Official Syllabus Item: 9 Commercial Mathematics

Percentage & Its Practical Applications

Master percentage fundamentals, conversions, percentage increase/decrease, successive changes, expenditure/consumption balancing, and marks/population problems for RIMC Entrance Exam.

Overview & Fundamental Concept

Percentage ('per centum' or parts per hundred) is one of the most critical commercial arithmetic tools in the RIMC Mathematics Paper (Paper I, 200 Marks, 90 Minutes). Prescribed under Item 9 of Chapter 8 in the official RIMC brochure, percentage problems appear routinely in Part A (5 Marks) testing rapid fractional conversions, percentage allocations, and error estimation, as well as in Part B (10 Marks) testing multi-tiered real-world scenarios: salary allocations, candidate pass/fail examination thresholds, price-consumption trade-offs, compound demographic growth, and successive market changes. Descriptive step marking requires candidates to clearly state the reference base ('Original Value'), formulate intermediate changes with fractional arithmetic, avoid premature rounding, and box final answers with standard percentage symbols or rupee amounts.

Official Syllabus Coverage Breakdown

Syllabus Item	Prescribed Scope	Exam Relevance & Pedagogical Objective
Item 9	Percentage: Calculations, conversions, percentage change, and applied word problems	Core commercial arithmetic syllabus item. Assesses base-value discipline, multi-step financial distributions, examination score modeling, and demographic compounding.

1. Definition, Conceptual Foundation & Inter-Conversions

The word percent is derived from the Latin phrase per centum, meaning 'by the hundred' or 'for every hundred'. A percentage is a fraction whose denominator is always $100$. The symbol for percentage is $\%$.

$$x\% = \frac{x}{100}$$

Three-Way Equivalence: Every percentage can be transformed seamlessly into a common fraction or a decimal numeral, and vice versa:

• Percentage to Fraction: Drop the $\%$ sign and divide by $100$, then reduce to irreducible terms: $35\% = \frac{35}{100} = \frac{7}{20}$.
• Fraction to Percentage: Multiply the fraction by $100\%$: $\frac{5}{8} = \frac{5}{8} \times 100\% = \frac{125}{2}\% = 62.5\%$.
• Decimal to Percentage: Multiply by $100$ (shift decimal point two places to the right) and attach the $\%$ sign: $0.075 = 0.075 \times 100\% = 7.5\%$.
• Percentage to Decimal: Divide by $100$ (shift decimal point two places to the left): $4.8\% = 0.048$.

High-Speed Cadet Fractional Equivalents (Must-Memorize)

$\frac{1}{2} = 50\%$, $\frac{1}{3} = 33\frac{1}{3}\%$, $\frac{1}{4} = 25\%$, $\frac{1}{5} = 20\%$, $\frac{1}{6} = 16\frac{2}{3}\%$
$\frac{1}{7} = 14\frac{2}{7}\%$, $\frac{1}{8} = 12\frac{1}{2}\%$, $\frac{1}{9} = 11\frac{1}{9}\%$, $\frac{1}{10} = 10\%$, $\frac{1}{12} = 8\frac{1}{3}\%$, $\frac{1}{16} = 6\frac{1}{4}\%$, $\frac{1}{20} = 5\%$.

2. Percentage Increase, Decrease & Reference Base Discipline

The most critical axiom in percentage arithmetic is: Percentage change is always calculated with respect to the ORIGINAL (initial) reference value, never the new value.

$$\text{Percentage Increase} = \left( \frac{\text{Absolute Increase}}{\text{Original Initial Value}} \times 100 \right)\%$$

$$\text{Percentage Decrease} = \left( \frac{\text{Absolute Decrease}}{\text{Original Initial Value}} \times 100 \right)\%$$

New Value Multipliers:

• If a quantity $Q$ increases by $r\%$, then $\text{New Value} = Q \times \left(1 + \frac{r}{100}\right)$.
• If a quantity $Q$ decreases by $r\%$, then $\text{New Value} = Q \times \left(1 - \frac{r}{100}\right)$.

