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.

Maturity value for the recurring deposits = Total Sum of Money deposited + Interest earned on it $\displaystyle P =$ Amount deposited every month $\displaystyle n =$ number of months the deposits were made $\displaystyle r\% =$ rate of interest $\displaystyle P = \text{ Rs. } 600, n = 20, r = 10\%$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle = 600 \times 20 + 600 \times \frac{20(20+1)}{2 \times 12} \times \frac{10}{100} = \text{ Rs. } 13050$ $\displaystyle \$

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. $\displaystyle P = \text{ Rs. } 640$, no of months $\displaystyle = 54, r = 12\%$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle = 640 \times 54 +640 \times \frac{54(54+1)}{2 \times 12} \times \frac{12}{100} = \text{ Rs. } 44064$ $\displaystyle \$

Question 3: Person A and Person B both opened recurring bank account in a bank, if A deposited Rs. 1200 per month for 3 years and B deposited Rs. 1500 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.

For Person A $\displaystyle P = \text{ Rs. } 1200$, no of months $\displaystyle = 36, r = 10\%$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle =1200 \times 36 +1200 \times \frac{36(36+1)}{2 \times 12} \times \frac{10}{100} = \text{ Rs. } 49860$

For Person B $\displaystyle P = \text{ Rs. } 1500$ , no of months $\displaystyle = 30, r = 10\%$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle =1500 \times 30 +1500 \times \frac{30(30+1)}{2 \times 12} \times \frac{10}{100} = \text{ Rs. } 50812.5$

B will get more amount. The difference is $\displaystyle 50812.5-49860 = \text{ Rs. } 952.5$ $\displaystyle \$

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 Rs. 12715 as the maturity value of this account, what sum of money did he pay every month? $\displaystyle P = \text{ Rs. } x,$, no of months $\displaystyle = 12, r = 11\%$ Maturity Amount $\displaystyle = \text{ Rs. } 12715$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 12715 =x \times 12 +x \times \frac{12(12+1)}{2 \times 12} \times \frac{11}{100}$ $\displaystyle x(12+ \frac{12 \times 13}{2 \times 12} \times \frac{11}{100} ) = 12715 \Rightarrow x = \text{ Rs. } 1000$ $\displaystyle \$

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 Rs. 10205 on maturity, find the value of monthly installment. $\displaystyle P = \text{ Rs. } x,$, no of months $\displaystyle = 42, r = 12\%$ Maturity Amount $\displaystyle = \text{ Rs. } 10205$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 10205 =x \times 42 +x \times \frac{42(42+1)}{2 \times 12} \times \frac{12}{100}$ $\displaystyle x(42+ \frac{42 \times 43}{2 \times 12} \times \frac{12}{100} ) = 10205 \Rightarrow x = \text{ Rs. } 200$ $\displaystyle \$

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. 8092 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. 7725 at the time of maturity, find the rate of interest per annum. 

i) $\displaystyle P = \text{ Rs. } 140$, no of months $\displaystyle = 48, rate = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 8092$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 8092 =140 \times 48 +140 \times \frac{48(48+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle r = \frac{(8092-140 \times 48) \times (2 \times 12) \times 100}{140 \times 48 \times 49} \Rightarrow r=10\%$

ii) $\displaystyle P = \text{ Rs. } 300$, no of months $\displaystyle = 24, rate = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 7725$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 7725 =300 \times 24 +300 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle r = \frac{(7725-300 \times 24) \times (2 \times 12) \times 100}{300 \times 24 \times 25} \Rightarrow r=7\%$ $\displaystyle \$

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] $\displaystyle P = \text{ Rs. } 150$, no of months $\displaystyle = 8, r = 8\%$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle =150 \times 8 +150 \times \frac{8(8+1)}{2 \times 12} \times \frac{8}{100} = \text{ Rs. } 1236$ $\displaystyle \$

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. 5565; find the rate of interest per annum. $\displaystyle P = \text{ Rs. } 350$, no of months $\displaystyle = 15, rate = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 5565$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 5565 =350 \times 15 +350 \times \frac{15(15+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle r = \frac{(5565-350 \times 15) \times (2 \times 12) \times 100}{350 \times 15 \times 16} \Rightarrow r=9\%$ $\displaystyle \$

Question 9: A recurring deposit account of Rs. 1200 per month has a maturity value of Rs. 12440. 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. $\displaystyle P = \text{ Rs. } 1200$, no of months $\displaystyle = n, rate = 8\%$ Maturity Amount $\displaystyle = \text{ Rs. } 12440$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 12440 =1200 \times n +1200 \times \frac{n(n+1)}{2 \times 12} \times \frac{8}{100}$ $\displaystyle 12440 = 1200n + 4n(n+1)$ $\displaystyle \text{or } n^2+301n-3110=0 \Rightarrow n = 10\text{ or }-311$

Hence $\displaystyle n = 10$ months. $\displaystyle \$

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. 8100; find the time (in years) of this recurring deposit account. $\displaystyle P = \text{ Rs. } 300$, no of months $\displaystyle = n, rate = 12\%$ Maturity Amount $\displaystyle = \text{ Rs. } 8000$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 8100 =300 \times n +300 \times \frac{n(n+1)}{2 \times 12} \times \frac{12}{100}$ $\displaystyle 8100 = 300n + \frac{3}{2} n(n+1)$ $\displaystyle \text{or } 1.5n^2+301.5n-8100=0 \Rightarrow n = 24\text{ or }-225$

Hence $\displaystyle n = 24$ months $\displaystyle \text{or } 2$ years. $\displaystyle \$

Question 11: 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. 67500. Find:

The total interest earned by Mr. Gupta

The rate of interest per annum. $\displaystyle P = \text{ Rs. } 2500$, no of months $\displaystyle = 24, rate = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 67500$ $\displaystyle \text{Maturity Value } = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle 67500 =2500 \times 24 +2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{r}{100}$ $\displaystyle r = \frac{(67500-2500 \times 24) \times (2 \times 12) \times 100}{2500 \times 24 \times 25} \Rightarrow r=12\%$ $\displaystyle \text{Interest } =2500 \times \frac{24(24+1)}{2 \times 12} \times \frac{12}{100} = \text{ Rs. } 7500$