NCERT Solutions Class 7 Maths Chapter 10 Operations with Integers

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 10: Operations with
Integers

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 25 – 45 20 76 English

Solutions, notes, sample papers & more at 79 pages

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 10: Operations with
Integers
Chapter 2 of Ganita Prakash Grade 7 Part II takes integers past addition and subtraction. A carrom coin on a
number line, green and red tokens in a bag, and a growing multiplication pattern are used to work out why a
negative times a negative is positive — and then the same ideas give the rules for division, the commutative,
associative and distributive properties, and a magic grid.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) Part II, 25 – 45

SECTIONS QUESTIONS

20 76

MEDIUM

English

In-text Questions — Page 25
Section 2.1 A Quick Recap of Integers — Rakesh’s Puzzle: A Number Game

Q1 Rakesh gives you a challenge. “I have thought of two numbers”, he says. “Their sum
is 25, and their difference is 11.” Can you tell me the two numbers?

The two numbers are 18 and 7.
Work down the table the way the book does. Keep the sum fixed at 25 and watch how the
difference changes.

FIRST NUMBER SECOND NUMBER SUM DIFFERENCE

10 15 25 –5

20 5 25 15

19 6 25 13

18 7 25 11

18 + 7 = 25 ✓

18 – 7 = 11 ✓

Page 1 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: every time you move 1 unit from the second number to the first,
the sum stays 25 but the difference goes up by 2. From (20, 5) the difference is 15;
move one step to (19, 6) and it drops to 13; one more step to (18, 7) and it is 11. So
the difference marches in steps of 2, and you only have to walk it to the number you
want.

Tip: you can also jump straight there. The bigger number is half of (sum +
difference) and the smaller is half of (sum – difference).

First = (25 + 11) ÷ 2 = 36 ÷ 2 = 18

Second = (25 – 11) ÷ 2 = 14 ÷ 2 = 7

Q2 “Think of two numbers whose sum is 25, but their difference is –11.” Use the same
method. Try different pairs of numbers and fill in the table again.

FIRST NUMBER SECOND NUMBER SUM DIFFERENCE

The same guess-table as on page 24 — first number, second number, their sum and their
difference — ready to be filled in again.

The numbers are 7 and 18 — the same pair as before, but swapped.

Page 2 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

FIRST NUMBER SECOND NUMBER SUM DIFFERENCE

15 10 25 5

10 15 25 –5

8 17 25 –9

7 18 25 –11

7 + 18 = 25 ✓

7 – 18 = –11 ✓

Why it happens: the difference means first number – second number. Swapping the
two numbers leaves the sum untouched, but it flips the difference to its additive
inverse. So a puzzle with difference –11 is the difference-11 puzzle read backwards.

Check it yourself: the shortcut works here too. First = (25 + (–11)) ÷ 2 = 14 ÷ 2 = 7,
and Second = (25 – (–11)) ÷ 2 = 36 ÷ 2 = 18.

Figure it Out — Page 26
Section 2.1 A Quick Recap of Integers

Q1 Let us try to find a few more pairs of numbers from their sums and differences: (a)
Sum = 27, Difference = 9 (b) Sum = 4, Difference = 12 (c) Sum = 0, Difference = 10 (d)
Sum = 0, Difference = –10 (e) Sum = –7, Difference = –1 (f) Sum = –7, Difference = –13

Use the same two half-steps every time.

First number = (sum + difference) ÷ 2

Second number = (sum – difference) ÷ 2

Page 3 of 79

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Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
m.
SUM DIFFERENCE FIRST SECOND CHECK

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NUMBER NUMBER

. a g l
m
ase
(a) 27 9 18 9 18 + 9 = 27, 18 – 9 = 9

a(b)gl 4 12 8 –4 8 + (–4) = 4, 8 – (–4) = 12

m
.co ag
(c) 0 10 5 –5 5 + (–5) = 0, 5 – (–5) = 10

asem
agl
(d) 0 –10 –5 5 –5 + 5 = 0, –5 – 5 = –10

(e) –7 –1 –4 –3 –4 + (–3) = –7, –4 – (–3) =
–1
co m
em.
m l a13s
.co
(f) –7 –13 –10 3 –10 + 3 = –7, –10 – 3 = –

a g
a s em
a l of them are worth walking through slowly.
gTwo
m a s
(e) First = (–7 + (–1)) ÷ 2 = (–8) ÷ 2 = –4
m .co agl
l a se
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Second = (–7 – (–1)) ÷ 2 = (–6) ÷ 2 = –3

co m
(f) First = (–7 + (–13)) ÷ 2 = (–20) ÷ 2 = –10
m .
m as e
.co
Second = (–7 – (–13)) ÷ 2 = 6 ÷ 2 = 3
a g l
se m
g l a
a Why it happens: add the two conditions and the second number cancels out. (first +

se m
comgiving second = (sum – difference) ÷ 2. That is a
second) + (first – second) = 2 × first, so first = (sum + difference) ÷ 2. Subtract them
. a g l
em
instead and the first number cancels,

a s
agl
why the trick never fails.

Tip: notice (c) and (d). They are the same two numbers written in the two possible
co m
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e
orders, so their differences are 10 and –10.
m l as
m .co a g
l a se
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.c
s e m
m a
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g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Q2 Choose a partner and take turns to play this game. In each turn, one of you can
think of two integers, and give their sum and difference; the other person must
then figure out the integers. After some practice, you can perform this magic trick
for your family members and surprise them!

Here is how to be unbeatable at the game.

1. Your partner announces the sum S and the difference D.
2. Add them and halve: that is the first number.
3. Subtract and halve: that is the second number.

First = (S + D) ÷ 2

Second = (S – D) ÷ 2

A sample turn — your partner says “sum 12, difference –30”.

First = (12 + (–30)) ÷ 2 = (–18) ÷ 2 = –9

Second = (12 – (–30)) ÷ 2 = 42 ÷ 2 = 21

Check: –9 + 21 = 12 ✓ and –9 – 21 = –30 ✓

Why it happens: S + D counts the first number twice and the second number once
positively and once negatively, so the second number vanishes. Halving what is left
leaves the first number alone. The same cancelling, with a subtraction, isolates the
second number.

Try This: when you set the puzzle, choose S and D that are both even or both odd. If
one is even and the other odd, S + D is odd and no pair of integers works — a good
way to catch out a careless partner!

In-text Questions — Pages 26–27

Page 5 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Section 2.1 A Quick Recap of Integers — Carrom Coin Integers

Q1 To begin with, the coin is at point 0. If the coin is struck twice, with the first strike
moving it by 4 units and the second strike moving it by 3 units, what will be the
final position of the coin?

The coin ends at 7.

Start at 0

After the first strike: 0 + 4 = 4

After the second strike: 4 + 3 = 7

So the coin is 4 + 3 = 7 units from 0, to the right.

+4
+3

0 1 2 3 4 5 6 7 8 9 10 11
Two rightward strikes from 0: 4 units, then 3 more. The coin stops at 7.

Why it happens: the second strike does not care where the coin started. It simply
pushes it 3 more units from wherever it is. So the two distances just pile up, and
adding them gives the final position.

Q2 If the coin is struck twice, and if the two movements are known, can you give a
formula for the final position of the coin?

Yes. If the first strike moves the coin a units to the right and the second strike moves it b units to
the right, then

P=a+b

Page 6 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

where P is the distance of the coin from the starting point 0.

Why it happens: after the first strike the coin sits at a. The second strike adds
another b units of travel to that. Nothing is lost or gained in between, so the total
distance from 0 is a + b. This is why the formula holds no matter what a and b are.

Check it yourself: put a = 4 and b = 3 into P = a + b. You get 7 — exactly the answer
found by walking the coin along the line.

Q3 Now, suppose the coin can be struck to move it in either direction — left or right.
The coin is at 0. If it is struck twice (the direction of the two strikes may be the
same or different) can you give a formula for the final position of the coin?

The formula is the same one: P = a + b. Only the meaning of a and b widens.

Take a rightward movement as positive.
Take a leftward movement as negative.

Then a is the first movement (positive if the strike is to the right, negative if to the left), b is the
second movement in the same sense, and

P=a+b

The four cases the book lists are all handled at once.

