Skip to content

NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers (PDF) – 2027

NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.

Chapter 4 of Ganita Prakash Part I introduces letter-numbers — letters such as a, n or w that stand for numbers. With them we write short algebraic expressions and formulas that capture a whole pattern in one line, and we learn to simplify them.

NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers PDF

All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.

PDFChapterDownload
Solutions PDFExpressions Using Letter NumbersView PDF

NCERT Solutions for Class 7 Maths Chapter 4 Expressions Using Letter Numbers – Questions and Answers

All 107 questions of the chapter (textbook pages 81 – 105) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.

In-text Questions — Page 81Section 4.1 The Notion of Letter-Numbers

Q1Example 1: Shabnam is 3 years older than Aftab. When Aftab’s age 10 years, Shabnam’s age will be 13 years. Now Aftab’s age is 18 years, what will Shabnam’s age be? _______
Answer

Shabnam is always 3 years older, so add 3 to Aftab’s age.

Shabnam’s age = Aftab’s age + 3
= 18 + 3
= 21 years
Q2Given Aftab’s age, how will you find out Shabnam’s age?
Answer

Add 3 to Aftab’s age.

Shabnam’s age = Aftab’s age + 3
Aftab’s ageShabnam’s age
44 + 3 = 7
1010 + 3 = 13
2323 + 3 = 26
Q3Can we write this as an expression?
Answer

Yes. Give Aftab’s age a short name and the rule becomes one line.

Let a = Aftab’s age, s = Shabnam’s age
Shabnam’s age = Aftab’s age + 3
s = a + 3
Q4If a is 23 (Aftab’s age in years), then what is Shabnam’s age?
Answer

Replace a by 23 in the expression a + 3.

s = a + 3
= 23 + 3
= 26 years

In-text Questions — Page 82Section 4.1 The Notion of Letter-Numbers

Q1Given the age of Shabnam, write an expression to find Aftab’s age.
Aftab’s ageExpression for Shabnam’s age
44 + 3
1010 + 3
2323 + 3
?? + 3
aa + 3
Fig. 4.1 (page 81) — Aftab’s age and the matching expression for Shabnam’s age.
Answer

Aftab is 3 years younger, so subtract 3.

Aftab’s age = Shabnam’s age – 3
a = s – 3
Q2Use this expression to find Aftab’s age if Shabnam’s age is 20.
Answer

Replace s by 20.

a = s – 3
= 20 – 3
= 17 years
Q3Example 2: Parthiv is making matchstick patterns. He repeatedly places Ls next to each other. Each L has two matchsticks as shown in Figure 4.2. How many matchsticks are needed to make 5 Ls? 7 Ls? 45 Ls? Now, what is the relation between the number of Ls and the number of sticks?
Fig. 4.2, page 82 — one L, two Ls and three Ls made from matchsticks.
Answer

Every L uses exactly 2 matchsticks, so the count is always double the number of Ls.

5 Ls → 5 × 2 = 10 matchsticks
7 Ls → 7 × 2 = 14 matchsticks
45 Ls → 45 × 2 = 90 matchsticks

In words: Number of matchsticks = 2 × Number of L’s. Taking n for the number of Ls, the expression is

2 × n

In-text Questions — Page 83Section 4.1 The Notion of Letter-Numbers

Q1Example 3: Ketaki prepares and supplies coconut-jaggery laddus. The price of a coconut is ₹35 and the price of 1 kg jaggery is ₹60. How much should she pay if she buys 10 coconuts and 5 kg jaggery?
Answer

Find the two costs separately and add them.

Cost of 10 coconuts = 10 × ₹35 = ₹350
Cost of 5 kg jaggery = 5 × ₹60 = ₹300
Total cost = ₹350 + ₹300 = ₹650
Q2How much should she pay if she buys 8 coconuts and 9 kg jaggery?
Answer

Find the two costs and add them.

Cost of 8 coconuts = 8 × ₹35 = ₹280
Cost of 9 kg jaggery = 9 × ₹60 = ₹540
Total = ₹280 + ₹540 = ₹820
Q3Write an algebraic expression to find the total amount to be paid for a given number of coconuts and quantity of jaggery.
Answer

Let c be the number of coconuts and j the number of kilograms of jaggery.

Quantity neededRelationshipExpression
Cost of coconutsNumber of coconuts × 35c × 35
Cost of jaggeryNumber of kgs of jaggery × 60j × 60
Total amount = Cost of coconuts + Cost of jaggery
c × 35 + j × 60
Q4Use this expression (or formula) to find the total amount to be paid for 7 coconuts and 4 kg jaggery.
Answer

Put c = 7 and j = 4.

c × 35 + j × 60
= 7 × 35 + 4 × 60
= 245 + 240
= ₹485
Q5Example 4: The perimeter of a square is 4 times the length of its side, written as 4 × q. What is the perimeter of a square with sidelength 7 cm? Use the expression to find out.
Answer

Put q = 7 in 4 × q.

Perimeter = 4 × q
= 4 × 7
= 28 cm

Figure it Out — Pages 84–85Section 4.1 The Notion of Letter-Numbers

Q1Write 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
Answer

In each case the perimeter is the sidelength added as many times as there are sides. Let the sidelength be a units.

ShapeNumber of equal sidesFormula for perimeter
Equilateral triangle33a
Regular pentagon55a
Regular hexagon66a
(a) Perimeter = a + a + a = 3a units
(b) Perimeter = a + a + a + a + a = 5a units
(c) Perimeter = 6a units
Q2Munirathna 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.
Answer

Joining two pipes end to end adds their lengths.

Combined length = 20 + k
= (20 + k) metres
Q3What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5? Complete the following table:
No. of ₹100 notesNo. of ₹20 notesNo. of ₹5 notesExpression and total amount
356 
   6 × 100 + 4 × 20 + 3 × 5 = 695
84z 
xyz 
Answer

Each row follows the same rule: (number of ₹100 notes × 100) + (number of ₹20 notes × 20) + (number of ₹5 notes × 5).

