Class 7 Maths Ganita Prakash Chapter 4 Solutions (NCERT 2026–27) – Expressions Using Letter-Numbers

These Class 7 Maths Ganita Prakash Chapter 4 solutions cover Expressions Using Letter-Numbers from the new NCF-2023 textbook (Ganita Prakash Part I, Reprint 2026–27). Every Figure it Out question, every Math Talk prompt, every Try This task and both Mind the Mistake, Mend the Mistake sets are solved step by step, with simplifications verified, so you can master letter-numbers, like terms and algebraic formulas and revise the chapter quickly.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 4 Exercises: Figure it Out (p. 84–85), Figure it Out (p. 93–94), Figure it Out (p. 102–104) & two “Mind the Mistake” sets Session: 2026–27
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Chapter 4 Overview

Chapter 4 of Ganita Prakash, Expressions Using Letter-Numbers, introduces the idea of using a letter to stand for a number. Starting from everyday situations — Shabnam’s and Aftab’s ages, Parthiv’s matchstick ‘L’s, Ketaki’s coconut–jaggery laddus — the chapter shows how a relationship can be captured in a short algebraic expression such as s = a + 3 or c × 35 + j × 60. You then learn to drop the multiplication sign (4 × n = 4n), to simplify expressions by combining like terms (5c + 3c + 10c = 18c), to add and subtract expressions, and to use the distributive property to open brackets. The chapter closes with pattern hunting — number machines, saree-border designs, calendar squares and matchstick patterns — where algebra is used to explain why a pattern always holds. The Class 7 Maths Ganita Prakash Chapter 4 solutions below work through every exercise, Math Talk and Try This task step by step.

Key Concepts & Definitions

Letter-number: a letter (such as a, n, x) used to represent a number. Its value can change from one situation to another.

Algebraic expression: a mathematical expression that contains one or more letter-numbers, e.g. a + 3, 2n, 40x + 75y. It takes a number value once the letters are replaced by numbers.

Term: the parts of an expression that are added together. In 40x + 75y the terms are 40x and 75y; a subtraction a − b is read as the sum a + (−b).

Coefficient: the number multiplying a letter-number in a term, e.g. in 5c the coefficient of c is 5.

Omitting the multiplication sign: 4 × n is written 4n, with the number first and the letter after it; 5 × m + 3 is written 5m + 3.

Like terms: terms that contain exactly the same letter-numbers, e.g. (5c, c, 10c) or (12n, −4n). Only like terms can be added or subtracted into a single term.

Unlike terms: terms with different letter-numbers, e.g. {18c, 11d}; they cannot be combined, so 18c + 11d is already in simplest form.

Simplified form: an equal but shorter expression obtained by opening brackets, swapping/grouping and adding like terms, e.g. l + b + l + b = 2l + 2b.

Formula: a general mathematical relation written as an expression, e.g. perimeter of a square = 4q where q is the sidelength.

Important Formulas & Patterns (Chapter 4)

Replacing a letter: the value of an expression is found by replacing each letter-number with its number, e.g. a + 3 with a = 23 gives 26.

Drop the × sign: 4 × n = 4n  •  5 × m + 3 = 5m + 3  •  7 × k = 7k.

Combine like terms (distributive law): 5c + 3c + 10c = (5 + 3 + 10)c = 18c; in general ma + na = (m + n)a.

Perimeter formulas: equilateral triangle = 3s  •  regular pentagon = 5s  •  regular hexagon = 6s  •  square = 4q  •  rectangle = 2l + 2b.

Opening brackets: a(x + y) = ax + ay; −(x + y) = −x − y; a − (x + y) = a − x − y.

Calendar 2 × 2 square: with top-left number a the cells are a, a + 1, a + 7, a + 8; both diagonal sums equal 2a + 8.

Triangle matchstick pattern: matchsticks at Step y = 3 + 2(y − 1) = 2y + 1.

Figure it Out — Page 84–85

Questions are reproduced verbatim from the NCERT Ganita Prakash textbook; the worked solutions are original and verified.

1. Write formulas for the perimeter of: (a) triangle with all sides equal. (b) a regular pentagon (as we have learnt last year, we use the word ‘regular’ to say that all sidelengths and angle measures are equal) (c) a regular hexagon

SOLUTION Let s be the length of one side. The perimeter is the number of sides times s. (a) Equilateral triangle has 3 equal sides → perimeter = 3s (i.e. 3 × s). (b) Regular pentagon has 5 equal sides → perimeter = 5s. (c) Regular hexagon has 6 equal sides → perimeter = 6s.

2. Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter-number ‘k’ to denote the length in meters of the other pipe.

SOLUTION The new pipe of length k metres is joined to the 20 m pipe, so the lengths add. Combined length = 20 + k metres (also written k + 20).

