. Write the final answer with % sign clearly.

Question 9. A recurring deposit account of Rs. 1,200 per month has a maturity value of Rs. 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.
Answer:
Given:
Installment per month (P) = Rs. 1,200
Number of months (n) = n
Rate of interest (r) = 8% p.a.
Maturity value = Rs. 12,440

Step 1: Find Simple Interest using formula
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 1,200 \times \frac{n(n + 1)}{24} \times \frac{8}{100} = Rs \, 4n(n+1) \)

Step 2: Use maturity value formula
Maturity value = Rs. (1,200 × n) + Rs. 4n(n+1)
Rs. 12,440 = Rs. (1200n + 4n² + 4n)

Step 3: Solve the quadratic equation
\( 1200n + 4n² + 4n = 12,440 \)
\( 4n² + 1204n - 12,440 = 0 \)
\( n² + 301n - 3110 = 0 \)
\( (n + 311)(n - 10) = 0 \)

Since n cannot be negative, n = 10 months

Time = 10 months
In simple words: We make an equation using the maturity value formula. Then we solve it like any math equation to find how many months the money was deposited.

📝 Teacher's Note: Teach students to always check if their answer makes sense. 10 months is reasonable for this amount. Negative values for time don't make sense in real life.

🎯 Exam Tip: When you get a quadratic equation, take only the positive value for time. Always write "months" or "years" after your numerical answer.

Question 10. Mr. Gulati has a Recurring Deposit Account of Rs. 300 per month. If the rate of interest is 12% and the maturity value of this account is Rs. 8,100; find the time (in years) of this Recurring Deposit Account.
Answer:
Given:
Installment per month (P) = Rs. 300
Number of months (n) = n
Rate of interest (r) = 12% p.a.
Maturity value = Rs. 8,100

Step 1: Find Simple Interest using formula
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 300 \times \frac{n(n + 1)}{24} \times \frac{12}{100} = Rs \, 1.5n(n+1) \)

Step 2: Use maturity value formula
Maturity value = Rs. (300 × n) + Rs. 1.5n(n+1)
Rs. 8,100 = Rs. (300n + 1.5n² + 1.5n)

Step 3: Solve the quadratic equation
\( 300n + 1.5n² + 1.5n = 8,100 \)
\( 1.5n² + 301.5n - 8100 = 0 \)
\( n² + 201n - 5400 = 0 \)
\( (n + 225)(n - 24) = 0 \)

Since n cannot be negative, n = 24 months

Time = 24 months = 2 years
In simple words: We solve an equation to find the number of months. Then we convert months to years by dividing by 12.

📝 Teacher's Note: Remind students that 12 months = 1 year. So 24 months = 2 years. Practice this conversion often as it's commonly asked.

🎯 Exam Tip: The question asks for time "in years", so convert your answer from months to years. Write both: "24 months = 2 years".

Question 11. Mr. Gupta opened a recurring deposit account in a bank. He deposited Rs. 2,500 per month for two years. At the time of maturity he got Rs. 67,500. Find:
(i) the total interest earned by Mr. Gupta
(ii) the rate of interest per annum.
Answer:
(i)
Maturity value = Rs. 67,500
Money deposited = Rs. 2,500 × 24 = Rs. 60,000
Total interest earned = Rs. 67,500 - Rs. 60,000 = Rs. 7,500

(ii)
Installment per month (P) = Rs. 2,500
Number of months (n) = 24
Let rate of interest (r) = r% p.a.

\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 2500 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 2500 \times \frac{600}{24} \times \frac{r}{100} = Rs(625)r \)

Since 625r = 7500
\( r = \frac{7500}{625} = 12\% \)

(i) Total interest = Rs. 7,500
(ii) Rate of interest = 12% per annum
In simple words: Interest is the extra money he got. We subtract what he put in from what he got back. Then we use the RD formula to find the interest rate.

📝 Teacher's Note: Show that interest = maturity value - principal. This is the most basic concept in banking that students must understand clearly.

🎯 Exam Tip: For part (i), always subtract total deposits from maturity value. For part (ii), use the interest amount you found to calculate the rate.

