Official Syllabus Item: 11 Commercial Mathematics

Simple Interest, Amount & Financial Growth

Master Principal, Rate p.a., Time conversion (months/days/years), Simple Interest, Amount, sum doubling/tripling rules, and multi-period amount differences for RIMC Entrance Exam.

Overview & Fundamental Concept

Simple Interest (SI) is one of the most foundational and frequently tested commercial arithmetic topics in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Item 11 of Chapter 8 in the official RIMC brochure, questions in this module appear regularly in Part A (5 Marks) testing parameter substitutions, time conversions (months and days to fractional years), and missing parameter evaluations, and in Part B (10 Marks) testing multi-year amount differences ($A_1$ in $T_1$ years vs. $A_2$ in $T_2$ years), sum $n$-folding rules, and division of a principal into multiple investment schemes with equal returns. RIMC examiners demand formal declarations of all five core financial variables ($P, R, T, SI, A$), explicit demonstration of time conversion to years ($T = \text{months}/12$ or $\text{days}/365$), step-wise algebraic transposition, and boxed final answers with appropriate rupee amounts and percentage symbols.

Official Syllabus Coverage Breakdown

Syllabus Item	Prescribed Scope	Exam Relevance & Pedagogical Objective
Item 11	Simple Interest: Principal, rate, time, interest, amount, and applied financial problems	Core brochure syllabus item. Tests temporal unit conversions, linear capital accumulation, capital splitting, and multi-year difference analysis.

1. Core Financial Terms & Conceptual Definition

When money is borrowed or lent, a fee is paid for the use of that capital. This charge is termed Interest. In the Simple Interest system, interest is calculated strictly and uniformly on the original principal throughout the entire duration of the loan, regardless of accumulated past interest.

The five fundamental parameters of the system are:

• Principal $(P)$: The original sum of money borrowed, lent, or deposited.
• Time $(T)$: The duration for which the principal is utilized. Crucial Law: $T$ MUST always be converted into YEARS in all formulas.
• Rate of Interest $(R)$: The interest paid on every ₹ $100$ of principal for one complete year, expressed as a percentage per annum ($R\%\text{ p.a.}$).
• Simple Interest $(SI)$: The total uniform interest accrued over the loan duration.
• Total Amount $(A)$: The combined total repaid at the end of the term:

$$A = P + SI$$

Mandatory Time Unit Conversion Table

RIMC questions routinely provide time in months or days. Never plug raw months or days directly into the formula!

Months to Years: Divide by $12$: $T = \frac{\text{Number of Months}}{12}\text{ years}$. (e.g., $3\text{ years } 4\text{ months} = 3 + \frac{4}{12} = 3 + \frac{1}{3} = \frac{10}{3}\text{ years}$).
Days to Years: Divide by $365$ (non-leap year): $T = \frac{\text{Number of Days}}{365}\text{ years}$.
High-Yield 73-Day Multiples (Must Memorize):
$73\text{ days} = \frac{73}{365} = \mathbf{\frac{1}{5}\text{ year}}$, $\quad 146\text{ days} = \mathbf{\frac{2}{5}\text{ year}}$, $\quad 219\text{ days} = \mathbf{\frac{3}{5}\text{ year}}$, $\quad 292\text{ days} = \mathbf{\frac{4}{5}\text{ year}}$.

2. The Master Simple Interest Formula & Direct Transpositions

The foundational equation governing simple interest is:

$$SI = \frac{P \times R \times T}{100}$$

By algebraic transposition, any of the three independent variables can be isolated directly when the other parameters are known:

Required Parameter	Transposed Formula	Units & Exam Note
Principal $(P)$	$$P = \frac{100 \times SI}{R \times T}$$	Expressed in Rupees (₹).
Rate of Interest $(R)$	$$R = \frac{100 \times SI}{P \times T}$$	Expressed as Percentage per annum ($\%\text{ p.a.}$).
Time Duration $(T)$	$$T = \frac{100 \times SI}{P \times R}$$	Expressed in Years.
Principal from Amount $(A)$	$$P = \frac{100 \times A}{100 + (R \times T)}$$	Used when Amount $A$ is given instead of Interest $SI$.

