NCERT Solutions Class 7 Maths Chapter 4 Expressions Using Letter Numbers

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 7 · M AT H S

NCERT Solutions

Chapter 4: Expressions Using
Letter-Numbers

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

81 – 105 15 106 English

Solutions, notes, sample papers & more at 72 pages

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 4: Expressions Using
Letter-Numbers
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.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) 81 – 105

SECTIONS QUESTIONS

15 106

MEDIUM

English

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

Q1 Example 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? _______

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

Shabnam's age = Aftab's age + 3

= 18 + 3

= 21 years

Why it happens: The gap between two people's ages never changes. Aftab was 10
when Shabnam was 13 — a gap of 3. That same gap of 3 stays for ever, so we can
add 3 to any age of Aftab.

Page 1 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q2 Given Aftab’s age, how will you find out Shabnam’s age?

Add 3 to Aftab's age.

Shabnam's age = Aftab's age + 3

AFTAB'S AGE SHABNAM'S AGE

4 4+3=7

10 10 + 3 = 13

23 23 + 3 = 26

Tip: Describe the rule in plain words first. The words tell you exactly what to write in
symbols.

Q3 Can we write this as an expression?

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

Why it happens: Writing ‘Aftab’s age’ again and again is slow. A single letter a stands
for that number, whatever it is. Letters used this way are called letter-numbers, and
an expression like a + 3 is an algebraic expression.

Page 2 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q4 If a is 23 (Aftab’s age in years), then what is Shabnam’s age?

Replace a by 23 in the expression a + 3.

s=a+3

= 23 + 3

= 26 years

Check it yourself: Put a = 40. Then s = 43 — and 43 – 40 = 3, the same gap. The
formula never breaks.

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

Q1 Given the age of Shabnam, write an expression to find Aftab’s age.

Aftab is 3 years younger, so subtract 3.

Aftab's age = Shabnam's age – 3

a=s–3

Why it happens: s = a + 3 and a = s – 3 say the same thing from the two ends.
Adding 3 and subtracting 3 undo each other.

Q2 Use this expression to find Aftab’s age if Shabnam’s age is 20.

Replace s by 20.

Page 3 of 72

as e
Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
e m.
a=s–3
m l as
= 20 – 3
m .co a g
l a se
g
= 17 years
a

com
. ag
Check it yourself: Now go back — s = a + 3 = 17 + 3 = 20. It matches.
e m
g l as
a
m
Example 2: Parthiv is making matchstick patterns. He repeatedly places Ls next to
co
Q3

m.
each other. Each L has two matchsticks as shown in Figure 4.2. How many

as e
com the number of Ls and the number of sticks? l
matchsticks are needed to make 5 Ls? 7 Ls? 45 Ls? Now, what is the relation
. a g
sem
between

a
agl
m a s
m.co agl
l a se
a g

co m
m .
m as e
.co l
Fig. 4.2, page 82 — one L, two Ls and three Ls made from matchsticks.

a g
se m
g l a
a
se m

com
Every L uses exactly 2 matchsticks, so the count is always double the number of Ls.
g l a
m . a
ase
5 Ls → 5 × 2 = 10 matchsticks agl
m
7 Ls → 7 × 2 = 14 matchsticks
. co
45 Ls → 45 × 2 = 90 matchsticks
em
m l as
.co a g
a s emIn words: Number of matchsticks = 2 × Number of L’s. Taking n for the number of Ls, the
agl expression is
.c
s e m
m a
. co agl
2×n
e m
g l as
a

co m
m .
m ase
.co


a g l Page 4 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Why it happens: The Ls do not share any matchstick — each one is separate. So the
total is simply 2 counted n times, which is 2 × n.

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

Q1 Example 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?

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

Q2 How much should she pay if she buys 8 coconuts and 9 kg jaggery?

Same two steps, new numbers.

Cost of 8 coconuts = 8 × ₹35 = ₹280

Cost of 9 kg jaggery = 9 × ₹60 = ₹540
Total = ₹280 + ₹540 = ₹820

Tip: 8 × 35 is easy as 8 × 35 = 8 × 30 + 8 × 5 = 240 + 40 = 280.

Page 5 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q3 Write an algebraic expression to find the total amount to be paid for a given
number of coconuts and quantity of jaggery.

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

QUANTITY NEEDED RELATIONSHIP EXPRESSION

Cost of coconuts Number of coconuts × 35 c × 35

Cost of jaggery Number of kgs of jaggery × 60 j × 60

Total amount = Cost of coconuts + Cost of jaggery
c × 35 + j × 60

Why it happens: The rates 35 and 60 never change; only the quantities c and j
change. That is exactly what a formula is built to handle.

Q4 Use this expression (or formula) to find the total amount to be paid for 7 coconuts
and 4 kg jaggery.

Put c = 7 and j = 4.

c × 35 + j × 60

= 7 × 35 + 4 × 60

= 245 + 240

= ₹485

Page 6 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q5 Example 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.

Put q = 7 in 4 × q.

Perimeter = 4 × q

=4×7

= 28 cm

Why it happens: A square has 4 equal sides, so its boundary is the same length
added 4 times: q + q + q + q = 4 × q.

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

Q1 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

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

SHAPE NUMBER OF EQUAL SIDES FORMULA FOR PERIMETER

Equilateral triangle 3 3a

Regular pentagon 5 5a

Regular hexagon 6 6a

Page 7 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

(a) Perimeter = a + a + a = 3a units

(b) Perimeter = a + a + a + a + a = 5a units

(c) Perimeter = 6a units

Why it happens: In a regular polygon every side is the same length. Adding the
same number several times is multiplication, so a polygon with k equal sides of
length a has perimeter ka.

