Mixture & Alligations (Quick revision & Practice Problems)
Duration: 1 hr 22 min
This video lesson is available to enrolled students.
AI Summary
An AI-generated summary of this video lecture.
This educational video provides a comprehensive tutorial on solving mixture and alligation problems, commonly found in competitive exams like TCS and Infosys. The instructor, Yash Jain, begins with a basic two-component problem involving badam nuts, demonstrating both the traditional algebraic method and the faster alligation rule. He then progresses to more complex scenarios, including problems with three components (alcohol varieties) and mixtures involving free ingredients like water or seeds. Each problem is solved step-by-step on a digital whiteboard, with clear diagrams and calculations. The instructor emphasizes the importance of identifying the mean value and calculating differences to find ratios. The lesson concludes with a percentage-based problem, reinforcing the versatility of the alligation method in various contexts.
Chapters
0:00 – 2:00 00:00-02:00
The video begins with a mixture problem involving Kaccha Badam and Pakka Badam. The instructor presents the question: Kaccha Badam is sold at Rs. 450 per kg, and Pakka Badam is sold at Rs. 510 per kg. The goal is to find the ratio in which they should be mixed so that the mixture costs Rs. 475 per kg. The instructor sets the stage for solving this using different methods, highlighting the key values on the screen and introducing the concept of finding a mixing ratio for a desired average price.
2:00 – 5:00 02:00-05:00
The instructor demonstrates the "Traditional Approach" to solve the badam mixture problem. He defines the quantities of Kaccha and Pakka badam as 'p' and 'q' respectively. He sets up the weighted average equation: $475 = (450p + 510q) / (p+q)$. He then cross-multiplies to simplify the equation, leading to $475p + 475q = 450p + 510q$. This algebraic method is shown step-by-step on the whiteboard, providing a foundational understanding before introducing the shortcut.
5:00 – 10:00 05:00-10:00
The instructor introduces the "Alligation Rule" as a faster alternative. He draws a diagram with the lower price (450) and higher price (510) on the top line, and the average price (475) in the center. He calculates the differences: $475 - 450 = 25$ and $510 - 475 = 35$. The required ratio is derived from these differences, showing that the ratio of Kaccha to Pakka badam is 35:25, which simplifies to 7:5. This visual method is highlighted as efficient for exams.
10:00 – 15:00 10:00-15:00
A new problem is introduced involving "Vimal Paan Masala". The text on screen states there are two qualities: Bina Kesar Wala Vimal costing Rs. 28.50 per kg and Kesar Wala Vimal costing Rs. 34.50 per kg. The mixture is sold at Rs. 30.50 per kg. The instructor asks to find the ratio of the two qualities of Vimal in the mixture, setting up the next example for the class to practice the alligation technique on decimal values.
15:00 – 20:00 15:00-20:00
The instructor solves the Vimal Paan problem using the alligation method. He places 28.50 and 34.50 on the sides and 30.50 in the middle. He calculates the differences: $30.50 - 28.50 = 2$ and $34.50 - 30.50 = 4$. The ratio of Bina Kesar to Kesar Wala is found to be 4:2, which simplifies to 2:1. He emphasizes the importance of identifying the correct order for the ratio to match the question's requirement.
20:00 – 25:00 20:00-25:00
The next problem involves two types of oil. The rates are Rs. 4 per kg and Rs. 5 per kg. They are mixed to produce a mixture having a rate of Rs. 4.6 per kg. The problem states that the amount of the first type of oil in the mixture is 40 kg. The task is to find the amount of the second type of oil. The instructor highlights the given values on the screen, preparing to apply the ratio concept to find an unknown quantity.
25:00 – 30:00 25:00-30:00
The instructor solves the oil problem using alligation. He places 4 and 5 on the sides and 4.6 in the center. The differences are $4.6 - 4 = 0.6$ and $5 - 4.6 = 0.4$. The ratio of the first oil to the second oil is 0.4:0.6, which simplifies to 2:3. Since the first oil (2 parts) is 40 kg, he calculates that 1 part is 20 kg, making the second oil (3 parts) equal to 60 kg. This demonstrates how to use ratios to find specific quantities.
30:00 – 35:00 30:00-35:00
A problem regarding average weights is presented. The average weight of boys in Gajjumal College is 40 kg, and the average weight of girls is 30 kg. The average weight of 80 students is 33 kg. The question asks to find the number of boys in the college. The instructor underlines the key numbers: 40, 30, 33, and 80, preparing to apply the alligation rule to find the ratio of boys to girls, treating weight as a "price" in this context.
