OP Malhotra Class 10 Maths Solutions Chapter 2 Banking Exercise 2

 

Question 1. Mr. Rajiv Anand has opened a recurring deposit account of Rs. 7400 per month for 20 months in a bank. Find the amount he will get at the time of maturity, if the rate of interest is 8.5% p.a., and if the interest is calculated at the end of each month.
Answer:
Monthly deposit (P) = Rs. 400 (as used in the provided solution)
Period (n) = 20 months
Rate (R) = 8.5% p.a.

First, we calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 400 \times \frac{20 \times(20+1)}{2} \)
\( = 400 \times \frac{20 \times 21}{2} \)
\( = 400 \times 10 \times 21 \)
\( = \) Rs. 84000

Next, we calculate the interest earned:
Interest \( = \frac{PRT}{100} \)
\( = \frac{84000 \times 8.5 \times 1}{100 \times 12} \)
\( = \frac{714000}{1200} \)
\( = \) Rs. 595

Finally, we find the maturity value:
Maturity value \( = (P \times n) + \text{Interest} \)
\( = (400 \times 20) + 595 \)
\( = 8000 + 595 \)
\( = \) Rs. 8595
In simple words: We first find the total principal that earns interest over the period. Then, we use this to calculate the simple interest. Finally, the total amount received at maturity is the sum of all monthly deposits and the earned interest.

🎯 Exam Tip: Always clearly list the given values (P, n, R) at the start of your solution to avoid errors. Ensure the time period for interest calculation is in years when using the standard formula.

 

Question 2. Mrs. Savita Khosla deposits 7900 per month in a recurring account for 2 years. If she gets 71800 as interest at the time of maturity, find the rate of interest if the interest is calculated at the end of each month.
Answer:
Monthly deposit (P) = Rs. 900 (as used in the provided solution)
Period (n) = 2 years = 24 months
Interest (I) = Rs. 1800 (as used in the provided solution)
Let Rate = R% p.a.

First, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 900 \times \frac{24 \times(24+1)}{2} \)
\( = 900 \times \frac{24 \times 25}{2} \)
\( = 900 \times 12 \times 25 \)
\( = \) Rs. 270000

Now, use the interest formula to find the rate (R):
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
Since the Principal for 1 month already accounts for `n(n+1)/2`, the `T` in this context represents 1/12 years for converting the monthly rate to annual.
\( 1800 = \frac{270000 \times R \times 1}{100 \times 12} \)
\( 1800 = \frac{2700 \times R}{12} \)
\( 1800 \times 12 = 2700 \times R \)
\( 21600 = 2700 \times R \)
\( R = \frac{21600}{2700} \)
\( R = 8 \) % p.a.
In simple words: We first calculate the total sum on which interest is charged for one month. Then, we use the given interest amount along with the principal and period to work backwards and find the percentage rate of interest per year. The rate tells us how quickly the money grows.

🎯 Exam Tip: When finding the rate of interest, ensure you use the correct total principal (calculated over the entire deposit period) and convert all time units to years or months consistently.

 

Question 3. Mr. Brown deposits Rs. 1100 per month in a cumulative time deposit account in a bank for 16 months. If at the end of maturity he gets Rs. 19096, find the rate of interest if interest is calculated at the end of each month.
Answer:
Monthly deposit (P) = Rs. 1100
Period (n) = 16 months
Maturity value (MV) = Rs. 19096
Let rate = R% p.a.

First, calculate the total amount deposited:
Total deposit \( = P \times n = 1100 \times 16 = \) Rs. 17600

Now, find the interest earned:
Interest (I) \( = \text{Maturity value} - \text{Total deposit} \)
\( = 19096 - 17600 \)
\( = \) Rs. 1496

Next, calculate the principal for one month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 1100 \times \frac{16 \times(16+1)}{2} \)
\( = 1100 \times \frac{16 \times 17}{2} \)
\( = 1100 \times 8 \times 17 \)
\( = \) Rs. 149600

Finally, calculate the rate of interest (R):
Rate \( = \frac{\text{Interest} \times 100 \times 12}{\text{Principal for 1 month} \times 1} \)
\( = \frac{1496 \times 100 \times 12}{149600 \times 1} \)
\( = \frac{149600 \times 12}{149600} \)
\( = 12 \) % p.a.
In simple words: First, we find how much extra money was earned (interest) by subtracting the total money deposited from the final amount. Then, we calculate the total amount of money that earned interest each month. Finally, we use these numbers to figure out the yearly interest rate.

