Get the most accurate RBSE Solutions for Class 6 Mathematics Chapter 12 Algebra here. Updated for the 2026-27 academic session, these solutions are based on the latest RBSE textbooks for Class 6 Mathematics. Our expert-created answers for Class 6 Mathematics are available for free download in PDF format.
Detailed Chapter 12 Algebra RBSE Solutions for Class 6 Mathematics
For Class 6 students, solving RBSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 6 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 12 Algebra solutions will improve your exam performance.
Class 6 Mathematics Chapter 12 Algebra RBSE Solutions PDF
Algebra Ex 12.1
Question 1. Make the matchstick pattern of the letters given below. Draw the figures of the patterns in your note book. Create rules to find out number of matchsticks required for each pattern. (You can use literals like a, b, x, y e.t.c. to create rules).
(i) T pattern of T, T, TT
(ii) N pattern of N, N, NN,
(iii) W pattern of W, W, WW,
Answer:
(i) Pattern of T
For the letter 'T', two matchsticks are needed to form one pattern. If we need to make 'n' patterns of 'T', the total number of matchsticks required will be given by the rule: \( 2 \times n = 2n \). For example, for two 'T' patterns, \( 2 \times 2 = 4 \) matchsticks are needed.
(ii) Pattern of N
For the letter 'N', three matchsticks are needed to form one pattern. If we need to make 'n' patterns of 'N', the total number of matchsticks required will be given by the rule: \( n \times 3 = 3n \). This pattern uses diagonal matchsticks to connect the vertical ones.
(iii) Pattern of W
For the letter 'W', four matchsticks are needed to form one pattern. If we need to make 'n' patterns of 'W', the total number of matchsticks required will be given by the rule: \( n \times 4 = 4n \). The 'W' shape is formed by two 'V' shapes joined together.
In simple words: First, count how many matchsticks are needed for just one letter shape (like T, N, W). Then, to find the total matchsticks for 'n' of those shapes, multiply that number by 'n'.
🎯 Exam Tip: When forming patterns with matchsticks, always start by clearly counting the number of sticks for a single unit of the pattern. This base number is crucial for creating the correct algebraic rule.
Question 2. Tree plantation program was held in a school. 4 Trees were planted in each row. Write the number of trees planted in terms of the number of rows.
Answer: In this tree plantation program, 4 trees were planted in every single row. Let \( x \) represent the total number of rows where trees were planted. To find the total number of trees, we multiply the number of trees in one row by the total number of rows. So, the total number of trees planted will be \( 4 \times x = 4x \). This algebraic expression helps us calculate trees for any number of rows.
In simple words: If 4 trees are in each row and there are 'x' rows, you just multiply 4 by 'x' to find all the trees.
🎯 Exam Tip: Always clearly define what your variable (like 'x' or 'n') stands for in your answer to avoid confusion and make your algebraic expression meaningful.
Question 3. Ranu is 5 years younger than Leela.
(i) Let Leela's age be years. Write the age of Ranu in terms of
(ii) Let Ranu's age of be P years. Write the age of Leela in terms of P.
Answer:
(i) If Leela's age is \( x \) years, and Ranu is 5 years younger than Leela, then Ranu's age can be written as \( (x - 5) \) years. This shows how Ranu's age relates to Leela's age with a simple subtraction.
(ii) If Ranu's age is \( P \) years, and Leela is 5 years older than Ranu (since Ranu is younger), then Leela's age can be written as \( (P + 5) \) years. This expresses Leela's age in terms of Ranu's age.
In simple words: If someone is younger, you subtract from the older person's age. If someone is older, you add to the younger person's age.
🎯 Exam Tip: Pay close attention to keywords like "younger than" (means subtract) and "older than" (means add) when forming algebraic expressions for age problems.
Question 4. Cost of a Pen is Rs.5 Madan has some money with him. He pays all of that money to buy those pens. Write the number of pens purchased in terms of the money he had.
Answer: The cost of one pen is Rs. 5. Let's say Madan has \( x \) rupees in total. He spends all his money to buy pens. If \( n \) is the number of pens Madan bought, then the total money spent must be equal to the cost of one pen multiplied by the number of pens. So, we can write this as \( x = 5 \times n \).
\( \implies \) To find the number of pens \( n \), we divide the total money \( x \) by the cost of one pen, which is Rs. 5.
\( \implies n = \frac{x}{5} \). This formula helps determine how many pens can be bought with any amount of money.
In simple words: To find out how many pens Madan bought, divide the total money he had by the price of one pen.
🎯 Exam Tip: When dealing with cost and quantity problems, remember that the total cost is always the unit price multiplied by the quantity. To find the quantity, you'll usually divide the total cost by the unit price.
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RBSE Solutions Class 6 Mathematics Chapter 12 Algebra
Students can now access the RBSE Solutions for Chapter 12 Algebra prepared by teachers on our website. These solutions cover all questions in exercise in your Class 6 Mathematics textbook. Each answer is updated based on the current academic session as per the latest RBSE syllabus.
Detailed Explanations for Chapter 12 Algebra
Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 6 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 6 students who want to understand both theoretical and practical questions. By studying these RBSE Questions and Answers your basic concepts will improve a lot.
Benefits of using Mathematics Class 6 Solved Papers
Using our Mathematics solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 6 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 12 Algebra to get a complete preparation experience.
FAQs
The complete and updated RBSE Solutions Class 6 Maths Chapter 12 Algebra Exercise 12.1 is available for free on StudiesToday.com. These solutions for Class 6 Mathematics are as per latest RBSE curriculum.
Yes, our experts have revised the RBSE Solutions Class 6 Maths Chapter 12 Algebra Exercise 12.1 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 RBSE language because RBSE marking schemes are strictly based on textbook definitions. Our RBSE Solutions Class 6 Maths Chapter 12 Algebra Exercise 12.1 will help students to get full marks in the theory paper.
Yes, we provide bilingual support for Class 6 Mathematics. You can access RBSE Solutions Class 6 Maths Chapter 12 Algebra Exercise 12.1 in both English and Hindi medium.
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