Exercise 2B

Question 1. Pramod deposits Rs. 600 per month in a Recurring Deposit Account for 4 years. If the rate of interest is 8% per year; calculate the maturity value of his account.
Answer:
Given:
Installment per month (P) = Rs. 600
Number of months (n) = 48
Rate of interest (r) = 8% p.a.

Step 1: Find Simple Interest
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 600 \times \frac{48(48 + 1)}{2 \times 12} \times \frac{8}{100} \)
\( = 600 \times \frac{2352}{24} \times \frac{8}{100} = Rs \, 4,704 \)

Step 2: Find maturity value
The amount that Pramod will get at the time of maturity
= Rs. (600 × 48) + Rs. 4,704
= Rs. 28,800 + Rs. 4,704
= Rs. 33,504

Maturity value = Rs. 33,504
In simple words: Maturity value is all the money put in plus the interest earned. It's like your savings plus bonus money from the bank.

📝 Teacher's Note: Emphasize that 4 years = 48 months. Students often forget this conversion. Show them that maturity value = principal + interest.

🎯 Exam Tip: Convert years to months first (4 years = 48 months). Use the RD interest formula carefully. Add principal and interest for final answer.

Question 2. Ritu has a Recurring Deposit Account in a bank and deposits Rs. 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of account is Rs. 1,554.
Answer:
Given:
Installment per month (P) = Rs. 80
Number of months (n) = 18
Let rate of interest (r) = r% p.a.
Maturity value = Rs. 1,554

Step 1: Find Simple Interest formula
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 80 \times \frac{18(18 + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 80 \times \frac{342}{24} \times \frac{r}{100} = Rs(11.4r) \)

Step 2: Use maturity value equation
Maturity value = Rs. (80 × 18) + Rs. (11.4r)
Rs. 1,554 = Rs. 1,440 + Rs. (11.4r)

Step 3: Solve for r
11.4r = 1,554 - 1,440 = 114
\( r = \frac{114}{11.4} = 10\% \)

Rate of interest = 10% per annum
In simple words: We find how much extra money (interest) was earned. Then we work backwards to find what interest rate gives that amount.

📝 Teacher's Note: Show students that we can work backwards from the final answer to find missing information. This reverse calculation is very useful in banking problems.

🎯 Exam Tip: First find the interest earned (maturity value - principal). Then use the interest formula to find the rate. Show all working clearly.

Question 3. The maturity value of a R.D. Account is Rs. 16,176. If the monthly installment is Rs. 400 and the rate of interest is 8%; find the time (period) of this R.D Account.
Answer:
Given:
Installment per month (P) = Rs. 400
Number of months (n) = n
Rate of interest (r) = 8% p.a.
Maturity value = Rs. 16,176

Step 1: Find Simple Interest formula
\( S.I. = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( = 400 \times \frac{n(n + 1)}{24} \times \frac{8}{100} = Rs \, \frac{4n(n + 1)}{3} \)

Step 2: Use maturity value equation
Maturity value = Rs.(400 × n) + Rs. \( \frac{4n(n + 1)}{3} \)
Rs. 16,176 = Rs.(400n) + Rs. \( \frac{4n(n + 1)}{3} \)

Step 3: Solve the equation
1200n + 4n² + 4n = 48,528
4n² + 1204n = 48,528
n² + 301n - 12132 = 0
(n + 337)(n - 36) = 0

Since n cannot be negative, n = 36

Time = 36 months = 3 years
In simple words: We make an equation using the given maturity value and solve it to find how many months the deposits were made.

📝 Teacher's Note: When solving quadratic equations in practical problems, always reject negative answers as they don't make sense in real life situations.

🎯 Exam Tip: Convert your final answer to years if it makes sense (36 months = 3 years). Write both forms to be safe.

