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. These Class 7 Mathematics Chapter 4 solutions are also useful as quick revision notes before exams.

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