Samacheer Kalvi Class 6 Maths Solutions Term 1 Chapter 6 Information Processing Exercise 6.3

Get the most accurate TN Board Solutions for Class 6 Maths Chapter 06 Information Processing here. Updated for the 2026-27 academic session, these solutions are based on the latest TN Board textbooks for Class 6 Maths. Our expert-created answers for Class 6 Maths are available for free download in PDF format.

Detailed Chapter 06 Information Processing TN Board Solutions for Class 6 Maths

For Class 6 students, solving TN Board textbook questions is the most effective way to build a strong conceptual foundation. Our Class 6 Maths solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 06 Information Processing solutions will improve your exam performance.

Class 6 Maths Chapter 06 Information Processing TN Board Solutions PDF

 

Question 1. How many Triangles are there in each of the following figures?
(i) 12 triangles
(ii) 16 triangles
(iii) 32 triangles
(iv) 35 triangles
Answer: We need to count the total number of triangles in each given figure. For the first figure (i), after careful observation, there are 12 distinct triangles. In the second figure (ii), there are 16 triangles. The third figure (iii) contains 32 triangles. Finally, the fourth figure (iv) has 35 triangles. Finding all triangles requires checking different sizes and combinations of lines.
In simple words: We need to count all the triangles in each picture. Figure (i) has 12, figure (ii) has 16, figure (iii) has 32, and figure (iv) has 35 triangles in total.

🎯 Exam Tip: When counting triangles in complex figures, break them down by size (small, medium, large) and orientation to ensure no triangle is missed or double-counted.

 

Question 2. Find the number of dots in the tenth figure of the following patterns.
Answer:
(i) For the first pattern, the number of dots forms a sequence of triangular numbers. The figures have 1, 3, 6, 10, 15 dots respectively. The formula for the \( n^{th} \) triangular number is \( T_n = \frac{n(n+1)}{2} \). Therefore, for the tenth figure \( (n=10) \), the number of dots will be \( T_{10} = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = \frac{110}{2} = 55 \). Triangular numbers are naturally formed by arranging dots in the shape of an equilateral triangle.
In simple words: The first pattern adds dots like triangles, so for the 10th shape, there are 55 dots.
(ii) For the second pattern, the number of dots forms a sequence of square numbers. The figures have 1, 4, 9, 16 dots respectively, which are \( 1^2, 2^2, 3^2, 4^2 \). The formula for the \( n^{th} \) square number is \( n^2 \). Therefore, for the tenth figure \( (n=10) \), the number of dots will be \( 10^2 = 10 \times 10 = 100 \). Square numbers are simply a number multiplied by itself.
In simple words: The second pattern adds dots like squares, so for the 10th shape, there are 100 dots.

🎯 Exam Tip: Always identify the type of pattern first (e.g., arithmetic, geometric, triangular, square) by looking at the differences between terms. This helps in finding the correct formula for any term.

 

Question 3.
(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Answer:
(i) The given patterns have 2, 5, 9, 14 dots. The differences between consecutive terms are 3, 4, 5. So, the next difference will be 6. The 5th pattern will have \( 14 + 6 = 20 \) dots. The pattern starts with 2 dots, then adds 3, then 4, then 5. So, the 5th pattern adds 6 dots to the 4th pattern. You would draw a similar arrangement of dots but with 20 dots in total.
In simple words: The pattern grows by adding 3, then 4, then 5 dots. So, the next pattern will add 6 dots, making it 20 dots in total.
(ii) The table showing the number of dots for each pattern is:

Pattern1234
Number of dots25914

Pattern1234
Number of dots22+32+3+42+3+4+5

In simple words: The tables show how many dots are in each pattern from 1 to 4.
(iii) The pattern is a quadratic sequence. This means the difference between consecutive terms is not constant, but the difference of these differences (the second difference) is constant. The first differences are 3, 4, 5, so the second difference is 1. The formula for the \( n^{th} \) term is \( \frac{n^2 + 3n}{2} \). This type of pattern grows faster as the number of terms increases.
In simple words: The pattern adds more dots each time. First it adds 3, then 4, then 5, and so on. It grows quickly.
(iv) To find the number of dots in the 25th pattern, we use the formula \( \frac{n^2 + 3n}{2} \).
For \( n = 25 \):
Number of dots \( = \frac{25^2 + 3 \times 25}{2} \)
\( = \frac{625 + 75}{2} \)
\( = \frac{700}{2} \)
\( = 350 \)
The 25th pattern will have 350 dots.
In simple words: Using the pattern's rule, we calculate that the 25th shape will have 350 dots.