🎯 Exam Tip: Remember that "cumulative time deposit" and "recurring deposit" accounts follow the same interest calculation principles. Be careful with unit conversions, especially months to years.

 

Question 4. Sandhya has a recurring deposit account in Vijaya Bank and deposits 400 per month for 3 years. If she gets Rs. 16,176 on maturity, find the rate of interest given by the bank.
Answer:
Monthly deposit (P) = Rs. 400
Period (n) = 3 years = 36 months
Maturity value (MV) = Rs. 16176
Let rate = R% p.a.

First, calculate the total amount deposited:
Total deposit \( = P \times n = 400 \times 36 = \) Rs. 14400

Now, find the interest earned:
Interest (I) \( = \text{Maturity value} - \text{Total deposit} \)
\( = 16176 - 14400 \)
\( = \) Rs. 1776

Next, calculate the principal for one month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 400 \times \frac{36 \times(36+1)}{2} \)
\( = 400 \times \frac{36 \times 37}{2} \)
\( = 400 \times 18 \times 37 \)
\( = \) Rs. 266400

Finally, calculate the rate of interest (R):
Rate \( = \frac{\text{Interest} \times 100 \times 12}{\text{Principal for 1 month} \times 1} \)
\( = \frac{1776 \times 100 \times 12}{266400 \times 1} \)
\( = \frac{177600 \times 12}{266400} \)
\( = \frac{2131200}{266400} \)
\( = 8 \) % p.a.
In simple words: We figure out the total extra money Sandhya earned (interest) and the overall amount of money that generated this interest. Using these figures, we can calculate the yearly percentage rate offered by the bank.

🎯 Exam Tip: When calculating interest rate, ensure the principal value reflects the sum over which the interest is truly earned, and the time factor is correctly applied (e.g., multiplying by 12 for annual rate if the formula uses monthly principal).

 

Question 5. A man deposits Rs. 600 per month in a bank for 12 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits if the rate of interest is 8% p.a., and interest is calculated at the end of every month?
Answer:
Monthly deposit (P) = Rs. 600
Period (n) = 12 months
Rate (R) = 8% p.a.

First, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 600 \times \frac{12 \times(12+1)}{2} \)
\( = 600 \times \frac{12 \times 13}{2} \)
\( = 600 \times 6 \times 13 \)
\( = \) Rs. 46800

Next, calculate the interest earned:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{46800 \times 8 \times 1}{100 \times 12} \)
\( = \frac{374400}{1200} \)
\( = \) Rs. 312

Finally, calculate the maturity value:
Maturity value \( = (P \times n) + \text{Interest} \)
\( = (600 \times 12) + 312 \)
\( = 7200 + 312 \)
\( = \) Rs. 7512
In simple words: First, we find the total amount of money that earned interest. Then, we calculate the interest based on the given rate and time. The final maturity value is the total money deposited each month plus the interest earned.

🎯 Exam Tip: Remember that for recurring deposits, the principal for interest calculation is not just `P`, but `P * n * (n+1) / 2` because each monthly deposit earns interest for a different period.

 

Question 6. Anil deposits 300 per month in a recurring deposit account for 2 years. If the rate of interest is 10% per year, calculate the amount that Anil will receive at the end of 2 years, i.e., at the time of maturity.
Answer:
Monthly deposit (P) = Rs. 300
Period (n) = 2 years = 24 months
Rate (R) = 10% p.a.