Q2 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.

Joining two pipes end to end adds their lengths.

Combined length = 20 + k

= (20 + k) metres

Check it yourself: If the second pipe is 15 m, k = 15 and the total is 20 + 15 = 35 m. If
it is 8 m, the total is 28 m — one expression covers every case.

Page 8 of 72

as e
Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
m.
What is the total amount Krithika has, if she has the following numbers of notes of
e
Q3

m l as
.co
₹100, ₹20 and ₹5? Complete the following table:

a g
s m OF ₹100
eNO.
a
gl NOTES
NO. OF ₹20 NO. OF ₹5 EXPRESSION AND TOTAL
a NOTES NOTES AMOUNT

m
.co ag
3 5 6

a sem
agl
6 × 100 + 4 × 20 + 3 × 5 = 695

8 4 z

co m
m.
x y z

m as e
.co a g l
se m
g l a
a Each row follows the same rule: (number of ₹100 notes × 100) + (number of ₹20 notes × 20) +
m a s
agl
(number of ₹5 notes × 5).

.co
m NO. OF ₹5
NO. OF ₹100
l
NO. OF ₹20
a se EXPRESSION AND TOTAL
NOTES g
NOTES
a NOTES AMOUNT

com
3 5 6 3 × 100 + 5 × 20 + 6 × 5 = 430

6 × 100 + 4e×m
.
a s
com agl
6 4 3 20 + 3 × 5 = 695

8m
.
a s e
agl
4 z 8 × 100 + 4 × 20 + z × 5 = 880 + 5z

m
x y z x × 100 + y × 20 + z × 5 = 100x + 20y +

a se
com l
5z

. a g
m
ase
Row 1: 300 + 100 + 30 = ₹430
agl
Row 2: 600 + 80 + 15 = ₹695

co m
Row 3: 800 + 80 + 5z = ₹(880 + 5z)
m .
m as e
.co
Row 4: ₹(100x + 20y + 5z)
a g l
se m
g l a
a c
Why it happens: The last row is the general formula. Put x = 8, y = 4 into it and you
m .
m a s e
co agl
get 800 + 80 + 5z — exactly row 3. The rows above are only special cases of the same

m .
e
expression.

g l as
a

co m
m .
m ase
.co


a g l Page 9 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q4 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 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

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

Why it happens: The 10 seconds of starting up happens once, no matter how much
grain there is, so it is added, not multiplied. The 8 seconds repeats for every
kilogram, so it is multiplied by y.

Check it yourself: For y = 5 kg, time = 10 + 40 = 50 s. Option (b) would give 90 s and
option (c) 400 s — both far too long.

Q5 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

Let the number be n.

IN WORDS EXPRESSION

(a) 5 more than a number n+5

(b) 4 less than a number n–4

(c) 2 less than 13 times a number 13n – 2

(d) 13 less than 2 times a number 2n – 13

Page 10 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Why it happens: (c) and (d) use the same two numbers but in opposite roles. For n =
10, 13n – 2 = 128 while 2n – 13 = 7. Read the sentence carefully — ‘times’ tells you
what to multiply, ‘less than’ tells you what to subtract at the end.

Q6 Describe situations corresponding to the following algebraic expressions: (a) 8 × x +
3 × y (b) 15 × j – 2 × k

(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.

Try This: Many different stories fit one expression. Write a cricket story for 8x + 3y —
say x runs for every boundary and y runs for every six.

Page 11 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q7 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’.

November 2024

Mon Tue Wed Thu Fri Sat Sun

1 2

3 4 5 6 7 8 9 w–1 w
10 11 12 13 14 15 16

17 18 19 20 21 22 23

24 25 26 27 28 29 30

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.

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–8 w–7 w–6

w–1 w w+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

Why it happens: A calendar row holds 7 days, so the cell just above any date is
exactly 7 less.

Check it yourself: In the picture the marked block is 12, 13, 14 over 19, 20, 21. Here
w = 20, so w – 1 = 19, w + 1 = 21, w – 7 = 13, w – 8 = 12 and w – 6 = 14. Every one
matches.

Page 12 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

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

Q1 Let 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)

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

Why it happens: In (2) we used swapping and grouping — 83 and –13 go together,
28 and 32 go together. In (5) the two ways shown in the book agree: 68 – (18 + 13) =
68 – 31 = 37, and also 68 – 18 – 13 = 50 – 13 = 37. A minus sign outside a bracket flips
the sign of every term inside.

Tip: A term that contains × must be turned into a single number before you add. That
is why 23 – 10 × 2 is 3 and not 26.

Q2 Find an algebraic expression to get the nth term of this sequence: 4, 8, 12, 16, 20, 24,
28, ...

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

Page 13 of 72

ase
Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
e m.
3rd term = 4 × 3 = 12
m l as
.co
29th term = 4 × 29 = 116
m a g
l a se
g
nth term = 4 × n = 4n
a

. com
Why it happens: The 1st term is 4 × 1, the 2nd is 4 × 2, the 3rd is 4 × 3 … The
ag
a s em so position n gives 4 × n. As a standard
position number is always the multiplier,

a l 4n, number first, letter after.
practice we drop the × sign andgwrite

co m
em.
c+o3mtakes when m = 2. l as
Find the value of the expression 7k when k = 4. Find the value that the expression
g
.
Q3
a
sem
5m

la
agANSWER
m a s
.co agl
Remember 7k means 7 × k and 5m means 5 × m.