3. Special High-Yield RIMC Exam Paradigms

In RIMC Stage-I, two classic problem archetypes appear with extreme regularity:

Paradigm A: Doubling / Tripling (n-Folding) of Principal

When a question states that a sum of money becomes $n$ times itself in $T$ years at $R\%$ p.a.:

Amount $A = n \times P$.
Simple Interest $SI = A - P = nP - P = (n - 1)P$.
Substitute into formula: $(n - 1)P = \frac{P \times R \times T}{100} \implies \mathbf{R \times T = 100(n - 1)}$.
• For Doubling $(n = 2)$: $R \times T = 100(2 - 1) = 100$. (e.g., at $12.5\%$, $T = \frac{100}{12.5} = 8\text{ years}$).
• For Tripling $(n = 3)$: $R \times T = 100(3 - 1) = 200$.
• For Quadrupling $(n = 4)$: $R \times T = 100(4 - 1) = 300$.

Paradigm B: Multi-Year Amount Differences ($A_1$ in $T_1$ vs. $A_2$ in $T_2$)

When a sum amounts to ₹ $A_1$ in $T_1$ years and to ₹ $A_2$ in $T_2$ years at the same rate:

Because principal remains unchanged, the difference between the two amounts is purely the interest accrued during the intervening period:
$$\text{Interest for } (T_2 - T_1) \text{ years} = A_2 - A_1$$
$$\text{Interest for } 1\text{ year} = \frac{A_2 - A_1}{T_2 - T_1}$$
$$\text{Principal } P = A_1 - [T_1 \times (\text{Interest for } 1\text{ year})]$$
$$\text{Rate } R = \left( \frac{\text{Interest for } 1\text{ year}}{P} \times 100 \right)\%\text{ p.a.}$$

4. Capital Splitting & Equal Return Investments

When a total principal $P$ is partitioned into two portions $P_1$ and $P_2$ ($P = P_1 + P_2$) such that the simple interest earned from the first scheme equals that from the second scheme ($SI_1 = SI_2$):

$$\frac{P_1 \times R_1 \times T_1}{100} = \frac{P_2 \times R_2 \times T_2}{100} \implies P_1 \times R_1 T_1 = P_2 \times R_2 T_2$$

$$\frac{P_1}{P_2} = \frac{R_2 T_2}{R_1 T_1} = \frac{1}{R_1 T_1} : \frac{1}{R_2 T_2}$$

The portions are in the inverse ratio of their products $(R \times T)$.

Illustrated Conceptual Diagrams & Visual Schematics

Simple Interest Linear Growth vs Principal Blueprint

Linear Capital Growth: In Simple Interest, interest earned remains constant every year because it is computed exclusively on the initial Principal ($P$). Formula: $I = rac{P imes R imes T}{100}$, Total Amount $A = P + I$.

Topper Strategy & Descriptive Step Breakdown

Mastering RIMC Mathematics Standard

In RIMC, clear step-by-step layout is essential. Write down all given values, declare formulas before substitution, and conclude with full sentences with proper units.

Official Descriptive Marking Scheme & Step Allocation

In RIMC Stage-I, answers without step justification receive only 20-30% of total marks even if the numeric answer is correct. Follow this strict 4-tier layout on every question:

Step 1: Declare Given Parameters & Convert Time to Years 1 - 2 Marks

State given values for P, R, and T clearly. Show explicit fractional conversion of months/days to years.

Booklet Layout:
Given:\nPrincipal $P = ₹ 4,500$\nRate of Interest $R = 7.5\% = \frac{15}{2}\%\text{ p.a.}$\nTime $T = 3\text{ years } 4\text{ months} = 3 + \frac{4}{12} = 3 + \frac{1}{3} = \frac{10}{3}\text{ years}$.

Step 2: Formula Citation & Symbolic Algebraic Setup 1 - 2 Marks

State the standard Simple Interest formula before plugging in numerical values.

Booklet Layout:
$$\text{Simple Interest } (SI) = \frac{P \times R \times T}{100}$$

Step 3: Stepwise Substitution & Arithmetic Reduction 2 - 4 Marks

Substitute fractions, show clean cross-cancellations, and evaluate SI and Amount step-by-step.

