Class 7 Maths Chapter 4 Solutions — Expressions Using Letter-Numbers

Learn algebra with these NCERT Ganita Prakash Class 7 Chapter 4 solutions. Follow worked simplifications, number machines, calendar patterns and original matchstick diagrams. Each pattern is read from the textbook drawing, with notes where the printed operations are inconsistent.

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. 102–105) & 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 worked solutions below explain the exercises and in-text patterns.

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 with the same variable part (the same letters raised to the same powers), 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

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.
For side length s, an equilateral triangle has perimeter 3s, a regular pentagon 5s and a regular hexagon 6s.
For side length s, an equilateral triangle has perimeter 3s, a regular pentagon 5s and a regular hexagon 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
In a two-row calendar block with bottom-middle w, the upper row is w − 8, w − 7, w − 6.
In a two-row calendar block with bottom-middle w, the upper row is w − 8, w − 7, w − 6.

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.

SOLUTIONThere is a mistake. Substitute s = 7: 3s − 2 = 3 × 7 − 2 = 21 − 2 = 19, not 15.

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

SOLUTIONThere is a mistake. 2r means 2 × r, so 2 × 8 + 1 = 16 + 1 = 17. Reading 2r as the two-digit number 28 would wrongly give 29.

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.

SOLUTIONThere is a mistake. 2t + b = 2 × 4 + 3 = 8 + 3 = 11. The expression 24 is not the required sum; multiply 2 by t and then add b.

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.

SOLUTIONPicture 1: two 5y terms, two x terms and −6 + 2 give 10y + 2x − 4. By rows: (5y − 6 + x) + (x + 2 + 5y).Picture 2: four 2p terms, four 3q terms and −2 + 3 + 3 − 2 give 4(2p) + 4(3q) + 2 = 8p + 12q + 2. By rows: (2p + 3q + 1) + (2p + 3q + 1) + (2p + 3q) + (2p + 3q).Picture 3: four corner terms −5g and twelve terms 5k give 60k − 20g. By rows: (10k − 10g) + 20k + 20k + (10k − 10g). Each method counts every entry once.
The middle picture has four 2p terms and four 3q terms, plus −2, 3, 3 and −2, so its total is 8p + 12q + 2.
The middle picture has four 2p terms and four 3q terms, plus −2, 3, 3 and −2, so its total is 8p + 12q + 2.

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)

SOLUTIONThe factored expression 3(j + 2k + 3h + 4) is equal to the original. However, this exercise specifies a simplest form with brackets removed. Under that convention use 3j + 6k + 9h + 12; there are no like terms left to combine.

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 102–105

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.

SOLUTIONUpper-left: w + 2 → subtract 5 → w − 3 → multiply by 3 → 3w − 9.Upper-right: w + 2 → add 3 → w + 5 → multiply by 4 → 4w + 20.Lower-left: w + 2 → subtract 8 → w − 6 → subtract 4 → w − 10.Lower-right printing issue: the supplied textbook shows +3 and ×4 again, so its missing oval is w + 5 and the output must be 4w + 20. Its printed output 3w − 6 is inconsistent. To obtain 3w − 6 instead, the operations would have to change, for example to −4 followed by ×3. Those are a possible correction, not the operations printed in the diagram.
The paths give 3w − 9, 4w + 20 and w − 10. The repeated +3 then ×4 path also gives 4w + 20; the printed 3w − 6 is inconsistent.
The paths give 3w − 9, 4w + 20 and w − 10. The repeated +3 then ×4 path also gives 4w + 20; the printed 3w − 6 is inconsistent.

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.
The train travels four legs of t minutes and halts for two minutes at each of three intermediate stations.
The train travels four legs of t minutes and halts for two minutes at each of three intermediate stations.

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?

SOLUTIONFollow the back-and-forth folds shown in the picture. Zero folds give one strand and 2 pieces; one fold gives two strands and 3 pieces; two folds give three strands and 4 pieces.Each additional bend adds one strand crossed by the cut. Thus r folds give r + 1 cut points along the original rope, and hence r + 2 pieces. For 10 folds: 12 pieces.This is different from repeatedly folding the entire bundle in half, which would double the strands. That is not the folding pattern pictured here.
The pictured pattern has 1, 2 and 3 strands, giving 2, 3 and 4 pieces after one cut. Ten such folds give 12 pieces.
The pictured pattern has 1, 2 and 3 strands, giving 2, 3 and 4 pieces after one cut. Ten such folds give 12 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.
One, two and three joined squares need 4, 7 and 10 sticks. Ten squares need 31 sticks.
One, two and three joined squares need 4, 7 and 10 sticks. Ten squares need 31 sticks.