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

co m
m .
m as e
.co a g l
Tip: 5m + 3 is not 5 × (m + 3). Multiply first, then add — 5m is one term and 3 is

se m
a
another.

a g l
se m
com — Page 87 g l a
. a
emSymbol
Mind the Mistake, Mend the Mistake
a s
agl
Section 4.3 Omission of the Multiplication

TRY THIS

co m
m .
as e
com
If a = – 4, then 10 – a = 6.
l
Q1

. a g
e m
as ANSWER
agl c
Mistake. The negative sign of a was ignored.
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 14 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

10 – a = 10 – (–4)

= 10 + 4

= 14

Why it happens: Subtracting a negative number is the same as adding its opposite.
Whoever solved this simply wrote 10 – 4.

Q2 If d = 6, then 3d = 36.

Mistake. 3d means 3 × d, not d × d.

3d = 3 × 6 = 18

Why it happens: 36 comes from 6 × 6. The number written in front of a letter-
number is the multiplier, and here it is 3.

Q3 If s = 7, then 3s – 2 = 15.

Mistake. The 2 was subtracted before multiplying.

3s – 2 = 3 × 7 – 2

= 21 – 2

= 19

Why it happens: 15 = 3 × (7 – 2). But 3s – 2 has two terms, 3s and –2. The
multiplication inside the term 3s must be done first.

Page 15 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q4 If r = 8, then 2r + 1 = 29.

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

2r + 1 = 2 × 8 + 1

= 16 + 1

= 17

Why it happens: 2r is a product, not a two-digit number. Replacing r by 8 gives 2 × 8,
never “28”.

Q5 If j = 5, then 2j = 10.

No mistake. This one is correct.

2j = 2 × 5 = 10

Q6 If m = –6, then 3 (m + 1) = 19.

Mistake. The bracket was not evaluated with the negative value.

3 (m + 1) = 3 × (–6 + 1)

= 3 × (–5)

= –15

Why it happens: –6 + 1 is –5, not 6 + 1. Substitute the whole value, sign included,
before working inside the bracket.

Page 16 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q7 If f = 3, g = 1 then 2f – 2g = 2.

Mistake. The subtraction was done before the multiplications.

2f – 2g = 2 × 3 – 2 × 1

=6–2

=4

Why it happens: 2 comes from 2 × (3 – 1). Here 2f and 2g are separate terms; each
must become a number first.

Q8 If t = 4, b = 3 then 2t + b = 24.

Mistake. The + was treated as a ×.

2t + b = 2 × 4 + 3

=8+3

= 11

Why it happens: 24 = 2 × 4 × 3. The expression has two terms, 2t and b, joined by
addition.

Q9 If h = 5, n = 6 then h – (3 – n) = 4.

Mistake. The bracket was handled wrongly.

Page 17 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

h – (3 – n) = 5 – (3 – 6)

= 5 – (–3)

=5+3
=8

Why it happens: 3 – 6 is –3, and subtracting –3 means adding 3. You can also open
the bracket first: h – 3 + n = 5 – 3 + 6 = 8. Both routes agree.

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

MATH TALK

Q1 Example 5: The price per pencil is c and the price per eraser is d. 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 ______.

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

Day 2: 3 pencils → 3c

Day 3: 10 pencils → 10c

DAY 1 DAY 2 DAY 3

Pencils (price c) 5 3 10

Money from pencils 5c 3c 10c

Q2 If c = ₹50, find the total amount earned by the sale of pencils.

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

Page 18 of 72

ase
Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
e m.
5c + 3c + 10c = (5 + 3 + 10) × c = 18c
m l as
.co
18c = 18 × ₹50
m a g
l a
= ₹900 se
a g

co m
. ag
Why it happens: Adding 5 lots of c, 3 lots of c and 10 lots of c gives 18 lots of c. That
is the distributive property at work. em
g l as
a

co m
m.
Write the expression for the total money earned by selling erasers. Then, simplify
e
Q3

m l as
.co
the expression.
a g
se m
g l a
a Erasers sold: 4 on Day 1, 6 on Day 2 and 1 on Day 3, each priced d.

m a s
m.co agl
se
Total from erasers = 4d + 6d + 1d
g l a
= (4 + 6 + 1) × d a
m
= 11d
. co
se m
o m l a
g altogether is 18c +
m .c1d is simply written as d. So the shop's three-day earning
Tip: a
ase11d.
agl
se m
com g l a
m . a
ase
Q4 Can the expression 18c + 11d be simplified further?

agl

co m
.
No. It is already in its simplest form.

em
m l as
.co a g
m terms. c is the price of a pencil and d is the price of an eraser; there is no way to
Why it happens: 18c and 11d contain different letter-numbers — they are unlike

l a se
ag
c
combine ₹-per-pencil with ₹-per-eraser into a single term.
m .
m a s e
e m . co
Did you know? Sets like (5c, 3c, 10c) are called like terms and can always be agl
g l as
merged into one term. Sets like {18c, 11d} are unlike terms and cannot.
a

com
m .
m ase
.co


a g l Page 19 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q5 Check that both expressions take the same value when c is replaced by different
numbers. (5c + 3c + 10c and 18c)

Try three different values of c.

C 5C + 3C + 10C 18C

2 10 + 6 + 20 = 36 36

5 25 + 15 + 50 = 90 90

50 250 + 150 + 500 = 900 900

Why it happens: Simplifying only rewrites an expression using the same rules that
hold for numbers. So the two forms must agree for every value of c — that is exactly
what ‘equal expressions’ means.

Q6 As earlier, a big rectangle is split into two smaller rectangles as shown below. Write
an expression to find the area of the rectangle AEFD.

4
D F C

n

A 12 E B

Rectangle ABCD, page 90 — length AB = 12, breadth AD = n. The dashed line FE splits it
into AEFD and EBCF, with FC = 4.