First, calculate the principal for one month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 300 \times \frac{24 \times(24+1)}{2} \)
\( = 300 \times \frac{24 \times 25}{2} \)
\( = 300 \times 12 \times 25 \)
\( = \) Rs. 90000

Next, calculate the interest earned:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{90000 \times 10 \times 1}{100 \times 12} \)
\( = \frac{900000}{1200} \)
\( = \) Rs. 750

Finally, calculate the maturity value:
Maturity value \( = (P \times n) + \text{Interest} \)
\( = (300 \times 24) + 750 \)
\( = 7200 + 750 \)
\( = \) Rs. 7950
In simple words: We first find the total amount on which interest will be calculated. Then, we figure out how much interest Anil earns over two years. The final amount he gets is the sum of all his deposits plus this earned interest.

🎯 Exam Tip: Always convert the time period to months for 'n' when calculating the principal for a recurring deposit. Ensure that the rate is an annual rate as well.

 

Question 7. Sudhir opened a recurring deposit account with a bank for \( 1\frac {1}{2} \) years. If the rate of interest is 10% and the bank pays 1554 on maturity, find how much did Sudhir deposit per month ?
Answer:
Maturity value (MV) = Rs. 1554
Rate (R) = 10% p.a.
Period (n) = \( 1\frac {1}{2} \) years = 18 months
Let P be the monthly deposit.

First, calculate the principal for one month in terms of P:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = P \times \frac{18 \times(18+1)}{2} \)
\( = P \times \frac{18 \times 19}{2} \)
\( = P \times 9 \times 19 \)
\( = 171 P \)

Next, calculate the interest earned in terms of P:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{171 P \times 10 \times 1}{100 \times 12} \)
\( = \frac{1710 P}{1200} \)
\( = \frac{171 P}{120} \)

Now, use the maturity value formula:
Maturity value \( = (P \times n) + \text{Interest} \)
\( 1554 = (P \times 18) + \frac{171 P}{120} \)
\( 1554 = 18P + \frac{171 P}{120} \)
To add the terms, find a common denominator:
\( 1554 = \frac{18P \times 120}{120} + \frac{171 P}{120} \)
\( 1554 = \frac{2160 P + 171 P}{120} \)
\( 1554 = \frac{2331 P}{120} \)
Now, solve for P:
\( P = \frac{1554 \times 120}{2331} \)
\( P = \frac{186480}{2331} \)
\( P = 80 \)

So, the monthly deposit = Rs. 80
In simple words: We set up equations using the known maturity amount and the formulas for principal and interest, all in terms of the unknown monthly deposit. By solving these equations, we can find out how much money Sudhir deposited each month.

🎯 Exam Tip: When the monthly deposit is unknown, express all other variables (principal for 1 month, interest) in terms of 'P' and then solve the maturity value equation for 'P'.

 

Question 8. Renu has a cumulative deposit account of Rs. 200 per month at 10% per annum. If she gets 6775 at the time of maturity, find the total time for which the account was held.
Answer:
Monthly deposit (P) = Rs. 200
Rate (R) = 10% p.a.
Maturity value (MV) = Rs. 6775
Let the period be 'n' months.

First, calculate the principal for 1 month in terms of 'n':
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 200 \times \frac{n(n+1)}{2} \)
\( = 100 n(n+1) \)

Next, calculate the interest earned in terms of 'n':
Interest (I) \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{100 n(n+1) \times 10 \times 1}{100 \times 12} \)
\( = \frac{10 n(n+1)}{12} \)
\( = \frac{5 n(n+1)}{6} \) ... (i)

Now, express the interest using the maturity value and total deposit:
Total deposit \( = P \times n = 200 \times n = 200n \)
Interest (I) \( = \text{Maturity value} - \text{Total deposit} \)
\( = 6775 - 200n \) ... (ii)

Equate the two expressions for interest (i) and (ii):
\( \frac{5 n(n+1)}{6} = 6775 - 200n \)
Multiply both sides by 6:
\( 5n(n+1) = 6 \times (6775 - 200n) \)
\( 5n^2 + 5n = 40650 - 1200n \)
Rearrange into a quadratic equation:
\( 5n^2 + 5n + 1200n - 40650 = 0 \)
\( 5n^2 + 1205n - 40650 = 0 \)
Divide the entire equation by 5 to simplify:
\( n^2 + 241n - 8130 = 0 \)

Solve the quadratic equation by factoring. We need two numbers that multiply to -8130 and add to 241. These numbers are 271 and -30.
\( n^2 + 271n - 30n - 8130 = 0 \)
Factor by grouping:
\( n(n+271) - 30(n+271) = 0 \)
\( (n-30)(n+271) = 0 \)

This gives two possible values for n:
\( n-30 = 0 \implies n = 30 \)
\( n+271 = 0 \implies n = -271 \)
Since the period (n) cannot be negative, we take \( n = 30 \) months.

