1 Apr - Aptitude - Ratio & Proportion
Duration: 1 hr 30 min
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AI Summary
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This educational video lecture, presented by Yash Jain, provides a comprehensive overview of Ratio and Proportion concepts tailored for competitive exams like GATE. The session begins with fundamental definitions and properties of ratios, including equivalent ratios and the principle that multiplying or dividing terms by the same number preserves the ratio. The instructor then systematically works through various types of ratio problems, such as finding compound ratios, combining multiple ratios (a:b:c, a:b:c:d), and solving algebraic expressions involving ratios. Key topics covered include duplicate, sub-duplicate, triplicate, and inverse ratios, as well as the properties of equivalent ratios where the sum of numerators divided by the sum of denominators equals the individual ratio. The lecture features numerous worked examples on a digital whiteboard, demonstrating step-by-step solutions for problems involving coins, salaries, and cricket scores. Towards the end, the instructor promotes a GATE Guidance course by Sanchit Sir, showcasing the course content and syllabus on a website interface before concluding the session.
Chapters
0:00 – 2:00 00:00-02:00
The video begins with a title slide displaying 'Ratio & Proportion - By Yash Jain' against a pink background with confetti graphics. The instructor introduces the topic, and the first content slide appears titled 'What is a Ratio?'. It defines a ratio as the comparison of two or more numbers in terms of 'how many times'. The slide explains that in certain situations, comparing quantities by division is efficient. It states that the simplified form of two quantities of the same kind is referred to as a ratio. The text emphasizes that the two numbers in a ratio can only be compared when they have the same unit. The symbol used to denote a ratio is shown as ':'. The instructor verbally reinforces these definitions, setting the stage for the mathematical concepts to follow.
2:00 – 5:00 02:00-05:00
The lecture transitions to the 'Properties of Ratio'. The instructor writes '3:4' and '3/4 = 0.75' on the board. He explains the first property: if you multiply or divide the numerator and denominator of a ratio by the same number, the ratio does not change. An example is provided: '3 * 2 / 4 * 2 = 6 / 8 = 0.75'. The text on the slide confirms that the numerator and denominator are multiplied by 2, and there is no change in the ratio. The instructor further clarifies that the ratio remains unchanged if both antecedent and consequent are multiplied by the same number. This foundational property is crucial for simplifying complex ratio problems later in the session.
5:00 – 10:00 05:00-10:00
A problem is presented: 'If p:q = r:s = t:u = 2:3 then what is the value of (p+r+t) / (q+s+u)?'. The instructor writes the ratios as fractions: p/q = r/s = t/u = 2/3. He then derives expressions for p, r, and t in terms of q, s, and u respectively: p = 2q/3, r = 2s/3, t = 2u/3. Substituting these into the target expression, he gets (2q/3 + 2s/3 + 2u/3) / (q+s+u). He factors out 2/3 to get (2/3 * [q+s+u]) / (q+s+u). The terms [q+s+u] cancel out, leaving the final answer as 2/3. This demonstrates the property that the ratio of the sum of numerators to the sum of denominators is equal to the individual ratio.
10:00 – 15:00 10:00-15:00
The next problem asks: 'If a:b = c:d = 2:3 then what is the value of (ab+cd) / (b^2+d^2)?'. The instructor rewrites the given ratios as a/b = c/d = 2/3. He manipulates the terms to match the denominator of the target expression. He writes ab/b^2 = 2/3 and cd/d^2 = 2/3. This implies ab = 2/3 * b^2 and cd = 2/3 * d^2. Substituting these into the numerator of the target expression, he gets (2/3 * b^2 + 2/3 * d^2) / (b^2+d^2). Factoring out 2/3 gives (2/3 * [b^2+d^2]) / (b^2+d^2). The terms cancel out, resulting in the answer 2/3. This example reinforces the technique of manipulating ratios to fit the structure of the required expression.
15:00 – 20:00 15:00-20:00
A more complex problem is introduced: 'If p:q = r:s = t:u = 2:3 then what is the value of (mp+nr+ot) : (mq+ns+ou)?'. The instructor writes the ratios as fractions: p/q = r/s = t/u = 2/3. He then multiplies each ratio by a corresponding variable to match the numerator terms: mp/mq = p/q = 2/3, nr/ns = r/s = 2/3, ot/ou = t/u = 2/3. This shows that mp = 2/3 * mq, nr = 2/3 * ns, and ot = 2/3 * ou. Substituting these into the expression, he gets (2/3 * mq + 2/3 * ns + 2/3 * ou) / (mq+ns+ou). Factoring out 2/3 leads to the final answer of 2/3. This generalizes the property that any linear combination of terms with the same ratio will maintain that ratio.