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

Page 20 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

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

D F C

n

A E B
12 4
Rectangle ABCD of sides 12 and n, split at EF. The shaded strip EBCF has area 4n.

Why it happens: Both routes must agree, and they do: 12n – 4n = (12 – 4)n = 8n. This
is the distributive property seen as areas.

Q7 Example 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.

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

Page 21 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

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)

Q8 Can we simplify this expression? If yes, how? If not, why not? [(40x + 75y) – (6x +
10y)]

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

Why it happens: The minus sign outside the bracket changes the sign of both terms
inside. After that, 40x and –6x are like terms, and so are 75y and –10y.

Check it yourself: For x = 2 chairs and y = 1 table: paid ₹155, returned ₹22, so ₹133
is kept. And 34 × 2 + 65 × 1 = 68 + 65 = ₹133. ✔

Page 22 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

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

MATH TALK

Q1 Could we have written the initial expression as (40x + 75y) + (– 6x – 10y)?

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

Why it happens: Every subtraction can be turned into an addition of the opposite
number. Written as a sum of terms, an expression becomes much easier to swap and
group.

Q2 Example 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?

ROUND EXPRESSION WHAT IT SAYS

1 7p – 3q 7 correct answers and 3 wrong answers

2 8p – 4q 8 correct answers and 4 wrong answers

3 6p – 2q 6 correct answers and 2 wrong answers

Why it happens: Each correct answer earns p marks, so 7 of them earn 7p. Each
wrong answer costs q marks, so 3 of them cost 3q — a subtraction.

Page 23 of 72

ase
Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
m.
If the score for a correct answer is 4 (p = 4) and the penalty for a wrong answer is 1
e
Q3

m l as
.co
(q = 1), find Charu’s score in the first round.

a g
se m
g l a
a

7p – 3q = 7 × 4 – 3 × 1
co m
em . ag
as
= 7 × 4 + (–3 × 1)

= 28 + (–3) a g l
= 25
co m
em.
m l as
.co a g
a s em
gl
What are her scores in the second and third rounds?
a
Q4

m a s
.co agl

Use p = 4 and q = 1 in each expression.
se m
g l a
a
Round 2: 8p – 4q = 8 × 4 – 4 × 1 = 32 – 4 = 28

co m
Round 3: 6p – 2q = 6 × 4 – 2 × 1 = 24 – 2 = 22
m .
m as e
.co a g l
s m it yourself: 25 + 28 + 22 = 75, and we will see below that her total is 21p – 9q
eCheck
gl a
a = 84 – 9 = 75. ✔

se m
com g l a
m . a
e
as What will be the value of q in that situation?
Q5
g l
What if there is no penalty?
a
co m
m .
No penalty means nothing is taken away for a wrong answer.
m as e
.co a g l
se mq=0
g l a
a c
Round 1 score becomes 7p – 3 × 0 = 7p = 28
m .
m a s e
. co agl
Total becomes 21p – 9 × 0 = 21p = 84
e m
g l as
a

co m
m .
m ase
.co


a g l Page 24 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Why it happens: Putting q = 0 makes every penalty term vanish, because any
number multiplied by 0 is 0.

Q6 What is her final score after the three rounds?

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

Q7 Give some possible scores for Krishita in the three rounds so that they add up to
give 23p – 7q.

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 ✔

Try This: Another set is 10p – 4q, 6p – 1q and 7p – 2q. Make up a third set of your
own — there are many.

Page 25 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q8 Can we say who scored more? Can you explain why? (Charu: 21p – 9q, Krishita: 23p –
7q)

Krishita scored more.

Krishita has 23p, Charu has 21p → Krishita gains 2p more

Krishita loses 7q, Charu loses 9q → Krishita loses 2q less

Why it happens: Krishita wins on both counts — more correct answers and fewer
wrong ones. Since p and q are positive numbers, her total must be larger, whatever
the marking scheme is.

Q9 How much more has Krishita scored than Charu? Simplify this expression further:
23p – 7q – (21p – 9q)

23p – 7q – (21p – 9q)

= 23p – 7q – 21p + 9q

= 23p – 21p + (–7q) + 9q

= (23 – 21)p + (9 – 7)q

= 2p + 2q

Check it yourself: With p = 4 and q = 1, Krishita has 92 – 7 = 85 and Charu has 84 – 9
= 75. The difference is 10, and 2p + 2q = 8 + 2 = 10. ✔

Page 26 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q10 Example 9: Simplify the expression 4 (x + y) – y

4 (x + y) – y

= 4x + 4y – y (distributive property)

= 4x + 4y + (– y)

= 4x + (4 – 1)y

= 4x + 3y

Why it happens: 4 times a sum is the sum of 4 times each part. After that, 4y and –y
are like terms, and 4y – y is 3y.

Q11 Example 10: Are the expressions 5u and 5 + u equal to each other?

No. They describe two different operations.

5u means 5 times the number u

5 + u means 5 more than the number u

Why it happens: Two expressions are equal only if they give the same value for
every value of the letter. Here they agree only at u = 1.25, and differ everywhere else
— so they are not equal expressions.

Page 27 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q12 Fill 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)

U 5U 5+U

11 55 16

2 10 7

8 40 13

5 25 10

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.

Q13 Are 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?