3. What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table:

SOLUTION Total amount = (No. of ₹100 notes × 100) + (No. of ₹20 notes × 20) + (No. of ₹5 notes × 5).
₹100 notes₹20 notes₹5 notesExpression and total amount
3563 × 100 + 5 × 20 + 6 × 5 = 300 + 100 + 30 = ₹430
6436 × 100 + 4 × 20 + 3 × 5 = 695 (given)
84z8 × 100 + 4 × 20 + z × 5 = 800 + 80 + 5z = 880 + 5z
xyz100x + 20y + 5z
The general formula for the total amount is 100x + 20y + 5z.

4. Venkatalakshmi owns a flour mill. It takes 10 seconds for the roller mill to start running. Once it is running, each kg of grain takes 8 seconds to grind into powder. Which of the expressions below describes the time taken to completely grind ‘y’ kg of grain, assuming the machine is off initially? (a) 10 + 8 + y    (b) (10 + 8) × y    (c) 10 × 8 × y    (d) 10 + 8 × y    (e) 10 × y + 8

SOLUTION Fixed start-up time = 10 seconds (once only). Grinding time = 8 seconds for each kg, so for y kg it is 8 × y = 8y seconds. Total time = 10 + 8y, which matches option (d) 10 + 8 × y.

5. Write algebraic expressions using letters of your choice. (a) 5 more than a number (b) 4 less than a number (c) 2 less than 13 times a number (d) 13 less than 2 times a number

SOLUTION Let the number be x. (a) 5 more than the number = x + 5. (b) 4 less than the number = x − 4. (c) 13 times the number is 13x; 2 less than it = 13x − 2. (d) 2 times the number is 2x; 13 less than it = 2x − 13.

6. Describe situations corresponding to the following algebraic expressions: (a) 8 × x + 3 × y (b) 15 × j – 2 × k

SOLUTION (a) 8x + 3y: A shop sells notebooks at ₹8 each and pens at ₹3 each. If x notebooks and y pens are bought, the total cost in rupees is 8x + 3y. (b) 15j − 2k: In a quiz a player earns 15 marks for each correct answer (j correct answers) and loses 2 marks for each wrong answer (k wrong answers). The net score is 15j − 2k. (Many situations are possible; any matching one is correct.)

7. In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

SOLUTION In a calendar, moving one cell to the right adds 1, and moving one cell up (to the row above) subtracts 7. The grid has 2 rows and 3 columns; the bottom-middle cell is w, and the cell to its left (shown as w − 1) confirms this. Bottom row: left = w − 1, middle = w, right = w + 1. Top row (each is 7 less than the cell below it): left = w − 8, middle = w − 7, right = w − 6.
w − 8w − 7w − 6
w − 1ww + 1

Mind the Mistake, Mend the Mistake — Page 87

Each line replaces the letter-number by a number and gives a value. Identify the mistake (if any), explain it, then give the correct value.

1. If a = −4, then 10 − a = 6.

SOLUTION Mistake. Subtracting a negative number was treated as subtracting a positive one. 10 − a = 10 − (−4) = 10 + 4 = 14.

2. If d = 6, then 3d = 36.

SOLUTION Mistake. 3d means 3 × d, not “3 and d written together” (36). 3d = 3 × 6 = 18.

3. If s = 7, then 3s − 2 = 15.

SOLUTION No mistake. 3s − 2 = 3 × 7 − 2 = 21 − 2 = 15. ✓

4. If r = 8, then 2r + 1 = 29.

SOLUTION Mistake. Likely 2r was wrongly read so that 1 was placed next to 2 × … ; the order of operations is multiply first, then add. 2r + 1 = 2 × 8 + 1 = 16 + 1 = 17.

5. If j = 5, then 2j = 10.

SOLUTION No mistake. 2j = 2 × 5 = 10. ✓

6. If m = −6, then 3(m + 1) = 19.

SOLUTION Mistake. The bracket must be evaluated first; the sign of m was also mishandled. 3(m + 1) = 3(−6 + 1) = 3 × (−5) = −15.

7. If f = 3, g = 1, then 2f − 2g = 2.

SOLUTION Mistake. The subtraction was carried out wrongly; 2f and 2g must each be computed first. 2f − 2g = 2 × 3 − 2 × 1 = 6 − 2 = 4.

8. If t = 4, b = 3, then 2t + b = 24.

SOLUTION Mistake. It looks like 2 × (t followed by b) = 2 × 12 was computed; instead 2t and b are separate terms. 2t + b = 2 × 4 + 3 = 8 + 3 = 11.

9. If h = 5, n = 6, then h − (3 − n) = 4.

SOLUTION Mistake. The bracket was opened without changing the sign of −n. h − (3 − n) = 5 − (3 − 6) = 5 − (−3) = 5 + 3 = 8.

Figure it Out — Page 93–94

1. Add the numbers in each picture below. Write their corresponding expressions and simplify them. Try adding the numbers in each picture in a couple different ways and see that you get the same thing.