Convert months to years:
\( n = 30 \text{ months} = \frac{30}{12} \text{ years} = 2.5 \text{ years} = 2\frac{1}{2} \text{ years} \)
Therefore, the total time for which the account was held is \( 2\frac{1}{2} \) years.
In simple words: We set up equations for the interest earned, one based on the rate and one based on the maturity amount. By making these equal, we get a quadratic equation involving the number of months. Solving this equation gives us the period of the account in months, which we then convert to years.

🎯 Exam Tip: When solving for 'n' (number of months/years), you'll often encounter a quadratic equation. Remember to discard any negative or impractical solutions, as time cannot be negative.

 

Question 1. 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 the interest is calculated at the end of every month. (ICSE 2001)
Answer:
Monthly deposit (P) = Rs. 150
Period (n) = 8 months
Rate (R) = 8% p.a.

First, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 150 \times \frac{8 \times(8+1)}{2} \)
\( = 150 \times \frac{8 \times 9}{2} \)
\( = 150 \times 4 \times 9 \)
\( = \) Rs. 5400

Next, calculate the interest earned:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{5400 \times 8 \times 1}{100 \times 12} \)
\( = \frac{43200}{1200} \)
\( = \) Rs. 36

Finally, calculate the maturity value:
Maturity value \( = (P \times n) + \text{Interest} \)
\( = (150 \times 8) + 36 \)
\( = 1200 + 36 \)
\( = \) Rs. 1236
In simple words: We first find the total sum on which interest is calculated over the period. Then, we calculate the simple interest. Finally, we add this interest to the total amount Amit deposited to find the maturity value.

🎯 Exam Tip: Ensure that the principal used in the interest calculation is the "principal for 1 month", which accounts for all deposits and their respective periods of interest accrual.

 

Question 2. Mr. R.K. Nair gets Rs. 6455 at the end of one year at the rate of 14% per annum in a Recurring Deposit Account. Find the monthly instalment. (ICSE 2005)
Answer:
Let monthly deposit = Rs. x
Rate (R) = 14% p.a.
Period (n) = 1 year = 12 months
Maturity value (MV) = Rs. 6455

First, calculate the principal for 1 month in terms of x:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = x \times \frac{12 \times(12+1)}{2} \)
\( = x \times \frac{12 \times 13}{2} \)
\( = x \times 6 \times 13 \)
\( = 78x \)

Next, calculate the interest earned in terms of x:
Interest (I) \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{78x \times 14 \times 1}{100 \times 12} \)
\( = \frac{1092x}{1200} \)
\( = \frac{91x}{100} \) ... (i)

Now, express the interest using the maturity value and total deposit:
Total deposit \( = P \times n = x \times 12 = 12x \)
Interest (I) \( = \text{Maturity value} - \text{Total deposit} \)
\( = 6455 - 12x \) ... (ii)

Equate the two expressions for interest (i) and (ii):
\( \frac{91x}{100} = 6455 - 12x \)
Multiply both sides by 100:
\( 91x = 100 \times (6455 - 12x) \)
\( 91x = 645500 - 1200x \)
Move terms with x to one side:
\( 91x + 1200x = 645500 \)
\( 1291x = 645500 \)
Solve for x:
\( x = \frac{645500}{1291} \)
\( x = 500 \)

Therefore, the monthly instalment = Rs. 500
In simple words: We set up equations for the interest earned, one in terms of the unknown monthly payment 'x' and another by subtracting total deposits from the final maturity amount. By solving these equations, we find the value of 'x', which is the monthly installment.

🎯 Exam Tip: When the monthly instalment is unknown, represent it as 'x' and set up an equation where the calculated interest (in terms of x) equals the difference between the maturity value and total deposits (also in terms of x).