Booklet Layout:
$$SI = \frac{4500 \times \frac{15}{2} \times \frac{10}{3}}{100} = \frac{4500 \times 15 \times 10}{100 \times 2 \times 3} = \frac{45 \times 150}{6} = 45 \times 25 = ₹ 1,125$$
$$\text{Total Amount } A = P + SI = 4500 + 1125 = ₹ 5,625$$

Step 4: Concluding Statement with Proper Units & Answer Box 1 Mark

State both requested answers (SI and Amount) in words and enclose in an answer box.

Booklet Layout:
$$\mathbf{\text{Answer: Simple Interest = } ₹\;1,125; \quad \text{Total Amount = } ₹\;5,625}$$

Examiner Traps & Common Blunders (Marks Penalties)

Using Raw Months or Days in the Formula -2 to -3 Marks

Common Mistake: Plugging $T = 4$ for 4 months, computing interest that is 12 times too large.

Cadet Antidote: Time $T$ must ALWAYS be expressed in years. Divide months by 12 and days by 365 before substituting.

Confusing Simple Interest with Total Amount -2 Marks

Common Mistake: Computing $SI = ₹ 1,125$ and boxing it as the final answer when the question asked for the Total Amount.

Cadet Antidote: Always re-read the prompt: Amount $A = P + SI$. Add Principal to Simple Interest to find Amount.

Sum Doubling Misinterpretation -2 to -3 Marks

Common Mistake: Setting $SI = 2P$ in doubling problems instead of $SI = P$.

Cadet Antidote: If a sum doubles, the final Amount is $2P$. Therefore, the Interest earned is $A - P = 2P - P = P$.

Calendar Day Counting Errors -1 Mark

Common Mistake: Including both the date of deposit and the date of withdrawal when counting exact days.

Cadet Antidote: In banking and commercial mathematics, only ONE of the boundary days is counted (either deposit day or withdrawal day, never both).

Cadet Topper Mental Verification Shortcuts

The 73-Day Divisor Shortcut

Whenever time is given in days, look for multiples of 73: $73 / 365 = 1/5$, $146 / 365 = 2/5$, $219 / 365 = 3/5$, $292 / 365 = 4/5$.

Find SI for 146 days: directly use $T = \frac{2}{5}\text{ years}$.

The R × T = 100(n - 1) n-Folding Identity

For any problem where a sum becomes $n$ times itself, use $R \times T = 100(n - 1)$ immediately.

A sum doubles in 5 years. Find rate: $R \times 5 = 100(2 - 1) = 100 \implies R = 20\%$ p.a. in 2 seconds!

Direct Amount-to-Principal Bridge Formula

When Amount $A$, Rate $R$, and Time $T$ are given, bypass computing SI first: $P = \frac{100 \times A}{100 + RT}$.

If ₹ 4,410 is the amount in 2.25 years at $10\%$: $P = \frac{100 \times 4410}{100 + 22.5} = \frac{441000}{122.5} = ₹ 3,600$.

Key Formulas & Conversion Cheat Sheet

Simple Interest Formula

$$SI = \frac{P \times R \times T}{100}$$

Uniform interest on principal P at rate R% p.a. for T years.

Total Amount Formula

$$A = P + SI = P \left( 1 + \frac{R \times T}{100} \right)$$

Sum of principal and simple interest.

Principal Formula from SI

$$P = \frac{100 \times SI}{R \times T}$$

Finds initial principal when interest is known.

Principal Formula from Amount

$$P = \frac{100 \times A}{100 + (R \times T)}$$

Calculates principal directly from accumulated amount.

Rate of Interest Formula

$$R = \frac{100 \times SI}{P \times T}$$

Finds annual interest percentage.

Time Duration Formula

$$T = \frac{100 \times SI}{P \times R}$$

Finds duration in years.

Sum n-Folding Identity

$$R \times T = 100(n - 1)$$

Equation governing a sum becoming n times itself.

Annual Interest from Multi-Year Amounts

$$\text{Interest per year} = \frac{A_2 - A_1}{T_2 - T_1}$$

Finds constant yearly interest from two different maturity amounts.