The Asymmetry Fallacy (A vs. B)

If $A$ is $25\%$ more than $B$, is $B$ also $25\%$ less than $A$? NO! Because the base changes from $B$ to $A$. If $B = 100$, then $A = 125$. The shortfall of $B$ relative to $A$ is $\frac{125 - 100}{125} \times 100\% = \frac{25}{125} \times 100\% = \mathbf{20\%}$.

3. Price, Consumption & Constant Expenditure Balance

In military and household logistics, expenditure on a commodity equals the product of its unit price and consumption:

$$\text{Expenditure} = \text{Price} \times \text{Consumption}$$

When the unit price of a commodity increases by $r\%$, to maintain total expenditure strictly constant, consumption must be reduced by:

$$\text{Percentage Reduction in Consumption} = \left( \frac{r}{100 + r} \times 100 \right)\%$$

Conversely, if the price decreases by $r\%$, consumption can increase by:

$$\text{Percentage Increase in Consumption} = \left( \frac{r}{100 - r} \times 100 \right)\%$$

4. Successive Percentage Changes & Compounded Growth

If a quantity undergoes two successive percentage changes of $a\%$ and $b\%$ (taking positive for increase and negative for decrease), the net effective single percentage change is given by the Net Successive Identity:

$$\text{Net Effective \%} = \left( a + b + \frac{a \times b}{100} \right)\%$$

• Equal Increase & Decrease: If a quantity is first increased by $x\%$ and subsequently decreased by $x\%$, there is always a net decrease given by: $\mathbf{-\frac{x^2}{100}\%}$.
• Demographic Growth / Depreciation: If initial population is $P$ and growth rate is $R\%$ per annum, population after $n$ years is $P_n = P\left(1 + \frac{R}{100}\right)^n$.

Illustrated Conceptual Diagrams & Visual Schematics

Master Percentage-Fraction-Decimal Equivalence Grid

Cadet Mental Speed Multipliers: Memorizing standard fraction equivalents eliminates scratch division in descriptive steps. For percentage increases, multiply by $(1 + rac{r}{100})$; for decreases, multiply by $(1 - rac{r}{100})$.

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: Reference Base & Parameter Declaration 1 - 2 Marks

State the original initial value, target change, and define variables clearly (e.g., 'Let the maximum marks be M').

Booklet Layout:
Given:\nOriginal Price $P_1 = ₹ 80\text{ per litre}$\nNew Price $P_2 = ₹ 92\text{ per litre}$

Step 2: Absolute Difference Calculation 1 - 2 Marks

Calculate the exact numeric increase or decrease before setting up the percentage fraction.

Booklet Layout:
$$\text{Absolute Increase in Price} = P_2 - P_1 = 92 - 80 = ₹ 12$$

Step 3: Percentage Formula Statement & Substitution 2 - 4 Marks

State the standard percentage formula with 'Original Value' in the denominator and show stepwise arithmetic reduction.

Booklet Layout:
$$\text{Percentage Increase} = \left( \frac{\text{Increase}}{\text{Original Price}} \times 100 \right)\% = \left( \frac{12}{80} \times 100 \right)\% = \left( \frac{3}{20} \times 100 \right)\% = 15\%$$

Step 4: Concluding Statement with Percentage Symbol & Box 1 Mark

State the final percentage clearly in words and enclose the answer in a neat box.

Booklet Layout:
$$\mathbf{\text{Answer: The price of petrol increased by } 15\%.}$$

Examiner Traps & Common Blunders (Marks Penalties)

Dividing by the New Value Instead of Original Base -2 to -3 Marks

Common Mistake: Computing percentage increase as $\frac{12}{92} \times 100\% \approx 13.04\%$.

Cadet Antidote: Always remember: the original, prior value goes into the denominator. The change began at $80$, so the divisor is strictly $80$.

Assuming +10% Followed by -10% Results in No Change -2 Marks

Common Mistake: Claiming an item increased by $10\%$ and then reduced by $10\%$ returns to its original price.

Cadet Antidote: The second change acts on a larger base. The result is always a net loss of $\frac{10^2}{100} = 1\%$.

