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#8 Compulsory Commercial Mathematics Math 8

Ratio & Proportion Fundamentalsअनुपात एवं समानुपात: मूल अवधारणाएं व अनुप्रयोग

Ratio and Proportion Fundamentals form the cornerstone of commercial and advanced arithmetic, carrying an essential weightage of 4-6 questions in the Super TET Junior Part 2C Mathematics syllabus. Mastering equivalent ratios, continued proportions, and proportional distribution is crucial to achieve a top rank.

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Syllabus Focus & High-Yield Strategy: Ratio and Proportion Fundamentals form the cornerstone of commercial and advanced arithmetic, carrying an essential weightage of 4-6 questions in the Super TET Junior Part 2C Mathematics syllabus. Mastering equivalent ratios, continued proportions, and proportional distribution is crucial to achieve a top rank.

1. Fundamental Axioms, Definitions & Foundational Properties

Definition of Ratio: A ratio is the comparison of two quantities of the same kind by division, expressed as $a : b$ where $b \neq 0$. It is a dimensionless pure number devoid of units.
Property of Scaling: Multiplying or dividing each term of a ratio by the same non-zero constant $k$ does not alter the ratio, i.e., $a : b = ka : kb$.
Definition of Proportion: An equality of two ratios is called a proportion, denoted as $a : b :: c : d$ or $\frac{a}{b} = \frac{c}{d}$.
Cross-Product Rule: In any valid proportion $\frac{a}{b} = \frac{c}{d}$, the product of the extremes equals the product of the means: $ad = bc$.
Exam Pro-Tip: Always convert all given quantities into identical units (e.g., paise to rupees, grams to kilograms) before formulating any ratio or proportion expression.
Memory Mnemonic: EXTREME — Extremes X Together Relate Equally Means Exactly ($ad = bc$).

2. Formulas, Standard Theorems & Algorithmic Procedures

Category / CaseFormula / Mathematical RuleTopper Calculation Shortcut
Fourth Proportional ($d$)If $a : b :: c : d$, then $d = \frac{b \cdot c}{a}$Multiply middle terms and divide by the first term instantly.
Third Proportional ($c$)If $a : b :: b : c$, then $c = \frac{b^2}{a}$Square the second term and divide by the single given term.
Mean Proportional ($b$)If $a : b :: b : c$, then $b = \sqrt{ac}$Take the geometric mean (square root of the product of outer terms).
Theorem of Compendendo & Dividendo: If $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.
Theorem on Equal Ratios: If $\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$, each ratio equals $\frac{pa + qc + re}{pb + qd + rf}$ for non-zero constants $p, q, r$.
Continued Proportion Rule: Quantities $a, b, c$ are in continued proportion if $\frac{a}{b} = \frac{b}{c} \implies b^2 = ac$.
15-Second Speed-Math Trick: When dividing a total sum $S$ in the ratio $a : b : c$, the individual shares are $\frac{a}{a+b+c} \times S$, $\frac{b}{a+b+c} \times S$, and $\frac{c}{a+b+c} \times S$. Use fractions directly without finding the common variable $x$ to save precious seconds.

3. Word Problem Modeling & Practical Applications

Problem Type 1 (Income & Expenditure): If incomes are in ratio $a : b$ and expenditures in ratio $c : d$ with equal savings $S$, then income of first person is $\frac{S(b-d)}{ad-bc}$.
Problem Type 2 (Mixture & Alloy): Mixing two solutions of concentration $c_1$ and $c_2$ in ratio $q_1 : q_2$ yields a final concentration $C = \frac{c_1 q_1 + c_2 q_2}{q_1 + q_2}$.
Problem Type 3 (Age Related Ratios): Represent ages as $ax$ and $bx$. After $t$ years, the new ratio becomes $\frac{ax+t}{bx+t} = \frac{p}{q}$, solved by cross-multiplication.
Problem Type 4 (Coin / Currency Problems): Total value = (Number of coins $\times$ Denomination value). Convert coin ratios to value ratios before applying total sum.
UP Exam High-Yield Tip: In coin-based word problems, always equate the total given money to the sum of (Number of coins $\times$ individual coin value). Do not equate raw coin counts to total monetary value.