No. of ₹100 notesNo. of ₹20 notesNo. of ₹5 notesExpression and total amount
3563 × 100 + 5 × 20 + 6 × 5 = 430
6436 × 100 + 4 × 20 + 3 × 5 = 695
84z8 × 100 + 4 × 20 + z × 5 = 880 + 5z
xyzx × 100 + y × 20 + z × 5 = 100x + 20y + 5z
Row 1: 300 + 100 + 30 = ₹430
Row 2: 600 + 80 + 15 = ₹695
Row 3: 800 + 80 + 5z = ₹(880 + 5z)
Row 4: ₹(100x + 20y + 5z)
Q4Venkatalakshmi 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 complete 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
Answer

The answer is (d) 10 + 8 × y.

Starting time = 10 seconds (only once)
Grinding time = 8 seconds for each kg = 8 × y
Total = 10 + 8y seconds
Q5Write 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
Answer

Let the number be n.

In wordsExpression
(a) 5 more than a numbern + 5
(b) 4 less than a numbern – 4
(c) 2 less than 13 times a number13n – 2
(d) 13 less than 2 times a number2n – 13
Q6Describe situations corresponding to the following algebraic expressions: (a) 8 × x + 3 × y (b) 15 × j – 2 × k
Answer

(a) 8 × x + 3 × y

A stationery shop sells a pen for ₹x and a notebook for ₹y. Abha buys 8 pens and 3 notebooks. Her bill is 8x + 3y rupees.

(b) 15 × j – 2 × k

A carpenter makes 15 stools every day and works for j days, so he makes 15j stools. On k of those days, 2 stools got damaged. The number of good stools left is 15j – 2k.

Q7In 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’.
November 2024MonTueWedThuFriSatSun123456789101112131415161718192021222324252627282930
w – 1w
The November 2024 calendar with a 2 × 3 grid of dates marked (left). The same grid to be filled in (right): the bottom-middle cell holds w.
Answer

In a calendar, moving one step right adds 1 and moving one step up subtracts 7. Starting from w in the bottom-middle cell:

w – 8w – 7w – 6
w – 1ww + 1
Left of w = w – 1, right of w = w + 1
Directly above w = w – 7
Above-left = w – 7 – 1 = w – 8
Above-right = w – 7 + 1 = w – 6

In-text Questions — Pages 85–86Section 4.2 Revisiting Arithmetic Expressions

Q1Let us revise these concepts and find the values of the following expressions: 1. 23 – 10 × 2 2. 83 + 28 – 13 + 32 3. 34 – 14 + 20 4. 42 + 15 – (8 – 7) 5. 68 – (18 + 13) 6. 7 × 4 + 9 × 6 7. 20 + 8 × (16 – 6)
Answer

Write each expression as a sum of terms, turn each term into a number, then add.

1. 23 – 10 × 2 = 23 + (–10 × 2) = 23 + (–20) = 3
2. 83 + 28 – 13 + 32 = (83 – 13) + (28 + 32) = 70 + 60 = 130
3. 34 – 14 + 20 = (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 + 8 × 10 = 20 + 80 = 100
Q2Find an algebraic expression to get the nth term of this sequence: 4, 8, 12, 16, 20, 24, 28, …
Answer

These are the multiples of 4 in increasing order, so the term at position n is 4 times n.

3rd term = 4 × 3 = 12
29th term = 4 × 29 = 116
nth term = 4 × n = 4n
Q3Find the value of the expression 7k when k = 4. Find the value that the expression 5m + 3 takes when m = 2.
Answer

Remember 7k means 7 × k and 5m means 5 × m.

7k when k = 4 → 7 × 4 = 28
5m + 3 when m = 2 → 5 × 2 + 3 = 10 + 3 = 13

Mind the Mistake, Mend the Mistake — Page 87Section 4.3 Omission of the Multiplication Symbol

Try This
Q1If a = – 4, then 10 – a = 6.
Answer

Mistake. The negative sign of a was ignored.

10 – a = 10 – (–4)
= 10 + 4
= 14
Q2If d = 6, then 3d = 36.
Answer

Mistake. 3d means 3 × d. The 3 and the 6 were written side by side as “36”.

3d = 3 × 6 = 18
Q3If s = 7, then 3s – 2 = 15.
Answer

Mistake. The 2 was subtracted before multiplying.

3s – 2 = 3 × 7 – 2
= 21 – 2
= 19
Q4If r = 8, then 2r + 1 = 29.
Answer

Mistake. The 2 and the 8 were written side by side as 28.

2r + 1 = 2 × 8 + 1
= 16 + 1
= 17
Q5If j = 5, then 2j = 10.
Answer

No mistake. This one is correct.

2j = 2 × 5 = 10
Q6If m = –6, then 3 (m + 1) = 19.
Answer

Mistake. Only m was multiplied by 3 (not the 1), and the minus sign of m was dropped: 3 × 6 + 1 = 19.

3 (m + 1) = 3 × (–6 + 1)
= 3 × (–5)
= –15
Q7If f = 3, g = 1 then 2f – 2g = 2.
Answer

Mistake. 2f and 2g were read as the two-digit numbers 23 and 21, and 23 – 21 = 2.

2f – 2g = 2 × 3 – 2 × 1
= 6 – 2
= 4
Q8If t = 4, b = 3 then 2t + b = 24.
Answer

Mistake. The + was treated as a ×.

2t + b = 2 × 4 + 3
= 8 + 3
= 11
Q9If h = 5, n = 6 then h – (3 – n) = 4.
Answer

Mistake. The bracket was handled wrongly.

h – (3 – n) = 5 – (3 – 6)
= 5 – (–3)
= 5 + 3
= 8

In-text Questions — Pages 88–90Section 4.4 Simplification of Algebraic Expressions

Math Talk
Q1Example 5: Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days. The money earned by selling pencils on Day 1 is 5c. Similarly, the money earned by selling pencils on Day 2 is _____, and Day 3 is ______.
Day 1Day 2Day 3
Pencils (Price ‘c’)5310
Erasers (Price ‘d’)461
Answer

Multiply the number sold that day by the price per pencil.

Day 2: 3 pencils → 3c
Day 3: 10 pencils → 10c
Day 1Day 2Day 3
Pencils (price c)5310
Money from pencils5c3c10c
Total money earned in the three days
= (5c + 3c + 10c) + (4d + 6d + d)
= 18c + 11d (pencils 18c, erasers 11d)
Q2If c = ₹50, find the total amount earned by the sale of pencils.
Day 1Day 2Day 3
Pencils (Price ‘c’)5310
Erasers (Price ‘d’)461
Answer

The total money from pencils is 5c + 3c + 10c, which simplifies to 18c.