Question 4. Mr. Bajaj needs Rs. 30,000 after 2 years. What least money (in multiple of 5) must he deposit every month in a recurring deposit account to get required money after 2 years, the rate of interest being 8% p.a.?
Answer:
Given:
Let installment per month = Rs. P
Number of months (n) = 24
Rate of interest = 8% p.a.
Required maturity value = Rs. 30,000

Step 1: Find Simple Interest formula
\( S.I. = P \times \frac{24(24 + 1)}{2 \times 12} \times \frac{8}{100} \)
\( = P \times \frac{600}{24} \times \frac{8}{100} = 2P \)

Step 2: Use maturity value equation
Maturity value = Rs.(P × 24) + Rs.(2P)
Rs. 30,000 = Rs.(24P + 2P) = Rs.(26P)

Step 3: Solve for P
\( P = \frac{30,000}{26} = Rs. \, 1,153.85 \)

Since the amount must be in multiple of 5, the least amount = Rs. 1,155

Monthly deposit = Rs. 1,155
In simple words: We work backwards from the target amount to find how much must be saved each month. Then we round up to the nearest multiple of 5.

📝 Teacher's Note: Explain that "multiple of 5" means the number must end in 0 or 5. Since we need "least money", we round UP to the next multiple of 5.

🎯 Exam Tip: When rounding to multiples, always round UP for "least money" questions because you need AT LEAST that much to reach your target.

Question 5. Rishabh has recurring deposit account in a post office for 3 years at 8% p.a. simple interest. If he gets Rs. 9,990 as interest at the time of maturity, find:
(i) The monthly installment.
(ii) The amount of maturity.
Answer:
Given:
Number of months (n) = 36
Rate of interest (r) = 8% p.a.
Interest = Rs. 9,990

Step 1: Find monthly installment using interest formula.
Simple Interest = \( P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \)
\( 9990 = P \times \frac{36(36+1)}{2 \times 12} \times \frac{8}{100} \)
\( 9990 = P \times \frac{1332}{24} \times \frac{8}{100} \)
\( 9990 = P \times 4.44 \)
\( P = \frac{9990}{4.44} = Rs. 2,250 \)

(i) Monthly installment = Rs. 2,250

Step 2: Find maturity value.
(ii) Maturity value = Rs. (2,250 × 36) + Rs. 9,990 = Rs. 90,990
In simple words: We use the interest formula to find how much money Rishabh puts in every month. Then we add all his monthly deposits plus the interest to get the final amount.

📝 Teacher's Note: Show students that recurring deposit interest is calculated differently from simple interest. Each month's deposit earns interest for different time periods. Use the special formula for RD interest.

🎯 Exam Tip: Always write the RD interest formula first. Show clear steps - find monthly installment first, then calculate maturity value. Include units (Rs.) in final answer.

Question 6. Gopal has a cumulative deposit account and deposits Rs. 900 per month for a period of 4 years he gets Rs. 52,020 at the time of maturity, find the rate of interest.
Answer:
Given:
Installment per month (P) = Rs. 900
Number of months (n) = 48
Maturity value = Rs. 52,020
Let rate of interest (r) = r% p.a.

Step 1: Use the RD interest formula.
Simple Interest = \( P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \)
\( = 900 \times \frac{48(48+1)}{2 \times 12} \times \frac{r}{100} \)
\( = 900 \times \frac{2352}{24} \times \frac{r}{100} = Rs. (882)r \)

Step 2: Find rate using maturity value formula.
Maturity value = Total deposits + Interest
\( 52,020 = (900 \times 48) + 882r \)
\( 52,020 = 43,200 + 882r \)
\( 882r = 52,020 - 43,200 = 8,820 \)
\( r = \frac{8820}{882} = 10\% \)

Rate of interest = 10% p.a.
In simple words: We know the final amount and monthly deposits. We use the formula to find what interest rate gives us this final amount.

📝 Teacher's Note: Explain that maturity value = all deposits + interest earned. Students often forget to subtract the total deposits from maturity value to get interest amount.

🎯 Exam Tip: Write "Let r = rate of interest" clearly. Show the equation setup step by step. Always check your answer by substituting back.

Question 7. Deepa has a 4-year recurring deposit account in a bank and deposits Rs. 1,800 per month. If she gets Rs. 1,08,450 at the time of maturity, find the rate of interest.
Answer:
Given:
Installment per month (P) = Rs. 1,800
Number of months (n) = 48
Maturity value = Rs. 1,08,450
Let rate of interest (r) = r% p.a.

