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Commercial Mathematics - Banking (Recurring Deposit Accounts)

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Recurring Deposit (RD) Account: A banking product where an individual deposits a fixed monthly installment PP for a predetermined number of months nn to earn interest at a fixed annual rate rr.

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Monthly Installment (PP): This is the constant amount deposited every month. It remains unchanged throughout the tenure of the account.

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Number of Months (nn): In ICSE banking problems, the time duration is always converted into months. For example, if a period is given as 1.5 years, it must be used as n=18n = 18 months.

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Equivalent Principal for One Month: Since the first installment earns interest for nn months and the last installment for 1 month, we calculate the interest on a cumulative principal. This is visually represented by the sum of an arithmetic progression: n+(n−1)+(n−2)+...+1n + (n-1) + (n-2) + ... + 1. The sum is calculated as n(n+1)2\frac{n(n+1)}{2}.

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Interest (II): The extra money earned on the total sum deposited. It is calculated using the simple interest formula applied to the equivalent principal for one month, which accounts for the varying time each installment stays in the bank.

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Maturity Value (MVMV): This is the final amount received by the depositor at the end of the term. It consists of the sum of all monthly installments (P×nP \times n) plus the total interest earned (II). Visually, it represents the accumulation of fixed 'blocks' of money plus an additional 'bonus' layer of interest.

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Annual Rate of Interest (rr): The interest rate is always quoted per annum (p.a.). In the calculations, it is adjusted for monthly deposits by dividing by 12 within the formula structure.

📐Formulae

Total Deposit=P×nTotal\ Deposit = P \times n

Interest (I)=P×n(n+1)2×12×r100Interest\ (I) = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100}

Maturity Value (MV)=(P×n)+IMaturity\ Value\ (MV) = (P \times n) + I

MV=(P×n)+(P×n(n+1)24×r100)MV = (P \times n) + \left( P \times \frac{n(n+1)}{24} \times \frac{r}{100} \right)

💡Examples

Problem 1:

Mrs. Goswami deposits 10001000 every month in a Recurring Deposit Account for 33 years at 8%8\% interest per annum. Find the maturity value of her account.

Solution:

Given: Monthly deposit P=1000P = 1000 Time n=3×12=36n = 3 \times 12 = 36 months Rate r=8%r = 8\%

Step 1: Calculate Interest (II) I=P×n(n+1)2×12×r100I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} I=1000×36×3724×8100I = 1000 \times \frac{36 \times 37}{24} \times \frac{8}{100} I=10×133224×8I = 10 \times \frac{1332}{24} \times 8 I=10×55.5×8=4440I = 10 \times 55.5 \times 8 = 4440

Step 2: Calculate Maturity Value (MVMV) MV=(P×n)+IMV = (P \times n) + I MV=(1000×36)+4440MV = (1000 \times 36) + 4440 MV=36000+4440=40440MV = 36000 + 4440 = 40440

Answer: The maturity value is 4044040440.

Explanation:

First, we convert the time from years to months (3636 months). Then we apply the interest formula for RD accounts. Finally, we add the total money deposited (3600036000) to the interest earned (44404440) to get the maturity value.

Problem 2:

Mohan has a Recurring Deposit Account in a bank for 22 years at 6%6\% p.a. simple interest. If he gets 12001200 as interest at the time of maturity, find the monthly installment.

Solution:

Given: Interest I=1200I = 1200 Time n=2×12=24n = 2 \times 12 = 24 months Rate r=6%r = 6\%

Step 1: Substitute values into the Interest formula I=P×n(n+1)2×12×r100I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} 1200=P×24×2524×61001200 = P \times \frac{24 \times 25}{24} \times \frac{6}{100}

Step 2: Simplify the equation 1200=P×25×61001200 = P \times 25 \times \frac{6}{100} 1200=P×1501001200 = P \times \frac{150}{100} 1200=P×1.51200 = P \times 1.5

Step 3: Solve for PP P=12001.5=800P = \frac{1200}{1.5} = 800

Answer: The monthly installment is 800800.

Explanation:

In this problem, the interest is already provided. We use the Interest formula and treat the monthly installment PP as the unknown variable. By rearranging the formula and solving the linear equation, we find the value of PP.