5c + 3c + 10c = (5 + 3 + 10) × c = 18c
18c = 18 × ₹50
= ₹900
Q3Write the expression for the total money earned by selling erasers. Then, simplify the expression.
Day 1Day 2Day 3
Pencils (Price ‘c’)5310
Erasers (Price ‘d’)461
Answer

Erasers sold: 4 on Day 1, 6 on Day 2 and 1 on Day 3, each priced d.

Total from erasers = 4d + 6d + 1d
= (4 + 6 + 1) × d
= 11d
Q4Can the expression 18c + 11d be simplified further?
Answer

No. It is already in its simplest form.

Q5Check that both expressions take the same value when c is replaced by different numbers. (5c + 3c + 10c and 18c)
Answer

Try three different values of c.

c5c + 3c + 10c18c
210 + 6 + 20 = 3636
525 + 15 + 50 = 9090
50250 + 150 + 500 = 900900
Q6Example 6: A big rectangle is split into two smaller rectangles as shown. Write an expression describing the area of the bigger rectangle.
v43
Page 89 — a rectangle of breadth v, split into two parts of lengths 4 and 3.
Answer

Area of the bigger rectangle = 7v square units.

Way 1 — use its sides: v × (4 + 3) = 7v
Way 2 — add the two smaller areas: 4v + 3v = 7v
So 4v + 3v = 7v
Q7As earlier, a big rectangle is split into two smaller rectangles as shown below. Write an expression to find the area of the rectangle AEFD.
n4DFCA12EB
Rectangle ABCD, page 90 — length AB = 12, breadth AD = n. The dashed line FE splits it into AEFD and EBCF, with FC = 4.
Answer

AEFD is what is left of ABCD after the strip EBCF is removed.

Method 1 — use its own sides:
AE = 12 – 4 = 8, AD = n
Area of AEFD = 8 × n = 8n

Method 2 — subtract:
Area of ABCD = 12n, Area of EBCF = 4n
Area of AEFD = 12n – 4n = 8n
Q8Example 7: A shop rents out chairs (₹40 each) and tables (₹75 each); on return it pays back ₹6 per chair and ₹10 per table. For x chairs and y tables, describe the procedure to get these amounts.
Answer

Amount paid at the beginning: multiply each rate by how many pieces are taken and add.

Chairs: x × ₹40 = 40x
Tables: y × ₹75 = 75y
Paid at the start = 40x + 75y

Amount returned: the same procedure with the return rates.

Chairs: x × ₹6 = 6x
Tables: y × ₹10 = 10y
Returned = 6x + 10y
So the total amount actually paid = (40x + 75y) – (6x + 10y)
Q9Can we simplify this expression? If yes, how? If not, why not? [(40x + 75y) – (6x + 10y)]
Answer

Yes. Open the brackets, then collect like terms.

(40x + 75y) – (6x + 10y)
= 40x + 75y – 6x – 10y
= 40x + (–6x) + 75y + (–10y) (swapping and grouping)
= (40 – 6)x + (75 – 10)y
= 34x + 65y

In-text Questions — Pages 91–93Section 4.4 Simplification of Algebraic Expressions

Math Talk
Q1Could we have written the initial expression as (40x + 75y) + (– 6x – 10y)?
Answer

Yes. Subtracting an expression is the same as adding its opposite.

– (6x + 10y) = – 6x – 10y
So (40x + 75y) – (6x + 10y) = (40x + 75y) + (– 6x – 10y)
= 34x + 65y either way
Q2Example 8: Charu’s scores in three rounds of a quiz are 7p – 3q, 8p – 4q, and 6p – 2q, where p is the score for a correct answer and q the penalty for an incorrect answer. What do each of the expressions mean?
Answer
RoundExpressionWhat it says
17p – 3q7 correct answers and 3 wrong answers
28p – 4q8 correct answers and 4 wrong answers
36p – 2q6 correct answers and 2 wrong answers
Q3If the score for a correct answer is 4 (p = 4) and the penalty for a wrong answer is 1 (q = 1), find Charu’s score in the first round.
Answer
7p – 3q = 7 × 4 – 3 × 1
= 7 × 4 + (–3 × 1)
= 28 + (–3)
= 25
Q4What are her scores in the second and third rounds?
Answer

Use p = 4 and q = 1 in each expression.

Round 2: 8p – 4q = 8 × 4 – 4 × 1 = 32 – 4 = 28
Round 3: 6p – 2q = 6 × 4 – 2 × 1 = 24 – 2 = 22
Q5What if there is no penalty? What will be the value of q in that situation?
Answer

No penalty means nothing is taken away for a wrong answer.

q = 0
Round 1 score becomes 7p – 3 × 0 = 7p = 28
Total becomes 21p – 9 × 0 = 21p = 84
Q6What is her final score after the three rounds?
Answer

Add the three round scores and simplify.

(7p – 3q) + (8p – 4q) + (6p – 2q)
= 7p + (–3q) + 8p + (–4q) + 6p + (–2q)
= 7p + 8p + 6p + (–3q) + (–4q) + (–2q) (swapping and grouping)
= (7 + 8 + 6)p – (3 + 4 + 2)q
= 21p – 9q
Q7Give some possible scores for Krishita in the three rounds so that they add up to give 23p – 7q.
Answer

Split 23 into three parts and 7 into three parts.

One possibility:
Round 1: 8p – 3q
Round 2: 7p – 2q
Round 3: 8p – 2q
Sum = (8 + 7 + 8)p – (3 + 2 + 2)q = 23p – 7q ✔
Q8Can we say who scored more? Can you explain why? (Charu: 21p – 9q, Krishita: 23p – 7q)
Answer

Krishita scored more.

Krishita has 23p, Charu has 21p → Krishita gains 2p more
Krishita loses 7q, Charu loses 9q → Krishita loses 2q less
Q9How much more has Krishita scored than Charu? Simplify this expression further: 23p – 7q – (21p – 9q)
Answer
23p – 7q – (21p – 9q)
= 23p – 7q – 21p + 9q
= 23p – 21p + (–7q) + 9q
= (23 – 21)p + (9 – 7)q
= 2p + 2q
Q10Example 9: Simplify the expression 4 (x + y) – y
Answer
4 (x + y) – y
= 4x + 4y – y (distributive property)
= 4x + 4y + (– y)
= 4x + (4 – 1)y
= 4x + 3y
Q11Example 10: Are the expressions 5u and 5 + u equal to each other?
Answer

No. They describe two different operations.

