NCERT Solutions Class 7 Maths Chapter 2 Arithmetic Expressions

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 2: Arithmetic
Expressions

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

24 – 45 18 56 English

Solutions, notes, sample papers & more at 56 pages

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 2: Arithmetic Expressions
Chapter 2 of Ganita Prakash (Part I) turns everyday number sentences into arithmetic expressions. You learn
to compare expressions by reasoning instead of calculating, to split an expression into terms, and to use
brackets so that everyone reads the same expression the same way.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) 24 – 45

SECTIONS QUESTIONS

18 56

MEDIUM

English

In-text Questions — Page 24
Section 2.1 Simple Expressions

Q1 Choose your favourite number and write as many expressions as you can having
that value.

Let us choose 24. Using two numbers and any one of +, –, × and ÷ we can write many
expressions whose value is 24.

20 + 4 = 24

30 – 6 = 24

4 × 6 = 24

48 ÷ 2 = 24

12 + 12 = 24

34 – 10 = 24

3 × 8 = 24

96 ÷ 4 = 24

We can also use more than two numbers:

Page 1 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

10 + 10 + 4 = 24

5 × 4 + 4 = 24

2 × (10 + 2) = 24

Why it happens: A number does not have one “correct” expression. Every pair of
numbers that adds, subtracts, multiplies or divides to 24 gives another expression
with the same value. That is why different-looking expressions can be joined by an ‘=’
sign.

Try This: Pick your own favourite number — your house number or your jersey
number — and write ten expressions for it. Try to use each of +, –, × and ÷ at least
twice.

Figure it Out — Page 25
Section 2.1 Comparing Expressions

Q1 Fill in the blanks to make the expressions equal on both sides of the = sign: (a) 13 + 4
= ____ + 6 (b) 22 + ____ = 6 × 5 (c) 8 × ____ = 64 ÷ 2 (d) 34 – ____ = 25

Find the value of the side that is complete, then work backwards on the other side.

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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

(a) 13 + 4 = 17, so ___ + 6 = 17 → ___ = 17 – 6 = 11

13 + 4 = 11 + 6

(b) 6 × 5 = 30, so 22 + ___ = 30 → ___ = 30 – 22 = 8

22 + 8 = 6 × 5

(c) 64 ÷ 2 = 32, so 8 × ___ = 32 → ___ = 32 ÷ 8 = 4

8 × 4 = 64 ÷ 2

(d) 34 – ___ = 25 → ___ = 34 – 25 = 9

34 – 9 = 25

Why it happens: The ‘=’ sign does not mean “write the answer here”. It means both
sides have the same value. So we first find the value of the complete side and then
choose the missing number that gives that same value.

Check it yourself: Put your answers back in — 11 + 6 = 17 ✓, 22 + 8 = 30 ✓, 8 × 4 =
32 ✓, 34 – 9 = 25 ✓.

Q2 Arrange the following expressions in ascending (increasing) order of their values.
(a) 67 – 19 (b) 67 – 20 (c) 35 + 25 (d) 5 × 11 (e) 120 ÷ 3

First find each value.

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Class 7 Maths Chapter 2 Arithmetic Expressions
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EXPRESSION WORKING VALUE

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(a) 67 – 19
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(c) 35 + 25 35 + 25 60

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Why it happens: Expressions are ordered by their values, not by how they look.
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In-text Questions — Page 26
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Section 2.1 Comparing Expressions

Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do
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it without complicated calculations? Explain your thinking in each case. (a) 245 +
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289 ___ 246 + 285 (b) 273 – 145 ___ 272 – 144 (c) 364 + 587 ___ 363 + 589 (d) 124 + 245 ___

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129 + 245 (e) 213 – 77 ___ 214 – 76

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Compare the two sides part by part, the way we compared Raja’s and m
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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

(a) 245 + 289 > 246 + 285

245 is 1 less than 246, but 289 is 4 more than 285. Net gain = 4 – 1 = 3, so the left side is 3

more.
(Check: 534 and 531.)

(b) 273 – 145 = 272 – 144

273 is 1 more than 272, but we also take away 1 more (145 instead of 144). The extra and

the loss cancel.

(Check: 128 and 128.)

(c) 364 + 587 < 363 + 589
364 is 1 more than 363, but 587 is 2 less than 589. Net = 1 – 2 = –1, so the left side is 1 less.

(Check: 951 and 952.)

(d) 124 + 245 < 129 + 245

The second number is the same on both sides; 124 < 129, so the left sum is 5 less.

(Check: 369 and 374.)

(e) 213 – 77 < 214 – 76

The right side starts 1 higher and takes away 1 less, so it is 2 more.

(Check: 136 and 138.)

Why it happens: Think of each expression as a story. In (b) Raja starts with 1 rupee
more than Joy but also spends 1 rupee more — so both end with the same amount.
Comparing the changes is far quicker than adding four-digit numbers.

Tip: When comparing a + b with c + d, ask “how much did the first number go up?”
and “how much did the second number go down?” The bigger movement decides
the sign.

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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

In-text Questions — Pages 28 – 29
Section 2.2 Terms in Expressions

TRY THIS

Q1 Check if replacing subtraction by addition in this way does not change the value of
the expression, by taking different examples.

Take a few pairs and compare the two values.

SUBTRACTION FORM ADDITION FORM VALUE

18 – 10 18 + (–10) 8

83 – 14 83 + (–14) 69

40 – 55 40 + (–55) –15

–7 – 6 –7 + (–6) –13

100 – 100 100 + (–100) 0

In every case both forms give the same value.

Why it happens: Subtracting 14 and adding –14 do exactly the same job — both
move you 14 steps to the left on the number line. So a – b = a + (–b) always.

Tip: This is why we may rewrite every subtraction as an addition. Once an expression
is a pure sum, its parts are its terms and we can shuffle them freely.

Q2 Can you explain why subtracting a number is the same as adding its inverse, using
the Token Model of integers that we saw in the Class 6 textbook of mathematics?

In the Token Model a positive token (+1) and a negative token (–1) together make zero. Such a
pair can be added or taken away at any time without changing the value.
Take 5 – 3. Start with 5 positive tokens and remove 3 of them:

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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

+ + + + + → remove 3 positives → + + = 2

Now take 5 + (–3). Start with 5 positive tokens and put in 3 negative tokens. Each negative
cancels one positive:

+ + + + + and – – –

(+ –) (+ –) (+ –) cancel to 0, leaving + + = 2

Both actions leave the same 2 positive tokens, so 5 – 3 = 5 + (–3).

Why it happens: Removing a positive token and adding a negative token have the
same effect on the total. If there are not enough positive tokens to remove, we first
put in as many zero-pairs as we need — the value does not change — and then
remove. That is exactly what adding the inverse does.

Check it yourself: Try 3 – 7 with tokens. Add four zero-pairs first, remove 7 positives,
and you are left with 4 negative tokens, i.e. –4 — the same as 3 + (–7).

Page 7 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q3 In the following table, some expressions are given. Complete the table.

EXPRESSION EXPRESSION AS THE SUM OF ITS TERMS TERMS

13 – 2 + 6 13, – 2, 6

13 + –2 + 6

5+6×3

5 + 6×3

4 + 15 – 9

+ +

23 – 2 × 4 + 16

+ +

28 + 19 – 8

+ +

Page 29 — the table as printed. The first row and part of the second are already filled in.

Turn every subtraction into “add the inverse”, then read off the terms.

EXPRESSION EXPRESSION AS THE SUM OF ITS TERMS TERMS

13 – 2 + 6 13 + (–2) + 6 13, –2, 6

5+6×3 5 + (6 × 3) 5, 6 × 3

4 + 15 – 9 4 + 15 + (–9) 4, 15, –9

23 – 2 × 4 + 16 23 + (–2 × 4) + 16 23, –2 × 4, 16

28 + 19 – 8 28 + 19 + (–8) 28, 19, –8

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Class 7 Maths Chapter 2 Arithmetic Expressions
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Their values are 17, 23, 10, 31 and 39 respectively.
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Why it happens: A term is a piece separated by a ‘+’ sign. A chain of × or ÷ carries no

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‘+’ inside it, so 6 × 3 and 2 × 4 stay together as single terms. The minus sign in front
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Tip: Count the ‘+’ signs after rewriting. Number of terms = number of ‘+’ signs + 1.
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Q4 Does changing the order in which the terms are added give different values?
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No. The value stays the same however we order the terms.
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14 + 10 + (–5) = 24 + (–5) = 19

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(–5) + 10 + 14 = 5 + 14 = 19

10 + (–5) + 14 = 5 + 14 = 19 ag l a

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Why it happens: Adding is like collecting things into one heap. The heap is the same
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Tip: This only works once every subtraction has been turned into “add a negative
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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

EXPRESSION TERMS SWAPPED VALUE

6 + (–4) (–4) + 6 2

(–4) + (–2) (–2) + (–4) –6

(–6) + (–8) (–8) + (–6) –14

(–15) + 9 9 + (–15) –6

25 + (–25) (–25) + 25 0

Why it happens: On the number line the two moves are the same two jumps — one
to the right and one to the left. Doing the left jump first and the right jump second
lands you at exactly the same point.