Step 1: Use the RD interest formula.
Simple Interest = \( P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \)
\( = 1800 \times \frac{48(48+1)}{2 \times 12} \times \frac{r}{100} \)
\( = 1800 \times \frac{2352}{24} \times \frac{r}{100} = Rs. (1764)r \)

Step 2: Find rate using maturity value formula.
Maturity value = Total deposits + Interest
\( 1,08,450 = (1800 \times 48) + 1764r \)
\( 1,08,450 = 86,400 + 1764r \)
\( 1764r = 1,08,450 - 86,400 = 22,050 \)
\( r = \frac{22,050}{1764} = 12.5\% \)

Rate of interest = 12.5% p.a.
In simple words: We find how much extra money Deepa got above her deposits. Then we use the interest formula to find what rate gives this extra money.

📝 Teacher's Note: Show students that the interest earned is the difference between maturity value and total deposits. This makes the concept clearer.

🎯 Exam Tip: Calculate total deposits first (P × n). Then subtract from maturity value to get interest. Use this in the formula to find rate.

Question 8. Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets Rs. 8,088 from the bank after 3 years, find the value of his monthly instalment.
Answer:
Given:
Rate of interest (r) = 8%
Number of months (n) = 3 × 12 = 36
Maturity value = Rs. 8,088
Let monthly installment = Rs. P

Step 1: Write the maturity value formula.
Maturity value = Total deposits + Interest
\( M.V. = P \times n + P \times \frac{n(n+1)}{2} \times \frac{r}{12 \times 100} \)

Step 2: Substitute values.
\( 8088 = P \times 36 + P \times \frac{36 \times 37}{2} \times \frac{8}{12 \times 100} \)
\( 8088 = 36P + 4.44P \)
\( 8088 = 40.44P \)
\( P = \frac{8088}{40.44} = 200 \)

Monthly installment = Rs. 200
In simple words: We know the final amount and interest rate. We work backwards to find how much money Mr. Britto puts in each month.

📝 Teacher's Note: This is a reverse problem. Students need to understand that we start with the final result and find the monthly deposit amount.

🎯 Exam Tip: Set up the equation with P as unknown. Combine like terms carefully. Always verify your answer by substituting back into the original formula.

Question 9. Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs. 800 per month for 1½ years. If he received Rs. 15,084 at the time of maturity, find the rate of interest per annum.
Answer:
Given:
Monthly deposit (P) = Rs. 800
Time (n) = 1½ years = 1.5 × 12 = 18 months
Maturity value = Rs. 15,084

Step 1: Use maturity value formula.
\( M.V. = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \)
\( 15,084 = 800 \times 18 + 800 \times \frac{18 \times 19}{24} \times \frac{r}{100} \)
\( 15,084 = 14,400 + 114r \)
\( 114r = 684 \)
\( r = \frac{684}{114} = 6\% \)

Rate of interest = 6% per annum
In simple words: We find the extra money earned above deposits. Then we use the formula to find what interest rate gives this extra amount.

📝 Teacher's Note: Be careful with time conversion. 1½ years = 18 months, not 15 months. Students often make this mistake.

🎯 Exam Tip: Convert time to months correctly first. Show the interest calculation step clearly. Write "per annum" in your final answer.

Question 10. Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly installment is Rs. 1,000, find the:
(i) interest earned in 2 years
(ii) maturity value
Answer:
Given:
Monthly installment (P) = Rs. 1,000
Number of installments (n) = 2 years = 2 × 12 = 24 months
Rate of interest (r) = 6%

(i) Interest earned:
Interest = \( P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \)
\( = 1000 \times \frac{24 \times 25}{24} \times \frac{6}{100} \)
\( = 1000 \times 25 \times \frac{6}{100} \times \frac{1}{24} \)
\( = Rs. 1,500 \)

(ii) Maturity value:
Total money deposited = 24 × Rs. 1,000 = Rs. 24,000
Maturity value = Total money deposited + Interest
\( = Rs. (24,000 + 1,500) = Rs. 25,500 \)

(i) Interest earned = Rs. 1,500
(ii) Maturity value = Rs. 25,500
In simple words: Katrina puts Rs. 1,000 every month for 24 months. The bank gives her extra money (interest) of Rs. 1,500. So she gets Rs. 25,500 in total.

📝 Teacher's Note: Show students that maturity value is always total deposits plus interest earned. Use simple addition to verify the final answer.

