Question 1: A person opens a recurring deposit account with a Bank  and deposited 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:

Maturity value for the recurring deposits = Total Sum of Money deposited + Interest earned on it

P = Amount \ deposited \ every \ month

n = number \ of \ months \ the \ deposits \ were \ made

r\% = rate \ of \ interest

P = Rs. \ 600, \ n = 20, \ r = 10\%

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

 = 600 \times 20 + 600 \times \frac{20(20+1)}{2 \times 12} \times \frac{10}{100} = Rs. 13050

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Question 2: A person  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:

P = Rs. \ 640, \ no \ of \ months = 54, \ r = 12\%

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

 =640 \times 54 +640 \times \frac{54(54+1)}{2 \times 12} \times \frac{12}{100} = Rs. 44064

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Question 3: Person A and Person B both opened recurring bank account in a bank, if A deposited 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 Person A

P = Rs. \ 1200, \ no \ of \ months = 36, \ r = 10\%

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

 =1200 \times 36 +1200 \times \frac{36(36+1)}{2 \times 12} \times \frac{10}{100} = Rs. 49860

For Person B

P = Rs. \ 1500, \ no of months = 30, \ r = 10\%

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

 =1500 \times 30 +1500 \times \frac{30(30+1)}{2 \times 12} \times \frac{10}{100} = Rs. 50812.5

B will get more amount. The difference is 50812.5-49860 = Rs. 952.5

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Question 4: A person deposited a certain sum of money every month in a recurring deposit account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and he gets 12,715 as the maturity value of this account, what sum of money did he pay every month?

Answer:

P = Rs. \ x, \ no \ of \ months = 12, \ r = 11\% Maturity Amount = Rs. 12715

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

 12715 =x \times 12 +x \times \frac{12(12+1)}{2 \times 12} \times \frac{11}{100}  

x(12+\frac{12 \times 13}{2 \times 12} \times \frac{11}{100}) = 12715 \Rightarrow x = Rs. 1000

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Question 5: A man has recurring deposit account in a bank for 3 ½ years. If the rate of interest is 12% p.a. and the man gets 10,205 on maturity, find the value of monthly installment.

Answer:

P = Rs. \ x, \ no \ of \ months = 42, \ r = 12\% Maturity Amount = Rs. 10205

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

 10205 =x \times 42 +x \times \frac{42(42+1)}{2 \times 12} \times \frac{12}{100}  

x(42+\frac{42 \times 43}{2 \times 12} \times \frac{12}{100}) = 10205 \Rightarrow x = Rs. 200

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Question 6: Explain the following:

i) Punnet has a recurring deposit account in 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. [2008]

Answer:

i)

P = Rs. \ 140, \ no \ of \ months = 48, \ rate = r\%  \ Maturity Amount = Rs. 8092

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

8092 =140 \times 48 +140 \times \frac{48(48+1)}{2 \times 12} \times \frac{r}{100}  

r = \frac{(8092-140 \times 48) \times (2 \times 12) \times 100}{140 \times 48 \times 49} \Rightarrow r=10\%

ii)

P = Rs. \ 300, \ no \ of \ months = 24, \ rate = r\%  \ Maturity Amount = Rs. 7725

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

7725 =300 \times 24 +300 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100}  

r = \frac{(7725-300 \times 24) \times (2 \times 12) \times 100}{300 \times 24 \times 25} \Rightarrow r=7\%

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Question 7: Amit deposited 150 per month in a bank for 8 month 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? [2001, 2007]

Answer:

P = Rs. \ 150, \ no \ of \ months = 8, \ r = 8\%

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

 =150 \times 8 +150 \times \frac{8(8+1)}{2 \times 12} \times \frac{8}{100} = Rs. 1236

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Question 8: A person 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:

P = Rs. \ 350, \ no \ of \ months = 15, \ rate = r\%  \ Maturity Amount = Rs. 5565

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

5565 =350 \times 15 +350 \times \frac{15(15+1)}{2 \times 12} \times \frac{r}{100}  

r = \frac{(5565-350 \times 15) \times (2 \times 12) \times 100}{350 \times 15 \times 16} \Rightarrow r=9\%

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Question 9: A recurring deposit account of 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:

P = Rs. \ 1200, \ no \ of \ months = n, \ rate = 8\%  \ Maturity Amount = Rs. 12440

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

12440 =1200 \times n +1200 \times \frac{n(n+1)}{2 \times 12} \times \frac{8}{100}  

12440 = 1200n + 4n(n+1)

or n^2+301n-3110=0 \Rightarrow n = 10 \ or -311

Hence n = 10 months.

Notes: Please refer to quadratic equations for solving this.

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Question 10: A person 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:

P = Rs. \ 300, \ no \ of \ months = n, \ rate = 12\%  \ Maturity Amount = Rs. 8000

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

8100 =300 \times n +300 \times \frac{n(n+1)}{2 \times 12} \times \frac{12}{100}  

8100 = 300n + \frac{3}{2}n(n+1)

or 1.5n^2+301.5n-8100=0 \Rightarrow n = 24 \ or -225

Hence n = 24 months or 2 years.

Notes: Please refer to quadratic equations for solving this.

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Question 11: 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:

  • The total interest earned by Mr. Gupta
  • The rate of interest per annum. [2010]

Answer:

P = Rs. \ 2500, \ no \ of \ months = 24, \ rate = r\%  \ Maturity Amount = Rs. 67500

Maturity \ Value = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

67500 =2500 \times 24 +2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100}  

r = \frac{(67500-2500 \times 24) \times (2 \times 12) \times 100}{2500 \times 24 \times 25} \Rightarrow r=12\%

Interest =2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{12}{100} = Rs. \ 7500

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