STRIKE 1 STRIKE 2 A B P=A+B

5 right 3 right 5 3 8

5 left 3 left –5 –3 –8

5 right 3 left 5 –3 2

5 left 3 right –5 3 –2

Why it happens: a signed number stores two things at once — the magnitude tells
how far, the sign tells which way. Once the direction is packed into the number itself,
the four separate cases stop being separate. One addition covers all of them, which
is exactly why integers are worth inventing.

Page 7 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Q4 Suppose the first strike moves the coin rightward by 5 units from 0, and the second
strike leftward by 7 units, then we take the First Movement = 5 units, Second
Movement = – 7 units. What is the final position of the coin?

The coin ends at –2.

P=a+b

= 5 + (–7)

= –2

So the coin is 2 units to the left of 0.

b = –7
a = +5

–5 –4 –3 –2 –1 0 1 2 3 4

A strike of +5 followed by a strike of –7 leaves the coin at P = –2.

Why it happens: the coin first travels 5 units right, reaching 5. The second strike is
stronger and points the other way, so it undoes those 5 units and carries the coin 2
units past 0. Adding a negative number is exactly this “undo, then overshoot”
movement.

Q5 If the first movement is – 4 and the final position is 5, what is the second
movement?

The second movement is +9, that is, 9 units to the right.

Page 8 of 79

as e
Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
e m.
P=a+b
m l as
5 = (–4) + b
m .co a g
l a se
g
b = 5 – (–4)
ab = 5 + 4

com
. ag
b=9
e m
g l as
Check: (–4) + 9 = 5 ✓ a

. com
Why it happens: the first strike carries the coin to –4, which is 4 units left of 0. To

m a
finish 5 units right of 0, the second strike must first bring it back across
s em those 4 units
. a gl so b = +9.
co push it 5 units further. That is 4 + 5 = 9 units, all rightward,
em
and then

a s
a gl
a s
com agl
If there are multiple strikes causing movements in the order 1, – 2, 3, – 4, …, – 10,
.
Q6

em
what is the final position of the coin?
a s
ANSWER agl
The coin ends at –5.
co m
m .
as
The ten movements are 1, –2, 3, –4, 5, –6, 7, –8, 9, –10. Pair each odd movement with the
m e
.co
negative one just after it.
a g l
a s em
agl (1 + (–2)) + (3 + (–4)) + (5 + (–6)) + (7 + (–8)) + (9 + (–10))
se m
com l a
= (–1) + (–1) + (–1) + (–1) + (–1)
. a g
m
ase
= 5 × (–1)

= –5 agl

co m
m .
e
Why it happens: in every pair the leftward strike is exactly 1 unit stronger than the
m l as
.co g
rightward strike before it. So each pair drags the coin 1 unit to the left. Five such

e m pairs drag it 5 units left of 0. a
la s
ag c
m .
Check it yourself: add them straight through — 1, –1, 2, –2, 3, –3, 4, –4, 5, –5. The
m a s e
. co
running total after each strike lands on –5 at the end.
em agl
g l as
a

co m
m .
m ase
.co


a g l Page 9 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Try This — Page 27

Page 10 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Section 2.1 A Quick Recap of Integers — Carrom Coin Integers

TRY THIS

Q1 From the figures below, what can you conclude about the magnitudes of a and b
compared to each other, and what are their directions? Remember to start from 0.
(1), (2), (3)

1.

b

a

–ve P 0 +ve

2.

a

b

–ve 0 P +ve

3.

a
b

–ve 0 +ve
P
The three pictures on page 27. In each one the coin starts at 0, the first strike is a and
the second is b; the dashed coin shows where it lands in between and the solid coin its
final position P.

Page 11 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

In every figure, a is the first strike and b is the second, and P = a + b is where the coin finally
stops. Read each arrow’s direction, then compare its length.
Figure 1

b

a

–ve P 0 +ve

Figure 1: a takes the coin right of 0; the longer b brings it back past 0 to P.

a is positive (rightward) and b is negative (leftward), and the b-arrow is clearly the longer one.

magnitude of b > magnitude of a

a is rightward (positive), b is leftward (negative)

P = a + b is negative — P lies to the left of 0

Figure 2

a

b

–ve 0 P +ve

Figure 2: the long a goes right; the shorter b comes back, but P stays right of 0.

Page 12 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

a is positive (rightward) and b is negative (leftward), but this time the a-arrow is the longer
one.

magnitude of a > magnitude of b

a is rightward (positive), b is leftward (negative)

P = a + b is positive — P lies to the right of 0

Figure 3

a

b

–ve 0, P +ve

Figure 3: a carries the coin left, b brings it exactly back. P sits on 0.

a is negative (leftward) and b is positive (rightward), and the two arrows are the same length
— P has landed back on 0.

magnitude of a = magnitude of b

a is leftward (negative), b is rightward (positive)

P = a + b = 0, so b = –a

Why it happens: the sign of P is decided by whichever movement has the greater
magnitude, because the smaller one is completely cancelled by part of the bigger
one. Only what is left over survives, and it keeps the direction of the bigger strike.
When the magnitudes are equal, nothing is left over at all — the two movements are
additive inverses and P = 0.

Tip: a picture like this never gives you exact numbers, only comparisons. Reading “b
is longer than a and points the other way” is enough to say P is negative.

Page 13 of 79

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Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
m.
In-text Questions — Page 28
m as e
.co
Section 2.1 A Quick Recap of Integers — The Token Model
a g l
se m
l a
MATH TALK

ag
Find (+7) – (+18). To subtract 18 from 7, i.e., (+7) – (+18), we need to remove 18

com
Q1

. ag
positives from 7 positives. But there are not enough tokens to remove 18 positives!

a s emwe can remove 18 positives. How many? What is
We put in enough zero pairs so that
left?
agl

co m
m.

m
You need 11 zero pairs, and what is left is –11.
o l a se
g there.
Start with.c7 green tokens. You must take away 18 greens, but only 7 are
a
se m
l a
ag Greens needed = 18
m a s
.co agl
Greens present = 7

se m
l a
Greens short = 18 – 7 = 11
ag
So drop in 11 zero pairs — 11 greens together with 11 reds. The bag still stands for 7, because

co m
each pair is worth 0.
m .
as e
. om holds 7 + 11 = 18 greens and 11 reds
cnow a g l
m
ase
Bag

agl Remove 18 greens

se m
com a
What is left: 11 reds = –11

. a g l
m
ase
So 7 – 18 = –11

agl
Why it happens: a green and a red cancel each other, so a zero pair is worth nothing

. com
at all. Adding zero pairs changes what the bag looks like without changing what it is
worth. That is the trick that lets you take 18 out of a bag that e
s m to hold only 7.
m a
seemed

m .co agl
l a se
ag Tip: the same idea in one line — 7 – 18 = 7 + (–18) = –11, because 18 is 11 more than
.c
m
7 and the 18 is the negative one.

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

co m
m .
m ase
.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Q2 Using tokens, argue out the following statements. (a) 7 – 18 = 7 + (– 18) (additive
inverse of 18 is – 18) (b) 4 – (– 12) = 4 + 12 (additive inverse of – 12 is 12)

(a) 7 – 18 = 7 + (–18)

Left side. Put 7 greens in the bag. To remove 18 greens, first add 11 zero pairs, then take out
the 18 greens. 11 reds remain, so the answer is –11.
Right side. Put 7 greens in the bag and then put in 18 reds. Cancel 7 zero pairs. 11 reds
remain, so the answer is –11.

Removing 18 greens ⇒ 11 reds left ⇒ –11

Adding 18 reds ⇒ 11 reds left ⇒ –11

Both bags hold the same thing, so 7 – 18 = 7 + (–18)

(b) 4 – (–12) = 4 + 12

Left side. Put 4 greens in the bag. There is no red to remove, so add 12 zero pairs — 12
greens and 12 reds. Now take out the 12 reds. What is left is 4 + 12 = 16 greens, that is, 16.
Right side. Put 4 greens in the bag and add 12 more greens. That is 16 greens, that is, 16.

4 – (–12) = 16

4 + 12 = 16

Why it happens: taking a red out of the bag and putting a green in have exactly the
same effect on the value — each raises the total by 1. In the same way, taking a
green out and putting a red in each lower the total by 1. So subtracting a number is
the same as adding its additive inverse, whatever the sign of that number.