35:00 – 40:00 35:00-40:00
The instructor solves the weight problem. Using alligation with 40 and 30, and the average 33, he calculates the differences: $40 - 33 = 7$ and $33 - 30 = 3$. The ratio of boys to girls is 3:7. Since the total number of students is 80, he divides 80 into 10 parts (3+7). Each part is 8 students. Therefore, the number of boys is 3 parts * 8 = 24. This shows the application of alligation to population problems.
40:00 – 45:00 40:00-45:00
The video moves to a problem with three varieties of alcohol. The costs are Rs. 10 per 100 ml, Rs. 12 per 100 ml, and Rs. 16 per 100 ml. The mixture costs Rs. 13 per 100 ml. The instructor explains that for three components, one must pair them up. He suggests pairing (10, 12) and (12, 16) to find the ratios relative to the mean of 13, introducing a more complex variation of the standard alligation rule.
45:00 – 50:00 45:00-50:00
The instructor breaks down the three-varieties problem. He pairs (10, 12) with mean 13, finding a ratio. Then he pairs (12, 16) with mean 13, finding another ratio. He combines these ratios to determine the final ratio of the three varieties. He writes down the intermediate calculations on the board, showing how the common term (12) is adjusted to match the final ratio, ensuring the middle component is consistent across both pairs.
50:00 – 55:00 50:00-55:00
Another three-varieties problem is introduced. Three types of alcohol cost Rs. 120, Rs. 150, and Rs. 160 per litre. All three are mixed in a particular ratio to produce an alcohol costing Rs. 140 per litre. The goal is to find the ratio in which the three were mixed. The instructor highlights the prices and the target mixture price on the screen, setting up another example for three-component mixtures.
55:00 – 60:00 55:00-60:00
The instructor solves the second three-varieties problem. He pairs (120, 150) and (150, 160) relative to the mean 140. For (120, 150), the ratio is 1:2. For (150, 160), the ratio is 2:1. He combines these to find the final ratio of the three alcohols as 2:2:1. He emphasizes the logic of balancing the common term (150) in both pairs to get the final combined ratio.
60:00 – 65:00 60:00-65:00
A problem about pure milk and water is presented. Pure milk costs Rs. 60 per litre. Pushpa Raj makes a mixture and sells it for Rs. 40 per litre. The quantity of pure milk is 21 litres. The task is to find the quantity of water mixed. The instructor notes that water has a cost of Rs. 0. He sets up the alligation diagram with 60 and 0, treating water as a free ingredient to lower the average cost.
65:00 – 70:00 65:00-70:00
The instructor solves the milk problem. Using alligation with 60 and 0, and the average 40, the differences are 40 and 20. The ratio of milk to water is 2:1. Since the quantity of pure milk is 21 litres (2 parts), he calculates that 1 part is 10.5 litres. Therefore, the quantity of water mixed is 10.5 litres. This example reinforces the concept of mixing a costly item with a free item.
70:00 – 75:00 70:00-75:00
A problem about black pepper and papaya seeds is shown. Pure black pepper costs Rs. 210 per kg. It is adulterated by mixing dry papaya seeds. The mixture is sold for Rs. 165 per kg. If there is 18 kg of papaya seeds in the mixture, the task is to find the amount of pure black pepper. The instructor highlights the cost of pepper and the mixture price, preparing to solve for the unknown quantity of the expensive ingredient.
75:00 – 80:00 75:00-80:00
The instructor solves the pepper problem. Using alligation with 210 and 0 (seeds), and the average 165, the differences are 165 and 45. The ratio of pepper to seeds is 11:3. Since the quantity of seeds is 18 kg (3 parts), he calculates that 1 part is 6 kg. Therefore, the amount of pure black pepper is 11 parts * 6 kg = 66 kg. This demonstrates finding the quantity of the main ingredient given the adulterant.
80:00 – 81:52 80:00-81:52
The final problem involves percentages. 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. Given x > y, the instructor asks how many integer values x can take. He explains that since the average is 60, x must be greater than 60 and y must be less than 60. He sets up the equation to find the relationship between x and y, concluding the lesson with a more abstract application of the rule.
The video provides a comprehensive tutorial on solving mixture and alligation problems commonly found in competitive exams like TCS and Infosys. The instructor, Yash Jain, begins with a basic two-component problem involving badam nuts, demonstrating both the traditional algebraic method and the faster alligation rule. He then progresses to more complex scenarios, including problems with three components (alcohol varieties) and mixtures involving free ingredients like water or seeds. Each problem is solved step-by-step on a digital whiteboard, with clear diagrams and calculations. The instructor emphasizes the importance of identifying the mean value and calculating differences to find ratios. The lesson concludes with a percentage-based problem, reinforcing the versatility of the alligation method in various contexts.