Solved Model Examples (Descriptive Format with Step Mark Distribution)

Model 1: Fractional Time with Months Conversion 5 Marks

Calculate the simple interest and the total amount on ₹ $4,500$ at $7\frac{1}{2}\%$ per annum for $3\text{ years and } 4\text{ months}$.

Step-by-Step Model Solution:

Step 1: Declare Parameters & Convert Time to Years 2 Marks

Principal $P = ₹ 4,500$.
Rate $R = 7\frac{1}{2}\% = \frac{15}{2}\%\text{ p.a.}$.
Time $T = 3\text{ years } 4\text{ months} = 3 + \frac{4}{12} = 3 + \frac{1}{3} = \frac{10}{3}\text{ years}$.

Step 2: Apply Simple Interest Formula 2 Marks

$$SI = \frac{P \times R \times T}{100} = \frac{4500 \times \frac{15}{2} \times \frac{10}{3}}{100}$$
$$= \frac{4500 \times 15 \times 10}{100 \times 2 \times 3} = \frac{45 \times 150}{6} = 45 \times 25 = ₹ 1,125$$

Step 3: Compute Total Amount & Conclude 1 Mark

$$\text{Amount } A = P + SI = 4500 + 1125 = ₹ 5,625$$
$$\mathbf{\text{Answer: Simple Interest = } ₹\;1,125; \quad \text{Total Amount = } ₹\;5,625}$$

Examiner Scoring Key: Converting 3 years 4 months to 10/3 years carries 2 marks. Correct calculation of SI = ₹ 1,125 and Amount = ₹ 5,625 earns full 5 marks.

Model 2: Doubling & Tripling Period Evaluation 5 Marks

In how many years will a certain sum of money double itself at $12\frac{1}{2}\%$ per annum simple interest? At the same rate, in how many years will it become three times itself?

Step-by-Step Model Solution:

Step 1: Part (a) Doubling Period 2.5 Marks

Let principal be $P$. When sum doubles, Amount $A = 2P$.
Simple Interest $SI = A - P = 2P - P = P$.
Rate $R = 12\frac{1}{2}\% = \frac{25}{2}\%\text{ p.a.}$.
$$T_1 = \frac{100 \times SI}{P \times R} = \frac{100 \times P}{P \times \frac{25}{2}} = \frac{100 \times 2}{25} = 4 \times 2 = \mathbf{8\text{ years}}$$

Step 2: Part (b) Tripling Period 2.5 Marks

When sum becomes three times, Amount $A = 3P$.
Simple Interest $SI = 3P - P = 2P$.
$$T_2 = \frac{100 \times 2P}{P \times \frac{25}{2}} = \frac{200 \times 2}{25} = 8 \times 2 = \mathbf{16\text{ years}}$$

Examiner Scoring Key: Setting SI = P for doubling and SI = 2P for tripling is the key logic. Concluding with 8 years and 16 years gets full marks.

Model 3: Multi-Year Amount Differences ($A_1$ vs. $A_2$) 7 Marks

A certain sum of money amounts to ₹ $1,008$ in $2\text{ years}$ and to ₹ $1,164$ in $3\frac{1}{2}\text{ years}$ at a uniform rate of simple interest. Find the principal sum and the rate of interest per annum.

Step-by-Step Model Solution:

Step 1: Compute Interest for the Intervening Period 2 Marks

Amount in $3\frac{1}{2}\text{ years} = ₹ 1,164$.
Amount in $2\text{ years} = ₹ 1,008$.
Difference in Time $= 3\frac{1}{2} - 2 = 1\frac{1}{2}\text{ years} = \frac{3}{2}\text{ years}$.
$$\text{Simple Interest for } \frac{3}{2}\text{ years} = 1164 - 1008 = ₹ 156$$

Step 2: Calculate Interest for 1 Year & 2 Years 2 Marks

$$\text{Interest for } 1\text{ year} = 156 \div \frac{3}{2} = 156 \times \frac{2}{3} = 52 \times 2 = ₹ 104$$
$$\text{Interest for } 2\text{ years} = 104 \times 2 = ₹ 208$$