Did you know? The book writes the additive inverse of a as –a. So –(18) = –18 and –(–
18) = 18. Taking the inverse twice brings you straight back to where you started.

In-text Questions — Pages 29–30

Page 15 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Section 2.2 Multiplication of Integers

MATH TALK

Q1 Suppose we put some positive tokens into an empty bag as shown in the figure.
How many positives are in the bag now?

+ + + +

+ + + +

The figure on page 29 — green (positive) tokens being put into the empty bag, shown
group by group.

There are 8 positives in the bag.
The picture shows 2 positive tokens being dropped in, and that is done 4 times.

2 positives, 4 times

=4×2

=8

Here 4 is the multiplier (how many times), 2 is the multiplicand (what is being put in each time),
and 8 is the product.

Page 16 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: multiplication is repeated addition of the same amount. Each
round adds the same 2 greens, so after 4 rounds the bag has gained 2 + 2 + 2 + 2 =
8. Naming the two numbers multiplier and multiplicand matters here, because soon
they will play very different roles.

Q2 We have seen this kind of multiplication of positive integers before. Can we use
tokens to give meaning to multiplications like 4 × (– 2)?

Yes. Just change the colour of the tokens you drop in.
Start with an empty bag. 4 × (–2) means placing 2 negatives into the bag, 4 times. Negatives
are red tokens.

2 reds, 4 times

= 8 reds

4 × (–2) = –8

4×2=8 4 × (–2) = –8

+ + – –
+ + – –
+ + – –
+ + – –
Two green tokens four times gives 8; two red tokens four times gives –8.

Page 17 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: the multiplier 4 only says “do it four times”. It does not care what is
being put in. So if each round puts in –2 instead of +2, the bag falls by 2 each round
instead of rising by 2. Four rounds of falling by 2 lands on –8.

Q3 Similarly find the values of 4 × (– 6) and 9 × (– 7)? How can we interpret (– 4) × 2?

4 × (–6) = –24 and 9 × (–7) = –63.

4 × (–6): put 6 reds in, 4 times → 24 reds = –24

9 × (–7): put 7 reds in, 9 times → 63 reds = –63

Now (–4) × 2. Here the multiplier is negative, and that changes the instruction.

When the multiplier is positive, you place tokens into the bag.
When the multiplier is negative, you remove tokens from the bag.

So (–4) × 2 means: remove 2 positives (2 green tokens) from the bag, 4 times.

(–4) × 2 = remove 2 greens, 4 times = –8

Why it happens: the multiplier is the instruction and the multiplicand is the
material. A positive multiplier repeats “put in”; a negative multiplier repeats “take
out”. Taking 8 greens out of an empty bag leaves it 8 short, and being 8 short is
exactly what –8 means.

Q4 Why are we trying to remove green tokens and not red tokens?

Because the multiplicand decides which colour is handled, and here the multiplicand is +2.
In (–4) × 2, read the two numbers separately.

Page 18 of 79

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Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
m.
NUMBER ROLE WHAT IT TELLS US

mMultiplier as e
–4
.co a g l
Remove (because it is negative), and do it 4 times

se m
g l a Multiplicand Remove 2 positives — greens, because 2 is positive
a
2

o m
c ag
Why it happens: if we removed reds instead, we would be modelling (–4) × (–2), a
completely different product. The signm .
s e of the multiplicand names the colour, the sign

a g laKeeping the two jobs apart is what makes all four
of the multiplier names the action.
sign cases come out right.

. c om
m a s em 4 times”;
Check it yourself: read 4 × (–2) the same way. Multiplier +4 says “place,

. co –2 says “2 reds”. Place 2 reds four times → –8. agl
multiplicand

a s em
a gl
m a s
we model (–4) × (–2) with tokens? .co agl
Q5 What happens when both the integers in the multiplication are negative? How do

a s em
ANSWER agl
The product is positive: (–4) × (–2) = 8.
co m
m .
as e
Read the instruction as before. The multiplier –4 says remove, 4 times. The multiplicand –2 says 2
m l
.co g
negatives — red tokens.
m a
l a se
But the bag is empty, so first stock it with zero pairs, exactly as in subtraction.

ag
Round 1: put in 2 zero pairs (2 greens + 2 reds), then take out the 2 reds → 2 greens stay
se m
com g l a
m . a
ase
Do this 4 times

agl
Left in the bag: 8 greens

(–4) × (–2) = +8

co m
m .
m as e
.co l
So the four basic results the tokens have now established are:

a g
se m
g l a
a
4×2=8
.c
4 × (–2) = –8
s e m
m a
(–4) × 2 = –8
em . co agl
g l as
a
(–4) × (–2) = 8

co m
m .
m as e
.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: a zero pair costs nothing, so you may create as many as you need.
Each time you pull a red out of a pair, the green half is left behind. Removing
something negative therefore leaves something positive — which is why two
negatives multiply to a positive.

Tip: in everyday words, taking away a debt makes you richer. Removing 8 units of
debt is the same as gaining 8 units of fortune.

Figure it Out — Page 31
Section 2.2 Multiplication of Integers

MATH TALK

Q1 Using the token interpretation, find the values of: (a) 3 × (– 2) (b) (– 5) × (– 2) (c) (– 4) ×
(– 1) (d) (– 7) × 3

Read the multiplier for the action and the multiplicand for the colour.

PRODUCT TOKEN READING VALUE

(a) 3 × (–2) Place 2 reds in the bag, 3 times → 6 reds –6

(b) (–5) × (–2) Remove 2 reds, 5 times (using zero pairs) → 10 greens stay 10

(c) (–4) × (–1) Remove 1 red, 4 times (using zero pairs) → 4 greens stay 4

(d) (–7) × 3 Remove 3 greens, 7 times (using zero pairs) → 21 reds stay –21

(a) 3 × (–2) = –6

(b) (–5) × (–2) = 10

(c) (–4) × (–1) = 4

(d) (–7) × 3 = –21

Page 20 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: in (a) the bag is only ever fed red tokens, so it sinks. In (b) and (c)
reds are pulled out of zero pairs, and the greens left behind pile up — a positive
answer. In (d) greens are pulled out of zero pairs, and the reds left behind pile up —
a negative answer.

Q2 If 123 × 456 = 56088, without calculating, find the value of: (a) (– 123) × 456 (b) (– 123)
× (– 456) (c) (123) × (– 456)

The magnitude never changes — only the sign does.

(a) (–123) × 456 = –56088 (one negative)

(b) (–123) × (–456) = 56088 (two negatives)

(c) 123 × (–456) = –56088 (one negative)

Why it happens: the magnitude of a product depends only on the magnitudes of
the two numbers, and those are 123 and 456 in all four statements. So 56088 is
fixed. All that is left to decide is the sign, and that is settled by counting negatives —
an even count gives a positive product, an odd count a negative one.

Tip: this is why questions like these say “without calculating”. Once you have done
one long multiplication, the other three come free.

Q3 Try to frame a simple rule to multiply two integers.

Here is a rule in two steps.

1. Magnitudes: multiply the two numbers as if both were positive.
2. Sign: if the two signs are the same, the product is positive. If they are different, the product
is negative.

Page 21 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

SIGNS PRODUCT EXAMPLE

+ and + positive 4×2=8

– and – positive (–4) × (–2) = 8

+ and – negative 4 × (–2) = –8

– and + negative (–4) × 2 = –8

Test on (–13) × (–6):

Magnitudes: 13 × 6 = 78
Signs: both negative → positive

(–13) × (–6) = 78

Why it happens: the token model builds exactly these four rows. Two same signs
mean either “put greens in” or “take reds out”, and both make the bag richer. Two
different signs mean either “put reds in” or “take greens out”, and both make it
poorer.

Math Talk — Page 31
Section 2.2 Multiplication of Integers — different token sets for the same integer

MATH TALK

Q1 Consider the numbers represented by the following tokens: (a), (b), (c). We can see
that all of them represent the number (– 2). Now, take 4 times each of these token
sets. That is, place each set into the empty bag 4 times. What integer do we get as
the final answer in each case? Do we get different answers because the sets look
different, or the same answer because they all represent – 2?

All three give the same answer, –8. The sets look different, but they are worth the same.
First check what each set is worth.

Page 22 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

SET TOKENS SHOWN VALUE

(a) 2 reds –2

(b) 4 reds and 2 greens –4 + 2 = –2

(c) 6 reds and 4 greens –6 + 4 = –2

Now place each set into an empty bag 4 times.