 

Question 3. Mohan deposits 80 per month in a cumulative deposit account for six years. Find the amount payable to him on maturity, if the rate of interest is 6% per annum.
Answer:
Monthly deposit (P) = Rs. 80
Rate (R) = 6% p.a.
Period (n) = 6 years = 72 months

First, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 80 \times \frac{72 \times(72+1)}{2} \)
\( = 80 \times \frac{72 \times 73}{2} \)
\( = 80 \times 36 \times 73 \)
\( = \) Rs. 210240

Next, calculate the interest earned:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{210240 \times 6 \times 1}{100 \times 12} \)
\( = \frac{1261440}{1200} \)
\( = \) Rs. 1051.20

Finally, calculate the maturity value:
Maturity value \( = (P \times n) + \text{Interest} \)
\( = (80 \times 72) + 1051.20 \)
\( = 5760 + 1051.20 \)
\( = \) Rs. 6811.20
In simple words: We first find the total amount that acts as the principal for interest over the entire deposit period. Then, we calculate the interest using the given rate. The final amount Mohan receives is his total deposits plus the interest earned.

🎯 Exam Tip: Be careful with decimal places when calculating interest, especially if the final answer needs to be precise for money amounts.

 

Question 4. Saloni deposited Rs. 150 per month in her bank for eight months under the Recurring Deposit Scheme. What will the maturity value of her deposit, if the rate of interest is 8% per annum and the interest is calculated at the end of every month.
Answer:
Monthly deposit (P) = Rs. 150
Period (n) = 8 months
Rate (R) = 8% p.a.

First, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 150 \times \frac{8 \times(8+1)}{2} \)
\( = 150 \times \frac{8 \times 9}{2} \)
\( = 150 \times 4 \times 9 \)
\( = \) Rs. 5400

Next, calculate the interest earned:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{5400 \times 8 \times 1}{100 \times 12} \)
\( = \frac{43200}{1200} \)
\( = \) Rs. 36

Finally, calculate the maturity value:
Maturity value \( = (P \times n) + \text{Interest} \)
\( = (150 \times 8) + 36 \)
\( = 1200 + 36 \)
\( = \) Rs. 1236
In simple words: We first find the total principal that earns interest over the deposit period. Then, we calculate the interest based on the given rate. The total amount Saloni receives at the end is her total deposits plus the interest earned.

🎯 Exam Tip: Always make sure to use the correct number of months for 'n' in the formula for principal, and the annual rate 'R' (with 1/12 for the time 'T' if using monthly principal).

 

Question 5. A man opened a Recurring Deposit Account in a bank and deposited Rs. 300 per month for two years. If he received Rs. 7725 at the time of maturity, find the rate of interest per annum.
Answer:
Monthly deposit (P) = Rs. 300
Period (n) = 2 years = 24 months
Maturity value (MV) = Rs. 7725
Let R be the rate % per annum.

First, calculate the total amount deposited:
Total deposit \( = P \times n = 300 \times 24 = \) Rs. 7200

Next, find the interest earned:
Interest (I) \( = \text{Maturity value} - \text{Total deposit} \)
\( = 7725 - 7200 \)
\( = \) Rs. 525

Now, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 300 \times \frac{24 \times(24+1)}{2} \)
\( = 300 \times \frac{24 \times 25}{2} \)
\( = 300 \times 12 \times 25 \)
\( = \) Rs. 90000

Finally, use the interest formula to find the rate (R):
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( 525 = \frac{90000 \times R \times 1}{100 \times 12} \)
\( 525 = \frac{900 \times R}{12} \)
\( 525 \times 12 = 900 \times R \)
\( 6300 = 900 \times R \)
\( R = \frac{6300}{900} \)
\( R = 7 \) % p.a.
In simple words: We first find the total extra money received (interest) by subtracting the total deposits from the maturity value. Then, we calculate the total principal that earned this interest. Using these, we can find the yearly interest rate.

🎯 Exam Tip: When calculating the interest rate, ensure the "Principal for 1 month" is correctly calculated, as it is the foundation for finding 'R' or 'T'.