In-text Questions — Page 30
Section 2.2 Swapping and Grouping

TRY THIS

Q1 Can you explain why this is happening using the Token Model of integers that we
saw in the Class 6 textbook of mathematics?

Swapping two terms does not change the value because the tokens on the table do not change
— only the order in which we put them down changes.
Take 6 + (–4). Put down 6 positive tokens, then 4 negative tokens. Four zero-pairs cancel, leaving
2 positives, i.e. 2.
Take (–4) + 6. Put down 4 negative tokens first, then 6 positive tokens. The same four zero-pairs
cancel, leaving 2 positives, i.e. 2.

6 + (–4) = 2 and (–4) + 6 = 2

Why it happens: The final pile depends only on which tokens are on the table, never
on the order they arrived. That is exactly the commutative property of addition.

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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q2 Will this also hold when there are terms having negative numbers as well? Take
some more expressions and check.

Yes. Grouping the terms in either way gives the same value, even with negative terms.

EXPRESSION FIRST TWO GROUPED LAST TWO GROUPED VALUE

(–7) + 10 + (–11) (3) + (–11) (–7) + (–1) –8

(–5) + (–6) + 20 (–11) + 20 (–5) + 14 9

12 + (–4) + (–8) (8) + (–8) 12 + (–12) 0

(–3) + (–9) + (–4) (–12) + (–4) (–3) + (–13) –16

Why it happens: This is the associative property of addition. Whichever pair you add
first, all three terms end up in the same total.

Q3 Can you explain why this is happening using the Token Model of integers that we
saw in the Class 6 textbook of mathematics?

Take (–7) + 10 + (–11) as tokens: 7 negatives, 10 positives, 11 negatives — that is 18 negatives
and 10 positives in all.

10 zero-pairs cancel

18 – 10 = 8 negative tokens are left

Value = –8

Group the first two first: (–7) + 10 gives 3 positives; then 3 positives with 11 negatives leaves 8
negatives.
Group the last two first: 10 + (–11) gives 1 negative; then 7 negatives with 1 negative gives 8
negatives.

Why it happens: The whole collection of tokens is the same in both routes — 10
positives and 18 negatives. Cancelling zero-pairs in a different order cannot change
what is finally left over.

Page 11 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Try This: Do the same with (–6) + (–7) + (–13). There are no positives at all, so nothing
cancels and the value is –26 whichever pair you add first.

In-text Questions — Page 31
Section 2.2 Swapping and Grouping

TRY THIS

Q1 Does adding the terms of an expression in any order give the same value? Take
some more expressions and check. Consider expressions with more than 3 terms
also.

Yes. Any order and any grouping gives the same value.
Three terms: 9 + 4 + 6

(9 + 4) + 6 = 13 + 6 = 19

9 + (4 + 6) = 9 + 10 = 19

6 + 9 + 4 = 15 + 4 = 19

Four terms (one negative): 15 + (–8) + 5 + 12

(15 + (–8)) + 5 + 12 = 7 + 5 + 12 = 24

15 + 5 + 12 + (–8) = 32 + (–8) = 24

(15 + 5) + (12 + (–8)) = 20 + 4 = 24

Five terms: 20 + (–7) + (–3) + 11 + (–1) = 20 in every order.

Why it happens: Commutativity lets us swap any two terms and associativity lets us
bracket any two together. Repeating these two moves turns any order into any other
order, so all orders give the same sum.

Tip: Use this to add faster. In 20 + (–7) + (–3) + 11 + (–1), first collect all the negatives:
–7 – 3 – 1 = –11, then 20 + 11 – 11 = 20.

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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q2 Can you explain why this is happening using the Token Model of integers that we
saw in the Class 6 textbook of mathematics?

Write every term as a pile of tokens and push all the piles onto one table.
For 15 + (–8) + 5 + 12 the table holds 15 + 5 + 12 = 32 positive tokens and 8 negative tokens.

8 zero-pairs cancel

32 – 8 = 24 positive tokens are left

Value = 24

Why it happens: The order in which the piles are pushed onto the table does not
change how many positive and negative tokens end up on it. Since the value is
decided only by the leftover tokens after cancelling, every order must give the same
value.

Note: The Hindi edition prints this Token-Model prompt on page 30; the English
edition repeats it here on page 31.

Q3 Manasa is adding a long list of numbers. It took her five minutes to add them all
and she got the answer 11749. Then she realised that she had forgotten to include
the fourth number 9055. Does she have to start all over again?

No, she does not. She only has to add the missing number to the total she already has.

Correct sum = 11749 + 9055

= 11749 + 9000 + 55

= 20749 + 55

= 20804

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Class 7 Maths Chapter 2 Arithmetic Expressions
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Why it happens: The terms of a sum may be added in any order and in any

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grouping (commutative and associative properties). So leaving 9055 for the very end
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Q4 Manasa is going outside to play. Her mother says, “Wear your hat and shoes!”

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a It does not matter — she can wear the hat first and then the shoes, or the shoes first and
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a Addition of terms commutes: 6 + (–4) = (–4) + 6. Subtraction does not: 10 – 4 is not 4 –
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Try This: List three pairs of everyday actions that commute (comb hair / drink water)

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In-textmQuestions — Page 32 as e
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Section 2.2 More Expressions and Their Terms

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Seven friends order one dosa each at ₹23, and the tip stays ₹5.

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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Expression = 7 × 23 + 5

As a sum of terms: (7 × 23) + (5)

= 161 + 5

= ₹166

Terms: 7 × 23 and 5 — two terms.

Why it happens: 7 × 23 has no ‘+’ inside it, so the whole product is a single term.
Evaluating each term first and then adding gives 161 + 5 = 166. Writing 7 × (23 + 5)
would be wrong — that would mean paying a ₹5 tip to each of the seven friends’
dosas.

Check it yourself: 7 × (23 + 5) = 7 × 28 = ₹196, which is ₹30 too much.

Q2 Think and discuss why she wrote this. (Ruby wrote 6 × 5 + 3 when the teacher called
out ‘5’.)

There were 33 students playing. When the teacher called out ‘5’, the children formed groups of
5.

33 = 6 groups of 5, with 3 children left over

6 × 5 = 30 children in complete groups

3 children could not join any group

Expression = 6 × 5 + 3 = 30 + 3 = 33

The terms are 6 × 5 and 3.

Why it happens: Ruby is describing the picture, not just the total. The term 6 × 5
records how many complete groups formed, and the term 3 records the children
who were out. “6 × 5 + 3” is read as “3 more than 6 × 5”.

Page 15 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Tip: This is exactly division with remainder: 33 ÷ 5 gives quotient 6 and remainder 3,
so 33 = 6 × 5 + 3.

In-text Questions — Page 33
Section 2.2 More Expressions and Their Terms

Q1 For each of the cases below, write the expression and identify its terms: If the
teacher had called out ‘4’, Ruby would write ____________ . If the teacher had called out
‘7’, Ruby would write ____________ . Write expressions like the above for your class size.

There are 33 students playing, so we split 33 into complete groups plus those left out.

Teacher calls ‘4’: 33 ÷ 4 → 8 groups, 1 left out

Ruby would write 8 × 4 + 1

Terms: 8 × 4 and 1 (value 32 + 1 = 33)

Teacher calls ‘7’: 33 ÷ 7 → 4 groups, 5 left out

Ruby would write 4 × 7 + 5

Terms: 4 × 7 and 5 (value 28 + 5 = 33)

For your own class: suppose 40 children are playing.

TEACHER CALLS EXPRESSION TERMS

3 13 × 3 + 1 13 × 3, 1

5 8×5+0 8×5

6 6×6+4 6 × 6, 4

7 5×7+5 5 × 7, 5

Why it happens: The first term always counts children inside the groups and the
second term counts children who are out. Their sum must come back to the class
size.

Page 16 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Check it yourself: Count the students in your own class and repeat the game on
paper for the numbers 2 to 9.

Q2 Identify the terms in the two expressions above. (432 = 4 × 100 + 1 × 20 + 1 × 10 + 2 ×
1 and 432 = 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1)

Each product is a single term, because none of them contains a ‘+’ sign.

WAY EXPRESSION TERMS CHECK

1 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1 4 × 100, 1 × 20, 1 × 10, 2 × 1 400 + 20 + 10 + 2 = 432

2 8 × 50 + 1 × 10 + 4 × 5 + 2 × 1 8 × 50, 1 × 10, 4 × 5, 2 × 1 400 + 10 + 20 + 2 = 432

Each expression has four terms.

Why it happens: In each term the first number is how many notes or coins and the
second is the value of one note or coin. Adding the four terms gives the total money
handed over.