(a) 4 × 2 reds = 8 reds → –8

(b) 4 × (4 reds + 2 greens) = 16 reds + 8 greens; cancel 8 zero pairs → 8 reds → –8

(c) 4 × (6 reds + 4 greens) = 24 reds + 16 greens; cancel 16 zero pairs → 8 reds → –8

Why it happens: sets (b) and (c) are just set (a) with extra zero pairs thrown in — one
extra pair in (b), two in (c). Repeating a set 4 times repeats its zero pairs 4 times as
well, and those cancel out at the end. So the answer depends only on the value of
the set, never on how many tokens it happens to contain.

Tip: this is an important check on any model. If two pictures stand for the same
number, they must behave identically in every calculation — otherwise the model
would not be describing numbers at all.

Q2 Check this for 5 × 4, by taking different token sets corresponding to 4.

Every token set worth 4 gives the same product, 20.

SET WORTH 4 PLACED 5 TIMES AFTER CANCELLING ZERO PAIRS

4 greens 20 greens 20

5 greens + 1 red 25 greens + 5 reds 20

6 greens + 2 reds 30 greens + 10 reds 20

10 greens + 6 reds 50 greens + 30 reds 20

Page 23 of 79

as e
Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
e m.
Take 6 greens + 2 reds (value 6 – 2 = 4)
m l as
.co
5 times: 30 greens and 10 reds
m a g
l a se
g
Cancel 10 zero pairs
a20 greens left = 20

com
m . ag
l a se
Why it happens: each extra zero pair inside the set is copied 5 times, giving 5 extra
g pairs are still worth 0. The padding is copied
zero pairs at the end — and 5azero
faithfully and then cancels itself, so it can never affect the product.

co m
em.
m l as
.co a g
a s em
We have seen that – 4 × 2 is the number obtained by removing 2 positive tokens

gl from the empty bag 4 times. We know that removing or subtracting a number is the
Q3

a same as adding its inverse. Using this, can – 4 × 2 be defined through a process of

m a s
.co agl
addition of tokens instead of removal of tokens?

se m
g l a
a

Yes. Removing 2 positives is the same as adding 2 negatives.

co m
m .
Remove 2 greens, 4 times
as e
. com a g l
se m
= Add 2 reds, 4 times
a = 8 reds
agl
(–4) × 2 = –8
se m
com g l a
m . a
ase
agl
Both routes empty into the same bag, so both descriptions of (–4) × 2 are correct.

Why it happens: taking one green out lowers the value by 1, and putting one red in

. com
also lowers the value by 1. The two moves have identical effects, so any sequence of

m a s em colour. This is
removals can be rewritten as a sequence of additions of the opposite

m
o token version of the rule a – b = a + (–b).
.cthe agl
l a se
ag
.c
m
Tip: the addition form is easier to work with, because you never have to stock the

m a s e
co agl
bag with zero pairs first. It also explains the neat statement (–a) × b = –(a × b).

m .
as e
a g l
In-text Questions — Pages 32–33
co m
m .
m ase
.co


a g l Page 24 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Section 2.2 Multiplication of Integers — Patterns in Integer Multiplication

Q1 Using this understanding, let us construct a sequence of multiplications and
observe the patterns. What do you notice in this pattern? Can you describe it?

4 × 3 = 12
–3
3×3=9
–3
2×3=6
–3
1×3=3
–3
0×3=0

The sequence of multiplications printed on page 32, with a positive multiplicand; the
multiplier drops by 1 at each step.

The pattern is: every unit decrease in the multiplier drops the product by the multiplicand.

Page 25 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

STATEMENT PRODUCT CHANGE

4×3 12

3×3 9 –3

2×3 6 –3

1×3 3 –3

0×3 0 –3

Multiplier goes 4, 3, 2, 1, 0 — down by 1 each time

Product goes 12, 9, 6, 3, 0 — down by 3 each time

and 3 is the multiplicand

Why it happens: the multiplier counts how many 3s are in the bag. Lowering it by 1
means one fewer 3, so the total falls by exactly 3. The multiplicand is the size of one
step, which is why the product steps down by the multiplicand and by nothing else.

Page 26 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Q2 Will this pattern continue when the multiplier goes below zero and becomes a
negative number?

4 × 3 = 12
–3
3×3=9
–3
2×3=6
–3
1×3=3
–3
0×3=0

The sequence of multiplications printed on page 32, with a positive multiplicand; the
multiplier drops by 1 at each step.

Yes. The steps of –3 simply carry on past zero.

0×3=0

(–1) × 3 = –3

(–2) × 3 = –6

(–3) × 3 = –9

Each line is still 3 less than the one above it.

Page 27 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: nothing special happens at zero. The rule “one fewer 3” keeps
making sense — going one step below zero means the bag owes one 3, which is –3.
This pattern gives the same answers the tokens gave, and that agreement is what
tells us the sign rules are not arbitrary but forced.

Tip: a pattern that has to continue is a strong reason for a definition. If (–1) × 3 were
anything but –3, the neat staircase of –3s would break for no reason.

Q3 What is the pattern when the multiplicand is a negative integer?

4 × (–3) = –12
+3
3 × (–3) = –9
+3
2 × (–3) = –6
+3
1 × (–3) = –3
+3
0 × (–3) = 0

The sequence of multiplications printed on page 32, with a negative multiplicand; the
multiplier drops by 1 at each step.

The staircase turns the other way: every unit decrease in the multiplier now raises the
product by the multiplicand’s magnitude.

Page 28 of 79

as e
Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
m.
STATEMENT PRODUCT CHANGE

m as e
4 × (–3)
.co
–12
a g l
se m
g l a +3
a
3 × (–3) –9

2 × (–3) –6 +3

com
1 × (–3) –3
e m . +3 ag
g l as
0 × (–3)
a 0 +3

co m
As the multiplier falls by 1, the product increases by 3
em.
m l as
m .co
This is the inverse of the earlier pattern
a g
l a se
a g
Why it happens: the step size is still the multiplicand, but the multiplicand is now –3.
m a s
c o agl
Removing one lot of –3 means removing a debt of 3, which lifts the total by 3. So the
.
m by the multiplicand” — is still true; it is just
s e
same sentence — “the product decreases
a up by 3.
that decreasing by –3 means lgoing
ag

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
em . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Q4 Will this pattern continue when the multiplier goes below zero and becomes a
negative integer?

4 × (–3) = –12
+3
3 × (–3) = –9
+3
2 × (–3) = –6
+3
1 × (–3) = –3
+3
0 × (–3) = 0

The sequence of multiplications printed on page 32, with a negative multiplicand; the
multiplier drops by 1 at each step.

Yes — and this is where the last sign rule appears.

0 × (–3) = 0

(–1) × (–3) = 3

(–2) × (–3) = 6

(–3) × (–3) = 9

Each line is 3 more than the one above it, so the products march up through the positive
numbers.

Page 30 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: the staircase has climbed by 3 at every single step so far, and there
is no reason for it to stop at zero. Continuing it forces (–1) × (–3) to be +3. The tokens
said the same thing when reds were pulled out of zero pairs. Two different models
agreeing is strong evidence that negative × negative = positive is the only sensible
choice.

Did you know? Whatever is true for multiplication when the integers are positive
stays true when they are negative. That is the whole point of these patterns.

Figure it Out — Pages 33–34
Section 2.2 Multiplication of Integers — Patterns in Integer Multiplication

Q1 Find the following products. (a) 4 × (– 3) (b) (– 6) × (– 3) (c) (– 5) × (– 1) (d) (– 8) × 4 (e) (–
9) × 10 (f) 10 × (– 17)

Multiply the magnitudes, then fix the sign by counting negatives.

(a) 4 × 3 = 12, one negative → –12

(b) 6 × 3 = 18, two negatives → 18

(c) 5 × 1 = 5, two negatives → 5

(d) 8 × 4 = 32, one negative → –32

(e) 9 × 10 = 90, one negative → –90

(f) 10 × 17 = 170, one negative → –170

Page 31 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

PRODUCT SIGNS ANSWER

(a) 4 × (–3) different –12

(b) (–6) × (–3) same 18

(c) (–5) × (–1) same 5

(d) (–8) × 4 different –32

(e) (–9) × 10 different –90

(f) 10 × (–17) different –170

Why it happens: part (c) is worth a second look. Multiplying by –1 keeps the
magnitude at 5 and only flips the sign, so (–5) × (–1) = 5. That is the rule –1 × a = –a in
action.