Forgetting to State Maximum Marks in Exam Problems -2 Marks

Common Mistake: Computing passing percentage without formally declaring $M$ as total maximum marks.

Cadet Antidote: Set up $x\% \times M + \text{shortfall} = \text{passing marks}$ explicitly.

Premature Decimal Rounding -1 Mark

Common Mistake: Rounding $16.666\%$ to $16.6\%$ or $17\%$ early in multi-step equations.

Cadet Antidote: Keep values as exact rational fractions (e.g., $\frac{50}{3}\%$ or $\frac{1}{6}$) until the final step.

Cadet Topper Mental Verification Shortcuts

Fractional Multiplier Shortcut for Rapid Calculation

Instead of writing $\frac{x}{100} \times Q$, use memorized fractions: $25\% = \frac{1}{4}$, $20\% = \frac{1}{5}$, $12.5\% = \frac{1}{8}$, $16\frac{2}{3}\% = \frac{1}{6}$.

Find $12.5\%$ of ₹ $6,400$: $\frac{1}{8} \times 6400 = ₹ 800$ in 1 second.

The r/(100 + r) Sugar-Consumption Formula

When price rises by $r\%$, the reduction in consumption to keep expenditure constant is $\frac{r}{100 + r} \times 100\%$. If price drops, use $\frac{r}{100 - r} \times 100\%$.

Price of sugar rises by $25\%$: Reduction in consumption $= \frac{25}{125} \times 100\% = \frac{1}{5} \times 100\% = 20\%$ instantly.

Direct Passing Marks Equality Principle

Equate two candidates' scores to passing marks: $\text{Marks}_1 + \text{Failed By} = \text{Marks}_2 - \text{Exceeded By}$.

If $30\%$ fails by 15 marks and $40\%$ gets 35 marks extra: $(40\% - 30\%) \text{ of } M = 15 + 35 = 50 \implies 10\% \text{ of } M = 50 \implies M = 500$.

Key Formulas & Conversion Cheat Sheet

Basic Percentage Formula

$$\text{Percentage} = \left( \frac{\text{Value}}{\text{Total Base}} \times 100 \right)\%$$

Fraction of whole expressed per hundred.

Percentage Increase

$$\text{Percentage Increase} = \left( \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100 \right)\%$$

Fractional growth relative to initial baseline.

Percentage Decrease

$$\text{Percentage Decrease} = \left( \frac{\text{Original Value} - \text{New Value}}{\text{Original Value}} \times 100 \right)\%$$

Fractional decay relative to initial baseline.

Constant Expenditure Consumption Reduction

$$\text{Consumption Reduction} = \left( \frac{r}{100 + r} \times 100 \right)\%$$

Reduction needed when price rises by r% to maintain constant spending.

Net Successive Percentage Change

$$\text{Net Change} = \left( a + b + \frac{ab}{100} \right)\%$$

Single equivalent percentage for two successive changes a% and b%.

Equal Increase and Decrease Net Loss

$$\text{Net Loss} = -\left( \frac{x^2}{100} \right)\%$$

Successive increase of x% followed by decrease of x% always produces a net loss.

Demographic Compounding Formula

$$P_n = P_0 \left( 1 \pm \frac{R}{100} \right)^n$$

Population or value after n years at compound annual rate R%.

Commodity Value Multiplier

$$V_{\text{new}} = V_{\text{original}} \times \left( 1 \pm \frac{r}{100} \right)$$

Direct multiplier for increased or decreased quantities.

Solved Model Examples (Descriptive Format with Step Mark Distribution)

Model 1: Examination Pass Threshold & Total Marks 5 Marks

A cadet who secures $35\%$ marks in an entrance examination fails by $14\text{ marks}$. Another cadet who scores $42\%$ marks gets $28\text{ marks}$ more than the minimum passing marks. Determine the maximum marks of the examination and the minimum percentage required to pass.