Q3 Can you think of some more ways of giving ₹432 to someone?

Yes — many. Here are four more, using ₹100, ₹50, ₹20, ₹10 notes and ₹5, ₹1 coins.

EXPRESSION MEANING TOTAL

4 × 100 + 6 × 5 + 2 × 1 4 notes of ₹100, 6 coins of ₹5, 2 coins of ₹1 ₹432

2 × 100 + 4 × 50 + 1 × 20 + 2 × 5 + 2 × 1 2 × ₹100, 4 × ₹50, 1 × ₹20, 2 × ₹5, 2 × ₹1 ₹432

8 × 50 + 3 × 10 + 2 × 1 8 notes of ₹50, 3 notes of ₹10, 2 coins of ₹1 ₹432

21 × 20 + 2 × 5 + 2 × 1 21 notes of ₹20, 2 coins of ₹5, 2 coins of ₹1 ₹432

Check the second one: 200 + 200 + 20 + 10 + 2 = 432 ✓

Page 17 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Why it happens: Only the total is fixed at ₹432. Any set of notes and coins that adds
up to 432 works, so there are very many correct expressions — and each one has as
many terms as the kinds of money used.

Try This: What is the smallest number of notes and coins that can make ₹432?
(Answer: 4 × 100 + 1 × 20 + 1 × 10 + 2 × 1 — that is 8 pieces.)

In-text Questions — Page 34
Section 2.2 More Expressions and Their Terms

Q1 What is the expression for the arrangement in the right making use of the number
of yellow and blue squares?

Arrangement on the left Arrangement on the right

Page 33 — the two arrangements of squares. The left one is described by 5 × 2 + 3.

The right-hand picture has two identical columns. Each column is made of 5 yellow squares on
top and 3 blue squares below.

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Class 7 Maths Chapter 2 Arithmetic Expressions
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One column = 5 + 3 squares
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Two such columns = 2 × (5 + 3)
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5 + 3 + 5 + 3 OR 5 × 2 + 3 × 2

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Why it happens: Brackets are needed here. Without them, 2 × 5 + 3 would mean “3

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more than 2 × 5” = 13, which describes the left picture (5 columns of 2 green squares
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Figure it Out — Pages 34 – 35
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Section 2.2 More Expressions and Their Terms
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First write each expression as a sum of terms, evaluate each term, then add.

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(a) 28 + (–7) + 8 28, –7, 8 29

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39 + (–2 × 6) + 11 39, –2 × 6, 11 38

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(e) (6 × 3) + (–4 × 8 × 5) 6 × 3, –4 × 8 × 5 –142

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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

(a) 28 – 7 + 8 = 28 + (–7) + 8 = 21 + 8 = 29

(b) –2 × 6 = –12, so 39 + (–12) + 11 = 27 + 11 = 38

(c) 40 + (–10) + 10 + 10 = 30 + 10 + 10 = 50

(d) –10 × 2 = –20 and 16 ÷ 2 = 8, so 48 + (–20) + 8 = 28 + 8 = 36

(e) 6 × 3 = 18 and 4 × 8 × 5 = 160, so 18 + (–160) = –142

Why it happens: A × or ÷ chain has no ‘+’ inside it, so it stays as one term. That is
why in (b) we do 2 × 6 before adding, and in (e) the whole product 4 × 8 × 5 carries
the minus sign.

Tip: In (c) do not read “40 – 10 + 10 + 10” as 40 – 30. The minus belongs only to the
first 10.

Q2 Write a story/situation for each of the following expressions and find their values.
(a) 89 + 21 – 10 (b) 5 × 12 – 6 (c) 4 × 9 + 2 × 6

(a) 89 + 21 – 10
Story: A school library had 89 books on its shelves. The librarian bought 21 new books. Later, 10
books were issued to students. How many books are left on the shelves?

89 + 21 + (–10) = 110 – 10 = 100 books

(b) 5 × 12 – 6
Story: Ravi buys 5 boxes of mangoes with 12 mangoes in each box. On the way home 6 mangoes
get spoilt. How many good mangoes does he have?

(5 × 12) + (–6) = 60 – 6 = 54 mangoes

Page 20 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

(c) 4 × 9 + 2 × 6
Story: In a rangoli, Meena uses 4 rows of 9 yellow dots and 2 rows of 6 red dots. How many dots
are there in all?

(4 × 9) + (2 × 6) = 36 + 12 = 48 dots

Why it happens: A good story must match the terms. In (b) the term 5 × 12 is “5
boxes of 12” and the term –6 is “6 lost”. If the story were “each box loses 6 mangoes”,
the expression would be 5 × (12 – 6) instead.

Page 21 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q3 For each of the following situations, write the expression describing the situation,
identify its terms and find the value of the expression. (a) Queen Alia gave 100 gold
coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa
used the coins to start a business and doubled her coins. Princess Anna bought
jewellery and has only half of the coins left. Write an expression describing how
many gold coins Princess Elsa and Princess Anna together have. (b) A metro train
ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total
cost of tickets: (i) for four adults and three children? (ii) for two groups having three
adults each? (c) Find the total height of the window by writing an expression
describing the relationship among the measurements shown in the picture.

Border
3 cm

Grill
2 cm
Total
Gap Height

5 cm

The window on page 35 — its border, grill bars and gaps, with the measurement of each
marked.

(a) The two princesses

Page 22 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Elsa doubled her 100 coins → 2 × 100

Anna has half of her 100 coins → 100 ⁄ 2

Expression = 2 × 100 + 100 ⁄ 2

Terms: 2 × 100 and 100 ⁄ 2

Value = 200 + 50 = 250 gold coins

(b) Metro tickets

(i) Four adults and three children

Expression = 4 × 40 + 3 × 20

Terms: 4 × 40 and 3 × 20

Value = 160 + 60 = ₹220

(ii) Two groups of three adults each

Expression = 2 × (3 × 40)
Term: 2 × (3 × 40) — a single term

Value = 2 × 120 = ₹240

(c) Height of the window
Reading the picture: the window has a border of 3 cm at the top and 3 cm at the bottom, 6 grill
bars each 2 cm thick, and 7 gaps each 5 cm tall (one above the first bar, one below the last bar
and one between every two bars).

Expression = 7 × 5 + 6 × 2 + 2 × 3

Terms: 7 × 5, 6 × 2, 2 × 3
Value = 35 + 12 + 6 = 53 cm

Page 23 of 56

as e
Class 7 Maths Chapter 2 Arithmetic Expressions
a g l AglaSem · NCERT Solutions

m
m .co
m l a se
. co border a3g cm
e m
g l as
a
grill 2 cm
co m
e m . ag
g l as
a
gap 5 cm
co m
e m.
m l as
m .co a g
l a se
a g
m Total = 53 cm a s
m .co agl
l a se
ag

. com
m emcm
borderas3
e m . co agl
g l as
a
m
The window as 2 borders (3 cm each), 6 grill bars (2 cm each) and 7 gaps (5 cm each).

a se
.com a g l
m
ase
Why it happens: Count carefully before writing the expression. With 6 bars there are
agl
always 7 gaps — one more gap than the number of bars — because a gap sits above
the first bar and below the last one.

co m
m .
m as e
.c+o5 + 2 + 5 + 3 = 53 cm ✓ g l
Check it yourself: Add the pieces one by one: 3 + 5 + 2 + 5 + 2 + 5 + 2 + 5 + 2 + 5 + 2
a
se m
g l a
a c
m .
a s e
In-text Questions — Page 37 com
e m. agl
g l as
a

com
m .
m ase
.co


a g l Page 24 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Tinker the Terms I

Q1 Some expressions are given in following three columns. In each column, one or
more terms are changed from the first expression. Go through the example (in the
first column) and fill the blanks, doing as little computation as possible.

Change one term by a little and the value changes by exactly the same little amount.
Column 1 — starting from 53 + (–16) = 37

54 + (–16) = 38 (54 is 1 more than 53, so the value is 1 more than 37)

53 + (–15) = 38 (–15 is 1 more than –16, so the value is 1 more than 37)

Column 2 — starting from 53 + (–16) = 37

52 + (–16) = 36 (52 is 1 less than 53, so the value is 1 less than 37)

53 + (–17) = 36 (–17 is 1 less than –16, so the value is 1 less than 37)

Column 3 — starting from –87 + (–16)

EXPRESSION REASONING VALUE

–87 + (–16) 87 + 16 = 103, both negative –103

–88 + (–15) first term 1 less, second term 1 more → no change –103

–86 + (–18) first term 1 more, second term 2 less → 1 less –104

–97 + (–26) first term 10 less, second term 10 less → 20 less than –103 –123

Why it happens: Careful with the negative terms. –15 is greater than –16 (it is one
step to the right on the number line), so replacing –16 by –15 increases the value.
Similarly –17 is one less than –16, so it decreases the value.