 

Question 6. Mrs. Goswami deposits Rs. 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.
Answer:
Monthly deposit (P) = Rs. 1000
Period (n) = 3 years = 36 months
Rate (R) = 8% p.a.

First, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 1000 \times \frac{36 \times(36+1)}{2} \)
\( = 1000 \times \frac{36 \times 37}{2} \)
\( = 1000 \times 18 \times 37 \)
\( = \) Rs. 666000

Next, calculate the interest earned:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{666000 \times 8 \times 1}{100 \times 12} \)
\( = \frac{5328000}{1200} \)
\( = \) Rs. 4440

Finally, calculate the maturity value:
Maturity value \( = (P \times n) + \text{Interest} \)
\( = (1000 \times 36) + 4440 \)
\( = 36000 + 4440 \)
\( = \) Rs. 40440
In simple words: We calculate the total amount deposited and the interest it earns over three years. The maturity value is simply the sum of all deposits plus the interest.

🎯 Exam Tip: Remember to express the time period 'n' in months for recurring deposit calculations, especially when using the principal formula `P * n * (n+1) / 2`.

 

Question 7. Mr. Gupta opened a recurring deposit account in a bank. He deposited Rs. 2500 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:
Monthly deposit (P) = Rs. 2500
Period (n) = 2 years = 24 months
Maturity value (MV) = Rs. 67500

First, calculate the total amount deposited:
Total deposited amount \( = P \times n = 2500 \times 24 = \) Rs. 60000

(i) To find the total interest earned by Mr. Gupta:
Interest (I) \( = \text{Maturity value} - \text{Total deposited amount} \)
\( = 67500 - 60000 \)
\( = \) Rs. 7500

(ii) To find the rate of interest per annum:
Let R be the rate % p.a.

Now, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 2500 \times \frac{24 \times(24+1)}{2} \)
\( = 2500 \times \frac{24 \times 25}{2} \)
\( = 2500 \times 12 \times 25 \)
\( = \) Rs. 750000

Use the interest formula to find the rate (R):
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( 7500 = \frac{750000 \times R \times 1}{100 \times 12} \)
\( 7500 = \frac{7500 \times R}{12} \)
\( 7500 \times 12 = 7500 \times R \)
\( R = 12 \) % p.a.
In simple words: First, we find the extra money Mr. Gupta received, which is the interest. Then, we use the total interest, the total principal that earned interest, and the time to calculate the annual interest rate.

🎯 Exam Tip: Break down multi-part questions into individual steps. Calculate total deposits and interest first, then use those values to find the rate of interest.

 

Question 8. Ahmed has a recurring deposit account in a bank. He deposits Rs. 2,500 per month for 2 years. If he get Rs. 66,250 at the time of maturity, find
(i) The interest paid by the bank.
(ii) The rate of interest.
Answer:
Monthly deposit (P) = Rs. 2500
Period (n) = 2 years = 24 months
Matured amount (MV) = Rs. 66250

First, calculate the total deposited amount:
Total deposited amount \( = P \times n = 2500 \times 24 = \) Rs. 60000

(i) To find the interest paid by the bank:
Interest (I) \( = \text{Matured amount} - \text{Total deposited amount} \)
\( = 66250 - 60000 \)
\( = \) Rs. 6250

(ii) To find the rate of interest:
Let R be the rate % p.a.

Now, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 2500 \times \frac{24 \times(24+1)}{2} \)
\( = 2500 \times \frac{24 \times 25}{2} \)
\( = 2500 \times 12 \times 25 \)
\( = \) Rs. 750000

Use the interest formula to find the rate (R):
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( 6250 = \frac{750000 \times R \times 1}{100 \times 12} \)
\( 6250 = \frac{7500 \times R}{12} \)
\( 6250 \times 12 = 7500 \times R \)
\( 75000 = 7500 \times R \)
\( R = \frac{75000}{7500} \)
\( R = 10 \) % p.a.
In simple words: We first find how much extra money Ahmed received as interest. Then, we use the total principal on which this interest was earned over the period to calculate the annual percentage rate.

🎯 Exam Tip: For problems with multiple parts, solve them sequentially. The result of one part (e.g., total interest) often becomes an input for the next part (e.g., rate of interest).