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

Y 10Y – 3 10 (Y – 3)

2 17 –10

0 –3 –30

7 67 40

10 97 70

Page 28 of 72

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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
e m.
y = 2: 10 × 2 – 3 = 17, but 10 × (2 – 3) = 10 × (–1) = –10
m l as
.co
y = 0: 0 – 3 = –3, but 10 × (–3) = –30
m a g
l a se
g
y = 7: 70 – 3 = 67, but 10 × 4 = 40
ay = 10: 100 – 3 = 97, but 10 × 7 = 70

co m
No, the two expressions are not equal. m. ag
l a se
ag
Why it happens: Opening the bracket shows the gap clearly: 10(y – 3) = 10y – 30,
which is 27 less than 10y – 3 for every y.
co m
em.
m l as
m .co a g
l a se
a g
m a s
m.co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
m . a
ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q14 Example 11: What is the sum of the numbers in the picture (unknown values are
denoted by letter-numbers)?

3 3 3 3

r s

r s

3 3 3 3

The picture for Example 11, page 93 — four 3s, then r and s, then r and s, then four 3s.

Three different routes, one answer.

Page 30 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

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

Why it happens: The picture is symmetric top-to-bottom, so the third method
works. All three expressions look different but simplify to the same thing — that is
what makes them equal expressions.

Figure it Out — Pages 93–94

Page 31 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Section 4.4 Simplification of Algebraic Expressions

MATH TALK

Q1 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.

2p 3q –2 3
5y –6 x

3q 2p 3 –2

x 2 5y
2p 3q

(i)
3q 2p

(ii)

–5g 5k 5k –5g

5k 5k 5k 5k

5k 5k 5k 5k

–5g 5k 5k –5g

(iii)

Pictures (i), (ii) and (iii), page 94 — the numbers in each are to be added.

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

Page 32 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

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

Why it happens: Terms can be added in any order and grouped in any way, so
however you sweep across the picture — by rows, by columns, or by collecting like
symbols — the simplified answer must come out the same.

Page 33 of 72

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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Simplify each of the following expressions: (a) p + p + p + p, p + p + p + q, (b) p + q + p
se
Q2

o m l a
– 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.–c d – d – c, (i) 2d – d – (d – c), (j) 2d – (d – d) – c, (k) 2d – d –acg– c

a s em
a gl

co m
. ag
EXPRESSION WORKING SIMPLEST FORM

e m
g l as
a
(a) p + p + p + p 4 lots of p 4p

3 lots of p, one q
m
(a) p + p + p + q 3p + q

co
(b) p + q + p – q (p + p) + (q – q) 2p
s e m.
(c) p – q +.cpo
m gla
2p –a2q
em
(p + p) + (–q – q)
s
–q

a
agl(d) p + q – p + q (p – p) + (q + q) 2q

m a s
.co agl
(e) p + q – (p + q) p+q–p–q 0

se m
l a
(f) p – q – p – q (p – p) + (–q – q) –2q

a g
(g) 2d – d – d – d (2 – 1 – 1 – 1)d –d

co m
(h) 2d – d – d – c (2 – 1 – 1)d – c = 0 – c –c
m .
m as e
.co
(i) 2d – d – (d – c) 2d – d – d + c
a g l
c

a s e(j)m2d – (d – d) – c
agl 2d – 0 – c 2d – c

(k) 2d – d – c – c (2 – 1)d – 2c d – 2c
se m
com g l a
m . a
Why it happens: Comparea(h) e
s and (i). In (h) we subtract d and then subtract c. In (i)
a g l
the bracket makes us subtract d but add c, because a minus sign outside flips every
sign inside. One bracket changes the whole answer.
co m
m .
m as e
.co a g l
Check it yourself: Put d = 5, c = 2. (h) gives 10 – 5 – 5 – 2 = –2 = –c ✔. (i) gives 10 – 5 –

s e m (5 – 2) = 5 – 3 = 2 = c ✔.
agla
.c
s e m
m a
o
Mind the Mistake, Mend the.cMistake — Pages 94–95
m agl
l a se
a g

co m
m .
m ase
.co


a g l Page 34 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Section 4.4 Simplification of Algebraic Expressions

Q1 3a + 2b → Simplest Form given: 5

Mistake. Unlike terms were added.

Correct simplest form: 3a + 2b

Why it happens: 3 and 2 were added while the letters were thrown away. But a and
b are different letter-numbers, so 3a and 2b are unlike terms and cannot be merged.

Q2 3b – 2b – b → Simplest Form given: 0

No mistake.

3b – 2b – b = (3 – 2 – 1)b = 0 × b = 0

Q3 6 (p + 2) → Simplest Form given: 6p + 8

Mistake. The 6 was multiplied only by p, and then added to 2.

6 (p + 2) = 6 × p + 6 × 2

= 6p + 12

Why it happens: The distributive property says the multiplier reaches every term
inside the bracket.

Page 35 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q4 (4x + 3y) – (3x + 4y) → Simplest Form given: x + y

Mistake. The sign of the y-term was not changed.

(4x + 3y) – (3x + 4y)

= 4x + 3y – 3x – 4y

= (4 – 3)x + (3 – 4)y

=x–y

Q5 5 – (2 – 6z) → Simplest Form given: 3 – 6z

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

5 – (2 – 6z) = 5 – 2 + 6z

= 3 + 6z

Check it yourself: For z = 1, 5 – (2 – 6) = 5 – (–4) = 9, and 3 + 6 = 9 ✔, while 3 – 6 = –3
✘.

Q6 2 + (x + 3) → Simplest Form given: 2x – 6

Mistake. The bracket was treated as a multiplication.

2 + (x + 3) = 2 + x + 3

=x+5

Page 36 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Why it happens: There is a plus sign before the bracket, so the bracket can simply
be dropped.

Q7 2y + (3y – 6) → Simplest Form given: – y + 6

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

2y + (3y – 6) = 2y + 3y – 6

= 5y – 6

Q8 7p – p + 5q – 2q → Simplest Form given: 7p + 3q

Mistake. The – p was left out.

