02. Banking (Recurring Deposit Accounts) Previous Year Questions

Maturity Value MV = Total Principal Deposited + Total Interest = (P × n) + I = ₹(500 × 18 + 570). [1 Mark]

P = ₹250, n = 3 × 12 = 36 months, MV = ₹10,110.
Total deposited = P × n = 250 × 36 = ₹9,000.
Interest earned I = MV − (P × n) = 10,110 − 9,000 = ₹1,110. [1 Mark]

(a) I = P × [n(n+1)/(2 × 12)] × (r / 100)
1110 = 250 × (36 × 37 / 24) × (r / 100) = 2.5 × 55.5 × r = 138.75 r ⇒ r = 8% per annum. [2 Marks]

(b) If deposit is ₹50 more ⇒ Extra interest = (50 / 250) × ₹1110 = ₹222. [1 Mark]

Monthly deposit P = ₹2000, R = 8% p.a., Interest I = ₹1040.
Formula: I = P × [n(n+1)/(2 × 12)] × (R/100) [1 Mark]
1040 = 2000 × [n(n+1)/24] × (8/100) = (20/3) × n(n+1)
n(n+1) = (1040 × 3)/20 = 156 ⇒ n² + n − 156 = 0 [1 Mark]
(n + 13)(n − 12) = 0 ⇒ n = 12 months.
Total time in years = 12/12 = 1 year. [1 Mark]

Total principal deposited $P \times n = 800 \times 6 = ₹4,800$.
Maturity Value $(MV) = ₹4,884$.
Interest earned = $MV - (P \times n) = 4884 - 4800 = ₹84$.
Correct Option: (a) [1 Mark]

Monthly deposit $P = ₹1,000$. Time period $n = 2\text{ years} = 24\text{ months}$.
Maturity value $MV = ₹26,000$.

Part (a) Total deposited amount $= P \times n = 1000 \times 24 = ₹24,000$.
Total Interest earned $I = MV - (P \times n) = 26,000 - 24,000 = \mathbf{₹2,000}$.2 Marks

Part (b) Recurring deposit interest formula:
$I = \frac{P \times n(n + 1)}{2 \times 12} \times \frac{r}{100}$
$2000 = \frac{1000 \times 24 \times 25}{24} \times \frac{r}{100}$1 Mark
$2000 = 10 \times 25 \times r \implies 2000 = 250r \implies r = \frac{2000}{250} = \mathbf{8\%}$.
Rate of interest = 8% per annum.1 Mark

Monthly deposit $P = ₹750$, Time $n = 2 \text{ years} = 24 \text{ months}$.
Step 1 Total deposit $= P \times n = 750 \times 24 = ₹18,000$.
Maturity Value $= ₹19,125 \implies$ Total Interest $I = 19,125 - 18,000 = \mathbf{₹1,125}$. [1 Mark]
Step 2 Apply RD interest formula:
$$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100} \implies 1125 = 750 \times \frac{24 \times 25}{24} \times \frac{R}{100}$$ [1 Mark]
$$1125 = 750 \times 25 \times \frac{R}{100} = 187.5 R \implies R = \frac{1125}{187.5} = \mathbf{6\% \text{ p.a.}}$$
Answer: Rate of interest = 6% per annum. [1 Mark]

Step 1 State parameters and formula:
Number of months $n = 20$.
Rate of interest $r = 9\% \text{ p.a.}$, Interest $I = \text{₹ } 441$.
Formula: $I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100}$. [1 Mark]
Step 2 Substitute given values into equation:
$441 = P \times \frac{20(21)}{24} \times \frac{9}{100}$
$441 = P \times \frac{420}{24} \times \frac{9}{100} = P \times 17.5 \times 0.09 = P \times 1.575$. [1 Mark]
Step 3 Solve for monthly deposit $P$:
$P = \frac{441}{1.575} = \frac{441000}{1575} = \mathbf{\text{₹ } 280}$.
Answer: Rekha deposited ₹ 280 each month. [1 Mark]

Solution Formula for RD interest: $I = P \times \frac{n(n + 1)}{24} \times \frac{r}{100}$.
Since $I \propto r$ when $P$ and $n$ remain constant, doubling the rate from $r\%$ to $2r\%$ doubles the interest earned.
Correct Answer: (d) double [1 Mark]

Step 1 (a) Interest earned by Anamika:
$P = \text{₹ } 300, n = 3 \times 12 = 36\text{ months}, r = 10\%$.
$I_{\text{Anamika}} = 300 \times \frac{36 \times 37}{24} \times \frac{10}{100} = 300 \times 55.5 \times 0.1 = \mathbf{\text{₹ } 1665}$. [2 Marks]
Step 2 Interest earned by Sanjeev:
$P = \text{₹ } 500, n = 2 \times 12 = 24\text{ months}, r = 10\%$.
$I_{\text{Sanjeev}} = 500 \times \frac{24 \times 25}{24} \times \frac{10}{100} = 500 \times 25 \times 0.1 = \text{₹ } 1250$. [1 Mark]
Step 3 (b) Compare interest earned:
$I_{\text{Anamika}} - I_{\text{Sanjeev}} = 1665 - 1250 = \text{₹ } 415$.
$\mathbf{\text{Anamika got more interest by ₹ 415}}$. [1 Mark]