5u means 5 times the number u
5 + u means 5 more than the number u
Q12Fill the blanks below by replacing the letter-numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take. (u = 11, 2, 8 and 5)
105uu = 11u = 2u = 8u = 575 + uu = 11u = 2u = 8u = 5
Page 92 — two diagrams. The middle box holds the expression; each arrow carries a value of u and points to a box for the value of the expression. One box in each diagram is filled in as the example.
Answer
u5u5 + u
115516
2107
84013
52510
5 × 11 = 55 but 5 + 11 = 16
5 × 2 = 10 but 5 + 2 = 7
5 × 8 = 40 but 5 + 8 = 13
5 × 5 = 25 but 5 + 5 = 10

The two columns never match, so 5u ≠ 5 + u.

Q13Are the expressions 10y – 3 and 10(y – 3) equal? Let us compare the values that these expressions take for different values of y (y = 2, 0, 7, 10). After filling in the two diagrams, do you think the two expressions are equal?
1710y – 3y = 2y = 0y = 10y = 7–1010 (y – 3)y = 2y = 0y = 10y = 7
Page 93 — two diagrams. The middle box holds the expression; each arrow carries a value of y and points to a box for the value of the expression. One box in each diagram is filled in as the example.
Answer

10y – 3 means ‘3 less than 10 times y’, while 10(y – 3) means ‘10 times (3 less than y)’.

y10y – 310 (y – 3)
217–10
0–3–30
76740
109770
y = 2: 10 × 2 – 3 = 17, but 10 × (2 – 3) = 10 × (–1) = –10
y = 0: 0 – 3 = –3, but 10 × (–3) = –30
y = 7: 70 – 3 = 67, but 10 × 4 = 40
y = 10: 100 – 3 = 97, but 10 × 7 = 70

No, the two expressions are not equal.

Q14Example 11: What is the sum of the numbers in the picture (unknown values are denoted by letter-numbers)?
33333333rsrs
The picture for Example 11, page 93 — four 3s, then r and s, then r and s, then four 3s.
Answer

The sum is 2r + 2s + 24. Three ways of adding:

1. Row-wise: (4 × 3) + (r + s) + (r + s) + (4 × 3)
= 12 + 12 + 2r + 2s = 2r + 2s + 24

2. Like terms together: (8 × 3) + (r + r) + (s + s)
= 24 + 2r + 2s = 2r + 2s + 24

3. Upper half, doubled: 2 × (4 × 3 + r + s)
= 2 × (12 + r + s) = 24 + 2r + 2s = 2r + 2s + 24

Figure it Out — Pages 93–94Section 4.4 Simplification of Algebraic Expressions

Math Talk
Q1Add 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.
5y–6xx25y
(i)
2p3q–233q2p3–22p3q3q2p
(ii)
–5g5k5k–5g5k5k5k5k5k5k5k5k–5g5k5k–5g
(iii)
Pictures (i), (ii) and (iii), page 94 — the numbers in each are to be added.
Answer

Picture (i) — two rows: 5y, –6, x and x, 2, 5y.

Way 1 (row-wise): (5y – 6 + x) + (x + 2 + 5y)
Way 2 (like terms): (5y + 5y) + (x + x) + (–6 + 2)
= 10y + 2x + (–4)
= 2x + 10y – 4

Picture (ii) — 2p and 3q appear 4 times each; the number tiles are –2, 3, 3, –2.

Way 1 (like terms): 4 × 2p + 4 × 3q + (–2 + 3 + 3 – 2)
= 8p + 12q + 2
Way 2 (column-wise): (2p + 3q) + (3q + 2p) + (–2 + 3 + 2p + 3q) + (3 – 2 + 3q + 2p)
= 8p + 12q + 2

Picture (iii) — the four corners are –5g and every other circle is 5k (12 of them).

Way 1 (like terms): 4 × (–5g) + 12 × 5k
= –20g + 60k
Way 2 (columns): first column (–5g + 5k + 5k – 5g) = –10g + 10k
middle two columns = 8 × 5k = 40k
last column = –10g + 10k
Total = 60k – 20g
Q2Simplify each of the following expressions: (a) p + p + p + p, p + p + p + q, (b) p + q + p – q, (c) p – q + p – q, (d) p + q – p + q, (e) p + q – (p + q), (f) p – q – p – q, (g) 2d – d – d – d, (h) 2d – d – d – c, (i) 2d – d – (d – c), (j) 2d – (d – d) – c, (k) 2d – d – c – c
Answer
ExpressionWorkingSimplest form
(a) p + p + p + p4 lots of p4p
(a) p + p + p + q3 lots of p, one q3p + q
(b) p + q + p – q(p + p) + (q – q)2p
(c) p – q + p – q(p + p) + (–q – q)2p – 2q
(d) p + q – p + q(p – p) + (q + q)2q
(e) p + q – (p + q)p + q – p – q0
(f) p – q – p – q(p – p) + (–q – q)–2q
(g) 2d – d – d – d(2 – 1 – 1 – 1)d–d
(h) 2d – d – d – c(2 – 1 – 1)d – c = 0 – c–c
(i) 2d – d – (d – c)2d – d – d + cc
(j) 2d – (d – d) – c2d – 0 – c2d – c
(k) 2d – d – c – c(2 – 1)d – 2cd – 2c

Mind the Mistake, Mend the Mistake — Pages 94–95Section 4.4 Simplification of Algebraic Expressions

Q13a + 2b → Simplest Form given: 5
Answer

Mistake. Unlike terms were added.

Correct simplest form: 3a + 2b
Q23b – 2b – b → Simplest Form given: 0
Answer

No mistake.

3b – 2b – b = (3 – 2 – 1)b = 0 × b = 0
Q36 (p + 2) → Simplest Form given: 6p + 8
Answer

Mistake. The 6 multiplied p, but it was added to the 2 (6 + 2 = 8) instead of multiplying it.

6 (p + 2) = 6 × p + 6 × 2
= 6p + 12
Q4(4x + 3y) – (3x + 4y) → Simplest Form given: x + y
Answer

Mistake. 3y – 4y was worked out as 4y – 3y = y; the minus sign before the bracket was not applied to 4y.