7p – p + 5q – 2q

= (7 – 1)p + (5 – 2)q
= 6p + 3q

Tip: A letter written alone, like p, means 1p. It must be counted.

Q9 5 (2w + 3x + 4w) → Simplest Form given: 10w + 15x + 20w

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

Page 37 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Tip: Add like terms inside the bracket first — 2w + 4w = 6w — and the work becomes
shorter.

Q10 3j + 6k + 9h + 12 → Simplest Form given: 3 (j + 2k + 3h + 4)

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

Why it happens: In this chapter ‘simplest form’ means brackets removed, like terms
added and number terms added. The given expression is already like that — nothing
needed to change.

Q11 4 (2r + 3s + 5) → Simplest Form given: – 20 – 8r – 12s

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

Page 38 of 72

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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Take a look at all the corrected simplest forms (i.e. brackets are removed, like
se
Q12

o m l a
terms are added, and terms with only numbers are also added). Is there any
g
m .c between the number of terms and the number of letter-numbers
relation a these
e
las
expressions have?
ag

com

.
em OF ag
NO. CORRECT SIMPLEST
a s
NUMBER NUMBER OF LETTER-
FORM
a gl TERMS NUMBERS

m
1 3a + 2b 2 2

co
e m.
m as
2 0 1 0

.6pco+ 12 a g l
m
ase
3 2 1

agl4 x–y 2 2

m a s
.co agl
5 3 + 6z 2 1

a s em
6
g l
x+5 2 1

7 5y – 6
a 2 1

co m
8
m .
e
6p + 3q 2 2

m l as
.co 30w + 15x a g
em
9 2 2

a s
agl 10 3j + 6k + 9h + 12 4 3

se m
com a
11 8r + 12s + 20 3 2

. a g l
m
ase
agl
Yes. In every row, the number of letter-numbers is never more than the number of terms:

m
Number of letter-numbers ≤ Number of terms

. co
em
m l as
.co a g
Why it happens: Once an expression is fully simplified, each letter-number can
m appear in only one term — all its like terms have already been merged. There may
l a se
ag
c
also be one extra term made of numbers alone, which carries no letter. So terms can
m .
be more than letters, but never fewer.
m a s e
e m . co agl
g l as
Formula Detective — Pages 95–96 a

co m
m .
m as e
.co


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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Section 4.5 Pick Patterns and Reveal Relationships

Q1 Look 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.

5 2 8 1 9 11 10 10

8 15 7 10

expression: expression: expression: expression:

6 4

expression:

The five number machines, page 95 — each takes the two numbers at the top of the ‘Y’
and produces the result at the bottom.

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

Page 40 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

INPUTS (A, B) EXPRESSION OUTPUT

5, 2 2×5–2 8

8, 1 2×8–1 15

9, 11 2 × 9 – 11 7

10, 10 2 × 10 – 10 10

6, 4 2×6–4 8

Why it happens: One example is never enough — 5 and 2 also fit “a + b + 1”. Testing
the rule on every pair is what confirms 2a – b.

Q2 Find the formulas of the number machines below and write the expression for each
set of inputs. (Machine 1: 5, 2 → 5; 8, 1 → 7; 9, 11 → 18; 10, 10 → 18; a, b → ? Machine
2: 4, 1 → 5; 6, 0 → 1; 3, 2 → 7; 10, 3 → ?; a, b → ?)

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

INPUTS EXPRESSION OUTPUT

5, 2 5+2–2 5

8, 1 8+1–2 7

9, 11 9 + 11 – 2 18

10, 10 10 + 10 – 2 18

a, b a+b–2 a+b–2

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

Page 41 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

INPUTS EXPRESSION OUTPUT

4, 1 4×1+1 5

6, 0 6×0+1 1

3, 2 3×2+1 7

10, 3 10 × 3 + 1 31

a, b a×b+1 ab + 1

Tip: The pair (6, 0) is the giveaway for Machine 2. Any rule that only added would
give 6 or 7; only multiplying can crush 6 down to 1.

Q3 Now, make a formula on your own. Write a few number machines as examples
using that formula. Challenge your classmates to figure it out!

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

INPUTS (A, B) WORKING OUTPUT

4, 1 3×4–2×1 10

5, 5 3×5–2×5 5

2, 3 3×2–2×3 0

7, 2 3×7–2×2 17

1, 4 3×1–2×4 –5

Try This: Include one pair that gives 0 and one that gives a negative answer, as
above. Those two cases make guessing much harder — and much more fun.

In-text Questions — Pages 96–97

Page 42 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Algebraic Expressions to Describe Patterns

Q1 Example 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.

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

DESIGN POSITIONS WHERE IT OCCURS FORMULA FOR THE NTH TIME

A 1, 4, 7, 10, 13, … 3n – 2

B 2, 5, 8, 11, 14, … 3n – 1

C 3, 6, 9, 12, 15, … 3n

Why it happens: D, E, F in the picture are just A, B, C repeating. Every third position
carries C, so C's positions are the multiples of 3. B always sits one place before C, and
A two places before.

Q2 Where would design C appear for the nth time?

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

nth occurrence of Design C is at position 3n

Check it yourself: The 20th C would be at position 3 × 20 = 60.

Q3 Similarly, find the formula that gives the position where the other Designs appear
for the nth time.