In-text Questions — Pages 34–35
Section 2.2 Multiplication of Integers — 1 × a, –1 × a and Commutativity

Q1 Consider the expression 1 × a. We know that the value of this expression is ‘a’ for all
positive integers. Is this true for all negative integers too?

Yes. 1 × a = a for every integer a, positive or negative.
Use the token model. The multiplier is 1, so you place the multiplicand into the empty bag just
once.

If a = –5, place 5 reds in the bag, once

The bag holds 5 reds

1 × (–5) = –5

So in general,

1 × a = a (for all integers a)

Page 32 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: doing something once changes nothing about it. The bag ends up
holding precisely the set you put in, whatever colour those tokens were. That is why
1 is called the multiplicative identity, and why the sign of a survives untouched.

Q2 What is the value of the expression – 1 × a?

–1 × a = –a — the additive inverse of a — for every integer a.

A –1 × A WHAT HAPPENED

7 –7 Same magnitude, sign flipped

–7 7 Same magnitude, sign flipped

0 0 0 is its own inverse

When a is positive, the product has magnitude a and is negative

When a is negative, the product has magnitude a and is positive

In both cases the product is the additive inverse of a

–1 × a = –a

Why it happens: the multiplier –1 says “remove the multiplicand once”. Removing a
from an empty bag leaves the bag owing a, and owing a is exactly –a. So multiplying
by –1 is the same as taking the additive inverse — it is a sign-flipper, nothing more.

Tip: this single fact settles a whole family of questions. For example, the number
whose product with –1 is –31 must be 31, because flipping the sign of 31 gives –31.

Q3 In the case of integers, is the product the same when we swap the multiplier and
the multiplicand? Try this for some numbers.

Yes. Swapping them never changes the product.

Page 33 of 79

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Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
e m.
3 × (–4) = –12 and (–4) × 3 = –12
m l as
.co
(–15) × (–8) = 120 and (–8) × (–15) = 120
m a g
l a se
g
7 × (–9) = –63 and (–9) × 7 = –63
a
The two sides of each pair agree every time.
co m
e m . ag
g l as
Why it happens: the token stories are different but the bags end up the same. For 3
a
× (–4) you put 4 reds in three times; for (–4) × 3 you take 3 greens out four times. The

m
first leaves 12 reds, the second leaves 12 reds. Different instructions, identical bag.
co
em.
m l as
.co a g
a s em
Observe the following pairs of multiplications (fill in the blanks where needed): 3 × –
gl 4 = –12, – 4 × 3 = –12; – 30 × 12 = _______, 12 × – 30 = _______; –15 × – 8 = 120, – 8 × –15 =
Q4

a 120; 14 × – 5 = –70, – 5 × _____ = – 70. What do you notice in these pairs of

m a s
.co agl
multiplication statements?

se m
g l a
a

The filled table is:

co m
m .
e
FIRST STATEMENT SWAPPED STATEMENT

co=m–12 g l as
. a
sem
3 × (–4) (–4) × 3 = –12

a (–30) × 12 = –360
agl 12 × (–30) = –360

se m
com l a
(–15) × (–8) = 120 (–8) × (–15) = 120

. a g
m
ase
agl
14 × (–5) = –70 (–5) × 14 = –70

(–30) × 12: 30 × 12 = 360, one negative → –360
co m
m .
se
12 × (–30): same magnitudes, one negative → –360
o m l a
m .(–5) ag → 14
c × ___ = –70: 70 ÷ 5 = 14, and one negative is already there
l a se
ag
.c
m
What we notice: the product is the same when the multiplier and multiplicand are swapped.

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

co m
m .
m ase
.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: a product carries two pieces of information — a magnitude and a
sign — and swapping disturbs neither. The magnitude is the product of the two
magnitudes, and that is unchanged because whole-number multiplication is
commutative. The sign depends only on how many of the two numbers are negative,
and swapping does not change that count.

Q5 The magnitude of the product does not change when the multiplier and the
multiplicand are swapped… Will this always happen?

Yes, always. The magnitude of a product depends only on the two magnitudes.

magnitude of (a × b) = (magnitude of a) × (magnitude of b)

and for positive numbers, m × n = n × m

so the magnitude cannot change when a and b are swapped

For example, both (–15) × (–8) and (–8) × (–15) have magnitude 15 × 8 = 120.

Why it happens: the sign of each number tells us the direction of the action, not its
size. Once the signs are set aside, only 15 and 8 are left, and a 15-by-8 array of
objects is the same array as an 8-by-15 one, just turned on its side. That old fact
about positive numbers is doing all the work here.

Q6 Does the sign of the product change if we swap the multiplier and multiplicand?

No. Whatever the signs are, they give the same result before and after swapping.

If both are positive, the product is positive before and after the swap.
If both are negative, the product is positive before and after the swap.
If one is positive and the other negative, the product is negative before and after the swap —
swapping only decides which of the two is written first.

So neither the magnitude nor the sign changes, and therefore

a × b = b × a for any two integers a and b

Page 35 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

That is, multiplication is commutative for integers.

Why it happens: the sign rule asks only one question — are the two signs the same
or different? “Same” and “different” are relationships between the two numbers, and
a relationship does not care which number you name first. That is why the sign is
untouched by swapping.

In-text Questions — Pages 35–36
Brahmagupta’s Rules for Multiplication and Division · Examples 1 and 2

Q1 An exam has 50 multiple choice questions. 5 marks are given for every correct
answer and 2 negative marks for every wrong answer. What are the maximum
possible marks in the exam? What are the minimum possible marks?

Maximum = 250 marks. Minimum = –100 marks.
The best case is all 50 answers correct.

50 × 5 = 250

The worst case is all 50 answers wrong.

50 × (–2) = –100

SITUATION CORRECT WRONG TOTAL MARKS

Best possible 50 0 250

Mala’s result 30 20 110

Worst possible 0 50 –100

Why it happens: every correct answer pushes the score up by 5 and every wrong
answer pulls it down by 2. To make the score as large as possible, use only pushes;
to make it as small as possible, use only pulls. The score can go below zero here,
which is exactly why the marks are written as integers.

Page 36 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Tip: if a student leaves a question blank instead of answering it wrongly, that
question is worth 0, not –2. So a blank paper scores 0 — better than a fully wrong
one.

Q2 There is an elevator in a mining shaft… If it begins to descend from 15 m above the
ground, what will be its position after 45 minutes? Find the solution to part (b)
using Method 1 described above.

The elevator will be at –120 m, that is, 120 metres below the ground.
Method 1 treats the journey as a subtraction — work out the distance travelled, then take it
away from the starting height.

Speed = 3 metres per minute

Time = 45 minutes

Distance travelled downwards = 45 × 3 = 135 metres

Starting position = 15

Position after 45 minutes = 15 – 135

= –120

So the elevator is 120 metres below ground level.

Why it happens: the cage drops 135 m in all. The first 15 m of that only brings it
down to ground level, and the remaining 120 m carry it below. Subtracting a bigger
number from a smaller one lands you on a negative, and here the negative is read
as “below the ground”.

Check it yourself: Method 2 writes the speed itself as the integer –3 and gives 15 +
(45 × (–3)) = 15 + (–135) = –120. Same answer, one line shorter.

A Magic Grid of Integers — Pages 37–38

Page 37 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Section 2.2 Multiplication of Integers — A Magic Grid of Integers

TRY THIS

Q1 A grid containing some numbers is given below. Follow the steps as shown until no
number is left. Circle any number; strike out the row and the column containing
that number; circle any unstruck number; when there are no more unstruck
numbers, stop. Multiply the circled numbers.

The product is –30240, and the book’s own example confirms it.

8 –4 12 –6

–28 14 –42 21

12 –6 18 –9

20 –10 30 –15

In the four rounds shown in the book the circled numbers are –6, 14, 20 and 18.

(–6) × 14 = –84

(–84) × 20 = –1680

(–1680) × 18 = –30240

Notice that the four circles use each row once and each column once — the striking-out rule
forces that.