Check it yourself: –86 + (–18) = –(86 + 18) = –104 ✓ and –97 + (–26) = –(97 + 26) = –
123 ✓

Figure it Out — Pages 37 – 38

Page 25 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Removing Brackets — I

Q1 Fill in the blanks with numbers, and boxes with operation signs such that the
expressions on both sides are equal. (a) 24 + (6 – 4) = 24 + 6 ☐ _____ (b) 38 + (_____ ☐
_____) = 38 + 9 – 4 (c) 24 – (6 + 4) = 24 ☐ 6 – 4 (d) 24 – 6 – 4 = 24 – 6 ☐ _____ (e) 27 – (8 + 3) =
27 ☐ 8 ☐ 3 (f) 27 – (_____ ☐ _____) = 27 – 8 + 3

A bracket with a plus before it keeps the signs inside; a bracket with a minus before it flips
them.

(a) 24 + (6 – 4) = 24 + 6 – 4

(b) 38 + (9 – 4) = 38 + 9 – 4

(c) 24 – (6 + 4) = 24 – 6 – 4

(d) 24 – 6 – 4 = 24 – 6 – 4

(e) 27 – (8 + 3) = 27 – 8 – 3

(f) 27 – (8 – 3) = 27 – 8 + 3

LEFT SIDE RIGHT SIDE BOTH EQUAL

(a) 24 + 2 24 + 6 – 4 26

(b) 38 + 5 38 + 9 – 4 43

(c) 24 – 10 24 – 6 – 4 14

(d) 24 – 6 – 4 24 – 6 – 4 14

(e) 27 – 11 27 – 8 – 3 16

(f) 27 – 5 27 – 8 + 3 22

Page 26 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Why it happens: In (c) and (e) the whole bracket is being taken away, so both
numbers inside must be taken away. In (f) we are asked to put the bracket back:
since 27 – 8 + 3 takes away 8 but gives back 3, the bracket must contain 8 – 3.

Q2 Remove the brackets and write the expression having the same value. (a) 14 + (12 +
10) (b) 14 – (12 + 10) (c) 14 + (12 – 10) (d) 14 – (12 – 10) (e) –14 + (12 – 10) (f) 14 – (–12 –
10)

GIVEN BRACKETS REMOVED VALUE

(a) 14 + (12 + 10) 14 + 12 + 10 36

(b) 14 – (12 + 10) 14 – 12 – 10 –8

(c) 14 + (12 – 10) 14 + 12 – 10 16

(d) 14 – (12 – 10) 14 – 12 + 10 12

(e) –14 + (12 – 10) –14 + 12 – 10 –12

(f) 14 – (–12 – 10) 14 + 12 + 10 36

(b) 14 – 12 – 10 = 2 – 10 = –8

(d) 14 – 12 + 10 = 2 + 10 = 12

(f) the minus flips both –12 and –10 to +12 and +10 → 14 + 12 + 10 = 36

Why it happens: A bracket after a ‘+’ changes nothing. A bracket after a ‘–’ flips the
sign of every term inside. So in (f), taking away (–12 – 10) means taking away –22,
which is the same as adding 22.

Did you know? (a) and (f) have the same value 36, even though (f) looks full of
minus signs.

Page 27 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q3 Find the values of the following expressions. For each pair, first try to guess
whether they have the same value. When are the two expressions equal? (a) (6 + 10)
– 2 and 6 + (10 – 2) (b) 16 – (8 – 3) and (16 – 8) – 3 (c) 27 – (18 + 4) and 27 + (–18 – 4)

(a) (6 + 10) – 2 = 16 – 2 = 14

6 + (10 – 2) = 6 + 8 = 14 → equal

Both are 6 + 10 + (–2).

(b) 16 – (8 – 3) = 16 – 5 = 11

(16 – 8) – 3 = 8 – 3 = 5 → not equal

First is 16 – 8 + 3, second is 16 – 8 – 3.

(c) 27 – (18 + 4) = 27 – 22 = 5

27 + (–18 – 4) = 27 – 22 = 5 → equal

Both are 27 + (–18) + (–4).

When are the two expressions equal? They are equal exactly when, after removing the
brackets, both expressions become the same list of terms with the same signs.

Why it happens: In (b) the bracket sits after a minus sign, so 16 – (8 – 3) becomes 16
– 8 + 3 while (16 – 8) – 3 stays 16 – 8 – 3. The sign of 3 is different, so the values differ
by 6.

Q4 In each of the sets of expressions below, identify those that have the same value.
Do not evaluate them, but rather use your understanding of terms. (a) 319 + 537,
319 – 537, –537 + 319, 537 – 319 (b) 87 + 46 – 109, 87 + 46 – 109, 87 + 46 – 109, 87 – 46 +
109, 87 – (46 + 109), (87 – 46) + 109

(a) Write each as a sum of terms:

Page 28 of 56

as e
Class 7 Maths Chapter 2 Arithmetic Expressions
a g l AglaSem · NCERT Solutions

co m
m.
EXPRESSION TERMS

m as e
319 + 537
.co
319, 537
a g l
se m
g l a
a
319 – 537 319, –537

–537 + 319 –537, 319

com
m . ag
se
537 – 319 537, –319

l a
g value — the same two terms, only swapped.
319 – 537 and –537 + 319 have theasame

m
(b) Again compare the terms:
co
em.
m l as
.co
EXPRESSION TERMS GROUP

a g
se m
l a
87 + 46 – 109 (three times) 87, 46, –109 Group 1

g
a 87 – 46 + 109 87, –46, 109 Group 2

m a s
.co 87, –46, –109 agl
87 – (46 + 109) on its own

se m
g l a
a
(87 – 46) + 109 87, –46, 109 Group 2

m
So the three copies of 87 + 46 – 109 are equal to one another, and 87 – 46 + 109 = (87 – 46) +

. co
m
109. The odd one out is 87 – (46 + 109).

o m l a se
.c it happens: Two expressions built only from additions
a gare equal when they
m
asecarry the same terms with the same signs, no matter what order they appear in.
Why

agl Removing the bracket in 87 – (46 + 109) flips 109 to –109, which puts it in a group of

se m
com a
its own.
. a g l
m
ase
agl
Check it yourself: Group 1 = 24, Group 2 = 150, and 87 – (46 + 109) = –68.

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

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m .
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.co


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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q5 Add brackets at appropriate places in the expressions such that they lead to the
values indicated. (a) 34 – 9 + 12 = 13 (b) 56 – 14 – 8 = 34 (c) –22 – 12 + 10 + 22 = –22

(a) 34 – (9 + 12) = 34 – 21 = 13

(without the bracket 34 – 9 + 12 would be 37)

(b) (56 – 14) – 8 = 42 – 8 = 34

(c) –22 – (12 + 10) + 22 = –22 – 22 + 22 = –22

Why it happens: In (a) the bracket makes 12 be taken away instead of added, a
change of 24 (37 → 13). In (c) the bracket after the minus flips +10 into –10, so the
middle part becomes –22, which cancels the last +22.

Tip: In (b) the value is already 34 without any bracket. The bracket only makes the
intended order explicit — that is allowed.

Page 30 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q6 Using only reasoning of how terms change their values, fill the blanks to make the
expressions on either side of the equality (=) equal. (a) 423 + ______ = 419 + ______ (b)
207 – 68 = 210 – ______

(a) 423 + 419 = 419 + 423

(commutative property — the same two terms, swapped)

Another answer: 423 + 6 = 419 + 10

(419 is 4 less than 423, so the second term must be 4 more)

(b) 207 – 68 = 210 – 71

210 is 3 more than 207, so we must subtract 3 more: 68 + 3 = 71

Check: 207 – 68 = 139 and 210 – 71 = 139 ✓

Why it happens: If the first term goes up by 3, the amount taken away must also go
up by 3 for the value to stay the same. Any change on one side must be balanced on
the other.

Q7 Using the numbers 2, 3 and 5, and the operators ‘+’ and ‘–’, and brackets, as
necessary, generate expressions to give as many different values as possible. For
example, 2 – 3 + 5 = 4 and 3 – (5 – 2) = 0.

Use each of 2, 3 and 5 once. Here are all the different values that can be reached.

Page 31 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

VALUE EXPRESSIONS

10 2 + 3 + 5; 5 + 3 + 2

6 3 – 2 + 5; 5 – 2 + 3; 5 + 3 – 2

4 2 – 3 + 5; 5 + 2 – 3; 5 – (3 – 2)

0 2 + 3 – 5; 3 – (5 – 2); 5 – 3 – 2; 5 – (3 + 2)

–4 3 – 2 – 5; 3 – (2 + 5)

–6 2 – 3 – 5; 2 – (3 + 5)

So we get six different values: –6, –4, 0, 4, 6 and 10.

Why it happens: With only + and –, the value is 2, 3 and 5 each taken with a plus or a
minus sign. Since the first number written is always added, the six sign patterns give
only these six sums. Brackets do not add new values here — they only give new ways
of writing the same six.

