NCERT Solutions for Class 7 Maths Chapter 4: Expressions Using Letter-Numbers – Ganita Prakash

Complete NCERT Solutions for Class 7 Maths Chapter 4 “Expressions Using Letter-Numbers” from the Ganita Prakash textbook, covering algebraic expressions, simplification, and pattern-based problems, with step-by-step working for every question.

Last Updated: September 23, 2026

4.1 The Notion of Letter-Numbers

In-Text Examples

Shabnam is 3 years older than Aftab. If Aftab’s age is 18 years, Shabnam’s age = 18 + 3 = 21 years. Using the expression “Aftab’s age = Shabnam’s age − 3”, if Shabnam’s age is 20, Aftab’s age = 20 − 3 = 17 years.

Coconuts and jaggery: 8 coconuts at Rs. 35 each and 9 kg jaggery at Rs. 60/kg: 8×35 = Rs. 280; 9×60 = Rs. 540; Total = Rs. 280 + Rs. 540 = Rs. 820. Using the general expression c×35 + j×60 for c=7 coconuts, j=4 kg jaggery: 7×35 + 4×60 = 245 + 240 = Rs. 485.

Perimeter of a square with side 7 cm = 4×7 = 28 cm.

Figure It Out (Page 84-85)

Q1. Perimeter formulas for regular polygons with side a: (a) Triangle = 3a (b) Pentagon = 5a (c) Hexagon = 6a.

Q2. Munirathna has a 20 m pipe and buys k more metres. Total length = (20 + k) m.

Q3. This question depends on a table image in the textbook and cannot be reproduced without the figure.

Q4. A flour mill charges a fixed cost of Rs. 10 plus Rs. 8 per kg (y kg): total charge = 10 + 8y.

Q5. Let n be the number. (a) 5 more than n = n + 5 (b) 4 less than n = n − 4 (c) 2 less than 13 times n = 13n − 2 (d) 13 less than twice n = 2n − 13.

Q6. Open-ended descriptive expression-writing exercise (answers vary based on the story chosen).

Q7. This question is based on a calendar grid image and cannot be reproduced without the figure.

4.2 Revisiting Arithmetic Expressions

Evaluate the following (using BODMAS/order of operations):

1. 23 − 10×2 = 23 − 20 = 3
2. 83 + 28 − 13 + 32 = 111 − 13 + 32 = 130
3. 34 − 14 + 20 = 20 + 20 = 40
4. 42 + 15 − (8 − 7) = 57 − 1 = 56
5. 68 − (18 + 13) = 68 − 31 = 37
6. 7×4 + 9×6 = 28 + 54 = 82
7. 20 + 8×(16 − 6) = 20 + 80 = 100

Mind the Mistake, Mend the Mistake (Page 87)

Each statement below is checked by substituting the given value; the corrected value is shown where the original was wrong:

1. a = −4: 10 − a = 10 − (−4) = 14 (not 6)
2. d = 6: 3d = 3×6 = 18 (not 36)
3. s = 7: 3s − 2 = 21 − 2 = 19 (not 15)
4. r = 8: 2r + 1 = 16 + 1 = 17 (not 29)
5. j = 5: 2j = 2×5 = 10 — correct as given
6. m = −6: 3(m+1) = 3×(−5) = −15 (not 19)
7. f = 3, g = 1: 2f − 2g = 6 − 2 = 4 (not 2)
8. t = 4, b = 3: 2t + b = 8 + 3 = 11 (not 24)
9. h = 5, n = 6: h − (3 − n) = 5 − (−3) = 8 (not 4)

4.4 Simplification of Algebraic Expressions

Pencils and erasers example: pencils sold over 3 days = 5, 3, 10 (total 18); cost per pencil = c. Total pencil sales = 18c. Checking c = Rs. 50: 5c + 3c + 10c = 250 + 150 + 500 = Rs. 900; and 18c = 18×50 = 900. Both methods match, confirming 5c+3c+10c = 18c.

Combining like terms: (40x + 75y) + (−6x − 10y) = (40x + 75y) − (6x + 10y) = 34x + 65y.

Charu’s game scores (p=4, q=1): Round 2 = 8p − 4q = 32 − 4 = 28; Round 3 = 6p − 2q = 24 − 2 = 22. Charu’s total = 7p − 3q + 8p − 4q + 6p − 2q = 21p − 9q.

Krishita’s total needs to equal 23p − 7q. Example: 8p−4q + 9p−2q + 6p−q = 23p − 7q ✓. The difference between Krishita’s and Charu’s totals: (23p−7q) − (21p−9q) = 2p + 2q = 2(p+q).

4.5 Pick Patterns and Reveal Relationships

Calendar diagonal sums (centre value a): a−8 + a = … general relation verified as 2a + 8 for one diagonal pair.

Plus-shaped 5-cell sums (centre value a): sum of the 5 cells (a−7, a−1, a, a+1, a+7) = 5a. Verified with centre=15: 8+14+15+16+22 = 75 = 15×5 ✓. Centre=20: 13+19+20+21+27 = 100 = 20×5 ✓. Centre=12: 5+11+12+13+19 = 60 = 12×5 ✓.

Matchstick pattern: at step y, horizontal sticks = y, diagonal sticks = y+1, so total = y + (y+1) = 2y + 1.

Figure It Out (Pages 102-105)

Q1. Jowar roti costs Rs. 30/plate, pulao costs Rs. 20/plate. For x plates of roti and y plates of pulao, total bill = 30x + 20y.