(4x + 3y) – (3x + 4y)
= 4x + 3y – 3x – 4y
= (4 – 3)x + (3 – 4)y
= x – y
Q55 – (2 – 6z) → Simplest Form given: 3 – 6z
Answer

Mistake. Only the first term inside the bracket had its sign changed.

5 – (2 – 6z) = 5 – 2 + 6z
= 3 + 6z
Q62 + (x + 3) → Simplest Form given: 2x – 6
Answer

Mistake. The + before the bracket was treated as ×, and the sign of 3 was also changed: 2(x – 3) = 2x – 6.

2 + (x + 3) = 2 + x + 3
= x + 5
Q72y + (3y – 6) → Simplest Form given: – y + 6
Answer

Mistake. The signs inside were flipped even though the bracket has a plus in front.

2y + (3y – 6) = 2y + 3y – 6
= 5y – 6
Q87p – p + 5q – 2q → Simplest Form given: 7p + 3q
Answer

Mistake. The – p was left out.

7p – p + 5q – 2q
= (7 – 1)p + (5 – 2)q
= 6p + 3q
Q95 (2w + 3x + 4w) → Simplest Form given: 10w + 15x + 20w
Answer

Mistake. The distribution was done correctly, but like terms were not collected — so it is not the simplest form.

5 (2w + 3x + 4w) = 5 (6w + 3x)
= 30w + 15x
Q103j + 6k + 9h + 12 → Simplest Form given: 3 (j + 2k + 3h + 4)
Answer

Mistake. The answer is correct as an equal expression, but it is not in simplest form because the bracket is still there.

Correct simplest form: 3j + 6k + 9h + 12
Q114 (2r + 3s + 5) → Simplest Form given: – 20 – 8r – 12s
Answer

Mistake. Every term was given a minus sign, though there is no minus anywhere.

4 (2r + 3s + 5) = 4 × 2r + 4 × 3s + 4 × 5
= 8r + 12s + 20
Q12Take a look at all the corrected simplest forms (i.e. brackets are removed, like terms are added, and terms with only numbers are also added). Is there any relation between the number of terms and the number of letter-numbers these expressions have?
Answer
No.Correct simplest formNumber of termsNumber of letter-numbers
13a + 2b22
2010
36p + 1221
4x – y22
53 + 6z21
6x + 521
75y – 621
86p + 3q22
930w + 15x22
103j + 6k + 9h + 1243
118r + 12s + 2032

Yes. In every row, the number of letter-numbers is never more than the number of terms:

Number of letter-numbers ≤ Number of terms

Formula Detective — Pages 95–96Section 4.5 Pick Patterns and Reveal Relationships

Q1Look at the picture given. In each case, the number machine takes in the 2 numbers at the top of the ‘Y’ as inputs, performs some operations and produces the result at the bottom. Find out the formula of this number machine.
528expression:
8115expression:
9117expression:
101010expression:
64expression:
The five number machines, page 95 — each takes the two numbers at the top of the ‘Y’ and produces the result at the bottom.
Answer

The formula is “two times the first number minus the second number”, that is 2a – b.

Inputs (a, b)ExpressionOutput
5, 22 × 5 – 28
8, 12 × 8 – 115
9, 112 × 9 – 117
10, 102 × 10 – 1010
6, 42 × 6 – 48
Q2Find the formulas of the number machines below and write the expression for each set of inputs.
525expression:
817expression:
91118expression:
101018expression:
abexpression:
415expression:
601expression:
327expression:
103expression:
abexpression:
Page 96 — two sets of number machines: the blue set (top row) and the purple set (bottom row). Each machine takes the two numbers at the top of the ‘Y’ and gives the result in the box below; empty boxes are to be found.
Answer

Machine 1 — “add the two numbers, then subtract 2”, that is a + b – 2.

InputsExpressionOutput
5, 25 + 2 – 25
8, 18 + 1 – 27
9, 119 + 11 – 218
10, 1010 + 10 – 218
a, ba + b – 2a + b – 2

Machine 2 — “multiply the two numbers, then add 1”, that is ab + 1.

InputsExpressionOutput
4, 14 × 1 + 15
6, 06 × 0 + 11
3, 23 × 2 + 17
10, 310 × 3 + 131
a, ba × b + 1ab + 1
Q3Now, make a formula on your own. Write a few number machines as examples using that formula. Challenge your classmates to figure it out!
Answer

Here is one you can use. Formula: 3a – 2b — “three times the first number minus twice the second”.

Inputs (a, b)WorkingOutput
4, 13 × 4 – 2 × 110
5, 53 × 5 – 2 × 55
2, 33 × 2 – 2 × 30
7, 23 × 7 – 2 × 217
1, 43 × 1 – 2 × 4–5

In-text Questions — Pages 96–97Algebraic Expressions to Describe Patterns

Q1Example 12: Somjit noticed a repeating pattern along the border of a saree (A B C D E F …). Somjit wonders if there is a way to describe all the positions where the (i) Design A occurs, (ii) Design B occurs, and (iii) Design C occurs.
A1B2C3D4E5F6…
Page 96 — the repeating border of Somjit’s saree, positions 1 to 6. The six motifs are redrawn here in simplified form; the order and the repeats are as printed.
Answer

The border repeats in blocks of three: A, B, C, A, B, C, …

DesignPositions where it occursFormula for the nth time
A1, 4, 7, 10, 13, …3n – 2
B2, 5, 8, 11, 14, …3n – 1
C3, 6, 9, 12, 15, …3n
Q2Where would design C appear for the nth time?
A1B2C3D4E5F6…
Page 96 — the repeating border of Somjit’s saree, positions 1 to 6. The six motifs are redrawn here in simplified form; the order and the repeats are as printed.
Answer

Design C first appears at position 3, then 6, then 9 — the multiples of 3.

nth occurrence of Design C is at position 3n
Q3Similarly, find the formula that gives the position where the other Designs appear for the nth time.
A1B2C3D4E5F6…
Page 96 — the repeating border of Somjit’s saree, positions 1 to 6. The six motifs are redrawn here in simplified form; the order and the repeats are as printed.
Answer

B occurs at 2, 5, 8, 11, 14, … — always one less than a multiple of 3.

nth occurrence of Design B: 3n – 1
nth occurrence of Design A: 3n – 2
Q4Given a position number can we find out the design that appears there? Which Design appears at Position 122?
A1B2C3D4E5F6…
Page 96 — the repeating border of Somjit’s saree, positions 1 to 6. The six motifs are redrawn here in simplified form; the order and the repeats are as printed.
Answer

Yes, and the design at position 122 is Design B.