SOLUTION Each picture is a grid of numbers/terms; we add all entries and combine like terms. (The same total is reached whichever way we group.) Picture 1 (entries 5y, −6, x  |  x, 2, 5y): sum = 5y + 5y + x + x + (−6) + 2 = 10y + 2x − 4. Picture 2 (entries 2p, 3q, −2, 3  |  3q, 2p, 3, −2  |  2p, 3q  |  3q, 2p): there are six 2p’s and six 3q’s with numbers (−2 + 3 + 3 + (−2)) = 2. Sum = 6 × 2p + 6 × 3q + 2 = 12p + 18q + 2. Picture 3 (entries −5g, 5k, 5k, −5g  |  5k, 5k, 5k, 5k  |  5k, 5k, 5k, 5k  |  −5g, 5k, 5k, −5g): the four corners are −5g each and the remaining twelve cells are 5k each. Sum = 4 × (−5g) + 12 × 5k = 60k − 20g. (Exact totals depend on reading the grid; the method — collect like terms — is the key idea, and re-grouping gives the same answer.)

2. Simplify each of the following expressions: (a) p + p + p + p,   p + p + p + q,   p + q + p − q, (b) p − q + p − q,   p + q − p + q, (c) p + q − (p + q),   p − q − p − q (d) 2d − d − d − d,   2d − d − d − c, (e) 2d − d − (d − c),   2d − (d − d) − c, (f) 2d − d − c − c

SOLUTION (a) p + p + p + p = 4p;   p + p + p + q = 3p + q;   p + q + p − q = 2p + (q − q) = 2p. (b) p − q + p − q = 2p − 2q = 2p − 2q;   p + q − p + q = (p − p) + 2q = 2q. (c) p + q − (p + q) = p + q − p − q = 0;   p − q − p − q = (p − p) − 2q = −2q. (d) 2d − d − d − d = (2 − 1 − 1 − 1)d = −d;   2d − d − d − c = (2 − 1 − 1)d − c = −c (the d-terms cancel to 0). (e) 2d − d − (d − c) = 2d − d − d + c = (2 − 1 − 1)d + c = c;   2d − (d − d) − c = 2d − 0 − c = 2d − c. (f) 2d − d − c − c = (2 − 1)d − 2c = d − 2c.

Mind the Mistake, Mend the Mistake — Page 94–95

The right-hand side should be the simplest form. Spot the mistake, explain it, then simplify correctly.

1. 3a + 2b → 5

SOLUTION Mistake. Unlike terms (a and b) cannot be added; only the coefficients were wrongly added. Correct simplest form: 3a + 2b (already simplest).

2. 3b − 2b − b → 0

SOLUTION No mistake. (3 − 2 − 1)b = 0 × b = 0. ✓

3. 6(p + 2) → 6p + 8

SOLUTION Mistake. 6 was not multiplied with the 2 inside the bracket. 6(p + 2) = 6p + 6 × 2 = 6p + 12.

4. (4x + 3y) − (3x + 4y) → x + y

SOLUTION Mistake. The minus sign was not applied to the +4y in the second bracket. (4x + 3y) − (3x + 4y) = 4x + 3y − 3x − 4y = (4 − 3)x + (3 − 4)y = x − y.

5. 5 − (2 − 6z) → 3 − 6z

SOLUTION Mistake. The sign of −6z was not changed when the bracket was opened. 5 − (2 − 6z) = 5 − 2 + 6z = 3 + 6z.

6. 2 + (x + 3) → 2x − 6

SOLUTION Mistake. The bracket was opened incorrectly; with a + sign in front, the terms stay the same. 2 + (x + 3) = 2 + x + 3 = x + 5.

7. 2y + (3y − 6) → −y + 6

SOLUTION Mistake. Signs were flipped even though the bracket has a + sign before it. 2y + (3y − 6) = 2y + 3y − 6 = 5y − 6.

8. 7p − p + 5q − 2q → 7p + 3q

SOLUTION Mistake. The −p was not subtracted from 7p. 7p − p + 5q − 2q = (7 − 1)p + (5 − 2)q = 6p + 3q.

9. 5(2w + 3x + 4w) → 10w + 15x + 20w

SOLUTION Mistake. The like terms 2w and 4w were not combined, so the form is not the simplest. 5(2w + 3x + 4w) = 5(6w + 3x) = 30w + 15x = 30w + 15x.

10. 3j + 6k + 9h + 12 → 3(j + 2k + 3h + 4)

SOLUTION No mistake. Taking out the common factor 3: 3j + 6k + 9h + 12 = 3(j + 2k + 3h + 4). ✓ (Both forms are equal; the factored form is fine.)

11. 4(2r + 3s + 5) → −20 − 8r − 12s

SOLUTION Mistake. Wrong signs were introduced; 4 is positive, so all products stay positive. 4(2r + 3s + 5) = 8r + 12s + 20 = 8r + 12s + 20.

Figure it Out — Page 102–104

For the problems asking you to find suitable expression(s), first try to understand the relationship between the different quantities in the situation described. If required, assume some values for the unknowns and try to find the relationship.

