Get the most accurate GSEB Solutions for Class 7 Mathematics Chapter 12 Algebraic Expressions here. Updated for the 2026-27 academic session, these solutions are based on the latest GSEB textbooks for Class 7 Mathematics. Our expert-created answers for Class 7 Mathematics are available for free download in PDF format.
Detailed Chapter 12 Algebraic Expressions GSEB Solutions for Class 7 Mathematics
For Class 7 students, solving GSEB textbook questions is the most effective way to build a strong conceptual foundation. Our Class 7 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 12 Algebraic Expressions solutions will improve your exam performance.
Class 7 Mathematics Chapter 12 Algebraic Expressions GSEB Solutions PDF
Question 1. Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
(a) For the digit 6, the pattern is 6, 11, 16, 21... which follows the rule \( (5n + 1) \).
(b) For the digit 4, the pattern is 4, 7, 10, 13... which follows the rule \( (3n + 1) \).
(c) For the digit 8, the pattern is 7, 12, 17, 22... which follows the rule \( (5n + 2) \).
If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 8.
Answer:
(a) For the digit 6:
The expression for the number of line segments needed to form 'n' figures of the digit 6 is \( 5n + 1 \).
For 5 figures, the number of line segments will be \( 5 \times 5 + 1 = 25 + 1 = 26 \).
For 10 figures, the number of line segments will be \( 5 \times 10 + 1 = 50 + 1 = 51 \).
For 100 figures, the number of line segments will be \( 5 \times 100 + 1 = 500 + 1 = 501 \).
(b) For the digit 4:
The expression for the number of line segments needed to form 'n' figures of the digit 4 is \( 3n + 1 \).
For 5 figures, the number of line segments will be \( 3 \times 5 + 1 = 15 + 1 = 16 \).
For 10 figures, the number of line segments will be \( 3 \times 10 + 1 = 30 + 1 = 31 \).
For 100 figures, the number of line segments will be \( 3 \times 100 + 1 = 300 + 1 = 301 \).
(c) For the digit 8:
The expression for the number of line segments needed to form 'n' figures of the digit 8 is \( 5n + 2 \).
For 5 figures, the number of line segments will be \( 5 \times 5 + 2 = 25 + 2 = 27 \).
For 10 figures, the number of line segments will be \( 5 \times 10 + 2 = 50 + 2 = 52 \).
For 100 figures, the number of line segments will be \( 5 \times 100 + 2 = 500 + 2 = 502 \).
In simple words: To find out how many segments are needed for any number of digits, just put the number of digits into the right formula for each digit type (6, 4, or 8) and calculate.
Exam Tip: Remember to use the correct algebraic expression for each digit type to avoid calculation errors. Always double-check your arithmetic.
Question 2. Use the given algebraic expression to complete the table of number patterns.
| S. No. | Expression | 1\(^{\text{st}}\) | 2\(^{\text{nd}}\) | 3\(^{\text{rd}}\) | 4\(^{\text{th}}\) | 5\(^{\text{th}}\) | ... | 10\(^{\text{th}}\) | ... | 100\(^{\text{th}}\) | ... |
|---|---|---|---|---|---|---|---|---|---|---|---|
| (i) | \( 2n - 1 \) | 1 | 3 | 5 | 7 | 9 | - | 19 | - | 199 | - |
| (ii) | \( 3n + 2 \) | 2 | 5 | 8 | 11 | 17 | - | 32 | - | 302 | - |
| (iii) | \( 4n + 1 \) | 5 | 9 | 13 | 17 | 21 | - | 41 | - | 401 | - |
| (iv) | \( 7n + 20 \) | 27 | 34 | 41 | 48 | 55 | - | 90 | - | 720 | - |
| (v) | \( n^2 + 1 \) | 2 | 5 | 10 | 17 | 26 | - | 101 | - | 10,001 | - |
Answer:
(i) For the expression \( 2n - 1 \):
The 100th term is calculated as \( 2 \times 100 - 1 = 200 - 1 = 199 \).
(ii) For the expression \( 3n + 2 \):
The 5th term is calculated as \( 3 \times 5 + 2 = 15 + 2 = 17 \).
The 10th term is calculated as \( 3 \times 10 + 2 = 30 + 2 = 32 \).
The 100th term is calculated as \( 3 \times 100 + 2 = 300 + 2 = 302 \).
(iii) For the expression \( 4n + 1 \):
The 5th term is calculated as \( 4 \times 5 + 1 = 20 + 1 = 21 \).
The 10th term is calculated as \( 4 \times 10 + 1 = 40 + 1 = 41 \).
The 100th term is calculated as \( 4 \times 100 + 1 = 400 + 1 = 401 \).
(iv) For the expression \( 7n + 20 \):
The 5th term is calculated as \( 7 \times 5 + 20 = 35 + 20 = 55 \).
The 10th term is calculated as \( 7 \times 10 + 20 = 70 + 20 = 90 \).
For the 100th term, it is calculated as \( 7 \times 100 + 20 = 700 + 20 = 720 \).
(v) For the expression \( n^2 + 1 \):
The 5th term is calculated as \( 5^2 + 1 = 25 + 1 = 26 \).
The 10th term is calculated as \( 10^2 + 1 = 100 + 1 = 101 \).
In simple words: To fill in the table, replace 'n' in each expression with the term number (like 5 for the 5th term, 10 for the 10th term, or 100 for the 100th term) and then do the math.
Exam Tip: Pay close attention to the order of operations (PEMDAS/BODMAS) when evaluating expressions, especially with exponents like \( n^2 \).
Note:
1. We can also create some patterns using dots, such as:
\( 1^2 = 1 \)
\( 2^2 = 4 \)
\( 3^2 = 9 \)
\( 4^2 = 16 \)
2. We can also have some geometrical patterns as:
| Polygon | Number of sides | Number of diagonals from a vertex. |
|---|---|---|
 | 4 | \( n - 3 = 4 - 3 = 1 \) |
 | 5 | \( n - 3 = 5 - 3 = 2 \) |
 | 6 | \( n - 3 = 6 - 3 = 3 \) |
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GSEB Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions
Students can now access the GSEB Solutions for Chapter 12 Algebraic Expressions prepared by teachers on our website. These solutions cover all questions in exercise in your Class 7 Mathematics textbook. Each answer is updated based on the current academic session as per the latest GSEB syllabus.
Detailed Explanations for Chapter 12 Algebraic Expressions
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