20:00 – 25:00 20:00-25:00
The instructor summarizes the general property of equivalent ratios. He writes p/q = r/s = t/u = v/w = x/y = k. He then demonstrates that (p+t)/(q+u) = k, (r+x)/(s+y) = k, and (p+r+t)/(q+s+u) = k. He extends this to more terms, showing (p+r+t+v)/(q+s+u+w) = k and (p+r+t+v+x)/(q+s+u+w+y) = k. This section solidifies the concept that if multiple ratios are equal to a constant k, then the ratio of the sum of the numerators to the sum of the denominators is also equal to k. This is a powerful tool for solving complex ratio problems quickly.
25:00 – 30:00 25:00-30:00
The lecture moves to 'Types of Ratio'. The instructor lists five types based on a ratio a:b. Type 1 is Duplicate Ratio, defined as a^2 : b^2 (e.g., 8^2 : 1^2 = 64:1). Type 2 is Sub-Duplicate Ratio, defined as sqrt(a) : sqrt(b) (e.g., sqrt(8) : sqrt(1) = 2sqrt(2):1). Type 3 is Triplicate Ratio, defined as a^3 : b^3 (e.g., 8^3 : 1^3 = 512:1). Type 4 is Sub-Triplicate Ratio, defined as cbrt(a) : cbrt(b) (e.g., cbrt(8) : cbrt(1) = 2:1). Type 5 is Inverse Ratio / Reciprocal Ratio, defined as 1/a : 1/b which simplifies to b:a (e.g., 1/8 : 1/1 = 1:8). The instructor writes these definitions and examples on the board for clarity.
30:00 – 35:00 30:00-35:00
The concept of 'Compound Ratio / Compounded Ratio / Mixed Ratio' is introduced. The definition states that for two or more ratios, if we take the antecedent as the product of antecedents and the consequent as the product of consequents, the resulting ratio is called a mixed or compound ratio. The formula is given as: compound ratio of m:n and p:q is mp:nq. In other words, when two or more ratios are multiplied term-wise, the ratio thus obtained is called a compound ratio. The instructor writes 'a:b' and 'p:q' and shows the resulting 'ap:bq' to illustrate the concept visually.
35:00 – 40:00 35:00-40:00
Two problems are solved. First, 'Find the compound ratio of 2/3, 7/9 and 12/35'. The instructor multiplies the numerators (2*7*12) and denominators (3*9*35). He simplifies the fraction to 8/45, so the answer is 8:45. Second, 'If a:b = 2:3 and b:c = 5:6, then find a:b:c?'. He writes a/b = 2/3 and b/c = 5/6. To combine them, he makes the 'b' term common by finding the LCM of 3 and 5, which is 15. He multiplies the first ratio by 5 (10:15) and the second by 3 (15:18). The combined ratio is a:b:c = 10:15:18. This demonstrates the method of equating common terms to combine ratios.
40:00 – 45:00 40:00-45:00
A more complex combination problem is solved: 'If a:b = 3:4 and b:c = 5:6 and c:d = 2:3, then find a:b:c:d?'. The instructor uses the LCM method again. First, he combines a:b and b:c. LCM of 4 and 5 is 20. a:b becomes 15:20, and b:c becomes 20:24. So a:b:c is 15:20:24. Next, he combines this with c:d. LCM of 24 and 2 is 24. The ratio a:b:c remains 15:20:24, and c:d becomes 24:36. The final combined ratio is a:b:c:d = 15:20:24:36. The instructor writes out the step-by-step multiplication factors to show how the terms are adjusted to make the common variables equal.
45:00 – 50:00 45:00-50:00
A word problem involving coins is presented: 'The number of coins of Rs 1, Rs 5 and Rs 10 denominations that a person has are in the ratio 5:3:13. Of the total amount, the percentage of money in Rs 5 coins is?'. The instructor sets up the number of coins as 5y, 3y, and 13y. He then calculates the total money for each denomination: Rs 1 coins = 5y * 1 = 5y, Rs 5 coins = 3y * 5 = 15y, Rs 10 coins = 13y * 10 = 130y. The total money is 5y + 15y + 130y = 150y. The percentage of money in Rs 5 coins is (15y / 150y) * 100 = 10%. This problem illustrates how to convert a ratio of quantities into a ratio of values.
50:00 – 55:00 50:00-55:00
A theoretical problem is discussed: 'If two distinct non-zero real variables x and y are such that (x+y) is proportional to (x-y) then the value of x/y'. The instructor writes (x+y) = k(x-y). He rearranges the equation to solve for x/y. x + y = kx - ky. Grouping x terms and y terms gives y + ky = kx - x, or y(1+k) = x(k-1). Therefore, x/y = (k+1)/(k-1). Since k is a constant of proportionality, the value of x/y is a constant. The instructor circles the answer 'is a constant' and explains that it depends only on the constant k, not on the specific values of x and y.