Why it happens: the very first circle removes its own row and column from play, so
the next circle must come from a different row and a different column, and so on.
With a 4 × 4 grid you make exactly 4 circles, one in each row and one in each column.

Q2 Try again, and choose different numbers this time. What product did you get? Was
it different from the first time? Try a few more times with different numbers!

The product is –30240 again — every single time.
Try three completely different sets of circles.

Page 38 of 79

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Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
m.
CIRCLED NUMBERS WORKING PRODUCT

m l a se –30240
.co ag
8, –42, –9, –10 8 × (–42) = –336; × (–9) = 3024; × (–10)

se m
g l a (–4) × (–28) = 112; × 18 = 2016; × (–15)
a
–4, –28, 18, –15 –30240

–6, –42, –6, 20 (–6) × (–42) = 252; × (–6) = –1512; × 20 –30240

. c om ag
s e m Every entry is a row number times a column
The reason is hidden in how the grid was built.
a
agl
number:

co m
× 4 –2 6 –3

e m–6.
com as
2
l
8 –4 12

. a g
m
ase
–7 –28 14 –42 21

agl3 12 –6 18 –9

m a s
.co agl
5 20 –10 30 –15

se m
g l a
a
Each circle contributes one row number and one column number

co m
.
All four row numbers get used: 2 × (–7) × 3 × 5 = –210
e m
m l as
.co
All four column numbers get used: 4 × (–2) × 6 × (–3) = 144
a g
s e m = (–210) × 144 = –30240
la
Product

ag
se m
com
Why it happens: because multiplication is commutative and associative, the sixteen
factors may be rearranged freely..Whichever g l a
m a
se column number exactly once — so the same eight
cells you circle, you pick up each row

l a
ag
number exactly once and each
numbers are always multiplied, only in a different order. The answer therefore

m
cannot change.

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

co m
m .
m as e
.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Q3 Play the same game with the grid below. What answer do you get?

8 –4 12 –6

–28 14 –42 21

12 –6 18 –9

20 –10 30 –15

The grid printed on page 38 for the second round of the game.

The answer is –30240 once more.
In the copy of the book being used here the second grid carries exactly the same sixteen entries
as the first — 8, –4, 12, –6 / –28, 14, –42, 21 / 12, –6, 18, –9 / 20, –10, 30, –15 — so its magic
number is the same –30240.

Row numbers: 2, –7, 3, 5 → product –210

Column numbers: 4, –2, 6, –3 → product 144

Magic number = (–210) × 144 = –30240

Check it yourself: whatever grid your copy shows, you do not have to play the game
to find its magic number. Read off the four row numbers and the four column
numbers, multiply all eight together, and you have the answer.

Why it happens: the game is really just a disguised way of multiplying the four row
numbers by the four column numbers. The circling only decides the order of the
multiplication, and order never affects a product.

Q4 What is so special about these grids? Is the magic in the numbers or the way they
are arranged or both? Can you make more such grids?

The magic is in the arrangement — the numbers are ordinary, but they are laid out as a
multiplication table.

Page 40 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Every cell is (its row number) × (its column number). Because the rules force one circle per row
and one per column, the final product is always

(r₁ × r₂ × r₃ × r₄) × (c₁ × c₂ × c₃ × c₄)

Making your own grid. Pick any four row numbers and any four column numbers, then fill
each cell with their product. Here is a fresh one built from rows 1, –2, 4, 3 and columns 5, –1, 2, –
4.

× 5 –1 2 –4

1 5 –1 2 –4

–2 –10 2 –4 8

4 20 –4 8 –16

3 15 –3 6 –12

Rows: 1 × (–2) × 4 × 3 = –24

Columns: 5 × (–1) × 2 × (–4) = 40

Magic number = (–24) × 40 = –960

Rub out the header row and column before you show it to a friend, and the grid looks like
sixteen unrelated integers.

Why it happens: the trick rests entirely on commutativity and associativity. Those
two properties let the eight hidden factors be gathered in any order, so the product
is fixed before the game even begins.

Tip: to make the answer positive, use an even number of negatives among your
eight chosen numbers. To make it 0, put a 0 in the list — then every game ends at 0.

In-text Questions — Pages 38–39

Page 41 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Section 2.2 Multiplication of Integers — Division of Integers

Q1 Can you summarise the rules for integer division looking at the above pattern?

WE KNOW THAT… …THEREFORE

25 × (– 4) = (– 100) (– 100) ÷ 25 = (– 4)

(– 4) × 25 = (– 100) (– 100) ÷ (– 4) = 25

(– 25) × (– 2) = 50 50 ÷ (– 25) = (– 2)

The worked division examples on page 38 that the question refers to.

Division follows exactly the same sign rule as multiplication.

1. Magnitudes: divide as if both numbers were positive.
2. Sign: same signs give a positive quotient, different signs a negative quotient.

In symbols, for positive integers a and b with b ≠ 0,

a ÷ (–b) = –(a ÷ b)

(–a) ÷ b = –(a ÷ b)

(–a) ÷ (–b) = a ÷ b

DIVISION REWRITTEN AS A MULTIPLICATION QUOTIENT

(–100) ÷ 25 25 × ? = –100, and 25 × (–4) = –100 –4

(–100) ÷ (–4) (–4) × ? = –100, and (–4) × 25 = –100 25

50 ÷ (–25) (–25) × ? = 50, and (–25) × (–2) = 50 –2

Why it happens: a division question is a multiplication question in disguise — “what
must I multiply the divisor by to reach the dividend?” Since the answer has to obey
the multiplication sign rule, the quotient inherits that very same rule. This is exactly
what Brahmagupta wrote in 628 CE: the product or quotient of two debts is a
fortune.

Page 42 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Figure it Out — Page 39
Section 2.2 Multiplication of Integers — Division of Integers

Q1 Find the values of: (a) 14 × (– 15) (b) – 16 × (– 5) (c) 36 ÷ (– 18) (d) (– 46) ÷ (– 23)

(a) 14 × 15 = 210, signs different → –210

(b) 16 × 5 = 80, signs same → 80

(c) 36 ÷ 18 = 2, signs different → –2

(d) 46 ÷ 23 = 2, signs same → 2

EXPRESSION MAGNITUDE SIGN ANSWER

(a) 14 × (–15) 210 different → – –210

(b) (–16) × (–5) 80 same → + 80

(c) 36 ÷ (–18) 2 different → – –2

(d) (–46) ÷ (–23) 2 same → + 2

Why it happens: for (c), ask “what times –18 gives 36?” Since (–18) × (–2) = 36, the
quotient is –2. Checking a division by turning it back into a multiplication is the safest
habit with integers.

Q2 A freezing process requires that the room temperature be lowered from 32°C at the
rate of 5°C every hour. What will be the room temperature 10 hours after the
process begins?

The room will be at –18 °C.
A fall of 5 °C each hour is a change of –5 °C per hour.

Page 43 of 79

as e
Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
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Change in 10 hours = 10 × (–5) = –50
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Final temperature = 32 + (–50)
m a g
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= 32 – 50
a= –18 °C

co m
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Why it happens: the first 32 °C of the drop only brings the room down to 0 °C, and
g remaining fall of 18 °C carries it below zero.
that takes a little over 6 hours.aThe
Writing the rate as –5 instead of 5 lets one multiplication handle the whole journey,

co m
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above and below zero alike.

m as e
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Check it yourself: after how many hours does the room reach 0 °C? You need a fall

a gl of 32 °C at 5 °C per hour, which is 6 hours and 24 minutes.
m a s
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a s
gl [Represent the profit/loss as integers.] (a) The
Q3 A cement company earns a profit of ₹8 per bag of white cement sold and a loss of ₹5
per bag of grey cementasold.
company sells 3,000 bags of white cement and 5,000 bags of grey cement in a
month. What is its profit or loss? (b) If the number of bags of grey cement sold is
co m
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6,400 bags, what is the number of bags of white cement the company must sell to
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have neither profit nor loss.
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Write profit as +8 per white bag and loss as –5 per grey bag.
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(a) The company makes a loss of ₹1,000.
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From white cement: 3000 × 8 = 24000 agl
From grey cement: 5000 × (–5) = –25000
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(b) The company must sell 4,000 bags of white cement.

m .
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co m
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.co


a g l Page 44 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Loss on grey cement = 6400 × (–5) = –32000

For no profit and no loss, the white cement must earn +32000

Number of bags = 32000 ÷ 8 = 4000

Check: 4000 × 8 + 6400 × (–5) = 32000 – 32000 = 0 ✓

Why it happens: “neither profit nor loss” means the two integers cancel exactly, so
their sum is 0. That is the same as saying the white-cement earning must be the
additive inverse of the grey-cement loss. Once you know the loss is ₹32,000, you only
need to ask how many ₹8 profits make ₹32,000.