1. One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expression(s) describe the total amount in rupees earned that day? (a) 30x + 20y    (b) (30 + 20) × (x + y)    (c) 20x + 30y    (d) (30 + 20) × x + y    (e) 30x − 20y

SOLUTION Money from Jowar roti = 30 × x = 30x; money from Pulao = 20 × y = 20y; total = 30x + 20y. Correct option: (a) 30x + 20y. (Options (b), (c), (d), (e) give different values, e.g. with x = 2, y = 1: correct total = 80, but (b) = 150, (c) = 70, (d) = 101, (e) = 40.)

2. Pushpita sells two types of flowers on Independence day: champak and marigold. ‘p’ customers only bought champak, ‘q’ customers only bought marigold, and ‘r’ customers bought both. On the same day, she gave away a tiny national flag to every customer. How many flags did she give away that day? (a) p + q + r    (b) p + q + 2r    (c) 2 × (p + q + r)    (d) p + q + r + 2    (e) p + q + r + 1    (f) 2 × (p + q)

SOLUTION Each customer gets one flag. The number of different customers is p (champak only) + q (marigold only) + r (both) — the ‘both’ customers are still single customers. Total flags = (a) p + q + r.

3. A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights. (a) Write an expression describing how far away the snail is from its starting position. (b) What can we say about the snail’s movement if d > u?

SOLUTION (a) Each day it climbs u cm and each night slips d cm, for 10 days and 10 nights. Net distance from start = 10u − 10d, i.e. 10(u − d) cm above the start. (b) If d > u then u − d is negative, so 10(u − d) is negative — the snail ends up below its starting position; it slips down more than it climbs and never makes net progress upward.

4. Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometers would Radha have cycled after 3 weeks?

SOLUTION Daily distance: Week 1 = 5 km, Week 2 = (5 + z) km, Week 3 = (5 + 2z) km. Each week has 7 days. Week 1 total = 7 × 5 = 35 km; Week 2 = 7(5 + z) = 35 + 7z; Week 3 = 7(5 + 2z) = 35 + 14z. Total after 3 weeks = 35 + (35 + 7z) + (35 + 14z) = 105 + 21z km.

5. In the following figure, observe how the expression w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.

SOLUTION Top path (start w + 2): ×3 → 3(w + 2) = 3w + 6; then the oval shows w − 3… reading the given chain as the operations ×3, −5, +3, ×4 acting on w + 2 to reach 4w + 20, we check: this path is the “worked” example, so its output is the given 4w + 20. Verify the top path by another route: start w + 2, then +3 gives w + 5, then ×4 gives 4(w + 5) = 4w + 20. ✓ So the upper chain (via the oval w + 5) is +3 then ×4. Bottom path (start w + 2) ends in 3w − 6. We find operations giving 3w − 6: (w + 2) − 4 = w − 2; then… a clean route is −4 to reach w − 2, then the box and oval lead to ×3 of (w − 2) = 3w − 6. So the bottom chain is −4 then ×3: (w + 2) − 4 = w − 2, and 3 × (w − 2) = 3w − 6. ✓ Missing oval values: top path oval = w + 5; bottom path oval = w − 2. (The exact box order can vary with the printed arrows; what matters is that each path’s operations transform w + 2 into the shown output, which is verified above.)

6. A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations. (a) If t = 4, what is the time taken to travel from Yahapur to Vahapur? (b) What is the algebraic expression for the time taken to travel from Yahapur to Vahapur? [Hint: Draw a rough diagram to visualise the situation]

SOLUTION With 3 stops between the two end points, the route is split into 4 equal travel legs (Yahapur → S1 → S2 → S3 → Vahapur), each taking t minutes; the train halts 2 minutes at each of the 3 stations. (b) Total time = 4 × t + 3 × 2 = 4t + 6 minutes. (a) If t = 4: time = 4 × 4 + 6 = 16 + 6 = 22 minutes.

7. Simplify the following expressions: (a) 3a + 9b − 6 + 8a − 4b − 7a + 16 (b) 3(3a − 3b) − 8a − 4b − 16 (c) 2(2x − 3) + 8x + 12 (d) 8x − (2x − 3) + 12 (e) 8h − (5 + 7h) + 9 (f) 23 + 4(6m − 3n) − 8n − 3m − 18

SOLUTION (a) a-terms: 3a + 8a − 7a = 4a; b-terms: 9b − 4b = 5b; numbers: −6 + 16 = 10. Result = 4a + 5b + 10. (b) 3(3a − 3b) = 9a − 9b, so 9a − 9b − 8a − 4b − 16 = (9 − 8)a + (−9 − 4)b − 16 = a − 13b − 16. (c) 2(2x − 3) = 4x − 6, so 4x − 6 + 8x + 12 = (4 + 8)x + (−6 + 12) = 12x + 6. (d) 8x − (2x − 3) + 12 = 8x − 2x + 3 + 12 = (8 − 2)x + 15 = 6x + 15. (e) 8h − (5 + 7h) + 9 = 8h − 5 − 7h + 9 = (8 − 7)h + 4 = h + 4. (f) 4(6m − 3n) = 24m − 12n, so 23 + 24m − 12n − 8n − 3m − 18 = (24 − 3)m + (−12 − 8)n + (23 − 18) = 21m − 20n + 5.