55:00 – 60:00 55:00-60:00
A cricket-themed problem is solved: 'Three cricket players - Virat Kohli, Rohit Sharma and MS Dhoni play for three different cricket teams... Virat's runs to Rohit's runs and Rohit's runs to Dhoni's runs are in the ratio 5:7. If the total runs scored by all the three players in the tournament is 327, find the total runs scored by Rohit Sharma in IPL 2020?'. The instructor writes Virat:Rohit = 5:7 and Rohit:Dhoni = 5:7. To combine them, he makes the Rohit term common (LCM of 7 and 5 is 35). Virat:Rohit becomes 25:35, and Rohit:Dhoni becomes 35:49. The combined ratio V:R:D is 25:35:49. The sum of parts is 25+35+49 = 109. Since total runs = 327, and 327/109 = 3, each part is 3. Rohit's runs = 35 * 3 = 105.
60:00 – 65:00 60:00-65:00
An algebraic expression problem is solved: 'If x/y = 7/8, then find the value of the expression (3x-5y)/(2x+7y)?'. The instructor substitutes x=7 and y=8 directly into the expression. The numerator becomes 3(7) - 5(8) = 21 - 40 = -19. The denominator becomes 2(7) + 7(8) = 14 + 56 = 70. The final value is -19/70. He also shows an alternative method by substituting x = 7y/8 into the expression, which leads to the same result after simplification. This demonstrates the direct substitution method for evaluating expressions given a ratio.
65:00 – 70:00 65:00-70:00
The instructor transitions to promoting a course. He shows a webpage for 'GATE Guidance by Sanchit Sir'. The page describes a GATE 2026 bundle covering all syllabus from basics to advanced level, containing recorded videos, PYQ video solutions, and test series. The rating is shown as 4.9 stars. The content section lists 'Welcome Guide' and 'Gate Live Classes'. The instructor points out that the course is valid for 486 days and includes 20 courses. This segment serves as an advertisement for the educational platform associated with the lecture.
70:00 – 75:00 70:00-75:00
The instructor continues to navigate the course website, showing the syllabus for 'Data Interpretation'. The syllabus includes topics like 'Ratio & Proportion - Variation (Definition & Types)', 'Ratio & Proportion - Componendo & Dividendo Property', 'Ratio & Proportion - Addendo & Equivalent Ratio Property', and 'Ratio & Proportion - Divide a Number in x ratio of y'. The instructor scrolls through the list, highlighting the comprehensive coverage of ratio and proportion topics within the Data Interpretation module. This reinforces the relevance of the lecture content to the broader course curriculum.
75:00 – 80:00 75:00-80:00
The video returns to the lecture content, specifically revisiting the problem about (x+y) being proportional to (x-y). The instructor writes the equation (x+y) = k(x-y) again. He rearranges it to y(1+k) = x(k-1) and solves for x/y = (k+1)/(k-1). He emphasizes that since k is a constant, x/y is also a constant. He underlines the answer 'is a constant' and explains that the value does not depend on x or y individually, but only on the constant of proportionality k. This repetition helps reinforce the concept for students.
80:00 – 85:00 80:00-85:00
The instructor continues to discuss the proportionality problem, writing 's = d/t' and 'O(n^2)' on the board, possibly drawing an analogy to algorithmic complexity or physics formulas, though the context is slightly ambiguous. He also writes 'log m + log n = log (mn)', likely as a side note or analogy for combining terms. He then circles a box with '4y' and 'O' inside, possibly referring to a specific variable or constant in a different context. The focus remains on the mathematical manipulation of the proportionality equation.
85:00 – 89:37 85:00-89:37
The video concludes with the instructor speaking directly to the camera. He is wearing a black shirt with a 'KG Knowledge' logo. He summarizes the key takeaways from the session, reiterating the importance of understanding the properties of ratios and proportions for solving complex problems. He encourages students to practice the examples shown and to check out the GATE Guidance course for further preparation. The video ends with a close-up of the instructor, providing a personal touch to the educational content.
The lecture provides a structured and detailed exploration of Ratio and Proportion, starting from basic definitions and progressing to complex algebraic manipulations. The instructor effectively uses a digital whiteboard to demonstrate step-by-step solutions for various problem types, including equivalent ratios, compound ratios, and combining multiple ratios. Key properties such as the sum of numerators over sum of denominators are highlighted as powerful tools. The session also covers specific types of ratios like duplicate and triplicate, and applies these concepts to real-world word problems involving coins, salaries, and sports scores. The theoretical aspect is reinforced with problems involving proportionality constants. The lecture concludes with a promotion of a comprehensive GATE preparation course, linking the specific topic to a broader curriculum. Overall, the video serves as a valuable resource for students preparing for competitive exams, offering clear explanations and practical problem-solving techniques.