Try This: Repeat with 1, 4 and 6 and see how many different values you get.

Q8 Whenever Jasoda has to subtract 9 from a number, she subtracts 10 and adds 1 to it.
For example, 36 – 9 = 26 + 1. (a) Do you think she always gets the correct answer?
Why? (b) Can you think of other similar strategies? Give some examples.

(a) Yes, she always gets the correct answer.

9 = 10 – 1

So, any number – 9 = number – (10 – 1)

= number – 10 + 1 (bracket after a minus flips the signs)

36 – 9 = 36 – 10 + 1 = 26 + 1 = 27 ✓

84 – 9 = 84 – 10 + 1 = 74 + 1 = 75 ✓

123 – 9 = 123 – 10 + 1 = 113 + 1 = 114 ✓

Page 32 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

(b) Other strategies of the same kind:

TO DO THIS DO THIS INSTEAD EXAMPLE

subtract 8 subtract 10, add 2 55 – 8 = 55 – 10 + 2 = 47

subtract 99 subtract 100, add 1 432 – 99 = 432 – 100 + 1 = 333

subtract 19 subtract 20, add 1 76 – 19 = 76 – 20 + 1 = 57

add 9 add 10, subtract 1 48 + 9 = 48 + 10 – 1 = 57

multiply by 9 multiply by 10, subtract the number 23 × 9 = 230 – 23 = 207

Why it happens: Round numbers like 10 and 100 are easy to subtract. Writing the
awkward number as a round number minus a small correction turns one hard step
into two easy ones.

Q9 Consider the two expressions: a) 73 – 14 + 1, b) 73 – 14 – 1. For each of these
expressions, identify the expressions from the following collection that are equal to
it. (a) 73 – (14 + 1) (b) 73 – (14 – 1) (c) 73 + (–14 + 1) (d) 73 + (–14 – 1)

Remove every bracket first, then compare the terms.

EXPRESSION WITHOUT BRACKETS TERMS VALUE

(a) 73 – (14 + 1) 73 – 14 – 1 73, –14, –1 58

(b) 73 – (14 – 1) 73 – 14 + 1 73, –14, 1 60

(c) 73 + (–14 + 1) 73 – 14 + 1 73, –14, 1 60

(d) 73 + (–14 – 1) 73 – 14 – 1 73, –14, –1 58

73 – 14 + 1 = 60 is equal to (b) and (c)
73 – 14 – 1 = 58 is equal to (a) and (d)

Why it happens: Everything turns on the sign of the 1. A bracket after a minus flips
it — that is why 73 – (14 – 1) has +1 inside. A bracket after a plus leaves it alone.

Page 33 of 56

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Class 7 Maths Chapter 2 Arithmetic Expressions
a g l AglaSem · NCERT Solutions

co m
m.
In-text Questions — Page 39
m as e
Removing Brackets — II
.co a g l
se m
l a
MATH TALK

a g
What about the total amount they have to pay? Can it be described by the
m
Q1
expression: 2 × 43 + 24?
. co ag
e m
g l as
a
No. Lhamo and Norbu each ordered one vegetable cutlet (₹43) and one rasgulla (₹24), so each

co m
m.
share is 43 + 24 and the total is double the whole share.

m as e
2 × 43 .+co a g l
m
ase
24 as a sum of terms = (2 × 43) + (24)

agl = 86 + 24 = ₹110
s
This means “24 more than 2 × 43” — two cutlets but only one rasgulla. ✗
m a
m .co agl
l
Correct expression = 2 × (43 +g24)ase
a
= 2 × 67 = ₹134

co m
m .
m as e
.co
Same thing as two cutlets and two rasgullas:
a g l
a s e2m× 43 + 2 × 24 = 86 + 48 = ₹134 ✓
agl
se m
com l a
Therefore 2 × (43 + 24) = 2 × 43 + 2 × 24.
. a g
s e m
Why it happens: Without aabracket,
g l ‘×’ binds only to the number next to it, so 2 × 43
+ 24 has the two termsa2 × 43 and 24. The bracket in 2 × (43 + 24) makes the whole
share the thing that is doubled — this is the distributive property.
co m
m .
m as e
.co a g l
se mQ2
a
If another friend, Sangmu, joins them and orders the same items, what will be the

ag l expression for the total amount to be paid?
.c
s e m
om a
. c agl

s e m (₹43) and a rasgulla (₹24).
a
Now three friends each have a cutlet

agl

co m
m .
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.co


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Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Expression = 3 × (43 + 24)

= 3 × 67

= ₹201

The same amount, counted item by item:

3 × 43 + 3 × 24 = 129 + 72 = ₹201 ✓

Why it happens: Three equal shares of ₹67 can be counted either as 3 whole shares
or as 3 cutlets plus 3 rasgullas. Both routes must give the same bill, so 3 × (43 + 24) =
3 × 43 + 3 × 24.

Tip: For n friends the bill is n × (43 + 24) = 67n rupees.

In-text Questions — Page 40
Removing Brackets — II

Q1 5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why?

5 × 4 + 3 = (5 × 4) + (3) = 20 + 3 = 23
5 × (4 + 3) = 5 × 7 = 35

23 ≠ 35, so 5 × 4 + 3 ≠ 5 × (4 + 3)

Why it happens: In 5 × 4 + 3 there are two terms — the product 5 × 4 and the lone 3
— so only the 4 gets multiplied by 5. In 5 × (4 + 3) the bracket makes 4 + 3 one single
quantity, so both the 4 and the 3 get multiplied by 5. The difference is 5 × 3 – 3 = 12.

Tip: Picture it: 5 rows of 4 dots plus 3 loose dots (23 dots) is clearly not the same as 5
rows of 7 dots (35 dots).

Page 35 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q2 Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5?

Yes — all three are equal.

5 × (4 + 3) = 5 × 7 = 35

5 × (3 + 4) = 5 × 7 = 35

(3 + 4) × 5 = 7 × 5 = 35

Why it happens: Two rules are at work. Inside the bracket, 4 + 3 = 3 + 4 because
addition is commutative. Outside, 5 × 7 = 7 × 5 because multiplication is also
commutative. So the order of neither the addition nor the multiplication matters.

Check it yourself: 7 rows of 5 chairs and 5 rows of 7 chairs both seat 35 people —
the hall is just turned sideways.

In-text Questions — Page 41
Tinker the Terms II

MATH TALK

Q1 Find this value. (97 × 25 = 100 × 25 – 3 × 25)

97 × 25 = (100 – 3) × 25

= 100 × 25 – 3 × 25

= 2500 – 75

= 2425

Why it happens: 97 twenty-fives are 100 twenty-fives with 3 twenty-fives taken
away. Multiplying by 100 is very easy, and taking away 75 is easy too — much
quicker than the usual long multiplication.

Page 36 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Check it yourself: By the usual method, 97 × 25 = 97 × 20 + 97 × 5 = 1940 + 485 =
2425 ✓

Q2 Use this method to find the following products: (a) 95 × 8 (b) 104 × 15 (c) 49 × 50. Is
this quicker than the multiplication procedure you use generally?

(a) 95 × 8 = (100 – 5) × 8
= 100 × 8 – 5 × 8

= 800 – 40 = 760

(b) 104 × 15 = (100 + 4) × 15

= 100 × 15 + 4 × 15

= 1500 + 60 = 1560

(c) 49 × 50 = (50 – 1) × 50

= 50 × 50 – 1 × 50

= 2500 – 50 = 2450

Yes, it is quicker — each product becomes one easy multiplication by a round number plus one
easy addition or subtraction, with no carrying.

Why it happens: This is the distributive property: the multiple of a sum (or
difference) is the sum (or difference) of the multiples. Splitting 95 as 100 – 5 keeps
the answer the same but makes the arithmetic mental.

Q3 Which other products might be quicker to find like the ones above?

Any product where one number sits just above or just below a round number — a multiple of
10, 50, 100 or 1000.

Page 37 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

PRODUCT REWRITTEN VALUE

98 × 7 (100 – 2) × 7 = 700 – 14 686

103 × 9 (100 + 3) × 9 = 900 + 27 927

49 × 5 (50 – 1) × 5 = 250 – 5 245

52 × 12 (50 + 2) × 12 = 600 + 24 624

999 × 6 (1000 – 1) × 6 = 6000 – 6 5994

18 × 25 (20 – 2) × 25 = 500 – 50 450

Why it happens: Multiplying by 10, 100 or 1000 only shifts digits, and 25 or 50 are
easy halves and quarters of 100. Splitting the awkward number as round ± small
pushes all the difficulty into one tiny product.

Try This: Use it while shopping — 6 packets at ₹99 each is 6 × 100 – 6 = ₹594.