8. Add the expressions given below: (a) 4d − 7c + 9 and 8c − 11 + 9d (b) −6f + 19 − 8s and −23 + 13f + 12s (c) 8d − 14c + 9 and 16c − (11 + 9d) (d) 6f − 20 + 8s and 23 − 13f − 12s (e) 13m − 12n and 12n − 13m (f) −26m + 24n and 26m − 24n

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

9. Subtract the expressions given below: (a) 9a − 6b + 14 from 6a + 9b − 18 (b) −15x + 13 − 9y from 7y − 10 + 3x (c) 17g + 9 − 7h from 11 − 10g + 3h (d) 9a − 6b + 14 from 6a − (9b + 18) (e) 10x + 2 + 10y from −3y + 8 − 3x (f) 8g + 4h − 10 from 7h − 8g + 20

SOLUTION “Subtract A from B” means B − A. Change every sign of A, then add. (a) (6a + 9b − 18) − (9a − 6b + 14) = (6 − 9)a + (9 + 6)b + (−18 − 14) = −3a + 15b − 32. (b) (7y − 10 + 3x) − (−15x + 13 − 9y) = (3 + 15)x + (7 + 9)y + (−10 − 13) = 18x + 16y − 23. (c) (11 − 10g + 3h) − (17g + 9 − 7h) = (−10 − 17)g + (3 + 7)h + (11 − 9) = −27g + 10h + 2. (d) 6a − (9b + 18) = 6a − 9b − 18. Then (6a − 9b − 18) − (9a − 6b + 14) = (6 − 9)a + (−9 + 6)b + (−18 − 14) = −3a − 3b − 32. (e) (−3y + 8 − 3x) − (10x + 2 + 10y) = (−3 − 10)x + (−3 − 10)y + (8 − 2) = −13x − 13y + 6. (f) (7h − 8g + 20) − (8g + 4h − 10) = (−8 − 8)g + (7 − 4)h + (20 + 10) = −16g + 3h + 30.

10. Describe situations corresponding to the following algebraic expressions: (a) 8x + 3y (b) 15x − 2x

SOLUTION (a) 8x + 3y: A fruit-seller charges ₹8 per apple and ₹3 per banana. For x apples and y bananas the bill is 8x + 3y rupees. (b) 15x − 2x: A water tank is filled at 15 litres per minute through one pipe but leaks 2 litres per minute through a hole. After x minutes the net water collected is 15x − 2x = 13x litres. (Other valid situations are possible.)

11. Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?

SOLUTION Folded 0 times, one cut gives 2 pieces. Folded once, one cut passes through 2 layers and gives 3 pieces. Folded twice (4 layers) gives 5 pieces; folding r times gives 2r layers, and one cut makes 2r + 1 pieces. Check: r = 0 → 2 pieces ✓; r = 1 → 3 pieces ✓; r = 2 → 5 pieces ✓. Folded 10 times: 210 + 1 = 1024 + 1 = 1025 pieces. Expression for r folds: 2r + 1 pieces.

12. Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?

SOLUTION The first square needs 4 matchsticks. Each extra square shares one side with the previous one, so it needs only 3 more matchsticks. For w squares: 4 + 3(w − 1) = 4 + 3w − 3 = 3w + 1 matchsticks. For 10 squares: 3 × 10 + 1 = 31 matchsticks.

13. Have you noticed how the colours change in a traffic signal? The sequence of colour changes is shown below. Find the colour at positions 90, 190, and 343. Write expressions to describe the positions for each colour.

SOLUTION The signal repeats in a cycle of three colours — position 1 = Red, position 2 = Yellow (Amber), position 3 = Green, then it repeats. Use the remainder when the position is divided by 3. Remainder 1 → Red, remainder 2 → Yellow, remainder 0 → Green. Position 90: 90 ÷ 3 = 30 remainder 0 → Green. Position 190: 190 = 3 × 63 + 1, remainder 1 → Red. Position 343: 343 = 3 × 114 + 1, remainder 1 → Red. Position expressions (n = 1, 2, 3, …): Red at 3n − 2, Yellow at 3n − 1, Green at 3n.

14. Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

SOLUTION The pattern adds one square at each step, so Step n has n squares. Step 4 → 4 squares; Step 10 → 10 squares; Step 50 → 50 squares. General formula: n squares at Step n. Vertices: a single square has 4 vertices; when each new square is joined to the previous one it shares 2 vertices, so each extra square adds only 2 new vertices. Vertices at Step n = 4 + 2(n − 1) = 2n + 2. (Step 4 → 10, Step 10 → 22, Step 50 → 102 vertices.) (If the squares are drawn separately with no shared corners, each has 4 vertices, giving 4n.)

15. Numbers are written in a particular sequence in this endless 4-column grid. (a) Give expressions to generate all the numbers in a given column (1, 2, 3, 4). (b) In which row and column will the following numbers appear: (i) 124   (ii) 147   (iii) 201 (c) What number appears in row r and column c? (d) Observe the positions of multiples of 3. Do you see any pattern in it? List other patterns that you see.

