Addendo Property & Equivalent Ratio Property
Duration: 16 min
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Module outline
- English Comprehension/Verbal Ability: Comprehension / Reading Comprehension / Unseen Passages (Critical Reasoning) (Paragraph Questions), Vocabulary, Sentence Correction (Error Correction), Sentence Completion (Fill in the blanks), Sentence Re-arrangements (Para Jumbles) (Narrative Sequencing), Subject Verb Agreement (Verb Noun Agreement), Tenses, Verb, Articles, Preposition, Conjunction, Pronoun, Noun, Adjectives, Adverb, Modals, Cloze Test (Cloze Passage), Active Passive Voice (Voice), Direct and Indirect Speech, Sentence Construction
- Logical Ability: Coding Decoding, Directions (Direction Test), Analogy, Inequality, Data Sufficiency, Series (Number and Letter Series) (Numerical Relations and Reasoning), Syllogisms, Seating Arrangements, Puzzle Solving (Puzzles), Blood Relations, Deductive and Inductive Reasoning (Logical Deduction and Induction) (Prepositional Reasoning), Ranking, Order and Sequence, Logical Sequence Of Words, Data Arrangement, Venn Diagram, Calendar, Clock
- Quantitative Ability: Logarithm, Permutation and Combination, Probability, Profit and Loss, Speed Time and Distance, Mixture and Alligation, Partnership, Simple Interest and Compound Interest, Percentage, Ratio and Proportion (Ratios), Average, Time and Work, Pipe and Cistern, Number System, Divisibility Rules, HCF LCM, Algebra, Powers and Exponents (Surds and Indices), Mensuration and Geometry, Age Problems, Data Interpretation, Sequence and Series (Series)
- Basic Analytical Ability: Non Verbal Reasoning (Spatial Aptitude) (Spatial Reasoning) (Visual Reasoning), Analytical Reasoning (Counting Figures Reasoning), Cubes & Dices, Selection Decision Table (Decision Making)
- Computer Programming: Pseudo Codes, Coding Questions by Yash Sir, C Programming by Yash Sir, Programming in C++, Java, Python for Placements by Yash Sir
- DBMS (Computer Science): Basics of DBMS, ER Diagram, Relational Model & Functional Dependencies, Keys & Integrity Constraints, Normalization (1NF - BCNF), Transaction Management, Concurrency Control, Database Security & Authorization
- SQL (Computer Science): SQL
- OS (Computer Science): Introduction to OS, Process Management, CPU Scheduling, Deadlock, Process Synchronization, Threads & Process Creation, Memory Management, Virtual Memory, File Management
- Computer Networks (Computer Science): Introduction to CN, Data Communication, DLL: Access Control, DLL: Flow Control, DLL: Error Control, DLL: Framing, Data Link Layer - Ethernet, Net Layer: IPv4 & Proto, Net Layer: IP Addressing, Net Layer:Routing Protocol, Transport Layer Services, TL: Congestion & UDP, Application Layer, Hardware basics, Network Security
- Data Structures (Computer Science): Introduction to DS, Array, Stack, Queue, Linked List, Tree, Graphs, Hashing
- OOPs (Computer Science): Object Oriented Programming (OOPS)
- Software Engineering (Computer Science): Fundamentals of Software Engineering, Software Requirements and Quality Assurance, Software Design, Estimation and Metrics, Software Testing, Software Maintenance and Configuration Management
- SVAR Conversational - Spoken English: Communication Assessment
- HR Interview: MR & HR Interview Preparation
AI summary & chapters
AI Summary
An AI-generated summary of this video lecture.