Q2. With p customers choosing only Champak flags, q choosing only Marigold, and r choosing both (getting 1 flag each regardless), total flags handed out = p + q + r.

Q3. A snail climbs u cm during the day and slips d cm at night. (a) Net distance covered in 10 days = 10(u − d) cm. (b) If d > u, the snail’s net progress each day-night cycle is negative, so the snail will never reach the top.

Q4. Radha cycles 5 km/day in week 1, increasing by z km/day each subsequent week. Week 1 total = 5×7 = 35 km. Week 2 total = (5+z)×7 = 35+7z km. Week 3 total = (5+2z)×7 = 35+14z km. Grand total = 35 + (35+7z) + (35+14z) = 105 + 21z km.

Q5. This number-machine path question is based on a diagram and cannot be reproduced without the figure.

Q6. A train makes 3 stops; t minutes between each stop, 2 minutes at each stop. (a) For t=4: travel time = 4×4 = 16 min, stoppage time = 3×2 = 6 min, total = 22 minutes. (b) General expression: 4t + 6.

Q7. Simplify:

1. 3a+9b−6+8a−4b−7a+16 = 4a + 5b + 10
2. 3(3a−3b)−8a−4b−16 = 9a−9b−8a−4b−16 = a − 13b − 16
3. 2(2x−3)+8x+12 = 4x−6+8x+12 = 12x + 6
4. 8x−(2x−3)+12 = 8x−2x+3+12 = 6x + 15
5. 8h−(5+7h)+9 = h+4 = h + 4
6. 23+4(6m−3n)−8n−3m−18 = 23+24m−12n−8n−3m−18 = 21m − 20n + 5

Q8. Add:

1. (4d−7c+9) + (8c−11+9d) = 13d + c − 2
2. (−6f+19−8s) + (−23+13f+12s) = 7f + 4s − 4
3. (8d−14c+9) + (16c−11−9d) = 2c − d − 2
4. (6f−20+8s) + (23−13f−12s) = −7f − 4s + 3
5. (13m−12n) + (12n−13m) = 0
6. (−26m+24n) + (26m−24n) = 0

Q9. Subtract (second expression from first):

1. (6a+9b−18) − (9a−6b+14) = −3a + 15b − 32
2. (7y−10+3x) − (−15x+13−9y) = 16y + 18x − 23
3. (11−10g+3h) − (17g+9−7h) = 10h − 27g + 2
4. (6a−9b−18) − (9a−6b+14) = −3a − 3b − 32
5. (−3y+8−3x) − (10x+2+10y) = −13y − 13x + 6
6. (7h−8g+20) − (8g+4h−10) = 3h − 16g + 30

Q10. Open-ended word problems built around the expressions 8x+3y (notebooks and pens) and 15x−2x (rotten apples removed from a basket) — answers vary based on the story context chosen.

Q11. A rope is folded repeatedly: with 0 folds there are 2 pieces (after 1 cut), and each additional fold adds 1 more piece. For r folds, number of pieces = r + 2. For 10 folds: 10+2 = 12 pieces.

Q12. A matchstick pattern of squares: Step 1 = 4 sticks, Step 2 = 7 sticks, Step 3 = 10 sticks (each new square adds 3 sticks after the first). General formula for w squares = 4 + 3(w−1) = 3w + 1. For w=10: 3×10+1 = 31 matchsticks.

Q13. A traffic signal cycles: red on positions of the form 4n−3 (1,5,9,…), green on 4n−1 (3,7,11,…), yellow on all even-numbered positions. Position 90 (even) = yellow. Position 190 (even) = yellow. Position 343: 343 = 4(86)−1, matching the green pattern, so position 343 = green.

Q14. A growing square pattern: Step 1 = 5 sticks, Step 2 = 9, Step 3 = 13, Step 4 = 17 (each step adds 4). General formula = 4n + 1. Step 10: 4×10+1 = 41. Step 50: 4×50+1 = 201.

Q15. A 4-column number grid filled row-wise starting at 1: (a) General formula for row r, column c: 4(r−1) + c. (b) Row/column of specific numbers: 124 → row 31, column 4; 147 → row 37, column 3; 201 → row 51, column 1. (c) Formula confirmed: 4(r−1)+c. (d) Patterns observed: every fourth number is a multiple of 4; columns 2 and 4 contain even numbers, columns 1 and 3 contain odd numbers; the sum of each row increases by 16 as you move down (Row 1 sum = 1+2+3+4 = 10, Row 2 sum = 5+6+7+8 = 26, Row 3 sum = 9+10+11+12 = 42).

Note: A small number of Figure It Out questions in this chapter (Page 84 Q3 and Q7, Page 93 Q1, the Page 94-95 Mind the Mistake set, and Page 102 Q5) depend entirely on textbook figures/tables with no extractable text, so they have been noted rather than guessed at above.

Extra Questions: Class 7 Maths Chapter 4
Revision Notes: Class 7 Maths Chapter 4

Frequently Asked Questions

What does “letter-numbers” mean in the context of this chapter?
It refers to using letters (like x or n) to represent unknown or variable numbers in an algebraic expression — this is the introductory step into algebra, where letters stand in for numbers that can change or are yet to be found.

What is the difference between an algebraic expression and an algebraic equation?
An algebraic expression is a combination of letter-numbers, constants and operations with no equals sign (e.g. 3x+5); an algebraic equation states that two expressions are equal using an equals sign (e.g. 3x+5=20), which can then be solved to find the unknown value.

Written by Satish

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