 

Question 9. Kiran deposited Rs. 200 per month for 36 months in a bank's recurring deposit account. If the rate of interest is 11 % per annum, find the amount she gets on maturity.
Answer:
Monthly deposit (P) = Rs. 200
Period (n) = 36 months
Rate (R) = 11% p.a.

First, calculate the total amount deposited:
Amount deposited \( = P \times n = 200 \times 36 = \) Rs. 7200

Next, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 200 \times \frac{36 \times(36+1)}{2} \)
\( = 200 \times \frac{36 \times 37}{2} \)
\( = 200 \times 18 \times 37 \)
\( = \) Rs. 133200

Now, calculate the simple interest (S.I.) earned:
S.I. \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{133200 \times 11 \times 1}{100 \times 12} \)
\( = \frac{1465200}{1200} \)
\( = \) Rs. 1221

Finally, calculate the amount Kiran will get on maturity:
Maturity amount \( = \text{Amount deposited} + \text{S.I.} \)
\( = 7200 + 1221 \)
\( = \) Rs. 8421
In simple words: We calculate the total money Kiran puts in over three years. Then, we figure out how much extra money (interest) she earns on those deposits. Her final maturity amount is the sum of her total deposits and the interest.

🎯 Exam Tip: Distinguish between the "amount deposited" (P * n) and the "principal for 1 month" (P * n * (n+1) / 2) as they are used in different parts of the maturity value calculation.

 

Question 10. Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is 8% per annum and Mr. Britto gets Rs. 8088 from the bank after 3 years, find the value of his monthly instalment.
Answer:
Let monthly instalment = Rs. x
Period (n) = 3 years = 36 months
Rate (R) = 8% p.a.
Maturity value (MV) = Rs. 8088

First, calculate the principal for 1 month in terms of x:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = x \times \frac{36 \times(36+1)}{2} \)
\( = x \times \frac{36 \times 37}{2} \)
\( = x \times 18 \times 37 \)
\( = 666x \)

Next, calculate the interest earned (I) in terms of x:
I \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
I \( = \frac{666x \times 8 \times 1}{100 \times 12} \)
I \( = \frac{5328x}{1200} \)
I \( = \frac{444x}{100} = 4.44x \)

Now, use the maturity value formula:
Maturity value \( = (P \times n) + \text{Interest} \)
\( 8088 = (x \times 36) + 4.44x \)
\( 8088 = 36x + 4.44x \)
\( 8088 = 40.44x \)
Solve for x:
\( x = \frac{8088}{40.44} \)
\( x = \frac{8088 \times 100}{4044} \)
\( x = 2 \times 100 \)
\( x = 200 \)

Therefore, the value of the monthly instalment = Rs. 200
In simple words: We represent the unknown monthly payment as 'x'. We calculate the interest in terms of 'x' and use the maturity value formula. By setting the total maturity amount equal to the sum of all deposits and the calculated interest, we can solve for 'x'.

🎯 Exam Tip: When dealing with decimals in calculations, especially with division, it's often helpful to convert to fractions or multiply by powers of 10 to clear decimals for easier calculation.

 

Question 11. Katrina opened a recurring deposit account in a bank and deposited Rs. 800 per month for \( 1\frac{1}{2} \) years. If he received Rs. 15,084 at the time of maturity, find the rate of interest per annum.
Answer:
Monthly deposit (P) = Rs. 800
Period (n) = \( 1\frac{1}{2} \) years = 18 months
Maturity amount (MV) = Rs. 15084
Let the rate of interest = R% p.a.

First, calculate the total amount deposited:
Total deposited amount \( = P \times n = 800 \times 18 = \) Rs. 14400

Now, find the interest earned:
Interest (I) \( = \text{Maturity amount} - \text{Total deposited amount} \)
\( = 15084 - 14400 \)
\( = \) Rs. 684

Next, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 800 \times \frac{18 \times(18+1)}{2} \)
\( = 800 \times \frac{18 \times 19}{2} \)
\( = 800 \times 9 \times 19 \)
\( = \) Rs. 136800

Finally, use the interest formula to find the rate (R):
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( 684 = \frac{136800 \times R \times 1}{100 \times 12} \)
\( 684 = \frac{1368 \times R}{12} \)
\( 684 \times 12 = 1368 \times R \)
\( 8208 = 1368 \times R \)
\( R = \frac{8208}{1368} \)
\( R = 6 \) % p.a.
In simple words: We first determine how much extra money Katrina received (interest). Then, we calculate the total amount of money that earned this interest over the period. Using these figures, we can find the yearly interest rate.