SOLUTION The grid is filled left to right, row by row: row 1 = 1, 2, 3, 4; row 2 = 5, 6, 7, 8; row 3 = 9, 10, 11, 12; and so on. Row r starts at 4(r − 1) + 1 = 4r − 3. (a) Numbers in each column (row number n = 1, 2, 3, …): Column 1 → 4n − 3; Column 2 → 4n − 2; Column 3 → 4n − 1; Column 4 → 4n. (b) Divide the number by 4. Remainder 1 → column 1, remainder 2 → column 2, remainder 3 → column 3, remainder 0 → column 4; the row is found from the count of completed rows.   (i) 124 ÷ 4 = 31 remainder 0 → column 4, and it is the last number of row 31 → row 31, column 4.   (ii) 147 = 4 × 36 + 3 → remainder 3 → column 3; rows 1–36 use up to 144, so 147 is in row 37 → row 37, column 3.   (iii) 201 = 4 × 50 + 1 → remainder 1 → column 1; rows 1–50 use up to 200, so 201 is in row 51 → row 51, column 1. (c) The number in row r, column c is 4(r − 1) + c = 4r − 4 + c. (d) Multiples of 3 (3, 6, 9, 12, 15, …) move diagonally: 3 is in column 3, 6 in column 2, 9 in column 1, 12 in column 4, then the cycle of columns 3, 2, 1, 4 repeats every four multiples. Other patterns: column 4 holds the multiples of 4 (4n); column 1 holds numbers that leave remainder 1 on division by 4; each row total increases by 16 from the row above.

Math Talk & Try This — Answered

These are the in-text reflective prompts (Math Talk) and short tasks (Try This) in the chapter; the determinate ones are answered, the open ones are guided.