4. Mathematical Edge Cases, Sign Conventions & Common Traps

Critical Distinction: Ratio $a : b$ is strictly distinct from fraction $\frac{a}{b}$ in contextual word problems; ratios represent relative parts of a whole rather than absolute quantities.
Edge Case / Exception: Division by zero is undefined. In proportional equations, denominators $b$, $d$, and term $a$ in denominators must strictly satisfy $a \neq 0, b \neq 0, d \neq 0$.
Common Calculation Trap: Inverting terms in inverse ratio problems. The inverse ratio of $a : b$ is $\frac{1}{a} : \frac{1}{b}$ or simplified as $b : a$, not $-a : -b$.
Trap Prevention Trick: When given a ratio like $2:3:5$ and asked to increase each by $10\%$, do not add $10$ to each ratio term. Convert to absolute variables $2x, 3x, 5x$, apply percentage increase, and simplify back.
Memory Mnemonic: RATIO — Reduce All Terms Into One.

Comprehensive Mathematical Matrix & Method Comparison

Parameter / DimensionMethod / Concept AMethod / Concept BExam Speed & Accuracy Takeaway
Proportional TypeDirect Proportion ($y = kx$)Inverse Proportion ($xy = k$)Direct uses cross-multiplication; Inverse uses constant product.
Mean ProportionalFormula $b = \sqrt{ac}$Arithmetic Mean $\frac{a+c}{2}$Always use geometric mean for proportions; arithmetic mean leads to -1 negative marking.
Ratio CompositionDuplicate Ratio ($a^2 : b^2$)Sub-duplicate Ratio ($\sqrt{a} : \sqrt{b}$)Duplicate squares terms; Sub-duplicate takes square roots.
Coin Word ProblemsEquating coin counts directlyEquating monetary value sumsAlways convert coin ratios to value ratios before solving.
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Question 1 of 10 Super TET Junior Mathematics

1. If $x : y = 3 : 4$, what is the value of the expression $\frac{5x + 2y}{3x + 4y}$?

Correct Answer: Option A

Substitute $x = 3k$ and $y = 4k$ into the expression: $\frac{5(3k) + 2(4k)}{3(3k) + 4(4k)} = \frac{15k + 8k}{9k + 16k} = \frac{23k}{25k} = \frac{23}{25}$. Distractors arise from incorrect substitution or adding coefficients wrongly.

Question 2 of 10 Super TET Junior Mathematics

2. What is the mean proportional between $0.04$ and $0.36$?

Correct Answer: Option A

The mean proportional $b$ between $a$ and $c$ is given by $b = \sqrt{ac}$. Here, $b = \sqrt{0.04 \times 0.36} = \sqrt{0.0144} = 0.12$. Option 2 (0.20) is a common error from calculating $\frac{0.04+0.36}{2}$. Thus, The answer is Option A.

Question 3 of 10 Super TET Junior Mathematics

3. Two numbers are in the ratio $5 : 7$. If $9$ is subtracted from each, the new ratio becomes $3 : 5$. What is the greater number?

Correct Answer: Option A

Let the numbers be 5x and 7x. Then (5x-9)/(7x-9)=3/5, so 25x-45=21x-27 and x=4.5. The greater number is 7x=31.5. Subtracting 9 gives 13.5:22.5=3:5, as required. The answer is Option A.

Question 4 of 10 Super TET Junior Mathematics

4. If $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = 2.5$, then what is the value of $\frac{2a^2 + 3c^2 + 4e^2}{2b^2 + 3d^2 + 4f^2}$?

Correct Answer: Option A

Given $\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k = 2.5$. Then $a = kb, c = kd, e = kf$. Substituting these into the expression gives $\frac{2k^2b^2 + 3k^2d^2 + 4k^2f^2}{2b^2 + 3d^2 + 4f^2} = k^2 = (2.5)^2 = 6.25$. The answer is Option A.

Question 5 of 10 Super TET Junior Mathematics

5. A bag contains 50 paise, 25 paise, and 10 paise coins in the ratio $5 : 9 : 4$, amounting to a total of Rs. 206. Find the total number of coins in the bag.

Correct Answer: Option B

Let the coin counts be 5x, 9x, and 4x. Their value in paise is 250x+225x+40x=20600, so 515x=20600 and x=40. The total number of coins is 18x=720. The answer is Option B.

Question 6 of 10 Super TET Junior Mathematics

6. What is the fourth proportional to $4$, $9$, and $12$?

Correct Answer: Option A

Let the fourth proportional be $x$. By definition of proportion, $\frac{4}{9} = \frac{12}{x} \implies 4x = 9 \times 12 = 108 \implies x = \frac{108}{4} = 27$. Distractors like 36 or 24 arise from incorrect cross-multiplication or taking third proportional formulas.

