Short Trick to convert given ratio into required ratio

Duration: 8 min

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AI summary & chapters

AI Summary

An AI-generated summary of this video lecture.

This lecture by Yash Jain teaches a shortcut for ratio and proportion problems asking what number must be added to or subtracted from each term of a given ratio a:b so that it becomes c:d. The lesson begins by defining a ratio as a comparison of two quantities and posing the general question: what should be added or subtracted to each term of a:b to convert it into c:d? The instructor first demonstrates the standard algebraic method using two examples. For subtraction, he solves what should be subtracted from each term of 15:19 to make it 3:4 by setting up (15-y)/(19-y) = 3/4, cross-multiplying to get 4(15-y) = 3(19-y), expanding to 60 - 4y = 57 - 3y, and solving for y. For addition, he solves what should be added to each term of 9:16 to make it 2:3 by setting up (9+y)/(16+y) = 2/3, cross-multiplying to get 3(9+y) = 2(16+y), expanding to 27 + 3y = 32 + 2y, and solving to find y = 5. The core of the lesson is then a direct shortcut formula that avoids solving equations entirely. The instructor introduces the general formula |ad - bc| / |(a+c) - (b+d)|, then simplifies the denominator to |a-c| (or equivalently the difference of the target ratio terms). The method is: cross-multiply the original and target ratios, subtract the two products to get a numerator difference, then divide by the absolute difference of the target ratio terms. Applying this to 9:16 becoming 2:3, he calculates 9*3 = 27 and 16*2 = 32, subtracts to get 5, divides by 3-2 = 1, yielding the answer 5. He notes this saves time compared to algebra, referencing a timer showing 40 seconds crossed out and replaced with 60 seconds. The lesson concludes with two practice examples: adding to 11:21 to get 3:5 (calculating 11*5 = 55, 21*3 = 63, difference 8, divided by 5-3 = 2 to get answer 4), and subtracting from 23:38 to get 4:7 (calculating 23*7 = 161, 38*4 = 152, difference 9, divided by 7-4 = 3 to get answer 3). The consistent rule is: cross-multiply, subtract products, divide by the difference of target ratio terms.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The opening slide displays the heading 'RATIO & PROPORTION - BY YASH JAIN' with a whiteboard panel reading 'Ratio a comparison of 2 quantities' illustrated with colored dots labeled part, whole and total. A pink slide poses the general question 'What should be added/subtracted to each term of a:b to convert them to c:d?' At 35s an orange slide introduces the first specific example: 'Que: What should be added to each term of 9 : 16 so that new ratio becomes 2 : 3?' The instructor writes the fraction 9/16 with an arrow pointing to 2/3, then sets up (9+y)/(16+y) = 2/3 and cross-multiplies to '3(9+y) = 2(16+y)', expanding to '27+3y = 32+2y' and solving to get y = 5.

  2. 2:00 – 5:00 02:00-05:00

    The instructor first solves a subtraction problem for the ratio 15:19 to become 3:4 by setting up (15-y)/(19-y) = 3/4 and cross-multiplying to get 4(15-y) = 3(19-y), expanding to 60 - 4y = 57 - 3y. He then transitions back to the addition problem for 9:16 becoming 2:3 and introduces a general shortcut formula using absolute values of cross-products: |ad-bc| / |(a+c)-(b+d)|. He simplifies the denominator to |a-c|, highlighting it with a red circle. Applying the shortcut, he calculates 9*3 = 27 and 16*2 = 32, subtracts to get 5, and divides by the difference of target ratio terms (3-2 = 1), writing the final answer 5/1 on the board. A timer drawn on the board shows '40 sec' crossed out and replaced with '60 sec', indicating a time-saving goal.

  3. 5:00 – 7:54 05:00-07:54

    The instructor applies the shortcut to two additional practice problems. For 'What should be added to each term of 11 : 21 so that new ratio becomes 3 : 5?', he calculates 11*5 = 55 and 21*3 = 63, finds the difference 63-55 = 8, and divides by 5-3 = 2 to get the answer 4. For 'What should be subtracted from each term of 23 : 38 so that new ratio becomes 4 : 7?', he calculates 23*7 = 161 and 38*4 = 152, finds the difference 161-152 = 9, and divides by 7-4 = 3 to get the answer 3. The consistent method is cross-multiplying original and target ratio terms, subtracting the products, and dividing by the difference of the new ratio terms.

The central teaching point is a direct shortcut formula for converting one ratio to another by adding or subtracting the same number from each term. The standard algebraic approach sets up (a±y)/(b±y) = c/d and solves for y through cross-multiplication. The shortcut bypasses this by computing |ad - bc| divided by the absolute difference of the target ratio terms. The numerator comes from cross-multiplying the original ratio's first term with the target's second term and vice versa, then subtracting. The denominator is simply the difference between the two terms of the target ratio. This works for both addition and subtraction problems, as demonstrated with 9:16 to 2:3 (answer 5), 15:19 to 3:4, 11:21 to 3:5 (answer 4), and 23:38 to 4:7 (answer 3). The instructor emphasizes time efficiency, noting the shortcut saves significant calculation steps compared to full algebraic solving.

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