🎯 Exam Tip: Clearly distinguish between the total money deposited and the principal that accumulates interest. The latter is crucial for correctly applying the interest rate formula.

 

Question 12. Katrina opened a recurring deposit account with a National Bank for a period of 2 years. If the bank pays interest at the rate of 6% per annum and the monthly instalment is Rs. 1,000, find the:
(i) interest earned in 2 years.
(ii) matured value.
Answer:
Monthly instalment (P) = Rs. 1000
Period (n) = 2 years = 24 months
Rate (R) = 6% p.a.

(i) To find the interest earned in 2 years:
First, calculate the principal for 1 month:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = 1000 \times \frac{24 \times(24+1)}{2} \)
\( = 1000 \times \frac{24 \times 25}{2} \)
\( = 1000 \times 12 \times 25 \)
\( = \) Rs. 300000

Now, calculate the interest:
Interest (I) \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( = \frac{300000 \times 6 \times 1}{100 \times 12} \)
\( = \frac{1800000}{1200} \)
\( = \) Rs. 1500
Thus, the interest earned in 2 years is Rs. 1500.

(ii) To find the matured value:
Total sum deposited in two years \( = P \times n = 1000 \times 24 = \) Rs. 24000
Maturity value \( = \text{Total sum deposited} + \text{Interest} \)
\( = 24000 + 1500 \)
\( = \) Rs. 25500
Thus, the maturity value is Rs. 25500.
In simple words: First, we find the total interest Katrina earns based on her monthly deposits and the bank's rate. Then, we add this interest to the total money she deposited over two years to get the final maturity amount.

🎯 Exam Tip: For problems with multiple questions, ensure you answer each part distinctly and label them clearly. Calculate interest first, as it's needed for the maturity value.

 

Question 13. Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets 1200 as interest at the time of maturity, find :
(i) the monthly instalment
(ii) the amount of maturity
Answer:
Period (n) = 2 years = 24 months
Rate (R) = 6% p.a.
Interest (I) = Rs. 1200

(i) To find the monthly instalment:
Let the monthly instalment = P.

First, calculate the principal for 1 month in terms of P:
Principal for 1 month \( = P \times \frac{n(n+1)}{2} \)
\( = P \times \frac{24 \times(24+1)}{2} \)
\( = P \times \frac{24 \times 25}{2} \)
\( = P \times 12 \times 25 \)
\( = 300P \)

Now, use the interest formula with the given interest:
Interest \( = \frac{\text{Principal for 1 month} \times R \times T}{100} \)
\( 1200 = \frac{300P \times 6 \times 1}{100 \times 12} \)
\( 1200 = \frac{1800P}{1200} \)
\( 1200 = \frac{3P}{2} \)
Solve for P:
\( P = \frac{1200 \times 2}{3} \)
\( P = \frac{2400}{3} \)
\( P = 800 \)
So, the monthly instalment is Rs. 800.

(ii) To find the amount of maturity:
Total sum deposited \( = P \times n = 800 \times 24 = \) Rs. 19200
The amount that Mohan will get at the time of maturity:
Maturity amount \( = \text{Total sum deposited} + \text{Interest} \)
\( = 19200 + 1200 \)
\( = \) Rs. 20400
Hence, the amount of maturity is Rs. 20400.
In simple words: First, we use the given interest and rate to calculate Mohan's monthly deposit. Then, we add this interest to the total amount he deposited over two years to find the final maturity amount.

🎯 Exam Tip: When the monthly deposit is unknown, represent it by 'P' and set up the interest formula in terms of 'P'. Solve for 'P' first, then use this value for any subsequent calculations like maturity amount.

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