This lesson on Ratio and Proportion, presented by Yash Jain, focuses on two key algebraic properties: the Addendo Property and the Equivalent Ratio Property. The video begins by defining a ratio as a comparison of two quantities before introducing the Addendo Property, which states that if multiple ratios are equal (a/b = c/d = e/f), their common value is preserved when the sums of their antecedents and consequents are taken. This is demonstrated numerically using 2/3 = 6/9 = 8/12, where (2+6+8)/(3+9+12) simplifies to 16/24, which reduces back to 2/3. The instructor then provides a formal algebraic proof by setting the equal ratios to a constant k, expressing numerators as multiples of denominators (a=bk, c=dk), and showing that the combined fraction simplifies to k. A generalized version using variables p, q, r, s, t, u is also presented to show that multiplying terms by constants maintains equivalence. The lesson applies the Addendo Property to solve a problem where p:q = r:s = t:u = 2:3, finding (p+r+t)/(q+s+u) to be 2/3. The second half introduces the Equivalent Ratio Property, which states that if a:b :: c:d, then (a±c):(b±d) :: a:b. This is proven by setting a/b = c/d = k and showing that (a+c)/(b+d) simplifies to k. The property is extended to forms like (pa+qc)/(pb+qd). A final problem asks for the value of (ab+cd)/(b²+d²) given a:b = c:d = 2:3, which is solved by recognizing the expression as an application of the Equivalent Ratio Property with p=b and q=d, yielding 2/3. The lesson concludes with a thank-you slide.
Chapters
0:00 – 2:00 00:00-02:00
The lesson opens with a title slide for Ratio and Proportion, defining ratio as the comparison of two quantities. The instructor introduces the Addendo Property with the rule: if a:b = c:d = e:f, then each ratio equals (a+c+e):(b+d+f). A numerical example using 2/3 = 6/9 = 8/12 is shown, where (2+6+8)/(3+9+12) = 16/24 simplifies to 2/3. Red annotations highlight corresponding terms and circle the final combined fraction, reinforcing the visual connection between individual ratios and their sum.
2:00 – 5:00 02:00-05:00
The instructor transitions from the numerical example to a formal algebraic proof of the Addendo Property. By setting a/b = c/d = e/f = k, the numerators are expressed as multiples of denominators: a=bk, c=dk, e=fk. Substituting these into (a+c+e)/(b+d+f) yields (bk+dk+fk)/(b+d+f), which factors to k(b+d+f)/(b+d+f) and cancels to k. This proves that the combined fraction equals the original ratio, establishing the extended proportion a/b = c/d = e/f = (a+c+e)/(b+d+f).
5:00 – 10:00 05:00-10:00
The proof is generalized using new variables p, q, r, s, t, u to represent terms in equivalent fractions ma/mb, nc/nd, oe/of. The combined fraction (p+r+t)/(q+s+u) is shown to equal (ma+nc+oe)/(mb+nd+of), confirming that multiplying terms by constants preserves the ratio. The Addendo Property is then applied to a problem: if p:q = r:s = t:u = 2:3, then (p+r+t)/(q+s+u) = 2/3. The lesson transitions to the Equivalent Ratio Property, stating that if a:b :: c:d, then (a±c):(b±d) :: a:b.
10:00 – 15:00 10:00-15:00
The Equivalent Ratio Property is proven by setting a/b = c/d = k (k ≠ 0) and showing that (a+c)/(b+d) simplifies to k(b+d)/(b+d) = k. The property is extended algebraically to forms like (pa+qc)/(pb+qd). A problem on an orange slide asks: if a:b = c:d = 2:3, what is (ab+cd)/(b²+d²)? The instructor re-expresses the target fraction as (ba+dc)/(bb+dd), identifying it with p=b and q=d, so the value is 2/3 by the Equivalent Ratio Property.
15:00 – 15:45 15:00-15:45
The lesson concludes with a white slide headed 'Equivalent Ratio Property' that recaps two examples, boxing (ab+cd)/(b²+d²) and ending with a generalized form (4a-3c+9e)/(4b-3d+9f). A final 'THANKS FOR WATCHING' slide closes the video, signaling the end of the instructional content on ratio properties.
The lesson systematically builds from concrete examples to abstract proofs, a pedagogical approach that helps students internalize ratio properties. The Addendo Property is first demonstrated numerically, then proven algebraically using a constant k, and finally generalized with additional variables. This three-step progression (example → proof → generalization) is a key teaching pattern worth noting for exam preparation. The Equivalent Ratio Property follows a similar structure, with the proof emphasizing that (a±c)/(b±d) preserves the common ratio k. The final problem on (ab+cd)/(b²+d²) is particularly instructive because it requires recognizing that the expression fits the (pa+qc)/(pb+qd) form with p=b and q=d, a non-obvious substitution that tests conceptual understanding rather than rote application. Students should focus on identifying when a given fraction can be rewritten to match the property's standard form, as this is the core skill being assessed.