🎯 Exam Tip: When dealing with patterns, first find the differences between terms. If the first differences are not constant, find the second differences. A constant second difference indicates a quadratic pattern, for which a specific formula can be derived.

 

Question 4. (i) (ii)
Answer: We need to count the total number of squares in each of the given figures.
(i) After carefully counting all the square shapes of different sizes and orientations in the first figure, there are 20 squares in total. This figure has both aligned and tilted squares.
In simple words: The first picture has 20 squares when you count them all.
(ii) For the second figure, by systematically counting all the squares formed by the lines, we find there are 10 squares. This figure includes an outer square and a rotated inner square.
In simple words: The second picture has 10 squares when you count them all.

🎯 Exam Tip: When counting squares or other shapes in complex diagrams, adopt a methodical approach. Count shapes of one size first, then move to the next size, and consider all possible orientations to ensure accuracy.

 

Question 5. How many circles are there in the following figure?
Answer: By carefully examining the provided figure, we can count the number of distinct circles that form the pattern. There is one large central circle, and six smaller circles surrounding it, each touching the central circle and its neighbors. This makes a total of 7 circles in the figure. Each circle is a complete, distinct circular shape.
In simple words: In the picture, there is 1 circle in the middle and 6 circles around it. So, there are 7 circles in total.

🎯 Exam Tip: When counting circles or other overlapping shapes, ensure you count each complete shape only once, even if parts of them overlap with others. Focus on distinct boundaries.

 

Question 6. Find the minimum number of straight lines used in forming the following figures.
Answer: We need to determine the fewest continuous straight lines required to draw each given figure.
(i) For the first figure, which is a 3x3 grid of squares, there are 4 distinct horizontal lines and 4 distinct vertical lines required. This makes a total of \( 4 + 4 = 8 \) straight lines. Each line segment must be counted as part of a longer continuous line if possible.
In simple words: For the first picture, count all the straight lines going across and down. There are 8 lines in total.
(ii) For the second figure, which is a large triangle divided into smaller ones, we count the unique straight lines. There are 4 horizontal lines, 4 lines slanting to the left, and 4 lines slanting to the right. This adds up to a total of \( 4 + 4 + 4 = 12 \) straight lines. This figure is made by drawing a large triangle and then dividing it with more straight lines.
In simple words: For the second picture, count all the straight lines, including the slanted ones. There are 12 lines in total.

🎯 Exam Tip: To find the minimum number of lines, identify each unique, continuous straight line segment in the figure. Do not count segments that are part of a longer, unbroken line as separate lines.

TN Board Solutions Class 6 Maths Chapter 06 Information Processing

Students can now access the TN Board Solutions for Chapter 06 Information Processing prepared by teachers on our website. These solutions cover all questions in exercise in your Class 6 Maths textbook. Each answer is updated based on the current academic session as per the latest TN Board syllabus.

Detailed Explanations for Chapter 06 Information Processing

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 6 Maths 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 TN Board Questions and Answers your basic concepts will improve a lot.

Benefits of using Maths Class 6 Solved Papers

Using our Maths 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 06 Information Processing to get a complete preparation experience.

FAQs

Where can I find the latest Samacheer Kalvi Class 6 Maths Solutions Term 1 Chapter 6 Information Processing Exercise 6.3 for the 2026-27 session?

The complete and updated Samacheer Kalvi Class 6 Maths Solutions Term 1 Chapter 6 Information Processing Exercise 6.3 is available for free on StudiesToday.com. These solutions for Class 6 Maths are as per latest TN Board curriculum.

Are the Maths TN Board solutions for Class 6 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the Samacheer Kalvi Class 6 Maths Solutions Term 1 Chapter 6 Information Processing Exercise 6.3 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Maths concepts are applied in case-study and assertion-reasoning questions.

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