Step-by-Step Model Solution:

Step 1: Declare Variable & Express Passing Marks 2 Marks

Let the maximum marks of the examination be $M$.
Passing marks from Cadet 1 $= 35\% \text{ of } M + 14 = \frac{35}{100}M + 14$.
Passing marks from Cadet 2 $= 42\% \text{ of } M - 28 = \frac{42}{100}M - 28$.

Step 2: Equate Passing Marks & Solve for M 2 Marks

Since passing marks are identical for both candidates:
$$\frac{42}{100}M - 28 = \frac{35}{100}M + 14$$
$$\frac{42 - 35}{100}M = 14 + 28 \implies \frac{7}{100}M = 42$$
$$M = 42 \times \frac{100}{7} = 6 \times 100 = 600\text{ marks}$$

Step 3: Calculate Passing Marks & Passing Percentage 1 Mark

$$\text{Minimum Passing Marks} = \frac{35}{100}(600) + 14 = 210 + 14 = 224\text{ marks}$$
$$\text{Passing Percentage} = \left( \frac{224}{600} \times 100 \right)\% = \frac{224}{6}\% = \frac{112}{3}\% = 37\frac{1}{3}\% \approx 37.33\%$$
$$\mathbf{\text{Maximum Marks: } 600; \quad \text{Passing Percentage: } 37\frac{1}{3}\%}$$

Examiner Scoring Key: Setting up the difference 7% of M = 42 carries 2 marks. Concluding with both total marks (600) and passing percentage (37 1/3%) earns full marks.

Model 2: Household Expenditure & Price Rise 5 Marks

If the price of edible oil increases by $25\%$, by what percentage must a military mess cut down its consumption so that the total monthly expenditure on oil remains unchanged?

Step-by-Step Model Solution:

Step 1: Declare Standard Baseline Parameters 2 Marks

Let original price of oil $= ₹ 100\text{ per unit}$, and original consumption $= 100\text{ units}$.
Original Total Expenditure $= 100 \times 100 = ₹ 10,000$.

Step 2: Compute New Price & New Required Consumption 2 Marks

Price increased by $25\% \implies \text{New Price} = ₹ 125\text{ per unit}$.
To maintain expenditure at ₹ $10,000$:
$$\text{New Consumption} = \frac{\text{Target Expenditure}}{\text{New Price}} = \frac{10000}{125} = 80\text{ units}$$

Step 3: Calculate Percentage Reduction in Consumption 1 Mark

$$\text{Reduction in Consumption} = 100 - 80 = 20\text{ units}$$
$$\text{Percentage Reduction} = \left( \frac{20}{100} \times 100 \right)\% = \mathbf{20\%}$$
Alternatively, using formula: $\frac{r}{100 + r} \times 100\% = \frac{25}{125} \times 100\% = 20\%$.

Examiner Scoring Key: Either the ₹ 100 baseline method or the direct r/(100+r) formula with explanation receives full 5 marks.

Model 3: Salary Allocation & Residual Savings 7 Marks

An army officer spends $20\%$ of his monthly income on house rent, $15\%$ of the remaining salary on food, and $10\%$ of the balance on his children's education. If he is left with a remaining balance of ₹ $48,960$, determine his total monthly salary.

Step-by-Step Model Solution:

Step 1: Declare Total Salary & Deduct First Expense 2 Marks

Let the total monthly salary of the officer be ₹ $S$.
Expenditure on house rent $= 20\% \text{ of } S$.
Remaining balance after rent $= S - 0.20S = 0.80S = \frac{4}{5}S$.

Step 2: Deduct Food and Education from Sequential Remainders 3 Marks

Expenditure on food $= 15\% \text{ of Remaining}$.
Balance after food $= (100\% - 15\%) \times 0.80S = 0.85 \times 0.80S = 0.68S = \frac{17}{25}S$.
Expenditure on education $= 10\% \text{ of Balance}$.
Final remaining balance $= (100\% - 10\%) \times 0.68S = 0.90 \times 0.68S = 0.612S$.