Step 1 Given parameters:
Monthly deposit $P = ₹600$
Time $n = 2\frac{1}{2}\text{ years} = 2.5 \times 12 = 30\text{ months}$
Rate of interest $R = 10\%$ p.a.
Step 2 Calculate total simple interest earned:
$$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100}$$ $$I = 600 \times \frac{30 \times 31}{24} \times \frac{10}{100} = 600 \times \frac{930}{24} \times \frac{1}{10} = 25 \times 93 = ₹2,325$$ [1.5 Marks]
Step 3 Total sum deposited and maturity value:
Total principal deposited $= P \times n = 600 \times 30 = ₹18,000$ [0.5 Mark]
$$\text{Maturity Value } (MV) = (P \times n) + I = 18,000 + 2,325 = \mathbf{₹20,325}$$ [1 Mark]
Answer: The maturity value of the account is ₹20,325.

Step 1 RD Formula for interest: $I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100}$.
Here $P = 1000, R = 10\%, I = ₹5550$. [1 Mark]
Step 2 Substitute values:
$$5550 = 1000 \times \frac{n(n + 1)}{24} \times \frac{10}{100} = \frac{100 \times n(n + 1)}{24}$$ $$5550 \times 24 = 100(n^2 + n) \implies 133200 = 100(n^2 + n)$$ $$n^2 + n - 1332 = 0$$ [1 Mark]
Step 3 Factorise the quadratic equation ($36 \times 37 = 1332$):
$$(n + 37)(n - 36) = 0$$ Since time cannot be negative, $n = 36\text{ months}$.
$$\text{Time} = \frac{36}{12} = \mathbf{3\text{ years}}$$ [1 Mark]
Answer: 36 months (or 3 years).

Year 1 Principal $P_1 = ₹50,000$, Rate $R_1 = 12\%$.
$$\text{Interest for Year 1 } (I_1) = \frac{50,000 \times 12 \times 1}{100} = ₹6,000$$ $$\text{Amount at end of Year 1} = 50,000 + 6,000 = ₹56,000$$ [1 Mark]
Repayment Repaid at end of Year 1 $= ₹33,000$.
$$\text{Principal for Year 2 } (P_2) = 56,000 - 33,000 = ₹23,000$$ [1 Mark]
Year 2 Rate $R_2 = 15\%$.
$$\text{Interest for Year 2 } (I_2) = \frac{23,000 \times 15 \times 1}{100} = ₹3,450$$ $$\text{Amount to clear debt} = 23,000 + 3,450 = \mathbf{₹26,450}$$ [1 Mark]
Answer: Jaya must pay ₹26,450 at the end of the second year.

Qualifying Minimum Balances (10th to end of month)
• January: ₹5,600
• February: ₹4,100
• March: ₹4,100
• April: ₹2,000
• May: ₹8,500
• June: ₹10,000
$$\text{Total Qualifying Balance } (\sum P) = 5600 + 4100 + 4100 + 2000 + 8500 + 10000 = ₹34,300$$ [2.5 Marks]
(i) Calculate Interest
$$\text{Interest} = \frac{\sum P \times R \times 1}{100 \times 12} = \frac{34,300 \times 6 \times 1}{1200} = \frac{34,300}{200} = \mathbf{₹171.50}$$ [1.5 Marks]
(ii) Amount Received on Closing Final balance in June was ₹10,000.
$$\text{Total Amount} = 10,000 + 171.50 = \mathbf{₹10,171.50}$$ [1 Mark]
Answers: (i) Interest $= ₹171.50$, (ii) Amount received $= ₹10,171.50$.

Step 1 Time $n = 3 \times 12 = 36\text{ months}$, Rate $R = 7.5\%$, Interest $I = ₹8,325$.
$$I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{R}{100}$$ $$8325 = P \times \frac{36 \times 37}{24} \times \frac{7.5}{100} = P \times 55.5 \times 0.075 = 4.1625 P$$ [1 Mark]
(i) Monthly Deposit P
$$P = \frac{8325}{4.1625} = \mathbf{₹2,000}$$ [1 Mark]
(ii) Maturity Value MV Total deposited $= 2000 \times 36 = ₹72,000$.
$$MV = (P \times n) + I = 72,000 + 8,325 = \mathbf{₹80,325}$$ [1 Mark]
Answers: (i) Monthly deposit $= ₹2,000$, (ii) Maturity value $= ₹80,325$.