Step 3: Determine Principal and Rate 3 Marks

$$\text{Principal } P = \text{Amount in } 2\text{ years} - \text{Interest for } 2\text{ years} = 1008 - 208 = \mathbf{₹ 800}$$
Now calculate Rate $R$ using $P = 800$ and $1\text{ year interest} = 104$:
$$R = \left( \frac{100 \times 104}{800 \times 1} \right)\% = \frac{104}{8}\% = \mathbf{13\%\text{ p.a.}}$$
$$\mathbf{\text{Answer: Principal = } ₹\;800; \quad \text{Rate of Interest = } 13\%\text{ p.a.}}$$

Examiner Scoring Key: Finding interest for 3/2 years = ₹ 156 carries 2 marks. Principal = ₹ 800 carries 2 marks. Rate = 13% carries 3 marks.

Model 4: Exact Calendar Days Calculation (73-Day Multiples) 7 Marks

Find the simple interest on ₹ $7,300$ at $10\%$ per annum from $5\text{th January } 2026$ to $31\text{st May } 2026$. (Note: 2026 is a non-leap year).

Step-by-Step Model Solution:

Step 1: Compute Exact Number of Days 3 Marks

Counting days excluding the day of deposit (Jan 5):
• January: $31 - 5 = 26\text{ days}$
• February: $28\text{ days}$ (2026 is not a leap year)
• March: $31\text{ days}$
• April: $30\text{ days}$
• May: $31\text{ days}$
$$\text{Total Days} = 26 + 28 + 31 + 30 + 31 = 146\text{ days}$$

Step 2: Convert Days to Fractional Years 2 Marks

$$T = \frac{146}{365}\text{ years}$$
Notice $146 = 73 \times 2$ and $365 = 73 \times 5$:
$$T = \frac{2}{5}\text{ year}$$

Step 3: Compute Simple Interest 2 Marks

$$SI = \frac{P \times R \times T}{100} = \frac{7300 \times 10 \times \frac{2}{5}}{100} = 73 \times 10 \times \frac{2}{5} = 73 \times 4 = ₹ 292$$
$$\mathbf{\text{Answer: Simple Interest = } ₹\;292.}$$

Examiner Scoring Key: Day count of 146 days carries 3 marks. Recognition of 146/365 = 2/5 year carries 2 marks. Final SI = ₹ 292 carries 2 marks.

Model 5: Partitioning Capital into Two Schemes 10 Marks

A sum of ₹ $12,000$ is divided into two parts such that the simple interest on the first part for $3\text{ years}$ at $12\%$ per annum is equal to the simple interest on the second part for $4\frac{1}{2}\text{ years}$ at $16\%$ per annum. Find the amount of each part.

Step-by-Step Model Solution:

Step 1: Declare Variables for the Two Portions 2 Marks

Let the first part be ₹ $x$.
Then the second part $= ₹ (12000 - x)$.

Step 2: Formulate Simple Interest for Each Part 3 Marks

For First Part: $P_1 = x$, $T_1 = 3\text{ years}$, $R_1 = 12\%$.
$$SI_1 = \frac{x \times 12 \times 3}{100} = \frac{36x}{100}$$
For Second Part: $P_2 = 12000 - x$, $T_2 = 4\frac{1}{2} = \frac{9}{2}\text{ years}$, $R_2 = 16\%$.
$$SI_2 = \frac{(12000 - x) \times 16 \times \frac{9}{2}}{100} = \frac{(12000 - x) \times 72}{100}$$

Step 3: Equate Interests (ATQ) and Solve for x 3 Marks

Given $SI_1 = SI_2$:
$$\frac{36x}{100} = \frac{72(12000 - x)}{100}$$
Multiply both sides by $100$ and divide by $36$:
$$x = 2(12000 - x)$$
$$x = 24000 - 2x \implies 3x = 24000 \implies x = 8,000$$

Step 4: Compute Both Parts & Conclude 2 Marks

First part $x = \mathbf{₹ 8,000}$.
Second part $= 12000 - 8000 = \mathbf{₹ 4,000}$.
Check: $SI_1 = \frac{8000 \times 36}{100} = ₹ 2,880$. $SI_2 = \frac{4000 \times 72}{100} = ₹ 2,880$ (Equal!).