Step 3: Solve for Monthly Salary S 2 Marks

Given final balance $= ₹ 48,960$:
$$0.612 S = 48960$$
$$S = \frac{48960}{0.612} = \frac{48960000}{612}$$
Notice $612 \times 8 = 4896$:
$$S = 80 \times 1000 = ₹ 80,000$$
$$\mathbf{\text{Answer: The officer's monthly salary is } ₹\;80,000.}$$

Examiner Scoring Key: Sequential deduction on the remainder (not on the gross salary) is the critical logic step carrying 3 marks.

Model 4: Demographic Compounded Growth 7 Marks

The population of a cantonment town was $1,60,000$ three years ago. If the population grew by $3\%$, $4\%$, and $5\%$ respectively in the three successive years, calculate the present population of the town.

Step-by-Step Model Solution:

Step 1: Declare Parameters & Successive Growth Rates 2 Marks

Initial population three years ago $P_0 = 1,60,000$.
Rate in Year 1: $R_1 = 3\%$.
Rate in Year 2: $R_2 = 4\%$.
Rate in Year 3: $R_3 = 5\%$.

Step 2: Formulate Successive Compounding Formula 2 Marks

$$P_3 = P_0 \times \left(1 + \frac{R_1}{100}\right) \times \left(1 + \frac{R_2}{100}\right) \times \left(1 + \frac{R_3}{100}\right)$$
$$P_3 = 160000 \times \left(\frac{103}{100}\right) \times \left(\frac{104}{100}\right) \times \left(\frac{105}{100}\right)$$

Step 3: Step-by-Step Cancellation & Final Multiplication 3 Marks

Cancel $100 \times 100 = 10000$ with $160000$: leaves $16$ in numerator.
$$P_3 = \frac{16 \times 103 \times 104 \times 105}{100}$$
Notice $105 / 5 = 21$ and $100 / 5 = 20$. Then $16 / 4 = 4$ and $20 / 4 = 5$. Wait:
$$16 \times 103 = 1648$$
$$1648 \times 104 = 171392$$
$$171392 \times 105 = 17996160$$
$$P_3 = \frac{17996160}{100} = 1,79,961.6 \approx 1,79,962$$
Let's verify: $160000 \times 1.03 \times 1.04 \times 1.05 = 160000 \times 1.12476 = 1,79,961.6 \implies \mathbf{1,79,962\text{ people}}$.

Examiner Scoring Key: Rounding to nearest integer for population counts is standard practice in RIMC marking.

Model 5: Successive Percentage Change & Net Impact 10 Marks

A merchant marks up the price of an article by $30\%$ above its cost price. Due to sluggish market conditions, he then offers a discount of $20\%$ on the marked price. Later, during a clearance sale, he offers an additional special festival discount of $10\%$.
(a) What is the overall net percentage change relative to the initial cost price?
(b) Does the merchant make a profit or a loss, and by what percentage?

Step-by-Step Model Solution:

Step 1: Set Cost Price Base as ₹ 100 2 Marks

Let Cost Price $(\text{CP}) = ₹ 100$.
Marked Price $(\text{MP}) = ₹ 100 + 30\% \text{ of } 100 = ₹ 130$.

Step 2: Apply First Discount of 20% 3 Marks

Discount $D_1 = 20\% \text{ on MP}$: $0.20 \times 130 = ₹ 26$.
Price after First Discount $= 130 - 26 = ₹ 104$.

Step 3: Apply Second Festival Discount of 10% 3 Marks

Discount $D_2 = 10\% \text{ on } ₹ 104$: $0.10 \times 104 = ₹ 10.40$.
Final Selling Price $(\text{SP}) = 104 - 10.40 = ₹ 93.60$.

Step 4: Determine Net Profit / Loss Percentage 2 Marks

Since $\text{SP} = ₹ 93.60 < \text{CP} = ₹ 100$, the merchant incurs a Net Loss.
$$\text{Loss} = \text{CP} - \text{SP} = 100 - 93.60 = ₹ 6.40$$
$$\text{Net Loss Percentage} = \left(\frac{6.40}{100} \times 100\right)\% = \mathbf{6.4\%}$$
$$\mathbf{\text{Answer: Overall net decrease of } 6.4\% \text{ (Net Loss of } 6.4\%).}$$

Examiner Scoring Key: Sequential calculation MP -> SP1 -> SP2 is evaluated step-wise. Stating that the result is a loss of 6.4% earns full 10 marks.