Q4 Replace the blank with an integer to make a true statement. (a) (– 3) × _____ = 27 (b) 5
× _____ = (– 35) (c) _____ × (– 8) = (– 56) (d) _____ × (– 12) = 132 (e) _____ ÷ (– 8) = 7 (f) _____ ÷
12 = – 11

Turn each one round into a division (or, for the last two, into a multiplication).

STATEMENT WORKING BLANK

(a) (–3) × ___ = 27 27 ÷ (–3), signs different –9

(b) 5 × ___ = –35 (–35) ÷ 5, signs different –7

(c) ___ × (–8) = –56 (–56) ÷ (–8), signs same 7

(d) ___ × (–12) = 132 132 ÷ (–12), signs different –11

(e) ___ ÷ (–8) = 7 7 × (–8) –56

(f) ___ ÷ 12 = –11 (–11) × 12 –132

Page 45 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

(a) (–3) × (–9) = 27 ✓

(b) 5 × (–7) = –35 ✓

(c) 7 × (–8) = –56 ✓
(d) (–11) × (–12) = 132 ✓

(e) (–56) ÷ (–8) = 7 ✓

(f) (–132) ÷ 12 = –11 ✓

Why it happens: for (a)–(d) the missing number is a factor, so divide the product by
the factor you know. For (e) and (f) the missing number is the dividend, so multiply
the quotient by the divisor. In every case, decide the magnitude first and then ask
what sign makes the statement true.

In-text Questions — Pages 39–41
Section 2.2 Multiplication of Integers — Expressions Using Integers

MATH TALK

Q1 What is the value of the expression 5 × – 3 × 4? Does it matter whether we multiply 5
× – 3 and then multiply the product with 4, or if we multiply – 3 × 4 first and then
multiply the product with 5?

The value is –60, and no, the grouping does not matter.

(5 × (–3)) × 4

= (–15) × 4

= –60

5 × ((–3) × 4)

= 5 × (–12)

= –60

Page 46 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: both routes multiply the same three magnitudes 5, 3 and 4, giving
60, and both contain exactly one negative number, so the sign is negative. Since
neither the magnitude nor the sign depends on how you bracket, the answer cannot
depend on it either.

Q2 Take a few more examples of multiplication of 3 integers and check this property.
What do you observe?

The grouping never changes the answer. Integer multiplication is associative.

EXPRESSION GROUPED ONE WAY GROUPED THE OTHER WAY

(–2) × 6 × (–5) (–12) × (–5) = 60 (–2) × (–30) = 60

(–4) × (–3) × (–2) 12 × (–2) = –24 (–4) × 6 = –24

7 × (–1) × (–9) (–7) × (–9) = 63 7 × 9 = 63

So for any three integers a, b and c,

a × (b × c) = (a × b) × c

Why it happens: associativity already holds for positive numbers, and it fixes the
magnitude. The sign is decided by counting how many of the three numbers are
negative — a count that brackets cannot alter. Both halves of the answer are
therefore untouched by regrouping.

Q3 In the expression 5 × – 3 × 4, try to multiply 5 and 4 first and then multiply the
product with – 3: (5 × 4) × – 3. Are there orders in which 5 × – 3 × 4 can be evaluated?
Will the product be the same in all these cases?

Yes — every order gives –60.

(5 × 4) × (–3) = 20 × (–3) = –60

Page 47 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

The three numbers can be arranged in 6 orders, and each arrangement can be bracketed in 2
ways.

ORDER FIRST PRODUCT FINAL PRODUCT

5, –3, 4 5 × (–3) = –15 –60

5, 4, –3 5 × 4 = 20 –60

–3, 5, 4 (–3) × 5 = –15 –60

–3, 4, 5 (–3) × 4 = –12 –60

4, 5, –3 4 × 5 = 20 –60

4, –3, 5 4 × (–3) = –12 –60

Why it happens: commutativity lets you reorder the three numbers and associativity
lets you rebracket them. Used together, they say that a product of three or more
integers has one value, no matter how it is written. That is why we may drop the
brackets and simply write 5 × (–3) × 4.

Q4 Multiply the expression 25 × – 6 × 12 in all the different orders and check if the
product is the same in all cases.

Every order gives –1800.

(25 × (–6)) × 12 = (–150) × 12 = –1800

25 × ((–6) × 12) = 25 × (–72) = –1800

(25 × 12) × (–6) = 300 × (–6) = –1800

GROUPING STEP 1 STEP 2 VALUE

(25 × –6) × 12 –150 –150 × 12 –1800

25 × (–6 × 12) –72 25 × (–72) –1800

(25 × 12) × –6 300 300 × (–6) –1800

(–6 × 25) × 12 –150 –150 × 12 –1800

Page 48 of 79

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Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the magnitude is 25 × 6 × 12 = 1800 however it is grouped, and

m l a se
there is exactly one negative factor, so the sign is negative. The third grouping is also
o g
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m
the easiest —

l a se a product to make the arithmetic simple.
ag
reorder

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Tip: multiplying 25 × 12 first is smart because 25 × 12 = 300 is a round number.
.
m is allowed precisely because the order does
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Look at the following series of multiplications: – 1 × – 1 = 1, – 1 × –e1
s m× – 1 = – 1, – 1 × –
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1 ×c–o1 × – 1 = 1, – 1 × – 1 × – 1 × – 1 × – 1 = 1. When –1 is multiplied
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product is positive. When it is multiplied 3 or 5 times the product is negative. Can

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(–1) × (–1) × (–1) = –1 (3 factors, odd)
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(–1) × (–1) × (–1) × (–1) = 1 (4 factors, even)
m × (–1) × (–1) × (–1) × (–1) = –1 (5 factors, odd) a g
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— the sentence just below it correctly

a l –1s multiply to –1.
gives a negative product.gFive

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o m l a se
count is even, every –1 finds a partner and the whole product collapses to 1. If the

m ag and it drags the answer to
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.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Q6 Using this understanding of multiplication of many integers, can you give a simple
rule to find the sign of the product of many integers?

Count the negative factors.

An even number of negative factors → the product is positive.
An odd number of negative factors → the product is negative.
If any factor is 0, the product is 0.

(–2) × (–1) × (–5) × (–3): four negatives, even → positive

= 2 × 1 × 5 × 3 = 30

(–7) × 4 × (–1) × (–2): three negatives, odd → negative

= –(7 × 4 × 1 × 2) = –56

Why it happens: pull a factor of –1 out of every negative number. If there are k
negatives, the product becomes (–1) multiplied k times, times a product of positive
magnitudes. The magnitudes give a positive number, and the k copies of –1 give +1
when k is even and –1 when k is odd. That is the whole rule.

Tip: you never have to work out the sign step by step. Glance at the expression,
count the minus signs on the factors, and you know the answer’s sign before doing
any arithmetic.

Q7 Now, consider the expression 5 × (4 + (– 2)). As in the case of positive integers, is this
expression equal to 5 × 4 + 5 × (– 2)?

Yes. Both sides come to 10.

Page 50 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

5 × (4 + (–2))

=5×2

= 10

5 × 4 + 5 × (–2)

= 20 + (–10)

= 10

This is the distributive property, and it works for integers just as it did for whole numbers.

Why it happens: in the token picture, 5 × (4 + (–2)) is five copies of a set holding 4
greens and 2 reds. You may count the bag in two ways — cancel inside each set first
and then repeat it five times, or gather all the greens together and all the reds
together and count them separately. Counting the same bag differently cannot give
a different total.

Q8 Check if the distributive property holds for (– 2) × (4 + (– 3)) (that is, if this expression
equals (– 2) × 4 + (– 2) × (– 3)), and for a few other such expressions of your choice.
What do you observe? Will this always happen?

It holds. Both sides give –2.