Examiner Scoring Key: Setting up 36x = 72(12000 - x) simplifies the equation elegantly. Award full 10 marks for x = 8000 and (12000 - x) = 4000.

Model 6: Incremental Rate & Marginal Interest 10 Marks

A sum of ₹ $5,000$ was lent out partly at $6\%$ per annum and the rest at $9\%$ per annum simple interest. The total annual interest received from both investments combined was ₹ $390$. Find the sum lent at each rate.

Step-by-Step Model Solution:

Step 1: Declare Variable for Scheme Allocations 2 Marks

Let the sum lent at $6\%$ p.a. be ₹ $x$.
Then the sum lent at $9\%$ p.a. $= ₹ (5000 - x)$.

Step 2: Express Annual Interests from Both Schemes 3 Marks

Annual interest from first scheme ($T = 1\text{ year}$):
$$I_1 = \frac{x \times 6 \times 1}{100} = \frac{6x}{100}$$
Annual interest from second scheme ($T = 1\text{ year}$):
$$I_2 = \frac{(5000 - x) \times 9 \times 1}{100} = \frac{45000 - 9x}{100}$$

Step 3: Set Up Total Interest Equation & Solve 3 Marks

Total annual interest $= ₹ 390$:
$$\frac{6x}{100} + \frac{45000 - 9x}{100} = 390$$
$$6x + 45000 - 9x = 390 \times 100$$
$$45000 - 3x = 39000$$
$$3x = 45000 - 39000 = 6000 \implies x = \frac{6000}{3} = 2,000$$

Step 4: State Both Amounts & Conclude 2 Marks

Sum lent at $6\% = x = \mathbf{₹ 2,000}$.
Sum lent at $9\% = 5000 - x = 5000 - 2000 = \mathbf{₹ 3,000}$.
Check: $6\% \text{ of } 2000 = 120$; $9\% \text{ of } 3000 = 270$. $120 + 270 = ₹ 390$ (Verified!).

Examiner Scoring Key: Linear equation formulation carries 3 marks. Solving for x = 2000 and stating both sums clearly carries 4 marks.

RIMC Practice & Drill Exercises

Attempt these questions on paper following proper step-wise conventions, then click Reveal Solution to verify your answers.

Question 1 5 Marks

In how many years will a sum of money double itself at $10\%$ per annum simple interest?

Model Descriptive Solution:

Step 1: Set Up Doubling Relation 2.5 Marks

Let $P = x$. Amount $A = 2x \implies SI = 2x - x = x$.

Step 2: Compute Time 2.5 Marks

$$T = \frac{100 \times SI}{P \times R} = \frac{100 \times x}{x \times 10} = 10\text{ years}$$

Final Answer: $10\text{ years}$

Question 2 5 Marks

Find the simple interest on ₹ $2,500$ for $2\text{ years } 6\text{ months}$ at $6\%$ per annum.

Model Descriptive Solution:

Step 1: Convert Time to Years 2 Marks

$T = 2 + \frac{6}{12} = 2\frac{1}{2} = \frac{5}{2}\text{ years}$.

Step 2: Calculate Interest 3 Marks

$$SI = \frac{2500 \times 6 \times \frac{5}{2}}{100} = 25 \times 15 = ₹ 375$$

Final Answer: ₹ $375$

Question 3 5 Marks

At what rate percent per annum will ₹ $3,600$ yield a simple interest of ₹ $540$ in $2\text{ years}$?

Model Descriptive Solution:

Step 1: State Rate Formula 2 Marks

$$R = \frac{100 \times SI}{P \times T}$$

Step 2: Substitute and Solve 3 Marks

$$R = \frac{100 \times 540}{3600 \times 2} = \frac{540}{72} = 7.5\%\text{ p.a.}$$

Final Answer: $7.5\%\text{ p.a.}$ (or $7\frac{1}{2}\%$)

Question 4 7 Marks

A sum of money amounts to ₹ $6,200$ in $2\text{ years}$ and to ₹ $7,400$ in $3\text{ years}$ at simple interest. Find the principal and the rate of interest.

Model Descriptive Solution:

Step 1: Interest for 1 Year 2.5 Marks

$\text{Interest for } (3 - 2) = 1\text{ year} = 7400 - 6200 = ₹ 1,200$.