Model 6: Complex Election Two-Candidate Voting Problem 10 Marks

In an election between two candidates, $10\%$ of the voters on the electoral roll did not cast their votes, and $60\text{ votes}$ cast were declared invalid. The winning candidate secured $47\%$ of the total voters on the electoral list and defeated his rival by $308\text{ votes}$. Find the total number of voters on the electoral list.

Step-by-Step Model Solution:

Step 1: Declare Variable for Total Electoral Roll 3 Marks

Let total number of registered voters on the electoral roll be $x$.
Voters who did not vote $= 10\% \text{ of } x = 0.10x$.
Votes cast $= x - 0.10x = 0.90x$.
Invalid votes $= 60$.
Valid votes polled $= 0.90x - 60$.

Step 2: Express Votes Secured by Each Candidate 3 Marks

Winner secured $47\%$ of total registered voters $= 0.47x$.
Loser's votes $= \text{Valid votes} - \text{Winner's votes}$
$$= (0.90x - 60) - 0.47x = (0.90 - 0.47)x - 60 = 0.43x - 60$$

Step 3: Set Up Winning Margin Equation (ATQ) 2 Marks

Winner defeated rival by $308\text{ votes}$:
$$\text{Winner's Votes} - \text{Loser's Votes} = 308$$
$$0.47x - (0.43x - 60) = 308$$
$$0.47x - 0.43x + 60 = 308$$
$$0.04x + 60 = 308$$

Step 4: Solve for Total Voters x 2 Marks

$$0.04x = 308 - 60 = 248$$
$$x = \frac{248}{0.04} = \frac{24800}{4} = 6,200$$
$$\mathbf{\text{Answer: The total number of voters on the electoral roll is } 6,200.}$$

Examiner Scoring Key: Noting that the winner's 47% is based on the TOTAL ELECTORAL ROLL (not valid votes) is the linchpin carrying 3 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

Convert the fraction $\frac{7}{16}$ into its equivalent percentage form, expressing the answer as a mixed fraction and as a decimal.

Model Descriptive Solution:

Step 1: Multiply Fraction by 100% 2.5 Marks

$$\frac{7}{16} \times 100\% = \frac{7 \times 25}{4}\% = \frac{175}{4}\%$$

Step 2: Convert to Mixed Fraction & Decimal 2.5 Marks

$$\frac{175}{4}\% = 43\frac{3}{4}\% = 43.75\%$$

Final Answer: $43\frac{3}{4}\%$ or $43.75\%$

Question 2 5 Marks

Chalk contains Calcium, Carbon, and Sand in the ratio $12 : 3 : 10$. Find the percentage of Carbon in the chalk stick.

Model Descriptive Solution:

Step 1: Sum of Ratio Terms 2 Marks

Sum of parts $= 12 + 3 + 10 = 25$.

Step 2: Compute Carbon Percentage 3 Marks

$$\text{Carbon \%} = \left(\frac{3}{25} \times 100\right)\% = 3 \times 4 = 12\%$$

Final Answer: $12\%$

Question 3 5 Marks

If $A$'s salary is $25\%$ more than $B$'s salary, by what percentage is $B$'s salary less than $A$'s salary?

Model Descriptive Solution:

Step 1: Set B as 100 2.5 Marks

Let $B = 100$. Then $A = 125$.

Step 2: Compute Percentage Shortfall 2.5 Marks

$$\frac{125 - 100}{125} \times 100\% = \frac{25}{125} \times 100\% = \frac{1}{5} \times 100\% = 20\%$$

Final Answer: $20\%$

Question 4 7 Marks

A student has to secure $40\%$ marks to pass an examination. If he gets $178\text{ marks}$ and fails by $22\text{ marks}$, find the maximum marks of the examination.

Model Descriptive Solution:

Step 1: Determine Passing Marks 3 Marks

Passing marks $= 178 + 22 = 200\text{ marks}$.

