Class 10: Banking – Exercise 4(b) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

$\displaystyle P =$ Amount deposited every month

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

$\displaystyle r\% =$ rate of interest

$\displaystyle P = \text{ Rs. } 600, n = 20, r = 10\%$

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

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

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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:

$\displaystyle P = \text{ Rs. } 640$ , no of months $\displaystyle = 54, r = 12\%$

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

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

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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.

Answer:

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 $\displaystyle 50812.5-49860 = \text{ Rs. } 952.5$

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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 Rs. 12715 as the maturity value of this account, what sum of money did he pay every month?

Answer:

$\displaystyle P = \text{ Rs. } x,$ , no of months $\displaystyle = 12, r = 11\%$ Maturity Amount $\displaystyle = \text{ Rs. } 12715$

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

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

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

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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 Rs. 10205 on maturity, find the value of monthly installment.

Answer:

$\displaystyle P = \text{ Rs. } x,$ , no of months $\displaystyle = 42, r = 12\%$ Maturity Amount $\displaystyle = \text{ Rs. } 10205$

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

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

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

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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. 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. [2008]

Answer:

$\displaystyle P = \text{ Rs. } 140$ , 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 $\displaystyle = \text{ Rs. } 7725$

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

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

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

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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:

$\displaystyle P = \text{ Rs. } 150$ , no of months $\displaystyle = 8, r = 8\%$

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

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

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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. 5565; find the rate of interest per annum.

Answer:

$\displaystyle P = \text{ Rs. } 350$ , no of months $\displaystyle = 15, rate = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 5565$

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

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

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

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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.

Answer:

$\displaystyle P = \text{ Rs. } 1200$ , 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.

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. 8100; find the time (in years) of this recurring deposit account.

Answer:

$\displaystyle P = \text{ Rs. } 300$ , 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.

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. 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. [2010]

Answer:

$\displaystyle P = \text{ Rs. } 2500$ , no of months $\displaystyle = 24, rate = r\%$ Maturity Amount $\displaystyle = \text{ Rs. } 67500$