122 = 3 × 41 – 1
So 122 is one less than 123, a multiple of 3
Positions of the form 3n – 1 carry Design B
Q5Can the remainder obtained by dividing the position number by 3 be used for this? Observe the table below.
Position no.Quotient on division by 3Remainder
99330
122402
148491
Answer

Yes. The remainder alone decides the design.

Remainder on dividing by 3Form of the positionDesign
03nC
13n + 1, i.e. 3(n+1) – 2A
23n + 2, i.e. 3(n+1) – 1B
Q6Use this to find what design appears at positions 99, 122, and 148.
A1B2C3D4E5F6…
Page 96 — the repeating border of Somjit’s saree, positions 1 to 6. The six motifs are redrawn here in simplified form; the order and the repeats are as printed.
Position no.Quotient on division by 3Remainder
99330
122402
148491
Answer
PositionDivision by 3RemainderDesign
9999 = 3 × 33 + 00C
122122 = 3 × 40 + 22B
148148 = 3 × 49 + 11A
99 = 3 × 33 → Design C
122 = 3 × 41 – 1 → Design B
148 = 3 × 50 – 2 → Design A

In-text Questions — Pages 97–99Patterns in a Calendar

Math Talk
Q1Let us take the marked 2 × 2 square, and consider the numbers lying on the diagonals; 12 and 20; 13 and 19. Find their sums; 12 + 20, 13 + 19. What do you observe?
1213
1920
The marked 2 × 2 square of the November 2024 calendar, page 98.
Answer

Both diagonals add up to the same number.

12 + 20 = 32
13 + 19 = 32
The two diagonal sums are equal.
Q2Will the diagonal sums be equal in every 2 × 2 square in this endless grid? How can we be sure?
November 2024MonTueWedThuFriSatSun1234567891011121314151617181920212223242526272829303132333435363738394041424344⋮⋮⋮⋮⋮⋮⋮
Page 98 — the November 2024 calendar with its rows continued beyond 30, without end.
Answer

Yes — but checking examples can never make us sure, because there are unlimited 2 × 2 squares.

Q3Given that we know the top left number, how do we find the other numbers in this 2 × 2 square?
November 2024MonTueWedThuFriSatSun1234567891011121314151617181920212223242526272829303132333435363738394041424344⋮⋮⋮⋮⋮⋮⋮
Page 98 — the November 2024 calendar with its rows continued beyond 30, without end.
a???
Page 98 — a 2 × 2 square from this calendar; its top left number is called a.
Answer

Describe them in words first, then in symbols.

  • the number to the right of a is 1 more than it
  • the number below a is 7 more than it
  • the number diagonal to a is 8 more than it
aa + 1
a + 7a + 8
Diagonal 1: a + (a + 8) = a + a + 8 = 2a + 8
Diagonal 2: (a + 1) + (a + 7) = a + a + 1 + 7 = 2a + 8
Both diagonal sums are equal — for every value of a.
Q4Verify this expression for diagonal sums by considering any 2 × 2 square and taking its top left number to be ‘a’. [This expression: the diagonal sums a + (a + 8) and (a + 1) + (a + 7) both simplify to 2a + 8.]
November 2024MonTueWedThuFriSatSun1234567891011121314151617181920212223242526272829303132333435363738394041424344⋮⋮⋮⋮⋮⋮⋮
Page 98 — the November 2024 calendar with its rows continued beyond 30, without end.
Answer

Take the square with top-left number 25 (so a = 25): the square is 25, 26 over 32, 33.

Diagonal 1: 25 + 33 = 58
Diagonal 2: 26 + 32 = 58
Formula: 2a + 8 = 2 × 25 + 8 = 58 ✔

Now take a = 40: the square is 40, 41 over 47, 48.

40 + 48 = 88 and 41 + 47 = 88
2a + 8 = 80 + 8 = 88 ✔
Q5Consider a set of numbers from the calendar (having endless rows) forming the plus shape shown below. Find the sum of all the numbers. Compare it with the number in the centre: 15. Repeat this for another set of numbers that forms this shape. What do you observe?
814151622
Page 99 — the plus shape of five numbers taken from the endless calendar.
Answer
8 + 14 + 15 + 16 + 22
= (8 + 22) + (14 + 16) + 15
= 30 + 30 + 15
= 75
And 75 = 5 × 15, five times the centre.

Another plus shape, centred at 26: the numbers are 19 / 25, 26, 27 / 33.

19 + 25 + 26 + 27 + 33 = 130 = 5 × 26 ✔

Observation: the total is always 5 times the number in the centre.

Q6Will this always happen? How do you show this? [Hint: Consider a general set of numbers that forms this shape. Take the number at the centre to be ‘a’. Express the other numbers in terms of ‘a’.]
814151622
Page 99 — the plus shape of five numbers taken from the endless calendar.
Answer

Yes, always. Let the centre be a.

a – 7
a – 1aa + 1
a + 7
Sum = (a – 7) + (a – 1) + a + (a + 1) + (a + 7)
= a + a + a + a + a + (–7 + 7) + (–1 + 1)
= 5a + 0 + 0
= 5a
Q7Find other shapes for which the sum of the numbers within the figure is always a multiple of one of the numbers.
814151622
Page 99 — the plus shape of five numbers taken from the endless calendar.
Answer

Any shape that is symmetric about one cell works. Taking the centre as a:

ShapeNumbers in terms of aSum
Row of 3 (horizontal)a – 1, a, a + 13a
Column of 3 (vertical)a – 7, a, a + 73a
Diagonal of 3 (↘)a – 8, a, a + 83a
The X of 5 (four corners + centre)a – 8, a – 6, a, a + 6, a + 85a
Full 3 × 3 blocknine numbers around a9a
3 × 3 block: (a – 8) + (a – 7) + (a – 6) + (a – 1) + a + (a + 1) + (a + 6) + (a + 7) + (a + 8)
= 9a

In-text Questions — Pages 100–101Matchstick Patterns

Q1Look at the picture below. It is a pattern using matchsticks. Can you identify what the pattern is?
1
2
3
4
The matchstick pattern, page 100 — Steps 1, 2, 3 and 4.
Answer

Each step adds one more triangle to the row, using only 2 extra matchsticks each time, because the new triangle shares one side with the previous one.