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

Page 43 of 72

as e
Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
e m.
nth occurrence of Design B: 3n – 1
m l as
.co
nth occurrence of Design A: 3n – 2
m a g
l a se
g
aWhy it happens: In each block of three, C sits last, B just before it and A just before
B. So their positions are 3n, 3n – 1 and 3n – 2.
co m
e m . ag
g l as
a
Q4 Given a position number can we find out the design that appears there? Which

co m
m.
Design appears at Position 122?

m as e
.co a g l
se m
g l a
a
Com a s
.3c agl
A B D E F
1 2
a s em 4 5 6

ag l
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.

co m
m .
as e
. com a g l
emand the design at position 122 is Design B.

a s
agl Yes,

se m
122 = 3 × 41 – 1
com g l a
m . a
ase
agl
So 122 is one less than 123, a multiple of 3

Positions of the form 3n – 1 carry Design B

. com
m a
Why it happens: Every position is a multiple of 3 (→ C), one less
s em than a multiple of 3
m .c(→o B), or two less (→ A). There is no fourth possibility. agl
l a se
ag
.c
s e m
m a
em . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q5 Can the remainder obtained by dividing the position number by 3 be used for this?
Observe the table below.

POSITION NO. QUOTIENT ON DIVISION BY 3 REMAINDER

99 33 0

122 40 2

148 49 1

Yes. The remainder alone decides the design.

REMAINDER ON DIVIDING BY 3 FORM OF THE POSITION DESIGN

0 3n C

1 3n + 1, i.e. 3(n+1) – 2 A

2 3n + 2, i.e. 3(n+1) – 1 B

Check it yourself: Position 1 leaves remainder 1 and holds A ✔. Position 2 leaves
remainder 2 and holds B ✔. Position 3 leaves remainder 0 and holds C ✔.

Q6 Use this to find what design appears at positions 99, 122, and 148.

POSITION DIVISION BY 3 REMAINDER DESIGN

99 99 = 3 × 33 + 0 0 C

122 122 = 3 × 40 + 2 2 B

148 148 = 3 × 49 + 1 1 A

Page 45 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

99 = 3 × 33 → Design C

122 = 3 × 41 – 1 → Design B

148 = 3 × 50 – 2 → Design A

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

MATH TALK

Q1 Let 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?

Both diagonals add up to the same number.

12 + 20 = 32

13 + 19 = 32

The two diagonal sums are equal.

Try This: Take the square 5, 6 over 12, 13 from the same November 2024 calendar. 5
+ 13 = 18 and 6 + 12 = 18. Equal again.

Page 46 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q2 Will the diagonal sums be equal in every 2 × 2 square in this endless grid? How can
we be sure?

November 2024

Mon Tue Wed Thu Fri Sat Sun

1 2

3 4 5 6 7 8 9

10 11 12 13 14 15 16

17 18 19 20 21 22 23

24 25 26 27 28 29 30

31 32 33 34 35 36 37

38 39 40 41 42 43 44

⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮
Page 98 — the November 2024 calendar with its rows continued beyond 30, without end.

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

Why it happens: To be sure we need an argument that covers all squares at once.
Algebra does exactly that: call the top-left number a and describe the other three in
terms of a. One calculation then settles every square.

Page 47 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q3 Given that we know the top left number, how do we find the other numbers in this 2
× 2 square?

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

a a+1

a+7 a+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.

Why it happens: A calendar week has 7 days, so dropping one row adds 7. Moving
one column right adds 1. The diagonal cell does both, so it adds 8.

Q4 Verify this expression for diagonal sums by considering any 2 × 2 square and taking
its top left number to be ‘a’.

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 ✔

Page 48 of 72

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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
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co m
m.
Did you know? This is the power of algebraic modelling — one short proof replaces

m as e
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an unlimited number of checks.

m .co a g
l a se
a g
Q5 Consider a set of numbers from the calendar (having endless rows) forming the plus

co m
ag
shape shown below. Find the sum of all the numbers. Compare it with the number

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e
in the centre: 15. Repeat this for another set of numbers that forms this shape.
What do you observe?
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a

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a g
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14 a g 15 16
co m
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se m
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asshape of five numbers taken from the endless calendar.
a g l
Page 99 — the plus

co m
m .
m as e
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se m 8 + 14 + 15 + 16 + 22
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= (8 + 22) + (14 + 16) + 15
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= 30 + 30 + 15
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= 75
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a
And 75 = 5 × 15, five times the centre.

co m
m .
m ase
.co


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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

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.

Q6 Will 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’.]

8

14 15 16

22

Page 99 — the plus shape of five numbers taken from the endless calendar.

Yes, always. Let the centre be a.

a–7

a–1 a a+1

a+7

Page 50 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Sum = (a – 7) + (a – 1) + a + (a + 1) + (a + 7)

= a + a + a + a + a + (–7 + 7) + (–1 + 1)

= 5a + 0 + 0

= 5a

Why it happens: The shape is symmetric about its centre. The number above
cancels the number below (–7 and +7), and the number on the left cancels the one
on the right (–1 and +1). Only five copies of the centre survive.

Q7 Find other shapes for which the sum of the numbers within the figure is always a
multiple of one of the numbers.

8

14 15 16

22

Page 99 — the plus shape of five numbers taken from the endless calendar.

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

Page 51 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

SHAPE NUMBERS IN TERMS OF A SUM

Row of 3 (horizontal) a – 1, a, a + 1 3a

Column of 3 (vertical) a – 7, a, a + 7 3a

Diagonal of 3 (↘) a – 8, a, a + 8 3a

The X of 5 (four corners + centre) a – 8, a – 6, a, a + 6, a + 8 5a

Full 3 × 3 block nine numbers around a 9a

3 × 3 block: (a – 8) + (a – 7) + (a – 6) + (a – 1) + a + (a + 1) + (a + 6) + (a + 7) + (a + 8)

= 9a

Why it happens: In every one of these shapes the cells pair up around the centre,
and each pair adds to 2a. With k such shapes the total is always a multiple of the
centre number.