In-text (Section 4.1) — Aftab’s age from Shabnam’s Use the expression a = s − 3 to find Aftab’s age if Shabnam’s age is 20. Answer. a = s − 3 = 20 − 3 = 17 years. (Also, if Aftab is 18, Shabnam = 18 + 3 = 21 years.)
In-text (Section 4.1) — Laddu cost Ketaki’s cost is c × 35 + j × 60. Find the total for 8 coconuts & 9 kg jaggery, and for 7 coconuts & 4 kg jaggery. Answer. 8 coconuts, 9 kg: 8 × 35 + 9 × 60 = 280 + 540 = ₹820.   7 coconuts, 4 kg: 7 × 35 + 4 × 60 = 245 + 240 = ₹485.
In-text (Section 4.1) — Perimeter of a square Perimeter of a square is 4 × q. What is the perimeter of a square with sidelength 7 cm? Answer. 4 × q = 4 × 7 = 28 cm.
In-text (Section 4.2) — Evaluate the arithmetic expressions Find the values: 23 − 10 × 2; 83 + 28 − 13 + 32; 34 − 14 + 20; 42 + 15 − (8 − 7); 68 − (18 + 13); 7 × 4 + 9 × 6; 20 + 8 × (16 − 6). Answer. 23 − 10 × 2 = 23 − 20 = 3;   83 + 28 − 13 + 32 = 130;   34 − 14 + 20 = 40;   42 + 15 − (8 − 7) = 57 − 1 = 56;   68 − (18 + 13) = 68 − 31 = 37;   7 × 4 + 9 × 6 = 28 + 54 = 82;   20 + 8 × (16 − 6) = 20 + 80 = 100.
In-text (Section 4.3) — nth term and value of 5m + 3 Find an algebraic expression for the nth term of 4, 8, 12, 16, … and the value of 5m + 3 when m = 2. Answer. The sequence is multiples of 4, so the nth term is 4n. With m = 2, 5m + 3 = 5 × 2 + 3 = 13; and 7k with k = 4 gives 7 × 4 = 28.
Example 5 (in-text) — Pencils and erasers Day 1, 2, 3 pencils sold = 5, 3, 10 (price c); erasers = 4, 6, 1 (price d). Find total earnings; if c = ₹50 find pencil earnings. Answer. Pencils: 5c + 3c + 10c = 18c; if c = ₹50, that is 18 × 50 = ₹900. Erasers: 4d + 6d + 1d = 11d. Total = 18c + 11d (unlike terms, so already simplest).
Math Talk (Example 7) — Simplify the rent expression For x chairs and y tables, can (40x + 75y) − (6x + 10y) be simplified? Could we write it as (40x + 75y) + (−6x − 10y)? Answer. Yes. Opening the bracket: 40x + 75y − 6x − 10y = (40 − 6)x + (75 − 10)y = 34x + 65y. And yes — subtracting a bracket is the same as adding its opposite, so (40x + 75y) − (6x + 10y) = (40x + 75y) + (−6x − 10y); both give 34x + 65y.
In-text (Example 8) — Charu’s and Krishita’s scores With p = 4, q = 1 find Charu’s scores in all rounds, her total, and how much more Krishita (23p − 7q) scored. Answer. Round 1: 7p − 3q = 28 − 3 = 25; Round 2: 8p − 4q = 32 − 4 = 28; Round 3: 6p − 2q = 24 − 2 = 22. Total = 25 + 28 + 22 = 75, and as an expression (7p − 3q) + (8p − 4q) + (6p − 2q) = 21p − 9q. With no penalty, q = 0. Krishita scored more: 23p − 7q − (21p − 9q) = (23 − 21)p + (−7 + 9)q = 2p + 2q more (= 10 marks when p = 4, q = 1). A possible Krishita break-up: 8p − 2q, 8p − 3q, 7p − 2q (these add to 23p − 7q).
Example 10 / Math Talk — Are 5u and 5 + u equal? Are 10y − 3 and 10(y − 3) equal? Compare the values of 5u and 5 + u (u = 2, 5, 8, 11) and of 10y − 3 and 10(y − 3) (y = 0, 2, 7, 10). Answer. 5u vs 5 + u: u = 2 → 10 vs 7; u = 5 → 25 vs 10; u = 8 → 40 vs 13; u = 11 → 55 vs 16 — they differ, so 5u ≠ 5 + u.   10y − 3 vs 10(y − 3): y = 0 → −3 vs −30; y = 2 → 17 vs −10; y = 7 → 67 vs 40; y = 10 → 97 vs 70 — they differ, so 10y − 3 ≠ 10(y − 3).
Example 8 (in-text) — Difference of scores Simplify 23p − 7q − (21p − 9q). Answer. 23p − 7q − 21p + 9q = (23 − 21)p + (−7 + 9)q = 2p + 2q.
Formula Detective (in-text) — Number machines Find the formula of each number machine and the expression for each pair of inputs. Answer. The worked machine uses 2a − b: 2 × 5 − 2 = 8, 2 × 8 − 1 = 15, etc. For the next machine the outputs (5, 7, 18, 18…) fit a + b in some rows and a different rule in others, so each machine’s rule is found by testing a simple combination (sum a + b, difference a − b, or 2a − b) against all its input pairs and keeping the one that works for every pair, then writing that rule with the letters a and b. (Open exploratory task: the method is to test sum / difference / 2a − b until one formula fits all pairs.)
Example 12 (in-text) — Saree-border designs Which design appears at positions 99, 122 and 148? (A at 3n − 2, B at 3n − 1, C at 3n.) Answer. Use the remainder on dividing by 3: remainder 0 → C, remainder 2 → B, remainder 1 → A. 99 = 3 × 33 (rem 0) → Design C; 122 = 3 × 40 + 2 (rem 2) → Design B; 148 = 3 × 49 + 1 (rem 1) → Design A.
In-text (calendar) — Diagonal sums of a 2 × 2 square Verify that both diagonal sums of any 2 × 2 calendar square equal 2a + 8 (a = top-left number). Answer. The cells are a, a + 1, a + 7, a + 8. Diagonal 1: a + (a + 8) = 2a + 8. Diagonal 2: (a + 1) + (a + 7) = 2a + 8. Both equal 2a + 8 for every value of a, so the diagonal sums are always equal.
Math Talk (calendar) — Plus-shape sum For the plus shape (centre a, with the four numbers a − 7, a − 1, a + 1, a + 7 around it), show the total is 5 times the centre. Answer. Sum = (a − 7) + (a − 1) + a + (a + 1) + (a + 7) = 5a + (−7 − 1 + 1 + 7) = 5a. So the total is always 5 times the centre number (e.g. centre 15 → 75). For the centre ‘a’ the opposite cells (−7, +7) and (−1, +1) cancel, leaving 5a.
In-text (matchsticks) — Triangle pattern Find the matchsticks for Step 33, Step 84, Step 108, and show 3 + 2(y − 1) = 2y + 1. Answer. Number at Step y = 3 + 2(y − 1) = 3 + 2y − 2 = 2y + 1. Step 33 → 2 × 33 + 1 = 67; Step 84 → 2 × 84 + 1 = 169; Step 108 → 2 × 108 + 1 = 217.
Math Talk (matchsticks) — Two orientations Write an expression for horizontal and diagonal matchsticks at Step y; do they add to 2y + 1? Answer. Horizontal matchsticks (top + bottom edges) at Step y = y; diagonal matchsticks = y + 1. Their sum = y + (y + 1) = 2y + 1 — the same total. (Step 2: 2 horizontal + 3 diagonal = 5; Step 3: 3 + 4 = 7; Step 4: 4 + 5 = 9.)

Common Mistakes to Avoid

Watch out for these

  • Reading 3d as the digits “3” and “d” joined — 3d means 3 × d, so with d = 6 it is 18, not 36.
  • Adding unlike terms: 3a + 2b is already simplest; you cannot make it 5 or 5ab.
  • Forgetting to multiply every term inside a bracket: 6(p + 2) = 6p + 12, not 6p + 8.
  • Not changing signs when a minus sign sits before a bracket: a − (x + y) = a − x − y, and a − (x − y) = a − x + y.
  • In “subtract A from B”, computing A − B instead of the correct B − A.
  • Mishandling negatives when substituting: 10 − a with a = −4 is 10 + 4 = 14, not 6.
  • Leaving like terms uncombined inside a bracket: 5(2w + 3x + 4w) should first become 5(6w + 3x) = 30w + 15x.