Step number123456
Triangles123456
No. of matchsticks35791113
Q2Can you tell how many matchsticks there will be in the next step, Step 5?
Answer

Step 4 uses 9 matchsticks, and every step adds 2.

Step 5 = 9 + 2 = 11 matchsticks
Q3How many matchsticks will there be in Step 33, Step 84, and Step 108? Of course, we can draw and count, but is there a quicker way to find the answers using the pattern present here?
1
2
3
4
Page 100 — the matchstick pattern, Steps 1 to 4.
Answer

Yes — count how many 2s are added. Step 1 starts with 3, and each later step adds one more 2.

Step 33 = 3 + 2 × 32 = 3 + 64 = 67
Step 84 = 3 + 2 × 83 = 3 + 166 = 169
Step 108 = 3 + 2 × 107 = 3 + 214 = 217
Q4The number of matchsticks needed to make 33 triangles (Step 33) is _________. Similarly, find the number of matchsticks needed for Step 84 and Step 108.
Answer

Step 33 needs 67 matchsticks.

StepNumber of 2s addedWorkingMatchsticks
33323 + 2 + 2 + … (32 twos)67
84833 + 2 × 83169
1081073 + 2 × 107217
Q5What could be an expression describing the rule/formula to find out the number of matchsticks at any step? For step y, what is the expression?
Answer

Step y has one 3 and (y – 1) twos.

Number of matchsticks in Step y = 3 + 2 × (y – 1)

Noticing that the first step also contains a 2 (since 3 = 1 + 2) gives a second form:

2y + 1
Q6Does the above expression also give the number of matchsticks at each step correctly? Are these expressions the same? [The two expressions for Step y: 3 + 2 × (y – 1) and 2y + 1]
1
2
3
4
The matchstick pattern, page 100 — Steps 1, 2, 3 and 4.
Answer

Yes to both. Simplify the first expression and it turns into the second.

3 + 2 × (y – 1) = 3 + 2y – 2
= 2y + (3 – 2)
= 2y + 1
y3 + 2(y – 1)2y + 1
133
51111
336767
Q7Matchsticks are placed in two orientations — (a) horizontal ones at the top and bottom, and (b) the ones placed diagonally in the middle. For example, in step 2 there are 2 matchsticks placed horizontally and 3 matchsticks placed diagonally. What are these numbers in Step 3 and Step 4?
Answer
StepHorizontalDiagonalTotal
1123
2235
3347
4459
Step 3: 3 horizontal and 4 diagonal (3 + 4 = 7 ✔)
Step 4: 4 horizontal and 5 diagonal (4 + 5 = 9 ✔)
Q8How does the number of matchsticks change in each orientation as the steps increase? Write an expression for the number of matchsticks at Step ‘y’ in each orientation. Do the two expressions add up to 2y + 1?
Answer

Horizontal sticks go 1, 2, 3, 4, … and diagonal sticks go 2, 3, 4, 5, … Each set grows by exactly 1 per step.

Horizontal at Step y = y
Diagonal at Step y = y + 1

Sum = y + (y + 1) = 2y + 1 ✔

Figure it Out — Pages 102–105Section 4.5 Pick Patterns and Reveal Relationships

Try ThisMath Talk
Q1One 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
Answer

The correct expression is (a) 30x + 20y.

Money from jowar roti = 30 × x = 30x
Money from pulao = 20 × y = 20y
Total = 30x + 20y
Q2Pushpita 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)
Answer

The correct expression is (a) p + q + r.

Total customers = p + q + r
One flag per customer
Flags given away = p + q + r
Q3A 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?
Answer

(a) In one full day-and-night the snail gains (u – d) cm. This repeats 10 times.

Distance from start = 10 × (u – d)
= 10(u – d) cm = 10u – 10d cm

(b) If d > u, then u – d is negative.

10(u – d) is negative → the snail ends up below its starting point

It slips down more each night than it climbs during the day, so it keeps losing ground and will never reach the top.

Q4Radha 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?
Answer

Work out each week separately — a week has 7 days.

WeekDaily distanceDistance that week
15 km7 × 5 = 35
2(5 + z) km7(5 + z) = 35 + 7z
3(5 + 2z) km7(5 + 2z) = 35 + 14z
Total = 35 + (35 + 7z) + (35 + 14z)
= (35 + 35 + 35) + (7z + 14z)
= 105 + 21z km
Q5In 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.
–5w – 3×3+3w + 5×44w + 20–8–4–4×33w – 6w + 2
The path diagram, page 102 — ovals hold expressions and boxes hold operations. Four paths lead out of w + 2; the blank ovals are to be filled in.
Answer

Follow each arrow away from the centre oval w + 2, applying one box at a time.

PathStepsResult
Top right (given)w + 2 → +3 → w + 5 → ×44w + 20
Top leftw + 2 → –5 → w – 3 → ×33w – 9
Bottom rightw + 2 → –4 → w – 2 → ×33w – 6
Bottom leftw + 2 → –8 → w – 6 → –4w – 10
Top left: (w – 3) × 3 = 3w – 9
Bottom right: (w + 2) – 4 = w – 2, and (w – 2) × 3 = 3w – 6 ✔
Bottom left: (w + 2) – 8 = w – 6, and (w – 6) – 4 = w – 10
Q6A 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]
Answer

Three stations lie between the two towns, so the journey is cut into 4 stretches, with 3 halts.

Yahapur Vahapur t t t t 2 2 2
Four travel stretches of t minutes each, with a 2-minute halt at each of the three middle stations.
(a) When t = 4: time = 4 × 4 + 3 × 2 = 16 + 6 = 22 minutes

(b) Total time = 4 × t + 3 × 2 = 4t + 6 minutes
Q7Simplify 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
Answer
(a) 3a + 8a – 7a = 4a; 9b – 4b = 5b; –6 + 16 = 10
= 4a + 5b + 10

(b) 3(3a – 3b) = 9a – 9b
9a – 8a = a; –9b – 4b = –13b
= a – 13b – 16

(c) 2(2x – 3) = 4x – 6
4x + 8x = 12x; –6 + 12 = 6
= 12x + 6

(d) 8x – (2x – 3) + 12 = 8x – 2x + 3 + 12
= 6x + 15

(e) 8h – (5 + 7h) + 9 = 8h – 5 – 7h + 9
= h + 4

(f) 4(6m – 3n) = 24m – 12n
24m – 3m = 21m; –12n – 8n = –20n; 23 – 18 = 5
= 21m – 20n + 5
Q8Add 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
Answer
(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
(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
Q9Subtract 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
Answer

“Subtract A from B” means work out B – A. Change the sign of every term of A, then add.