(–2) × (4 + (–3))

= (–2) × 1

= –2

(–2) × 4 + (–2) × (–3)

= (–8) + 6

= –2

Two more checks of your own choosing:

Page 51 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

EXPRESSION BRACKET FIRST DISTRIBUTE FIRST

(–5) × (7 + (–10)) (–5) × (–3) = 15 (–35) + 50 = 15

6 × ((–4) + (–1)) 6 × (–5) = –30 (–24) + (–6) = –30

Yes, this will always happen. For any integers a, b and c,

a × (b + c) = (a × b) + (a × c)

Why it happens: the rectangular token arrangement shown in the book proves it. A
block of a rows, each holding b greens and c reds, can be read as a copies of (b + c),
or split down the middle into a × b and a × c. The same tokens are being counted, so
the two readings must agree — for any signs of b and c.

Try This — Page 41
Section 2.2 Multiplication of Integers — the distributive property with tokens

TRY THIS

Q1 Can you visually show the distributive property for an expression like –4 × (2 + (–3))?
[Hint: Use the fact that multiplying a number by –4 is adding the inverse of the
number 4 times.]

Yes. Use the hint: multiplying by –4 means adding the inverse of the multiplicand, 4 times.
The multiplicand here is the set “2 greens and 3 reds”. Its inverse is “2 reds and 3 greens”. Lay
out 4 rows of that inverse set.

Page 52 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

(–4) × 2 (–4) × (–3)

– – + + +

– – + + + total
=4
– – + + +

– – + + +
= –8 = +12
Four rows of the inverse set: 8 red tokens on the left and 12 green tokens on the right, giving –8 + 12
= 4.

Now read the picture in the two ways.

Whole block: (–4) × (2 + (–3)) = (–4) × (–1) = 4

Split block: (–4) × 2 + (–4) × (–3)

= (–8) + 12

=4

Why it happens: multiplying by a negative number turns every token in the set to
the opposite colour before it is repeated. Turning colours does not disturb which
column a token sits in, so the block can still be split down the middle exactly as
before. Both readings count the same 20 tokens, and 12 greens against 8 reds leave
4.

Tip: the picture also shows why the answer is positive even though the multiplier is
negative. The bracket 2 + (–3) is itself negative, and a negative times a negative is
positive.

Page 53 of 79

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Class 7 Maths Chapter 10 Operations with Integers
a g l AglaSem · NCERT Solutions

co m
m.
Pick the Pattern — Pages 41–42
m as e
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Section 2.2 Multiplication of Integers — Pick the Pattern
a g l
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aQ1gl Two pattern machines are given below. Each machine takes 3 numbers, does some
operations and gives out the result. Find the operations being done by Machine 1.

co m
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Machine 1 works out (first number) + (second number) – (third number), that is, a + b – c.

A B C A+B–C MACHINE’S RESULT
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5 8 3 5+8–3 10

a g
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10 12 10 + 11 – 12 9

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–3 10 2 (–3) + 10 – 2 5

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Row 3 is the useful one: 5 + 8 – (–3) = 13 + 3 = 16
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Row 5 checks it: (–4) + (–1) – (–6) = (–5) + 6 = 1
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a Why it happens: the first two rows only use positive numbers, so several guesses

se m
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would fit them. The rows with negative inputs are what pin the rule down — if the
machine were doing a + b + c, row.c3owould
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So, the result of the last group will be, (– 10) + (– 12) – (– 9) = _______.
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The answer is –13.

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.co


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Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

(–10) + (–12) – (–9)

= (–22) – (–9)

= (–22) + 9

= –13

Why it happens: the first two inputs pile up 22 units of debt. Subtracting –9 is the
same as adding 9, which pays back 9 of those units. What is left is a debt of 13, so
the star shows –13.

Q3 Find the operations being done by Machine 2 and fill in the blank.

Machine 2 works out –(a × b) – c, which can also be written as (–a) × b – c. The blank is –111.

A B C –(A × B) – C MACHINE’S RESULT

4 8 –3 –(32) – (–3) = –32 + 3 –29

6 –11 12 –(–66) – 12 = 66 – 12 54

5 3 7 –(15) – 7 –22

–3 9 –8 –(–27) – (–8) = 27 + 8 35

–7 4 6 –(–28) – 6 = 28 – 6 22

–10 –12 –9 –(120) – (–9) = –120 + 9 –111

Last group:

(–10) × (–12) = 120

–(120) – (–9)

= –120 + 9
= –111

Page 55 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Why it happens: the third row, 5, 3, 7 → –22, is the giveaway. All three inputs are
positive yet the answer is negative, so the machine must be flipping a sign
somewhere. Trying –(a × b) – c on that row gives –15 – 7 = –22, and the same rule
then fits all five rows.

Tip: a rule can be written in more than one way. Here (–a) × b – c and –(a × b) – c are
the same machine, because –1 × a × b = –(a × b).

Q4 Make your own machine and challenge your peers in finding its operations.

Here is a machine you can copy into your notebook. Show only the three inputs and the star,
and keep the rule hidden.

A B C RESULT

3 5 2 13

4 –2 6 –14

–5 3 1 –16

–6 –4 2 22

7 0 3 –3

The hidden rule is (a × b) – c.

3 × 5 – 2 = 15 – 2 = 13 ✓

4 × (–2) – 6 = –8 – 6 = –14 ✓

(–5) × 3 – 1 = –15 – 1 = –16 ✓

(–6) × (–4) – 2 = 24 – 2 = 22 ✓

7 × 0 – 3 = 0 – 3 = –3 ✓

Page 56 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

Tip for setting a good puzzle: include at least two rows with negative inputs and
one row where the answer’s sign is a surprise. Rows made only of positive numbers
can usually be explained by several rules, so they give your friend too much
freedom.

Why it happens: a machine is really an expression with three blanks in it. Guessing
the machine is guessing the expression, and the fastest way is to look for a row
whose answer has an unexpected sign or an unexpectedly large size — such a row
rules out most wrong guesses in one go.

Figure it Out — Pages 42–44
Section 2.2 Multiplication of Integers — Expressions Using Integers

MATH TALK TRY THIS

Q1 Find the values of the following expressions: (a) (– 5) × (18 + (– 3)) (b) (– 7) × 4 × (– 1)
(c) (– 2) × (– 1) × (– 5) × (– 3)

(a) (–5) × (18 + (–3))

= (–5) × 15

= –75

(b) (–7) × 4 × (–1)

= (–28) × (–1)

= 28

(c) (–2) × (–1) × (–5) × (–3)

= 2 × (–5) × (–3)
= (–10) × (–3)

= 30

Page 57 of 79

Class 7 Maths Chapter 10 Operations with Integers AglaSem · NCERT Solutions

NEGATIVE FACTORS MAGNITUDE ANSWER

(a) 1 (odd) 5 × 15 = 75 –75

(b) 2 (even) 7 × 4 × 1 = 28 28

(c) 4 (even) 2 × 1 × 5 × 3 = 30 30

Why it happens: in (a) the bracket must be simplified first, and 18 + (–3) = 15 is
positive, so only the –5 supplies a minus sign. In (b) and (c) simply count the minus
signs — two and four, both even — so both answers are positive.

Check it yourself: part (a) also works by distributing — (–5) × 18 + (–5) × (–3) = –90 +
15 = –75. Same answer, and a useful check.

Q2 Find the values of the following expressions: (a) (– 27) ÷ 9 (b) 84 ÷ (– 4) (c) (– 56) ÷ (– 2)

(a) (–27) ÷ 9: 27 ÷ 9 = 3, signs different → –3

(b) 84 ÷ (–4): 84 ÷ 4 = 21, signs different → –21

(c) (–56) ÷ (–2): 56 ÷ 2 = 28, signs same → 28

Turn each one into a multiplication to check it.

9 × (–3) = –27 ✓

(–4) × (–21) = 84 ✓

(–2) × 28 = –56 ✓

Why it happens: division inherits the multiplication sign rule, because a ÷ b is
answering “b × ? = a”. The unknown must carry whatever sign makes the product
come out right, and that is precisely the like-signs/unlike-signs rule.

Page 58 of 79

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Class 7 Maths Chapter 10 Operations with Integers
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co m
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Find the integer whose product with (– 1) is: (a) 27 (b) – 31 (c) – 1 (d) 1 (e) 0
e
Q3

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(b) –31 31
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