Figure it Out — Pages 41 – 42
The Distributive Property

Q1 Fill in the blanks with numbers, and boxes by signs, so that the expressions on both
sides are equal. (a) 3 × (6 + 7) = 3 × 6 + 3 × 7 (b) (8 + 3) × 4 = 8 × 4 + 3 × 4 (c) 3 × (5 + 8) = 3
×5 ☐ 3 × ____ (d) (9 + 2) × 4 = 9 × 4 ☐ 2 × ____ (e) 3 × (____ + 4) = 3 ____ + ____ (f) (____ + 6) × 4
= 13 × 4 + ____ (g) 3 × (____ + ____) = 3 × 5 + 3 × 2 (h) (____ + ____) × ____ = 2 × 4 + 3 × 4 (i) 5 × (9
☐ 5 × ____ (l) (8 – 3)
– 2) = 5 × 9 – 5 × ____ (j) (5 – 2) × 7 = 5 × 7 – 2 × ____ (k) 5 × (8 – 3) = 5 × 8
× 7 = 8 × 7 ☐ 3 × 7 (m) 5 × (12 – ____) = ____ ☐ 5 × ____ (n) (15 – ____) × 7 = ____ ☐ 6 × 7 (o) 5 ×
(____ – ____) = 5 × 9 – 5 × 4 (p) (____ – ____) × ____ = 17 × 7 – 9 × 7

Multiply the outside number into every term inside the bracket, keeping each sign.

Page 38 of 56

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Class 7 Maths Chapter 2 Arithmetic Expressions
a g l AglaSem · NCERT Solutions

co m
e m.
(a) 3 × (6 + 7) = 3 × 6 + 3 × 7 (given)
m l as
.co
(b) (8 + 3) × 4 = 8 × 4 + 3 × 4 (given)
m a g
l a se
g
(c) 3 × (5 + 8) = 3 × 5 + 3 × 8
a(d) (9 + 2) × 4 = 9 × 4 + 2 × 4

co m
. ag
(e) 3 × (10 + 4) = 3 × 10 + 3 × 4
e m
(f) (13 + 6) × 4 = 13 × 4 + 6 × 4
g l as
(g) 3 × (5 + 2) = 3 × 5 + 3 × 2 a

co m
m.
(h) (2 + 3) × 4 = 2 × 4 + 3 × 4

m as e
.co l
(i) 5 × (9 – 2) = 5 × 9 – 5 × 2
a g
a s em
(j) (5 – 2) × 7 = 5 × 7 – 2 × 7

a gl (k) 5 × (8 – 3) = 5 × 8 – 5 × 3
m a s
agl
(l) (8 – 3) × 7 = 8 × 7 – 3 × 7

m .co
se
(m) 5 × (12 – 3) = 5 × 12 – 5 × 3

×g
(n) (15 – 6) × 7 = 15 × 7 – 6 a 7 l a

m
(o) 5 × (9 – 4) = 5 × 9 – 5 × 4
. co
e m
as
(p) (17 – 9) × 7 = 17 × 7 – 9 × 7
m l
m .co a g
a s eWhy
agl it happens: The distributive property says a × (b + c) = a × b + a × c and a × (b –
c) = a × b – a × c. In (e) and (m) any number may be used in the first blank, as long as
se m
com
the same number is repeated on the right — for example 3 × (7 + 4) = 3 × 7 + 3 × 4
g l a
m . a
ase
also works.

a gl
Check it yourself: (m) 5 × 9 = 45 and 60 – 15 = 45 ✓ ; (n) 9 × 7 = 63 and 105 – 42 = 63

co m
.
✓

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

co m
m .
m ase
.co


a g l Page 39 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q2 In the boxes below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and
RHS. Use reasoning and understanding of terms and brackets to figure this out and
not by evaluating the expressions. (a) (8 – 3) × 29 ☐ (3 – 8) × 29 (b) 15 + 9 × 18 ☐ (15 +
9) × 18 (c) 23 × (17 – 9) ☐ 23 × 17 + 23 × 9 (d) (34 – 28) × 42 ☐ 34 × 42 – 28 × 42

(a) (8 – 3) × 29 > (3 – 8) × 29

8 – 3 is positive (5) while 3 – 8 is negative (–5); the same multiplier 29 keeps the order.

(b) 15 + 9 × 18 < (15 + 9) × 18

LHS multiplies only the 9 by 18; RHS multiplies both 15 and 9 by 18, adding 15 × 17

extra.

(c) 23 × (17 – 9) < 23 × 17 + 23 × 9

LHS = 23 × 17 – 23 × 9, so it subtracts 23 × 9 while the RHS adds it.

(d) (34 – 28) × 42 = 34 × 42 – 28 × 42

This is exactly the distributive property.

Why it happens: No multiplication is actually needed. In (c) both sides share the
piece 23 × 17; one side then takes away 23 × 9 and the other adds it, so the side that
adds must be bigger.

Check it yourself: (a) 145 > –145 ✓ (b) 177 < 432 ✓ (c) 184 < 598 ✓ (d) 252 = 252
✓

Q3 Here is one way to make 14: _2_ × ( _1_ + _6_ ) = 14. Are there other ways of getting
14? Fill them out below: (a) ____ × (____ + ____) = 14 (b) ____ × (____ + ____) = 14 (c) ____ × (____
+ ____) = 14 (d) ____ × (____ + ____) = 14

Split 14 as a product first, then split one factor as a sum.

Page 40 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

14 = 2 × 7 = 7 × 2 = 1 × 14 = 14 × 1

EXPRESSION CHECK

(a) 2 × (5 + 2) = 14 2 × 7 = 14

(b) 2 × (3 + 4) = 14 2 × 7 = 14

(c) 7 × (1 + 1) = 14 7 × 2 = 14

(d) 14 × (0 + 1) = 14 14 × 1 = 14

More answers: 1 × (7 + 7), 2 × (6 + 1), 7 × (0 + 2).

Why it happens: The bracket must add up to a number that divides 14 exactly —
that is 1, 2, 7 or 14. Once the bracket total is fixed, the two numbers inside can be
split in any way you like.

Page 41 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q4 Find out the sum of the numbers given in each picture below in at least two
different ways. Describe how you solved it through expressions.

4 8 4

8 4 8

4 8 4
Picture 1

5 6 6 5

6 5 5 6

6 5 5 6

5 6 6 5
Picture 2

Page 42, Question 4 — the two pictures.

Picture (I) — a 3 × 3 arrangement: five yellow squares showing 4 and four blue circles showing
8, placed like the dots on a die.

Page 42 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

4 8 4

8 4 8

4 8 4

Way 1 (by colour): 5 × 4 + 4 × 8

= 20 + 32 = 52

Way 2 (row by row): (4 + 8 + 4) + (8 + 4 + 8) + (4 + 8 + 4)

= 16 + 20 + 16 = 52

Way 3: 2 × (4 + 8 + 4) + (8 + 4 + 8)

= 2 × 16 + 20 = 32 + 20 = 52

Picture (II) — a 4 × 4 arrangement of circles: eight blue circles showing 5 and eight red circles
showing 6.

5 6 6 5

6 5 5 6

6 5 5 6

5 6 6 5

Way 1 (by colour): 8 × 5 + 8 × 6

= 40 + 48 = 88

Way 2 (distributive): 8 × (5 + 6)

= 8 × 11 = 88

Way 3 (row by row): each row adds to 22, and there are 4 rows

4 × 22 = 88

Page 43 of 56

as e
Class 7 Maths Chapter 2 Arithmetic Expressions
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: Way 1 and Way 2 of Picture (II) are the two sides of the distributive

m l a se
property: 8 × 5 + 8 × 6 = 8 × (5 + 6). Grouping by colour or by row simply counts the
o
m .c in a different order, so the total cannot change. ag
same numbers

l a se
g
aTip: Look for a pattern before adding. In Picture (I) the four corners and the centre

co m
ag
are all 4s, and the four edge-middles are all 8s — that gives 5 × 4 + 4 × 8 straight
m .
away.
ase
a g l

co m
m.
Figure it Out — Pages 42 – 44
m as e
.co l
Chapter Review Exercise

a g
a s em
aglQ1 Read the situations given below. Write appropriate expressions for each of them
and find their values. (a) The district market in Begur operates on all seven days of

m a s
agl
a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam
.co
em
supplies 11 kg of mangoes each day from his orchard to this market. Find the

a s
gl She spends ₹5,000 on rent, ₹5,000 on food, and ₹2,000
amount of mangoes supplied by them in a week to the local district market. (b)
Binu earns ₹20,000 peramonth.
on other expenses every month. What is the amount Binu will save by the end of a

c o m
.
year? (c) During the daytime a snail climbs 3 cm up a post, and during the night
while asleep, accidentally slips down by 2 cm. The post is 10 cmm
a s e high, and a delicious

. om is on its top. In how many days will the snail get athe
ctreat gltreat?
s em
aANSWER
a g l
(a) Mangoes at the Begur market
se m
com g l a
m . a
ase
agl
Both together supply (9 + 11) kg each day, for 7 days:
Expression = 7 × (9 + 11)

co m
.
= 7 × 20 = 140 kg

em
c o m g l as
m .The other way round: 7 × 9 + 7 × 11 = 63 + 77 = 140 kg ✓ a
l a se
ag
.c
(b) Binu’s yearly saving
s e m
m a
em . co agl
g l as
a

com
m .
m ase
.co


a g l Page 44 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Expression = 12 × 20000 – 12 × (5000 + 5000 + 2000)

= 240000 – 12 × 12000

= 240000 – 144000

= ₹96,000

Shorter route: monthly saving = 20000 – 12000 = ₹8000

Yearly saving = 12 × 8000 = ₹96,000 ✓

(c) The snail on the 10 cm post

In one full day-and-night the snail gains 3 – 2 = 1 cm.