(a) (6a + 9b – 18) – (9a – 6b + 14)
= 6a + 9b – 18 – 9a + 6b – 14 = –3a + 15b – 32

(b) (7y – 10 + 3x) – (–15x + 13 – 9y)
= 7y – 10 + 3x + 15x – 13 + 9y = 18x + 16y – 23

(c) (11 – 10g + 3h) – (17g + 9 – 7h)
= 11 – 10g + 3h – 17g – 9 + 7h = –27g + 10h + 2

(d) 6a – (9b + 18) = 6a – 9b – 18
(6a – 9b – 18) – (9a – 6b + 14) = –3a – 3b – 32

(e) (–3y + 8 – 3x) – (10x + 2 + 10y)
= –3y + 8 – 3x – 10x – 2 – 10y = –13x – 13y + 6

(f) (7h – 8g + 20) – (8g + 4h – 10)
= 7h – 8g + 20 – 8g – 4h + 10 = 3h – 16g + 30
Q10Describe situations corresponding to the following algebraic expressions: (a) 8x + 3y (b) 15x – 2x
Answer

(a) 8x + 3y

A fruit seller sells mangoes at ₹x each and bananas at ₹y each. A customer buys 8 mangoes and 3 bananas, so the bill is 8x + 3y rupees.

(b) 15x – 2x

A shopkeeper has a packet of 15 pencils, each costing ₹x. Two of them are damaged and cannot be sold. The money received for the packet is 15x – 2x = 13x rupees.

Q11Imagine 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?
no fold
folded once
folded twice
The rope cut once, page 104 — unfolded, folded once, and folded twice. The red dashed line is the cut.
Answer

Each extra fold gives one extra piece.

Number of folds0123…10
Pieces after one cut2345…12
Folded 10 times and cut → 10 + 2 = 12 pieces
Folded r times and cut → r + 2 pieces
Q12Look 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?
Page 104 — the matchstick pattern: one, two and three squares in a row.
Answer

The first square needs 4 matchsticks; every square after that shares one side, so it needs only 3 more.

Squares1234…10
Matchsticks471013…31
10 squares = 4 + 3 × 9 = 4 + 27 = 31 matchsticks
w squares = 4 + 3 × (w – 1)
= 4 + 3w – 3
= 3w + 1 matchsticks
Q13Have 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.
12345………
The traffic-signal colour sequence, page 104 — positions 1 to 5, continuing in the same way.
Answer

The printed sequence is red, yellow, green, yellow, red, … so the block red – yellow – green – yellow repeats every 4 positions.

Position12345678
ColourRedYellowGreenYellowRedYellowGreenYellow
90 = 4 × 22 + 2 → remainder 2 → Yellow
190 = 4 × 47 + 2 → remainder 2 → Yellow
343 = 4 × 85 + 3 → remainder 3 → Green

Expressions for the positions:

ColourPositionsExpression
Red1, 5, 9, 13, …4n – 3
Green3, 7, 11, 15, …4n – 1
Yellow2, 4, 6, 8, … (every even position)2n
Q14Observe 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?
Step 1
Step 2
Step 3
Page 104 — the first three steps of the pattern.
Answer

Each step is an X with one centre square and four arms. Step 1 has arms of length 1, Step 2 arms of length 2, and so on.

StepCentreFour armsSquares
114 × 15
214 × 29
314 × 313
414 × 417
1014 × 1041
5014 × 50201
Number of squares in Step n = 4n + 1
Step 4 → 17, Step 10 → 41, Step 50 → 201

Counting vertices: every square has 4 corners, so

Number of vertices = 4 × (4n + 1)
= 16n + 4

(This counts 4 corners for every square. Neighbouring squares touch at one corner; if each shared corner point is counted only once, there are 4n fewer: 16n + 4 – 4n = 12n + 4.)

Q15Numbers 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.
123412345678910111213141516⋮⋮⋮⋮
Page 105 — the endless 4-column grid: column numbers 1 to 4 on top, and the rows go on downwards without end.
Answer

(a) Column expressions (with r the row number):

ColumnNumbersExpression
11, 5, 9, 13, …4(r – 1) + 1 = 4r – 3
22, 6, 10, 14, …4(r – 1) + 2 = 4r – 2
33, 7, 11, 15, …4(r – 1) + 3 = 4r – 1
44, 8, 12, 16, …4(r – 1) + 4 = 4r

(b) Divide by 4: the remainder is the column (remainder 0 means column 4).

(i) 124 = 4 × 31 → row 31, column 4
(ii) 147 = 4 × 36 + 3 → row 37, column 3
(iii) 201 = 4 × 50 + 1 → row 51, column 1

(c) The number in row r and column c is

4(r – 1) + c

(d) Multiples of 3. Their columns run 3, 2, 1, 4, 3, 2, 1, 4, … repeating.

Multiple of 33691215182124
Column32143214

Other patterns:

  • Every number in column 4 is a multiple of 4.
  • Even numbers appear only in columns 2 and 4; odd numbers only in columns 1 and 3.
  • Going down any column, the numbers rise by 4 each time.
  • Along a row the numbers rise by 1; along a downward diagonal they rise by 5.

Chapter 4 of 8 – Maths

Also for This Chapter

NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter Numbers – Overview

ClassClass 7
SubjectMaths
Chapter4 – Expressions Using Letter Numbers
Textbook pages81 – 105
Questions solved107
Session2027
FormatFull answers on this page + PDF

NCERT Solutions Class 7 Maths Chapter 4 – FAQs

Where can I download NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter Numbers PDF?

Read every answer on this page, or download the free PDF from the table at the top.

Is this for the 2027 session?

Yes, it follows the latest NCERT Class 7 Maths book for 2027.

How many questions are solved in NCERT Class 7 Maths Chapter 4?

All 107 questions of the chapter (textbook pages 81 – 105) – the in-text questions, activities and the end-of-chapter exercises.

How should I use these solutions?

Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.

Official website: ncert.nic.inNational Council of Educational Research and Training publishes date sheets, admit cards and results here
Visit

Ask a question

?

Loading comments…