Question 7 of 10 Super TET Junior Mathematics

7. If $x$ is subtracted from each of the numbers $23, 30, 57$, and $78$, the numbers so obtained in this order are in proportion. What is the value of $x$?

Correct Answer: Option B

Proportion gives (23-x)/(30-x)=(57-x)/(78-x). Cross-multiplying and cancelling x^2 gives 1794-101x=1710-87x, so 84=14x and x=6. The answer is Option B.

Question 8 of 10 Super TET Junior Mathematics

8. The incomes of A and B are in the ratio $3 : 4$ and their expenditures are in the ratio $5 : 7$. If each saves Rs. 5,000, what is the income of A?

Correct Answer: Option A

Let incomes be $3x$ and $4x$, and expenditures be $5y$ and $7y$. Then $3x - 5y = 5000$ and $4x - 7y = 5000$. Equating savings: $3x - 5000 = 5y \implies y = \frac{3x-5000}{5}$. Substituting into second equation: $4x - 7\left(\frac{3x-5000}{5}\right) = 5000 \implies 20x - 21x + 35000 = 25000 \implies x = 10,000$. Income of A = $3x = 3 \times 10,000 = \text{Rs. } 30,000$. The answer is Option A.

Question 9 of 10 Super TET Junior Mathematics

9. What is the duplicate ratio of $5 : 7$ compounded with the inverse ratio of $25 : 42$ and the sub-duplicate ratio of $36 : 49$?

Correct Answer: Option A

Duplicate ratio of $5:7$ is $25:49$. Inverse ratio of $25:42$ is $42:25$. Sub-duplicate ratio of $36:49$ is $6:7$. Compounding them means multiplying: $\frac{25}{49} \times \frac{42}{25} \times \frac{6}{7} = \frac{42 \times 6}{49 \times 7} = \frac{252}{343} = \frac{36}{49}$ (since $252 = 36 \times 7$ and $343 = 49 \times 7$). The answer is Option A.

Question 10 of 10 Super TET Junior Mathematics

10. Divide Rs. 1,275 among A, B, and C such that A gets 2/3 of B's share and B gets 1/4 of C's share. What is B's share?

Correct Answer: Option A

The ratios are A:B=2:3 and B:C=1:4=3:12, so A:B:C=2:3:12. The 17 parts total Rs. 1,275, so each part is Rs. 75. B receives 3 parts, or Rs. 225. The answer is Option A.

5-Minute Exam Simulation Mode

Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.

Time Limit:

5 Minutes (300s)

Marking Scheme:

+3 Marks | -1 Negative

Passing Benchmark:

21+ Marks (70%)

High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)

#1
A ratio $a : b$ has no units and exists only between quantities of the same physical dimension.
#2
In a proportion $\frac{a}{b} = \frac{c}{d}$, the product of extremes equals the product of means ($ad = bc$).
#3
Third proportional to $a$ and $b$ is calculated as $\frac{b^2}{a}$.
#4
Mean proportional between $a$ and $c$ is given by the geometric mean $\sqrt{ac}$.
#5
Compounded ratio of $a:b$ and $c:d$ is $ac:bd$.
#6
Duplicate ratio of $a:b$ is $a^2:b^2$, and sub-duplicate ratio is $\sqrt{a}:sqrt{b}$.
#7
Inverse ratio of $a:b$ is $b:a$, representing reciprocal scaling.
#8
Dividendo and Componendo theorem states that if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a+b}{a-b} = \frac{c+d}{c-d}$.

Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)

Proportion Rule Mnemonic

Memory Key: EXTREME

E - Extremes, X - X-multiply, T - Terms, R - Relate, E - Equally, M - Means, E - Exactly.

Ratio Distribution Mnemonic

Memory Key: SHARE

S - Sum of parts, H - Holder ratio, A - Actual total, R - Ratio multiply, E - Evaluate.

Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)

Negative Marking Alert #1:

Negative Marking Trap 1: Failing to convert disparate units (e.g., meters and centimeters) before writing down a ratio.

Negative Marking Alert #2:

Negative Marking Trap 2: Confusing third proportional formula ($\frac{b^2}{a}$) with mean proportional formula ($\sqrt{ac}$).

Negative Marking Alert #3:

Negative Marking Trap 3: Directly adding numbers to ratio terms instead of introducing a common variable $x$.

Negative Marking Alert #4:

Negative Marking Trap 4: Forgetting the -1 negative marking penalty when guessing uncertain proportional word problems.

Self-Assessment Checklist

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