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.

SOLUTIONRead the pictured cycle: red, yellow, green, yellow, then repeat. Its period is 4.Position 90 leaves remainder 2 on division by 4: yellow. Position 190 also leaves remainder 2: yellow. Position 343 leaves remainder 3: green.For n = 1, 2, 3, …: red positions are 4n − 3; green positions are 4n − 1; yellow positions are all even numbers, 2n (equivalently the two lists 4n − 2 and 4n).
The textbook sequence is red, yellow, green, yellow, then repeats. Yellow occupies every even position.
The textbook sequence is red, yellow, green, yellow, then repeats. Yellow occupies every even position.

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?

SOLUTIONThe X-shaped arrangements contain 5, 9, 13, … squares. There is one central square and n squares on each of four arms at Step n. Hence the number is 4n + 1.Step 4: 17 squares. Step 10: 41. Step 50: 201.Vertices of all squares, counted separately: each square contributes four corners, giving 4(4n + 1) = 16n + 4. This gives 68, 164 and 804 for Steps 4, 10 and 50.If counting distinct corner points in the touching drawing: every added square shares one corner with its predecessor. There are 4n shared corner points, so subtract these duplicate counts: 16n + 4 − 4n = 12n + 4. This gives 52, 124 and 604. State which interpretation of “vertices” you are using.
Steps 1, 2 and 3 contain 5, 9 and 13 squares. Step n contains 4n + 1 squares.
Steps 1, 2 and 3 contain 5, 9 and 13 squares. Step n contains 4n + 1 squares.

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.

Terms and letter-numbers after simplification (page 95)

For the linear expressions in this exercise, each surviving letter-number contributes one term, and there may be one additional constant term. For example, 8r + 12s + 20 has two letter-numbers and three terms. Terms may disappear when their coefficients add to zero. This observation is specific to these expressions, not a rule for every possible algebraic expression.

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. For u = 2, 5, 8, 11, the values of 5u are 10, 25, 40, 55; the values of 5 + u are 7, 10, 13, 16. Thus these are not identical expressions. They can agree at a particular value: 5u = 5 + u when u = 1.25.

For y = 0, 2, 7, 10, 10y − 3 gives −3, 17, 67, 97, while 10(y − 3) gives −30, −10, 40, 70. Their difference is always 27, so these two never agree.
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. Worked machine: 2a − b. For (6, 4), the output is 2 × 6 − 4 = 8.

First new machine: a + b − 2. The checks are 5 + 2 − 2 = 5; 8 + 1 − 2 = 7; 9 + 11 − 2 = 18; 10 + 10 − 2 = 18.

Second new machine: ab + 1. The checks are 4 × 1 + 1 = 5; 6 × 0 + 1 = 1; 3 × 2 + 1 = 7. For (10, 3), the missing output is 31.

Make your own: choose a + 2b. Inputs (1, 2), (3, 4) and (5, 1) give 5, 11 and 7. Ask a classmate to find a simple rule fitting every example.
The first machine follows a + b − 2. The second follows ab + 1, so inputs 10 and 3 give 31.
The first machine follows a + b − 2. The second follows ab + 1, so inputs 10 and 3 give 31.
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.
The five entries are a − 7, a − 1, a, a + 1 and a + 7. Their total is 5a.
The five entries are a − 7, a − 1, a, a + 1 and a + 7. Their total is 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.
The triangle chain has 3, 5, 7 and 9 sticks. At step y, y horizontal and y + 1 diagonal sticks total 2y + 1.
The triangle chain has 3, 5, 7 and 9 sticks. At step y, y horizontal and y + 1 diagonal sticks total 2y + 1.
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 3 multiplied by 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 use substitution as an error check. Agreement at one value does not prove an identity; justify equality by valid algebraic steps. 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.

Textbook reference: NCERT Ganita Prakash, Chapter 4 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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