🎯 Exam Tip: Calculate interest first using the RD formula. Then add it to total deposits to get maturity value. Show both parts clearly.

Question 11. Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs. 1200 as interest at the time of maturity, find
(i) the monthly installment
(ii) the amount of maturity
Answer:
Given:
Interest = Rs. 1,200
Time (n) = 2 years = 2 × 12 = 24 months
Rate (r) = 6%

(i) To find monthly installment:
Interest = \( P \times \frac{n(n-1)}{2 \times 12} \times \frac{r}{100} \)
\( 1,200 = P \times \frac{24 \times 25}{24} \times \frac{6}{100} \)
\( 1,200 = P \times \frac{3}{2} \)
\( P = \frac{1,200 \times 2}{3} = Rs. 800 \)

(ii) Amount of maturity:
Total sum deposited = P × n = Rs. 800 × 24 = Rs. 19,200
Amount of maturity = Total sum deposited + Interest
\( = Rs. (19,200 + 1,200) = Rs. 20,400 \)

(i) Monthly installment = Rs. 800
(ii) Maturity amount = Rs. 20,400
In simple words: We know the interest earned. We use the formula backwards to find how much Mohan deposits each month. Then we add his total deposits plus interest to get final amount.

📝 Teacher's Note: This is another reverse problem. Students need to understand that we work backwards from interest to find monthly installment.

🎯 Exam Tip: Use the interest amount given to find monthly installment first. Then calculate total deposits and add interest to get maturity value.

Question 11. Peter has a recurring deposit account in Punjab National Bank at Sadar Bazar, Delhi for 4 years at 10% p.a. He will get Rs. 6,370 as interest on maturity. Find:
(i) monthly installment,
(ii) the maturity value of the account.
Answer:
Given:
Time = 4 years = 48 months
Rate = 10% p.a.
Interest = Rs. 6,370

(i) Monthly installment:
Using interest formula: Interest = \( P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \)
\( 6,370 = P \times \frac{48 \times 49}{24} \times \frac{10}{100} \)
\( 6,370 = P \times 98 \times \frac{10}{100} \)
\( 6,370 = P \times 9.8 \)
\( P = \frac{6,370}{9.8} = Rs. 650 \)

(ii) Maturity value:
Total deposits = 48 × 650 = Rs. 31,200
Maturity value = Total deposits + Interest
\( = Rs. (31,200 + 6,370) = Rs. 37,570 \)

(i) Monthly installment = Rs. 650
(ii) Maturity value = Rs. 37,570
In simple words: Peter saves Rs. 650 every month for 4 years. The bank gives him Rs. 6,370 extra as interest. So he gets Rs. 37,570 in total.

📝 Teacher's Note: Students should remember that maturity value includes both the money deposited and the interest earned. Show this clearly as addition.

🎯 Exam Tip: Work systematically - find monthly installment from interest first, then calculate maturity value. Show each calculation step clearly for full marks.

Solution:

Answer:
(i) Let the monthly instalment be Rs. P.
n = 4 years = 4 × 12 months = 48 months
Rate of interest, r = 10%
Interest = Rs. 6370

Now, Interest = \( P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \)
\( \Rightarrow 6370 = P \times \frac{48 \times 49}{24} \times \frac{10}{100} \)
\( \Rightarrow 6370 = P \times \frac{49}{5} \)
\( \Rightarrow P = \frac{6370 \times 5}{49} = \text{Rs. 650} \)
Thus, the monthly instalment is Rs. 650.

(ii) Total money deposited in the bank = 48 × Rs. 650 = Rs. 31200
∴ Maturity value = Total money deposited + Interest
= Rs. (31200 + 6370)
= Rs. 37570

In simple words: We used a special formula to find how much money to pay each month. Then we added all the monthly payments to get the total money put in the bank. Finally, we added interest to get the final amount.

📝 Teacher's Note: Show students that the instalment formula uses n(n+1)/2 because interest is calculated on reducing amounts each month. Use simple numbers like 3 months to show the pattern first.

🎯 Exam Tip: Always write the instalment formula clearly first. Calculate step by step. Write "Rs." with every amount. Check your answer - multiply monthly amount by number of months to verify.