Get the most accurate ISC Solutions for Class 11 Mathematics Chapter 30 Index Numbers here. Updated for the 2026-27 academic session, these solutions are based on the latest ISC textbooks for Class 11 Mathematics. Our expert-created answers for Class 11 Mathematics are available for free download in PDF format.
Detailed Chapter 30 Index Numbers ISC Solutions for Class 11 Mathematics
For Class 11 students, solving ISC textbook questions is the most effective way to build a strong conceptual foundation. Our Class 11 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 30 Index Numbers solutions will improve your exam performance.
Class 11 Mathematics Chapter 30 Index Numbers ISC Solutions PDF
Question 1. Construct the consumer price index number for 1990, taking 1989 as the base year and using simple average of price relative method for the following data :
| Commodities | Price in 1989 | Price in 1990 |
|---|---|---|
| Butter | 20 | 21 |
| Cheese | 16 | 12 |
| Milk | 3 | 3 |
| Egg | 2.80 | 2.80 |
Answer: First, we construct a table to organize the given prices for each commodity. Then, we calculate the price relative for each item by dividing its 1990 price (\( P_1 \)) by its 1989 price (\( P_0 \)) and multiplying the result by 100. The sum of all these price relatives is 380. We use this to find the Consumer Price Index for 1990 by dividing the total sum of price relatives by the number of commodities, which is 4. This calculation gives us an index number of 95. An index number provides a clear picture of how prices have changed over a period.
| Commodities | Price in 1989 \( P_0 \) | Price in 1990 \( P_1 \) | Price relative \( \frac{P_1}{P_0} \times 100 \) |
|---|---|---|---|
| Butter | 20 | 21 | \( \frac{21}{20} \times 100 = 105 \) |
| Cheese | 16 | 12 | \( \frac{12}{16} \times 100 = 75 \) |
| Milk | 3 | 3 | \( \frac{3}{3} \times 100 = 100 \) |
| Egg | 2.80 | 2.80 | \( \frac{2.80}{2.80} \times 100 = 100 \) |
| \( \Sigma \frac{P_1}{P_0} \times 100 = 380 \) | |||
Using the simple average of price relatives method:
Required index number \( P_{01} = \frac{1}{N} \Sigma \left(\frac{P_1}{P_0} \times 100\right) \)
\( \implies P_{01} = \frac{1}{4} \times 380 \)
\( \implies P_{01} = 95 \)
In simple words: First, we make a table with the prices. Then, for each item, we find how much its price changed from 1989 to 1990. We add up all these changes and divide by the number of items to get the final index number, which is 95.
🎯 Exam Tip: Remember to clearly identify the base year and current year prices before calculating price relatives. Ensure you sum the correct column for the final average.
Question 2. A small industrial concern used three raw materials A, B and C in the manufacturing using 1995 as the base year, calculate a simple aggregate price index for 2005.
| Commodity | Price (in Rs) in the year 1995 | Price (in Rs) in the year 2005 |
|---|---|---|
| A | 4 | 5 |
| B | 60 | 57 |
| C | 36 | 42 |
Answer: To calculate the simple aggregate price index for 2005 with 1995 as the base year, we first construct a table listing the commodities and their prices for both years. We then sum up all the prices for 1995 to get \( \Sigma P_0 = 100 \) and all prices for 2005 to get \( \Sigma P_1 = 104 \). Using the simple aggregate method, the required price index is calculated by dividing the total price of the current year by the total price of the base year, and multiplying by 100. This calculation results in a price index of 104. This shows a 4% increase in the aggregate prices from 1995 to 2005.
| Commodities | Price in year 1995 base price \( P_0 \) | Price in year 2005 current price \( P_1 \) |
|---|---|---|
| A | 4 | 5 |
| B | 60 | 57 |
| C | 36 | 42 |
| \( \Sigma P_0 = 100 \) | \( \Sigma P_1 = 104 \) |
By simple aggregate method, we have
Required price index for 2005 \( P_{01} = \frac{\Sigma P_1}{\Sigma P_0} \times 100 \)
\( \implies P_{01} = \frac{104}{100} \times 100 \)
\( \implies P_{01} = 104 \)
In simple words: We list the prices of materials for 1995 and 2005. We add up prices for each year. Then, we divide the total price of 2005 by the total price of 1995 and multiply by 100. The answer is 104.
🎯 Exam Tip: For simple aggregate methods, accurately summing \( \Sigma P_1 \) and \( \Sigma P_0 \) is key. Always express the final index as a percentage for easy interpretation.
