Class 10: Banking – Exercise 4(b)

$\displaystyle \textbf{Question 1: }\text{A person opens a Recurring Deposit Account with a bank and}$
$\displaystyle \text{deposits Rs. }600\text{ per month for }20\text{ months. Calculate the maturity value}$
$\displaystyle \text{of the account if the bank pays interest at }10\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Maturity Value}=\text{Total amount deposited}+\text{Interest earned}$
$\displaystyle \text{Here, }P=\text{Rs. }600,\ n=20,\ r=10\%$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle =600\times20+600\times\frac{20(20+1)}{2\times12}\times\frac{10}{100}$
$\displaystyle =12000+1050=\text{Rs. }13050$
$\\$

$\displaystyle \textbf{Question 2: }\text{A person opened a Recurring Deposit Account in a bank and}$
$\displaystyle \text{deposited Rs. }640\text{ per month for }4\frac{1}{2}\text{ years. Find the maturity}$
$\displaystyle \text{value of the account if the bank pays interest at }12\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }P=\text{Rs. }640,\ n=54,\ r=12\%$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle =640\times54+640\times\frac{54(54+1)}{2\times12}\times\frac{12}{100}$
$\displaystyle =34560+9504=\text{Rs. }44064$
$\\$

$\displaystyle \textbf{Question 3: }\text{Person A and Person B both opened Recurring Deposit Accounts in a}$
$\displaystyle \text{bank. If A deposited Rs. }1200\text{ per month for }3\text{ years and B deposited}$
$\displaystyle \text{Rs. }1500\text{ per month for }2\frac{1}{2}\text{ years, find who will get more amount}$
$\displaystyle \text{on maturity and by how much. The rate of interest is }10\%\text{ per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For Person A:}$
$\displaystyle P=\text{Rs. }1200,\ n=36,\ r=10\%$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle =1200\times36+1200\times\frac{36(36+1)}{2\times12}\times\frac{10}{100}$
$\displaystyle =\text{Rs. }49860$
$\displaystyle \text{For Person B:}$
$\displaystyle P=\text{Rs. }1500,\ n=30,\ r=10\%$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{30(30+1)}{2\times12}\times\frac{10}{100}$
$\displaystyle =1500\times30+1500\times\frac{30(30+1)}{2\times12}\times\frac{10}{100}$
$\displaystyle =\text{Rs. }50812.50$
$\displaystyle \text{Difference}=50812.50-49860=\text{Rs. }952.50$
$\displaystyle \therefore \text{Person B gets more by Rs. }952.50.$
$\\$

$\displaystyle \textbf{Question 4: }\text{A person deposited a certain sum every month in a Recurring Deposit}$
$\displaystyle \text{Account for }12\text{ months. If the bank pays interest at }11\%\text{ per annum}$
$\displaystyle \text{and the maturity value is Rs. }12715,\text{ find the monthly deposit.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the monthly deposit be Rs. }x.$
$\displaystyle P=\text{Rs. }x,\ n=12,\ r=11\%,\ \text{Maturity Value}=\text{Rs. }12715$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 12715=12x+x\times\frac{12(12+1)}{2\times12}\times\frac{11}{100}$
$\displaystyle 12715=x\left(12+\frac{13\times11}{200}\right)$
$\displaystyle 12715=x(12.715)$
$\displaystyle x=\frac{12715}{12.715}=\text{Rs. }1000$
$\displaystyle \therefore \text{The monthly deposit is Rs. }1000.$
$\\$

$\displaystyle \textbf{Question 5: }\text{A man has a Recurring Deposit Account in a bank for }3\frac{1}{2}\text{ years.}$
$\displaystyle \text{If the rate of interest is }12\%\text{ per annum and he gets Rs. }10205\text{ on}$
$\displaystyle \text{maturity, find the monthly installment.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the monthly installment be Rs. }x.$
$\displaystyle P=\text{Rs. }x,\ n=42,\ r=12\%,\ \text{Maturity Value}=\text{Rs. }10205$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 10205=42x+x\times\frac{42(42+1)}{2\times12}\times\frac{12}{100}$
$\displaystyle x\left(42+\frac{42\times43}{2\times12}\times\frac{12}{100}\right)=10205$
$\displaystyle x=\text{Rs. }200$
$\\$

$\displaystyle \textbf{Question 6: }\text{Explain the following:}$
$\displaystyle \text{(i) Punnet has a Recurring Deposit Account in Bank of Baroda and deposits}$
$\displaystyle \text{Rs. }140\text{ per month for }4\text{ years. If he gets Rs. }8092\text{ on maturity,}$
$\displaystyle \text{find the rate of interest given by the bank.}$
$\displaystyle \text{(ii) David opened a Recurring Deposit Account in a bank and deposited}$
$\displaystyle \text{Rs. }300\text{ per month for }2\text{ years. If he received Rs. }7725\text{ on maturity,}$
$\displaystyle \text{find the rate of interest per annum.}\hfill \text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Here, }P=\text{Rs. }140,\ n=48,\ r=r\%,\ \text{Maturity Value}=\text{Rs. }8092$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 8092=140\times48+140\times\frac{48(48+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle r=\frac{(8092-140\times48)\times(2\times12)\times100}{140\times48\times49}$
$\displaystyle r=10\%$
$\displaystyle \text{(ii) Here, }P=\text{Rs. }300,\ n=24,\ r=r\%,\ \text{Maturity Value}=\text{Rs. }7725$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 7725=300\times24+300\times\frac{24(24+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle r=\frac{(7725-300\times24)\times(2\times12)\times100}{300\times24\times25}$
$\displaystyle r=7\%$
$\\$