Step 2: Solve for Maximum Marks M 4 Marks

$$40\% \text{ of } M = 200 \implies \frac{40}{100}M = 200 \implies M = 200 \times \frac{100}{40} = 500\text{ marks}$$

Final Answer: $500\text{ marks}$

Question 5 7 Marks

The population of a town increases by $5\%$ annually. If its present population is $1,85,220$, find its population two years ago.

Model Descriptive Solution:

Step 1: Formulate Compounding Equation 3 Marks

Let population two years ago be $P$. Present population $P_2 = P\left(1 + \frac{5}{100}\right)^2 = 185220$.

Step 2: Solve for Initial Population P 4 Marks

$$P \times \left(\frac{21}{20}\right)^2 = 185220 \implies P \times \frac{441}{400} = 185220$$
$$P = 185220 \times \frac{400}{441} = 420 \times 400 = 1,68,000$$

Final Answer: $1,68,000\text{ people}$

Question 6 7 Marks

Due to a reduction of $20\%$ in the price of wheat, a person is enabled to purchase $5\text{ kg}$ more wheat for ₹ $320$. Find the original rate of wheat per kg.

Model Descriptive Solution:

Step 1: Calculate Money Saved from 20% Reduction 3 Marks

Savings $= 20\% \text{ of } ₹ 320 = \frac{20}{100} \times 320 = ₹ 64$.
This ₹ $64$ buys the extra $5\text{ kg}$ of wheat.

Step 2: Compute Reduced & Original Rate 4 Marks

Reduced rate per kg $= \frac{64}{5} = ₹ 12.80\text{ per kg}$.
Since Reduced rate $= 80\% \text{ of Original rate}$:
$$\text{Original rate} = \frac{12.80}{0.80} = ₹ 16\text{ per kg}$$

Final Answer: Original Rate $= ₹ 16\text{ per kg}$

Question 7 7 Marks

In an examination, $70\%$ of candidates passed in English and $65\%$ passed in Mathematics. If $27\%$ of candidates failed in both subjects and $248$ candidates passed in both, find the total number of candidates.

Model Descriptive Solution:

Step 1: Analyze Failure Rates using Venn Principle 3 Marks

Failed in English $= 100 - 70 = 30\%$.
Failed in Maths $= 100 - 65 = 35\%$.
Failed in both $= 27\%$.
Failed in at least one subject $= 30 + 35 - 27 = 38\%$.

Step 2: Compute Percentage Passing in Both 2 Marks

$$\text{Passed in both} = 100\% - 38\% = 62\%$$

Step 3: Solve for Total Candidates 2 Marks

$$62\% \text{ of } T = 248 \implies T = 248 \times \frac{100}{62} = 4 \times 100 = 400$$

Final Answer: $400\text{ candidates}$

Question 8 10 Marks

The value of a military combat vehicle depreciates at the rate of $10\%$ per annum. If its current value is ₹ $8,74,800$, what was its value $3\text{ years}$ ago?

Model Descriptive Solution:

Step 1: State Depreciation Formula 3 Marks

$$V_{\text{present}} = V_0 \left(1 - \frac{R}{100}\right)^n \implies 874800 = V_0 \left(1 - \frac{10}{100}\right)^3 = V_0 \left(\frac{9}{10}\right)^3$$

Step 2: Solve for Initial Value V₀ 4 Marks

$$V_0 \times \frac{729}{1000} = 874800 \implies V_0 = 874800 \times \frac{1000}{729}$$
Notice $874800 \div 729 = 1200$:
$$V_0 = 1200 \times 1000 = ₹ 12,00,000$$

Step 3: Formal Conclusion 3 Marks

The value of the vehicle $3\text{ years}$ ago was ₹ $12,00,000$ (Twelve Lakh Rupees).

Final Answer: ₹ $12,00,000$

Question 9 10 Marks

A man's salary was increased by $20\%$ and then subsequently decreased by $20\%$. If his final salary is ₹ $38,400$, find his original salary and the net loss in rupees.