Question 3. Find the consumer price index number for the year 2010 as the base year by using method of weighted aggregates.
| A | B | C | D | E | |
|---|---|---|---|---|---|
| Year 2000 price (in Rs) per unit | 16 | 40 | 0.50 | 5.12 | 2.00 |
| Year 2010 price (in Rs) per unit | 20 | 60 | 0.50 | 6.25 | 1.50 |
| weights | 40 | 25 | 5.00 | 20.00 | 10.00 |
Answer: To find the consumer price index using the weighted aggregate method, we first build a comprehensive table. This table includes prices for the base year (2000, \( P_0 \)), current year (2010, \( P_1 \)), and the assigned weights (\( w \)) for each commodity. Next, we calculate the product of the base year price and its weight (\( P_0 w \)), and the product of the current year price and its weight (\( P_1 w \)) for each commodity. After summing these columns, we get \( \Sigma P_0 w = 1764.9 \) and \( \Sigma P_1 w = 2442.5 \). Finally, we apply the formula: divide \( \Sigma P_1 w \) by \( \Sigma P_0 w \) and multiply by 100. The calculated index number is approximately 138.39, which indicates a significant price increase considering the different importance of items.
| Commodity | Price per unit in 2000 \( P_0 \) | Price per unit in 2010 \( P_1 \) | Weights \( w \) | \( P_0 w \) | \( P_1 w \) |
|---|---|---|---|---|---|
| A | 16 | 20 | 40 | 640 | 800 |
| B | 40 | 60 | 25 | 1000 | 1500 |
| C | 0.50 | 0.50 | 5 | 2.5 | 2.5 |
| D | 5.12 | 6.25 | 20 | 102.4 | 125 |
| E | 2 | 1.50 | 10 | 20 | 15 |
| \( \Sigma P_0 w = 1764.9 \) | \( \Sigma P_1 w = 2442.5 \) | ||||
Thus by weighted aggregate method,
Required index number \( = \frac{\Sigma P_1 w}{\Sigma P_0 w} \times 100 \)
\( \implies = \frac{2442.5}{1764.9} \times 100 \)
\( \implies = 138.39 \)
In simple words: We make a table with prices for two years and how important each item is (weight). We multiply each price by its weight for both years and then add these up. We divide the total for the current year by the total for the base year and multiply by 100. This gives us the index number, which is 138.39.
🎯 Exam Tip: Always double-check your calculations for \( P_0 w \) and \( P_1 w \) to avoid errors, as these products are crucial for the weighted aggregate formula.
Question 4. The price of six different commodities for years 2009 and year 2011 are as follows :
| Commodities | Price in 2009 \( p_0 \) | Price in 2011 \( p_1 \) |
|---|---|---|
| A | 35 | 50 |
| B | 80 | Y |
| C | 25 | 45 |
| D | 30 | 70 |
| E | 80 | 120 |
| F | X | 105 |
| \( \Sigma p_0 = 250 + x \) | \( \Sigma p_1 = 390 + y \) |
Answer: First, we construct a table with the commodity prices for 2009 (base year, \( p_0 \)) and 2011 (current year, \( p_1 \)), including the unknown values x and y. The problem states that the total price in 2009 is Rs 360. We sum the known 2009 prices (\( 35 + 80 + 25 + 30 + 80 = 250 \)) and add x to set up the equation: \( 250 + x = 360 \). Solving this gives us \( x = 110 \). Next, we use the given index number (125) and the simple aggregate method formula, substituting the total prices for 2009 (360) and 2011 (\( 390 + y \)). We then solve the equation \( 125 = \frac{390+y}{360} \times 100 \) to find y. This calculation leads to \( y = 60 \). Finding missing values like this is a key part of price index analysis.
In simple words: We have prices for items in 2009 and 2011, with some missing. We know the total price for 2009, so we find 'x' using that. Then, we use the given index number (125) and the formula to find the missing 'y' value. We get x = 110 and y = 60.
🎯 Exam Tip: When given a total sum, use it to find the missing component first. Then, apply the index number formula carefully to solve for any remaining unknowns.
Free study material for Mathematics
ISC Solutions Class 11 Mathematics Chapter 30 Index Numbers
Students can now access the ISC Solutions for Chapter 30 Index Numbers prepared by teachers on our website. These solutions cover all questions in exercise in your Class 11 Mathematics textbook. Each answer is updated based on the current academic session as per the latest ISC syllabus.
Detailed Explanations for Chapter 30 Index Numbers
Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 11 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 11 students who want to understand both theoretical and practical questions. By studying these ISC Questions and Answers your basic concepts will improve a lot.
Benefits of using Mathematics Class 11 Solved Papers
Using our Mathematics solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 11 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 30 Index Numbers to get a complete preparation experience.
FAQs
The complete and updated OP Malhotra Class 11 Maths Solutions Chapter 30 Index Numbers Chapter Test is available for free on StudiesToday.com. These solutions for Class 11 Mathematics are as per latest ISC curriculum.
Yes, our experts have revised the OP Malhotra Class 11 Maths Solutions Chapter 30 Index Numbers Chapter Test as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.
Toppers recommend using ISC language because ISC marking schemes are strictly based on textbook definitions. Our OP Malhotra Class 11 Maths Solutions Chapter 30 Index Numbers Chapter Test will help students to get full marks in the theory paper.
Yes, we provide bilingual support for Class 11 Mathematics. You can access OP Malhotra Class 11 Maths Solutions Chapter 30 Index Numbers Chapter Test in both English and Hindi medium.
Yes, you can download the entire OP Malhotra Class 11 Maths Solutions Chapter 30 Index Numbers Chapter Test in printable PDF format for offline study on any device.