$\displaystyle \textbf{Question 7: }\text{Amit deposited Rs. }150\text{ per month in a bank for }8\text{ months}$
$\displaystyle \text{under the Recurring Deposit Scheme. What will be the maturity value}$
$\displaystyle \text{of his deposits, if the rate of interest is }8\%\text{ per annum and}$
$\displaystyle \text{interest is calculated at the end of every month?}\hfill \text{[ICSE 2001, 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }150,\ n=8,\ r=8\%$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle =150\times8+150\times\frac{8(8+1)}{2\times12}\times\frac{8}{100}$
$\displaystyle =1200+36=\text{Rs. }1236$
$\\$

$\displaystyle \textbf{Question 8: }\text{A person deposited Rs. }350\text{ per month in a bank for }1\text{ year}$
$\displaystyle \text{and }3\text{ months under the Recurring Deposit Scheme. If the maturity}$
$\displaystyle \text{value of the deposits is Rs. }5565,\text{ find the rate of interest per annum.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }350,\ n=15,\ r=r\%,\ \text{Maturity Value}=\text{Rs. }5565$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 5565=350\times15+350\times\frac{15(15+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle r=\frac{(5565-350\times15)\times(2\times12)\times100}{350\times15\times16}$
$\displaystyle r=9\%$
$\\$

$\displaystyle \textbf{Question 9: }\text{A Recurring Deposit Account of Rs. }1200\text{ per month has}$
$\displaystyle \text{a maturity value of Rs. }12440.\text{ If the rate of interest is }8\%\text{ and}$
$\displaystyle \text{interest is calculated at the end of every month, find the time in months.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }1200,\ n=n,\ r=8\%,\ \text{Maturity Value}=\text{Rs. }12440$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 12440=1200n+1200\times\frac{n(n+1)}{2\times12}\times\frac{8}{100}$
$\displaystyle 12440=1200n+4n(n+1)$
$\displaystyle 12440=1200n+4n^2+4n$
$\displaystyle 4n^2+1204n-12440=0$
$\displaystyle n^2+301n-3110=0$
$\displaystyle (n-10)(n+311)=0$
$\displaystyle n=10\text{ or }n=-311$
$\displaystyle \text{Since the number of months cannot be negative, }n=10.$
$\displaystyle \therefore \text{The time is }10\text{ months.}$
$\\$

$\displaystyle \textbf{Question 10: }\text{A person has a Recurring Deposit Account of Rs. }300\text{ per month.}$
$\displaystyle \text{If the rate of interest is }12\%\text{ and the maturity value is Rs. }8100,$
$\displaystyle \text{find the time in years of this Recurring Deposit Account.}$
$\displaystyle \text{Answer:}$
$\displaystyle P=\text{Rs. }300,\ n=n,\ r=12\%,\ \text{Maturity Value}=\text{Rs. }8100$
$\displaystyle \text{Maturity Value}=P\times n+P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 8100=300n+300\times\frac{n(n+1)}{2\times12}\times\frac{12}{100}$
$\displaystyle 8100=300n+\frac{3}{2}n(n+1)$
$\displaystyle 8100=300n+\frac{3}{2}n^2+\frac{3}{2}n$
$\displaystyle 1.5n^2+301.5n-8100=0$
$\displaystyle n^2+201n-5400=0$
$\displaystyle (n-24)(n+225)=0$
$\displaystyle n=24\text{ or }n=-225$
$\displaystyle \text{Since the number of months cannot be negative, }n=24.$
$\displaystyle 24\text{ months}=2\text{ years}$
$\displaystyle \therefore \text{The time is }2\text{ years.}$
$\\$

$\displaystyle \textbf{Question 11: }\text{Gupta opened a Recurring Deposit Account in a bank. He deposited}$
$\displaystyle \text{Rs. }2500\text{ per month for }2\text{ years. At the time of maturity, he received}$
$\displaystyle \text{Rs. }67500.\text{ Find:}$
$\displaystyle \text{(i) The total interest earned}\qquad\text{(ii) The rate of interest per annum.}\hfill \text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }P=\text{Rs. }2500,\ n=24,\ \text{Maturity Value}=\text{Rs. }67500$
$\displaystyle \text{Total amount deposited}=2500\times24=\text{Rs. }60000$
$\displaystyle \text{Interest earned}=67500-60000=\text{Rs. }7500$
$\displaystyle \text{Let the rate of interest be }r\%\text{ per annum.}$
$\displaystyle \text{Interest}=P\times\frac{n(n+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle 7500=2500\times\frac{24(24+1)}{2\times12}\times\frac{r}{100}$
$\displaystyle r=\frac{7500\times(2\times12)\times100}{2500\times24\times25}$
$\displaystyle r=12\%$
$\displaystyle \therefore \text{(i) Interest earned}=\text{Rs. }7500$
$\displaystyle \therefore \text{(ii) Rate of interest}=12\%\text{ per annum}$
$\\$