Check it yourself: Take the 3 × 3 block centred on 15 (that is 7 to 23 in the
November calendar rows). Its nine numbers add to 135 = 9 × 15 ✔

In-text Questions — Pages 100–101

Page 52 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Matchstick Patterns

Q1 Look at the picture below. It is a pattern using matchsticks. Can you identify what
the pattern is?

4
3
2
1
The matchstick pattern, page 100 — Steps 1, 2, 3 and 4.

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 NUMBER 1 2 3 4 5 6

Triangles 1 2 3 4 5 6

No. of matchsticks 3 5 7 9 11 13

Why it happens: The first triangle needs all 3 sides. After that, the slanting side of
the previous triangle is reused, so only 2 new sticks are needed.

Q2 Can you tell how many matchsticks there will be in the next step, Step 5?

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

Step 5 = 9 + 2 = 11 matchsticks

Page 53 of 72

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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
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co m
m.
How many matchsticks will there be in Step 33, Step 84, and Step 108? Of course, we
e
Q3

m here?
copresent
can draw and count, but is there a quicker way to find the answers using the
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pattern

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Page 100 — the matchstick pattern, Steps 1 to 4.

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

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Step 33 = 3 + 2 × 32 = 3 + 64a=g67
a

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Step 84 = 3 + 2 × 83 = 3 + 166 = 169
. co
Step 108 = 3 + 2 × 107 = 3 + 214 = 217
e m
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a s eTip:
agl Drawing 108 triangles would take a whole page. The pattern gives the answer in
one line.
se m
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Q4
a
The number of matchsticks needed to make 33 triangles (Step 33) is _________.
Similarly, find the number of matchsticks needed for Step 84 and Step 108.

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Step 33 needs 67 matchsticks.

a s
agl STEP NUMBER OF 2S ADDED WORKING MATCHSTICKS
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33 32 3 + 2 + 2 + … (32 twos) 67

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84 83 3 + 2 × 83 169

108 107 a g 3 + 2 × 107 217

co m
m .
m as e
.co


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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Why it happens: Step 2 has one 2 added, Step 3 has two 2s, Step 4 has three 2s. So
Step y has (y – 1) twos added to the starting 3.

Q5 What 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?

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

Q6 Does the above expression also give the number of matchsticks at each step
correctly? Are these expressions the same?

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

Y 3 + 2(Y – 1) 2Y + 1

1 3 3

5 11 11

33 67 67

Page 55 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Why it happens: Two people can count the same picture in different ways and still
be right. Their expressions look different but simplify to the same thing.

Q7 Matchsticks 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?

STEP HORIZONTAL DIAGONAL TOTAL

1 1 2 3

2 2 3 5

3 3 4 7

4 4 5 9

Step 3: 3 horizontal and 4 diagonal (3 + 4 = 7 ✔)

Step 4: 4 horizontal and 5 diagonal (4 + 5 = 9 ✔)

Q8 How 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?

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 ✔

Page 56 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Why it happens: Each new triangle adds one horizontal stick and one slanting stick
— that is the “2 more each time”. The extra diagonal comes from the very first
triangle, which has two slanting sides.

Check it yourself: Step 6 → 6 horizontal + 7 diagonal = 13, and 2 × 6 + 1 = 13 ✔

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

TRY THIS MATH TALK

Q1 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

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

Money from jowar roti = 30 × x = 30x

Money from pulao = 20 × y = 20y

Total = 30x + 20y

Why it happens: (b) would be right only if every customer took one of each. (c)
swaps the two rates. (d) multiplies only x by 50. (e) subtracts the pulao money
instead of adding it.

Check it yourself: For x = 3 and y = 2, the true earning is ₹90 + ₹40 = ₹130, and 30 ×
3 + 20 × 2 = 130 ✔

Page 57 of 72

Class 7 Maths Chapter 4 Expressions Using Letter-Numbers AglaSem · NCERT Solutions

Q2 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)

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

Total customers = p + q + r

One flag per customer

Flags given away = p + q + r

Why it happens: Each flag goes to a customer, not to a flower. A customer who
bought both kinds is still one person and still gets one flag — which is why (b) and (c)
are wrong.

Q3 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?

(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.

Page 58 of 72

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Class 7 Maths Chapter 4 Expressions Using Letter-Numbers
a g l AglaSem · NCERT Solutions

co m
m.
Check it yourself: If u = 8 cm and d = 3 cm, the snail is 10 × 5 = 50 cm up. If u = 3 cm

m as e
l
and d = 8 cm, it is 10 × (–5) = –50 cm, i.e. 50 cm below where it began.

m .co a g
l a se
a g
Q4 Radha is preparing for a cycling race and practices daily. The first week she cycles 5

com
ag
km every day. Every week she increases the daily distance cycled by ‘z’ km. How

m .
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many kilometers would Radha have cycled after 3 weeks?

g l as
ANSWER a
Work out each week separately — a week has 7 days.
co m
e m.
m l as
.co g
WEEK DAILY DISTANCE DISTANCE THAT WEEK

m a
l a se
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1 5 km 7 × 5 = 35

a
(5 + z) km
s
2 7(5 + z) = 35 + 7z

m a
m.co agl
se
3 (5 + 2z) km 7(5 + 2z) = 35 + 14z

g l a
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Total = 35 + (35 + 7z) + (35 + 14z)

co m
= (35 + 35 + 35) + (7z + 14z)
m .
m as e
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= 105 + 21z km
a g l
se m
g l a
a Check it yourself: If z = 1 km, the totals are 35 + 42 + 49 = 126 km, and 105 + 21 × 1
se m
= 126 ✔
com g l a
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agl

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