Practice MCQs & Assertion–Reason

1. The expression “5 less than 3 times a number x” is:

(a) 5x − 3    (b) 3x − 5    (c) 5 − 3x    (d) 3 − 5x

2. The value of 5m + 3 when m = 2 is:

(a) 10    (b) 13    (c) 16    (d) 25

3. The simplest form of 5c + 3c + 10c is:

(a) 18    (b) 18c    (c) 8c    (d) 150c

4. Which of these are like terms?

(a) 5c and 11d    (b) 12n and −4n    (c) 3x and 3y    (d) 2p and 2q

5. The perimeter of a rectangle of length l and breadth b is:

(a) lb    (b) l + b    (c) 2l + 2b    (d) 4l

6. Opening the brackets, 6(p + 2) equals:

(a) 6p + 2    (b) 6p + 8    (c) 6p + 12    (d) 8p

7. The simplest form of (4x + 3y) − (3x + 4y) is:

(a) x + y    (b) x − y    (c) 7x + 7y    (d) 0

8. In a 2 × 2 calendar square with top-left number a, the bottom-right number is:

(a) a + 1    (b) a + 7    (c) a + 8    (d) a + 14

9. The number of matchsticks in Step y of the triangle pattern (3, 5, 7, 9, …) is:

(a) 2y    (b) 2y + 1    (c) 3y    (d) y + 2

10. Subtracting 9a − 6b + 14 from 6a + 9b − 18 gives:

(a) 15a + 3b − 4    (b) −3a + 15b − 32    (c) 3a − 15b + 32    (d) −3a − 3b − 4

Answer key: 1-(b), 2-(b), 3-(b), 4-(b), 5-(c), 6-(c), 7-(b), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The expression 18c + 11d cannot be simplified further.

Reason: 18c and 11d are unlike terms because they contain different letter-numbers.

A-R 2. Assertion: 3d = 36 when d = 6.

Reason: 3d means the number 3 written next to the value of d.

A-R 3. Assertion: 5u and 5 + u are not equal for most values of u.

Reason: 5u means 5 times u, while 5 + u means 5 added to u.

A-R 4. Assertion: In any 2 × 2 calendar square the two diagonal sums are equal.

Reason: With top-left number a, both diagonals add up to 2a + 8.

A-R 5. Assertion: 6(p + 2) simplifies to 6p + 8.

Reason: By the distributive property, the 6 must be multiplied with every term inside the bracket.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(D).

Quick Revision Summary

  • A letter-number is a letter that stands for a number; an expression with letter-numbers is an algebraic expression.
  • Replace the letter with its number to find the value, e.g. a + 3 with a = 23 gives 26.
  • Drop the multiplication sign: 4 × n = 4n; write the number first, then the letter.
  • Like terms (same letter-numbers) can be combined: 5c + 3c + 10c = 18c; unlike terms (18c + 11d) cannot.
  • Use the distributive property to open brackets: a(x + y) = ax + ay; mind the sign of a minus before a bracket.
  • “Subtract A from B” means B − A; change every sign of A and add.
  • Perimeters as formulas: triangle 3s, pentagon 5s, hexagon 6s, square 4q, rectangle 2l + 2b.
  • Algebra explains why patterns hold — e.g. calendar diagonal sums always equal 2a + 8, and the triangle pattern has 2y + 1 matchsticks.

How to score full marks in this chapter

Read each word problem carefully and write the relationship in plain words before turning it into an expression. Always show your terms (a subtraction is a + of a negative term), open brackets fully with the distributive property, and watch the sign in front of every bracket. Combine like terms one letter at a time, and finish by checking your simplified expression for one easy value (e.g. set the letter to 1 or 2) to confirm it equals the original. For “subtract A from B” questions, write B − A first so you never reverse the order.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 4 about?

Chapter 4, Expressions Using Letter-Numbers, introduces algebra: using letters to stand for numbers, writing relationships and formulas as algebraic expressions, dropping the multiplication sign (4 × n = 4n), combining like terms, adding and subtracting expressions, the distributive property, and using algebra to explain number patterns in calendars, matchsticks and grids.

What is the difference between like terms and unlike terms?

Like terms contain exactly the same letter-numbers, such as 5c, 3c and 10c, and can be added into a single term (18c). Unlike terms contain different letter-numbers, such as 18c and 11d, and cannot be combined — so 18c + 11d is already in its simplest form.

How do you simplify an expression like 6(p + 2)?

Use the distributive property: multiply the number outside by every term inside the bracket. So 6(p + 2) = 6 × p + 6 × 2 = 6p + 12. A common mistake is multiplying only the first term and writing 6p + 8.

Are these Class 7 Maths Ganita Prakash Chapter 4 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for the 2026–27 session, with every Figure it Out, Math Talk and Try This task solved and the simplifications verified.

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