Step 2: Compute Principal 2.5 Marks

$$\text{Principal } P = 6200 - (2 \times 1200) = 6200 - 2400 = ₹ 3,800$$

Step 3: Compute Rate 2 Marks

$$R = \left(\frac{100 \times 1200}{3800 \times 1}\right)\% = \frac{1200}{38}\% = \frac{600}{19}\% = 31\frac{11}{19}\%$$

Final Answer: Principal: ₹ $3,800$; Rate: $31\frac{11}{19}\%\text{ p.a.}$

Question 5 7 Marks

What sum will amount to ₹ $5,520$ in $3\text{ years}$ at $4\%$ per annum simple interest?

Model Descriptive Solution:

Step 1: State Amount Relation 3 Marks

$$A = P\left(1 + \frac{RT}{100}\right) \implies 5520 = P\left(1 + \frac{4 \times 3}{100}\right) = P\left(1 + \frac{12}{100}\right) = \frac{112}{100}P$$

Step 2: Solve for P 4 Marks

$$P = 5520 \times \frac{100}{112} = \frac{552000}{112} = ₹ 4,928.57$$

Final Answer: ₹ $4,928.57$

Question 6 7 Marks

A sum of money doubles itself in $8\text{ years}$ at simple interest. In how many years will it become $4$ times itself at the same rate?

Model Descriptive Solution:

Step 1: Find Rate from Doubling Condition 3.5 Marks

$R \times T = 100(2 - 1) \implies R \times 8 = 100 \implies R = 12.5\%$.

Step 2: Find Time to Quadruple 3.5 Marks

$$R \times T_{\text{new}} = 100(4 - 1) = 300 \implies 12.5 \times T_{\text{new}} = 300 \implies T_{\text{new}} = \frac{300}{12.5} = 24\text{ years}$$

Final Answer: $24\text{ years}$

Question 7 7 Marks

If the simple interest on a certain sum of money for $3\text{ years}$ at $8\%$ per annum is ₹ $120$ less than the simple interest on the same sum for $5\text{ years}$ at the same rate, find the sum.

Model Descriptive Solution:

Step 1: Formulate Difference Equation 3.5 Marks

Let sum be $P$.
$$SI_2 - SI_1 = 120 \implies \frac{P \times 8 \times 5}{100} - \frac{P \times 8 \times 3}{100} = 120$$
$$\frac{40P - 24P}{100} = 120 \implies \frac{16P}{100} = 120$$

Step 2: Solve for P 3.5 Marks

$$P = \frac{120 \times 100}{16} = \frac{12000}{16} = ₹ 750$$

Final Answer: ₹ $750$

Question 8 10 Marks

Divide ₹ $15,600$ into two parts such that the interest on one part at $5\%$ for $5\text{ years}$ may be equal to the interest on the other part at $4\frac{1}{2}\%$ for $6\text{ years}$.

Model Descriptive Solution:

Step 1: Define Parts 3 Marks

Let parts be $x$ and $(15600 - x)$.

Step 2: Equate Interests 4 Marks

$$\frac{x \times 5 \times 5}{100} = \frac{(15600 - x) \times \frac{9}{2} \times 6}{100}$$
$$25x = 27(15600 - x) \implies 25x + 27x = 27 \times 15600$$
$$52x = 421200 \implies x = \frac{421200}{52} = 8,100$$

Step 3: State Parts 3 Marks

First part $= ₹ 8,100$.
Second part $= 15600 - 8100 = ₹ 7,500$.

Final Answer: ₹ $8,100$ and ₹ $7,500$

Question 9 10 Marks

A man borrows ₹ $8,000$ at $10\%$ per annum simple interest. At the end of $2\text{ years}$, he pays ₹ $5,000$ in cash and gives a cow to clear the debt completely. What is the value of the cow?

Model Descriptive Solution:

Step 1: Calculate Total Debt in 2 Years 4 Marks

$$SI = \frac{8000 \times 10 \times 2}{100} = ₹ 1,600$$
$$\text{Total Debt Amount } A = 8000 + 1600 = ₹ 9,600$$

Step 2: Deduct Cash Payment to Find Cow Value 4 Marks

$$\text{Value of Cow} = \text{Total Debt} - \text{Cash Paid} = 9600 - 5000 = ₹ 4,600$$

Step 3: Conclude 2 Marks

The value of the cow is ₹ $4,600$.

