Banking

A. The Summation of Monthly Principals:

Suppose a customer deposits a fixed monthly installment of ₹$P$ at an annual interest rate of $r\%$ for a tenure of $n$ months:

• The 1st installment remains in the bank for $n$ months.
• The 2nd installment remains in the bank for $(n - 1)$ months.
• The 3rd installment remains in the bank for $(n - 2)$ months.
• ...
• The $n^{\text{th}}$ (last) installment remains in the bank for $1$ month.

The total equivalent principal for one single month is the arithmetic progression sum:

B. The Master ICSE Banking Formulas:

Since time $T = \frac{1}{12}$ year, applying the simple interest formula $I = \frac{\text{Principal} \times R \times T}{100}$ yields:

where:
• $P$ = Monthly installment amount (in ₹)
• $n$ = Total number of months (Tenure in years $\times 12$)
• $r$ = Annual rate of simple interest (%)
• $I$ = Total interest earned (in ₹)
• $MV$ = Maturity Value paid to customer at the end of tenure (in ₹)

Model 1: Direct Calculation of Maturity Value:

Problem: Manish opens a recurring deposit account with a bank for $2$ years at $10\%$ per annum simple interest. If he deposits ₹$1,000$ per month, find the amount he will receive upon maturity.

Solution:
Monthly deposit $P = 1000$; Rate $r = 10\%$; Tenure $n = 2 \times 12 = 24$ months.
$$I = \frac{1000 \times 24 \times (24 + 1)}{24} \times \frac{10}{100} = 1000 \times 25 \times 0.10 = ₹2,500$$ $$\text{Total Qualifying Deposit} = P \times n = 1000 \times 24 = ₹24,000$$ $$\text{Maturity Value } MV = 24,000 + 2,500 = \mathbf{₹26,500}.$$

Model 2: Reverse Calculation — Finding Monthly Installment $P$:

Problem: Rekha opened a recurring deposit account for $20$ months. The bank pays interest at $9\%$ p.a. If Rekha received ₹$441$ as interest at maturity, find her monthly installment.

Solution:
Given: $I = 441$, $n = 20$, $r = 9\%$.
$$441 = \frac{P \times 20 \times (21)}{24} \times \frac{9}{100} = \frac{P \times 420 \times 9}{2400} = P \times 1.575$$ $$P = \frac{441}{1.575} = \mathbf{₹280} \text{ per month}.$$

Determining Tenure $n$ Given Maturity Value:

Problem: Katrina opened a recurring deposit account with a bank for a certain number of months. If she deposited ₹$800$ per month at $10\%$ p.a. and received ₹$17,400$ at the time of maturity, find the total tenure of the account in years.

Step-by-Step Solution:

Given: $P = 800$, $r = 10$, $MV = 17400$. Let the tenure be $n$ months.
Total Qualifying Deposit = $800n$.
$$I = \frac{800 \times n(n+1)}{24} \times \frac{10}{100} = \frac{8000 \cdot n(n+1)}{2400} = \frac{10}{3} n(n+1) = \frac{10n^2 + 10n}{3}$$ Since $MV = Pn + I$:
$$17400 = 800n + \frac{10n^2 + 10n}{3}$$ Multiply the entire equation by $3$:
$$52200 = 2400n + 10n^2 + 10n \implies 10n^2 + 2410n - 52200 = 0$$ Divide by $10$:
$$n^2 + 241n - 5220 = 0$$ Factorizing the quadratic equation (finding two factors of $-5220$ whose sum is $+241$):
$261 \times (-20) = -5220$ and $261 - 20 = 241$.
$$(n + 261)(n - 20) = 0$$ Since time $n$ cannot be negative, $n = 20$ months.
$$\text{Tenure in Years} = \frac{20}{12} = 1\frac{8}{12} = \mathbf{1 \text{ year and } 8 \text{ months}} \text{ (or } 1\frac{2}{3} \text{ years)}.$$

Problem 1: Complex Tenure with Fraction of Year:

Question: Deepa has a recurring deposit account in a bank for $3\frac{1}{2}$ years at $9\%$ per annum. If she gets ₹$10,248$ as interest at the time of maturity, find: (i) the monthly installment; (ii) the total amount she receives at maturity.

Solution:
Tenure $n = 3.5 \times 12 = 42$ months. Rate $r = 9\%$. Interest $I = ₹10,248$.
$$I = \frac{P \cdot n(n+1)}{24} \times \frac{r}{100} \implies 10248 = \frac{P \times 42 \times 43}{24} \times \frac{9}{100}$$ $$10248 = \frac{P \times 1806 \times 9}{2400} = \frac{P \times 16254}{2400} = P \times 6.7725$$ $$P = \frac{10248}{6.7725} \approx \mathbf{₹1,513.18} \text{ (or taking exact fraction: } P = \frac{10248 \times 2400}{16254} = \mathbf{₹1,513}).$$ Total deposited = $P \times n = 1513 \times 42 = ₹63,546$.
Maturity Value $MV = 63,546 + 10,248 = \mathbf{₹73,794}$.

Problem 2: Reverse Rate of Interest Calculation:

Question: A man has a recurring deposit account in a post office for $3$ years at a certain rate of simple interest per annum. If the monthly installment is ₹$600$ and he gets ₹$25,122$ on maturity, find the rate of interest per annum.

Solution:
$P = 600$; $n = 3 \times 12 = 36$ months; $MV = 25,122$.
Total Deposit = $P \times n = 600 \times 36 = ₹21,600$.
Interest $I = MV - Pn = 25,122 - 21,600 = ₹3,522$.
$$I = \frac{P \cdot n(n+1)}{24} \times \frac{r}{100} \implies 3522 = \frac{600 \times 36 \times 37}{24} \times \frac{r}{100}$$ $$3522 = \frac{600 \times 1332}{2400} \times r = \frac{799200}{2400} \times r = 333 \times r$$ $$r = \frac{3522}{333} \approx \frac{1174}{111} = 10.57\% \approx \mathbf{10.6\% \text{ p.a.}}.$$

Rigorous Step-by-Step Solutions for Top-Band Scores:

Below is a curated compendium of standard ICSE board-level examination questions designed to test mathematical derivation, algebraic accuracy, unit precision, and rigorous geometric justifications.

How ICSE Examiners Award Marks in this Topic:

In the official CISCE evaluation rubrics, marks are systematically distributed across three distinct cognitive dimensions:

• Method Marks (M): Awarded for writing the correct formula, establishing the proper geometric theorem, setting up the correct equation, or constructing the proper table columns. Even if a careless calculation error occurs later, method marks are fully preserved!
• Accuracy Marks (A): Awarded for correct intermediate arithmetic steps, accurate factorization, algebraic simplification, and correct radical reduction.
• Final Statement & Unit Marks (B/A): Awarded for stating the final numerical answer with correct units, proper rounding (e.g. two decimal places), and answering all sub-parts explicitly.

Top 5 Practical Recommendations to Maximize Scores:

1. Never skip writing the standard formula: Always write out the formula in algebraic terms before substituting numerical values.
2. Keep rough calculations neatly organized: Draw a dedicated 2-inch rough margin on the right side of your answer sheet for long divisions and square root extractions.
3. Check boundary conditions: Ensure your final solutions belong strictly to the specified domain (e.g., natural numbers, positive lengths, valid quadrants).
4. State geometric reasons in parentheses: In geometry proofs, every equality must be accompanied by its supporting theorem name (e.g., '[angles in the same segment are equal]').
5. Proofread before moving to the next question: Spend 30 seconds verifying signs, basic arithmetic, and decimal point placements.