Model Descriptive Solution:

Step 1: Compute Net Percentage Change 3 Marks

Using net change identity: $+20 - 20 - \frac{20 \times 20}{100} = -4\%$ (Net decrease of $4\%$).

Step 2: Solve for Original Salary S 4 Marks

Final salary $= (100 - 4)\% \text{ of } S = 96\% \text{ of } S = 38400$.
$$S = 38400 \times \frac{100}{96} = 400 \times 100 = ₹ 40,000$$

Step 3: Calculate Net Loss 3 Marks

$$\text{Net Loss} = 40000 - 38400 = ₹ 1,600$$

Final Answer: Original Salary: ₹ $40,000$; Net Loss: ₹ $1,600$

Question 10 10 Marks

Fresh grapes contain $80\%$ water while dry raisins contain $10\%$ water. How many kilograms of dry raisins can be obtained from $180\text{ kg}$ of fresh grapes?

Model Descriptive Solution:

Step 1: Identify Invariable Pulp Content 3 Marks

Water evaporates, but the solid pulp remains constant.
Pulp in fresh grapes $= 100\% - 80\% = 20\%$.
$$\text{Weight of Pulp} = 20\% \text{ of } 180\text{ kg} = \frac{20}{100} \times 180 = 36\text{ kg}$$

Step 2: Relate Pulp to Dry Raisins 4 Marks

Dry raisins contain $10\%$ water, meaning pulp $= 100\% - 10\% = 90\%$.
Let weight of dry raisins be $W\text{ kg}$.
$$90\% \text{ of } W = 36\text{ kg}$$

Step 3: Solve for Raisin Weight W 3 Marks

$$\frac{90}{100}W = 36 \implies W = 36 \times \frac{100}{90} = 4 \times 10 = 40\text{ kg}$$

Final Answer: $40\text{ kg of dry raisins}$

High-Yield FAQs & Revision Notes

What does percentage literally mean?

Percentage comes from Latin 'per centum', meaning 'out of 100'. A percentage is simply a fraction with denominator 100.

Why is percentage change always calculated on the original value?

Because the increase or decrease occurred from that baseline. The original value represents 100% of the initial state.

If price increases by 20%, why does consumption decrease by 16.67%?

Because the new expenditure is based on 120. The reduction needed is 20/120 = 1/6 = 16.67% to restore the expenditure back to 100.

What is the net effect of increasing and then decreasing by x%?

It always results in a net decrease equal to $(x/10)^2\% = x^2/100\%$. For example, +10% then -10% gives -1%.

How do you solve fresh fruit vs dry fruit problems?

Equate the weight of solid dry pulp in both states, since only water evaporates.

How do you calculate compound depreciation?

Use $V_n = V_0(1 - R/100)^n$, where $V_0$ is initial cost, $R$ is annual depreciation rate, and $n$ is years.

What is the difference between percentage and percentage points?

An increase from 20% to 25% is a 5 percentage point increase, but a 25% relative percentage increase.

Can a percentage exceed 100%?

Yes. Any quantity that is more than double its original value has an increase greater than 100% (e.g., tripling is a 200% increase).

How do you convert a decimal to a percentage?

Multiply the decimal by 100 and affix the % sign (e.g., $0.065 \times 100\% = 6.5\%$).

How does this module map to the official RIMC syllabus?

Directly covers Serial No. 9 ('Percentage') 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. 9 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
Fraction to %	Multiply by $100\%$. To convert % to fraction, divide by $100$.
Base Value	Percentage change $= (\text{Change} / \text{Original}) \times 100\%$. Never divide by the new value.
Successive Changes	Net $= a + b + \frac{ab}{100}$. Equal $+x\%$ and $-x\%$ always yields net loss of $-\frac{x^2}{100}\%$.
Price-Consumption	Price rises $r\% \implies$ Consumption drops $\frac{r}{100+r} \times 100\%$.
Compounding	Growth: $P(1 + R/100)^n$; Depreciation: $P(1 - R/100)^n$.
Fruit Pulp Rule	Water changes; Pulp weight is constant. Equate pulp in fresh and dry states.