After 7 such cycles: 7 × (3 – 2) = 7 cm

On the 8th day it climbs 3 cm more: 7 + 3 = 10 cm — the top!

Expression = 7 × (3 – 2) + 3 = 7 + 3 = 10

Answer: 8 days

Why it happens: The snail does not take 10 days. Once it reaches the top during the
day it eats the treat and never slips back, so we only count full cycles until it is within
3 cm of the top — that is 7 cm after 7 days — and the last climb finishes the job on
day 8.

Check it yourself: Day-ends are 1, 2, 3, 4, 5, 6, 7 cm. On day 8 it is at 7 + 3 = 10 cm
before night falls.

Q2 Melvin reads a two-page story every day except on Tuesdays and Saturdays. How
many stories would he complete reading in 8 weeks? Which of the expressions
below describes this scenario? (a) 5 × 2 × 8 (b) (7 – 2) × 8 (c) 8 × 7 (d) 7 × 2 × 8 (e) 7 × 5 –
2 (f) (7 + 2) × 8 (g) 7 × 8 – 2 × 8 (h) (7 – 5) × 8

Melvin reads on 7 – 2 = 5 days a week, one story each of those days.

Page 45 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Stories in 8 weeks = (7 – 2) × 8 = 5 × 8 = 40 stories

The expressions that describe the scenario are (b) and (g).

EXPRESSION VALUE CORRECT?

(a) 5 × 2 × 8 80 No — this counts pages, not stories

(b) (7 – 2) × 8 40 Yes — 5 reading days × 8 weeks

(c) 8 × 7 56 No — all 56 days of 8 weeks

(d) 7 × 2 × 8 112 No

(e) 7 × 5 – 2 33 No

(f) (7 + 2) × 8 72 No

(g) 7 × 8 – 2 × 8 40 Yes — all 56 days minus the 16 skipped days

(h) (7 – 5) × 8 16 No — this counts the days he skips

Why it happens: (b) and (g) are the two sides of the distributive property: (7 – 2) × 8
= 7 × 8 – 2 × 8. One counts the reading days first; the other counts all days and then
removes the Tuesdays and Saturdays. Both give 40.

Q3 Find different ways of evaluating the following expressions: (a) 1 – 2 + 3 – 4 + 5 – 6 + 7
– 8 + 9 – 10 (b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1

(a) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10

Page 46 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Way 1 — pair them up:

(1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)

= (–1) + (–1) + (–1) + (–1) + (–1) = –5

Way 2 — collect positives and negatives:

(1 + 3 + 5 + 7 + 9) + (–2 – 4 – 6 – 8 – 10)

= 25 + (–30) = –5

Way 3 — left to right:

1 – 2 = –1; –1 + 3 = 2; 2 – 4 = –2; –2 + 5 = 3; 3 – 6 = –3;

–3 + 7 = 4; 4 – 8 = –4; –4 + 9 = 5; 5 – 10 = –5

(b) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1

Way 1 — pair them up:

(1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) = 0 + 0 + 0 + 0 + 0 = 0

Way 2 — collect signs:

(1 + 1 + 1 + 1 + 1) + (–1 –1 –1 –1 –1) = 5 + (–5) = 0

Way 3 — count the terms: five terms of +1 and five terms of –1 cancel exactly → 0

Why it happens: Once every subtraction is rewritten as adding a negative term, the
terms may be grouped and ordered any way we like. Pairing is fastest here because
each pair has a neat value.

Tip: Be careful — 1 – 1 + 1 – 1 + … is not 1 – (1 + 1 – 1 + …). The brackets you insert
must not change the signs.

Page 47 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q4 Compare the following pairs of expressions using ‘<’, ‘>’ or ‘=’ or by reasoning. (a) 49
–7+8 ☐ 49 – 7 + 8 (b) 83 × 42 – 18 ☐ 83 × 40 – 18 (c) 145 – 17 × 8 ☐ 145 – 17 × 6 (d) 23 ×
48 – 35 ☐ 23 × (48 – 35) (e) (16 – 11) × 12 ☐ –11 × 12 + 16 × 12 (f) (76 – 53) × 88 ☐ 88 × (53
– 76) (g) 25 × (42 + 16) ☐ 25 × (43 + 15) (h) 36 × (28 – 16) ☐ 35 × (27 – 15)

(a) 49 – 7 + 8 = 49 – 7 + 8 — identical expressions.

(b) 83 × 42 – 18 > 83 × 40 – 18 — 83 × 42 is 83 × 2 = 166 more than 83 × 40, and both lose

18.

(c) 145 – 17 × 8 < 145 – 17 × 6 — the left side takes away more (136 against 102).

(d) 23 × 48 – 35 > 23 × (48 – 35) — the right side is 23 × 48 – 23 × 35, which removes far

more than 35.

(e) (16 – 11) × 12 = –11 × 12 + 16 × 12 — the distributive property, terms just swapped.

(f) (76 – 53) × 88 > 88 × (53 – 76) — the first bracket is +23, the second is –23.

(g) 25 × (42 + 16) = 25 × (43 + 15) — both brackets equal 58.

(h) 36 × (28 – 16) > 35 × (27 – 15) — both brackets equal 12, and 36 > 35.

Page 48 of 56

as e
Class 7 Maths Chapter 2 Arithmetic Expressions
a g l AglaSem · NCERT Solutions

co m
m.
LHS SIGN RHS

m 50 as e
(a)
.co
=
a g l
50

se m
g
(b)
l a >
a
3468 3302

(c) 9 < 43

co m
m . ag
se
(d) 1069 > 299

l a
(e) 60
ag = 60

m
.co
(f) 2024 > –2024

1450em
s
com gla
(g) 1450 =

m . a
ase
(h) 432 > 420

agl
Why it happens: In every part it is enough to look at how the two sides differ. In (g)
m a s
.co
the bracket totals are the same (42 + 16 = 43 + 15 = 58), so the products must be
m agl
l a se
equal — one number gained exactly what the other lost.
ag

co m
m .
se
Q5 Identify which of the following expressions are equal to the given expression

o m l a
agis equal to the given
.cbrackets. There can be more than one expression which
without computation. You may rewrite the expressions using terms or removing

se m expression. (a) 83 – 37 – 12 : (i) 84 – 38 – 12 (ii) 84 – (37 + 12) (iii) 83 – 38 – 13 (iv) –37 +
g l a
a 83 – 12. (b) 93 + 37 × 44 + 76 : (i) 37 + 93 × 44 + 76 (ii) 93 + 37 × 76 + 44 (iii) (93 + 37) × (44

se m
a
+ 76) (iv) 37 × 44 + 93 + 76

. com a g l
m
ase
agl

(a) 83 – 37 – 12 has terms 83, –37, –12.

co m
OPTION TERMS
m . EQUAL?

m as e
.co
(i) 84 – 38 – 12
a g l
84, –38, –12 — both first numbers up by 1, so 84 – 38 = 83 – 37 Yes

a s em (ii) 84 – (37 + 12)
agl
84, –37, –12 — one more than the given No

.c
(iii) 83 – 38 – 13 83, –38, –13 — takes away 2 more No
s e m
m a
m . co agl
se
(iv) –37 + 83 – 12 –37, 83, –12 — same terms, reordered Yes

l a
tog83 – 37 – 12 (all three have value 34; (ii) is 35 and (iii) is 32).
So (i) and (iv) are equal a

co m
m .
m ase
.co


a g l Page 49 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

(b) 93 + 37 × 44 + 76 has terms 93, 37 × 44, 76.

OPTION TERMS EQUAL?

(i) 37 + 93 × 44 + 76 37, 93 × 44, 76 — the product is different No

(ii) 93 + 37 × 76 + 44 93, 37 × 76, 44 — the product is different No

(iii) (93 + 37) × (44 + 76) a single product 130 × 120 No

(iv) 37 × 44 + 93 + 76 37 × 44, 93, 76 — same terms, reordered Yes

So only (iv) is equal to 93 + 37 × 44 + 76 (both are 1797).

Why it happens: Two expressions are equal when they have the same terms,
whatever the order. In (b) swapping 93 and 37 changes which numbers sit inside the
product, so the term itself changes — that is why (i) fails.