Final Answer: Value of cow $= ₹ 4,600$

Question 10 10 Marks

A person deposits ₹ $50,000$ in a bank for $3\text{ years}$. The rate of interest is $6\%$ p.a. for the first year, $7\%$ p.a. for the second year, and $8\%$ p.a. for the third year. Calculate the total simple interest and the total maturity amount received by him at the end of the third year.

Model Descriptive Solution:

Step 1: Compute Year-wise Simple Interests 4 Marks

Interest for Year 1: $\frac{50000 \times 6 \times 1}{100} = ₹ 3,000$.
Interest for Year 2: $\frac{50000 \times 7 \times 1}{100} = ₹ 3,500$.
Interest for Year 3: $\frac{50000 \times 8 \times 1}{100} = ₹ 4,000$.

Step 2: Sum the Interests 3 Marks

$$\text{Total SI} = 3000 + 3500 + 4000 = ₹ 10,500$$

Step 3: Calculate Maturity Amount 3 Marks

$$\text{Total Maturity Amount} = 50000 + 10500 = ₹ 60,500$$

Final Answer: Total Simple Interest: ₹ $10,500$; Total Maturity Amount: ₹ $60,500$

High-Yield FAQs & Revision Notes

What is the fundamental difference between Simple Interest and Compound Interest?

In Simple Interest, the principal remains completely unchanged throughout the duration. In Compound Interest, interest is added to the principal at periodic intervals, causing the interest base to grow.

Why must time always be converted to years?

Because the rate of interest is defined 'per annum' (per year). The temporal units of Rate and Time must match.

How do you count days between two dates for interest?

Exclude the day of deposit and include the day of withdrawal (or vice versa). Never count both boundary days.

Why is 73 days significant in Simple Interest?

Because $365 = 73 \times 5$. Therefore, $73\text{ days} = 1/5\text{ year}$, simplifying fractional arithmetic without long decimals.

How does the doubling formula work?

When a sum doubles, the interest equals the principal ($SI = P$). Substituting into the formula yields $R \times T = 100$.

How do you find the annual interest from two amounts in different years?

Subtract the smaller amount from the larger amount and divide by the difference in years: $\text{Interest per year} = (A_2 - A_1) / (T_2 - T_1)$.

What is the formula for Principal when Amount is given?

$P = \frac{100 \times A}{100 + RT}$. This avoids having to compute interest separately.

Can the simple interest ever exceed the principal?

Yes. Whenever $R \times T > 100$, the accrued interest exceeds the original principal (e.g., at 10% for 12 years, SI = 120% of P).

What does 'per annum' mean?

'Per annum' (p.a.) is a Latin term meaning 'for each year'.

How does this module map to the official RIMC syllabus?

Directly covers Serial No. 11 ('Simple Interest') of Chapter 8 broader syllabus in the official RIMC admission brochure.

How does this topic relate to the official RIMC syllabus?

This module directly covers Serial No. 11 of the broader syllabus published in Chapter 8 of the official RIMC admission brochure, ensuring 100% syllabus alignment.

Last-Minute Cadet High-Yield Revision Matrix

Topic / Concept	High-Yield Exam Rules & Memory Anchor
SI Formula	$SI = \frac{PRT}{100}$. Time $T$ must strictly be in years.
Time Conversion	Months: divide by 12. Days: divide by 365. Multiples of 73 days $= k/5$ years.
Amount Formula	$A = P + SI$. Principal from amount: $P = \frac{100A}{100 + RT}$.
Doubling / n-Folding	$R \times T = 100(n - 1)$. For doubling: $RT = 100$; for tripling: $RT = 200$.
Two Amounts	$\text{Interest per year} = \frac{A_2 - A_1}{T_2 - T_1}$. Principal $= A_1 - T_1 \times (\text{yearly interest})$.
Capital Splitting	$P_1 : P_2 = \frac{1}{R_1 T_1} : \frac{1}{R_2 T_2}$. Inverse proportionality of returns.