Q6 Choose a number and create ten different expressions having that value.

Let us choose 36.

# EXPRESSION VALUE

1 30 + 6 36

2 40 – 4 36

3 6×6 36

4 72 ÷ 2 36

5 4×9 36

6 50 – 14 36

7 108 ÷ 3 36

8 12 × 3 36

9 2 × (15 + 3) 36

10 5×8–4 36

Page 50 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Why it happens: A value can be built in endless ways — as a sum, a difference, a
product, a quotient, or a mixture with brackets. All ten of these are joined by ‘=’
because they share the same value, not the same look.

Try This: Do the same for your age, and make sure at least three of your ten
expressions use brackets.

Expression Engineer! — Page 45
Chapter Challenge

TRY THIS

Q1 Using four 4’s, create expressions to get all values from 1 to 20.

Use exactly four 4s each time, with +, –, ×, ÷, brackets and the two-digit number 44.

Page 51 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

VALUE EXPRESSION

1 44 ÷ 44

2 4÷4+4÷4

3 (4 + 4 + 4) ÷ 4

4 4 + 4 × (4 – 4)

5 (4 × 4 + 4) ÷ 4

6 4 + (4 + 4) ÷ 4

7 44 ÷ 4 – 4

8 4+4+4–4

9 4+4+4÷4

10 (44 – 4) ÷ 4

12 (44 + 4) ÷ 4

15 4×4–4÷4

16 4+4+4+4

17 4×4+4÷4

20 (4 + 4 ÷ 4) × 4

With only +, –, × and ÷ (and the two-digit 44), the values 11, 13, 14, 18 and 19 cannot be reached
with exactly four 4s. They become possible once a decimal point is allowed:

11 = 4 ÷ .4 + 4 ÷ 4

13 = 4 ÷ .4 + 4 – 4 ÷ 4 (needs a fifth 4 — see the note)

14 = 4 + 4 + 4 + 4 ÷ .4 (also a five-4 answer)

18 = 4 ÷ .4 + 4 + 4

19 = 4 + 4 ÷ .4 + 4 + ...

The honest answer is this: 11, 13, 14, 18 and 19 have no four-4 expression using only +, –, ×
and ÷. Puzzle books allow the square root and the factorial for these, for example 19 = 4! – 4 – 4
÷ 4 and 14 = 4 × 4 – 4 ÷ √4.

Page 52 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Why it happens: Four 4s and four operations reach only a limited set of numbers,
because every 4 must be used once and the results of +, –, × and ÷ on 4s land on a
coarse grid. Puzzle-makers therefore allow extra tools — the decimal point (.4), the
square root and the factorial — to complete the run from 1 to 20.

Try This: How many of 1 to 20 can you reach with four 4s if 44 and 444 are allowed
but nothing else? Compare your list with a friend’s.

Q2 Using the numbers 1, 2, 3, 4, and 5 exactly once in any order get as many values as
possible between –10 and +10.

Use each of 1, 2, 3, 4, 5 once, with +, –, ×, ÷ and brackets. Every whole number from –10 to 10
can be reached.

VALUE EXPRESSION VALUE EXPRESSION

10 5+4+3–2×1 –1 1+2–3+4–5

9 1+2–3+4+5 –2 2+4–3–5×1

8 4+5+2–3×1 –3 1+2+3–4–5

7 1+2+3–4+5 –4 1×2+3–4–5

6 2×3+4–5+1 –5 1–2–3+4–5

5 1+2+3+4–5 –6 (4 – 5) × (1 + 2 + 3)

4 1×2+3+4–5 –7 1–2+3–4–5

3 1–2+3–4+5 –8 (4 – 5) × (1 + 3) × 2

2 5 × 2 – 4 × (3 – 1) –9 1+2–3–4–5

1 1+2–3–4+5 –10 2–3–4–5×1

0 5–4–3+2×1

Page 53 of 56

as e
Class 7 Maths Chapter 2 Arithmetic Expressions
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: Using only + and – on 1, 2, 3, 4, 5 always gives an odd total,

m l a se
because 1 + 2 + 3 + 4 + 5 = 15 is odd and flipping a sign changes the sum by an even
o
m
amount. To.c land on an even value we must bring in × or ÷ — thatagis why the even
a e use multiplication.
sabove
ag l
rows

co m
ag
Check it yourself: 5 × 2 – 4 × (3 – 1) = 10 – 8 = 2 ✓ and (4 – 5) × (1 + 3) × 2 = (–1) × 4 ×
m .
2 = –8 ✓
as e
a g l

co m
m.
Q3 Using the numbers 0 to 9 exactly once in any order, make an expression with a

m
value 100.
as e
.co a g l
se m
g l a

a Here is one neat solution using each of 0, 1, 2, …, 9 exactly once.
m a s
m .co agl
se
0+1+2+3+4+5+6+7+8×9

g l a
a
Terms: 0, 1, 2, 3, 4, 5, 6, 7 and 8 × 9
co m
m .
e
= (0 + 1 + 2 + 3 + 4 + 5 + 6 + 7) + 72
m l as
.co
= 28 + 72
a g
a s e=m100
agl
se m
Two more, if two-digit numbers are allowed:
com g l a
m . a
e
s 0 = 100
80 + 19 + 3 + 4 – 5 – 6 + 7 –l2a+
a g
91 + 5 + 8 + 4 – 7 – 6 + 3 + 2 + 0 = 100

co m
m .
m as e
.co a g l
Why it happens: The trick is to make one big term that carries most of the value.

s e m Here 8 × 9 = 72 does the heavy lifting, and the remaining digits 0 to 7 add up to
agla exactly 28 — the amount still needed to reach 100.
.c
s e m
m a
. co
Check it yourself: 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 and 8 × 9 = 72, so 28 + 72 = 100 ✓
e m agl
l as
Every digit from 0 to 9 has been used once.
g
a

co m
m .
m ase
.co


a g l Page 54 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Q4 What other similar interesting questions can you ask?

Here are some puzzles of the same family that you can set for your class.

Using three 5s and any operations, make every value from 1 to 10.
Using four 2s exactly once each, which numbers from 1 to 15 can you make?
Using the digits of the current year (for example 2, 0, 2 and 6) once each, make the numbers
1 to 12 — one for each month.
Using 1, 2, 3, 4 exactly once each, what is the largest value you can make? And the smallest?
Place brackets in 8 – 3 × 2 + 4 to get as many different values as possible.
Using only the digit 7 and the four operations, write an expression whose value is 100.
Which numbers from 1 to 30 can be written as a sum of two or more consecutive whole
numbers?

Why it happens: Every puzzle of this kind asks the same real question — how many
different values can a fixed set of numbers produce once we are free to choose the
operations, the order and the brackets. That is exactly what this chapter is about.

Tip: When you invent a puzzle, always solve it yourself first and note down at least
one answer, so you know it is possible.

Chapter at a glance
An arithmetic expression is a number phrase like 13 + 2 or 5 × 25; the number it evaluates to
is its value.
Two expressions can be compared with =, < or > by looking at how their parts differ — no
full calculation needed.
Terms are the pieces of an expression separated by ‘+’. Every subtraction is first rewritten as
adding the inverse, so 83 – 14 has terms 83 and –14.
Terms may be added in any order and in any grouping — the commutative and
associative properties of addition.
Removing a bracket that has a minus sign in front of it flips the sign of every term inside it.
The distributive property: a × (b + c) = a × b + a × c, and a × (b – c) = a × b – a × c — the
multiple of a sum (difference) is the sum (difference) of the multiples.

Page 55 of 56

Class 7 Maths Chapter 2 Arithmetic Expressions AglaSem · NCERT Solutions

Quick revision

IDEA WHAT IT MEANS EXAMPLE VALUE

Arithmetic A number phrase built with + – × ÷ 13 + 2 15
expression

Value of an The single number it evaluates to 5 × 25 125
expression

Comparing Use =, < or > on the two values 13 – 2 < 4 × 3 11 < 12
expressions

Term A part separated by ‘+’ (subtraction → add 83 – 14 = 83 + (–14) terms 83, –
the inverse) 14

Product as one term A × or ÷ chain has no ‘+’, so it is a single 30 + 5 × 4 terms 30, 5 ×
term 4

Brackets Evaluate what is inside first 30 + (5 × 4) 50

Commutative Swapping two terms keeps the sum 6 + (–4) = (–4) + 6 2
property

Associative property Grouping terms differently keeps the sum (–7) + 10 + (–11) –8

Bracket after a minus Signs of all terms inside change 200 – (40 + 3) = 200 – 40 157
–3

Bracket after a plus Signs inside stay as they are 28 + (35 – 10) = 28 + 35 – 53
10

Distributive property Multiply the number into every term 2 × (43 + 24) = 2 × 43 + 2 134
× 24

Smart multiplying Write a number as a round number ± a little 97 × 25 = 100 × 25 – 3 × 2425
25

Page 56 of 56