NCERT Solutions Class 7 Maths Chapter 11 Finding Common Ground

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 11: Finding Common
Ground

NCERT Textbook — Ganita Prakash

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 47 – 66 25 85 English

Solutions, notes, sample papers & more at 74 pages

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

CLASS 7 · MATHS · GANITA PRAKASH

NCERT Solutions — Chapter 11: Finding Common
Ground
Chapter 3 of Ganita Prakash Grade 7 Part II starts with Sameeksha tiling a 12 ft × 16 ft room and ends with a
relation that ties two numbers to their HCF and LCM. On the way it builds prime factorisation into a tool:
every factor of a number is a subpart of its prime factorisation, and that one idea gives both the HCF and the
LCM.

TEXTBOOK BOOK PAGES

Ganita Prakash (Class 7) Part II, 47 – 66

SECTIONS QUESTIONS

25 85

MEDIUM

English

In-text Questions — Page 47
Section 3.1 The Greatest of All

Q1 Sameeksha is building her new house. The main room of the house is 12 ft by 16 ft.
She feels that the room would look nice if the floor is covered with square tiles of
the same size. She also wants to use as few tiles as possible, and for the length of
the tile to be a whole number of feet. What size tiles should she buy?

She should buy square tiles of side 4 ft.
The tiles must fit the room exactly, with no cutting.

Across the breadth: the side of the tile must be a factor of 12

Along the length: the side of the tile must be a factor of 16

Factors of 12 = 1, 2, 3, 4, 6, 12

Factors of 16 = 1, 2, 4, 8, 16

Common factors = 1, 2, 4

Largest common factor = 4

So the possible tile sizes are 1 ft, 2 ft and 4 ft. The largest of these is 4 ft.

Page 1 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

16 ft

4 ft

12 ft

The 12 ft × 16 ft floor covered by 4 ft tiles — 3 rows of 4 tiles, 12 tiles in all, with nothing left over.

Why it happens: a tile of side s fits the 12 ft breadth only if 12 is a whole number of
tiles, that is, only if s divides 12. The same argument along the length forces s to
divide 16. So s has to be a common factor. Bigger tiles cover more floor each, so the
fewest tiles come from the biggest common factor — the HCF.

Tip: 4 is the Highest Common Factor (HCF) of 12 and 16. It is also called the
Greatest Common Divisor (GCD).

Q2 So, the square tiles can have sides 1 ft, 2 ft, and 4 ft. Among these, she should use
the largest sized square tile. Can you explain why?

Because a bigger tile covers more floor, so fewer tiles are needed.

Page 2 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Area of the room = 12 × 16 = 192 sq ft

With 1 ft tiles: 192 ÷ 1 = 192 tiles

With 2 ft tiles: 192 ÷ 4 = 48 tiles
With 4 ft tiles: 192 ÷ 16 = 12 tiles

Sameeksha wants as few tiles as possible, so she picks the 4 ft tile.

Why it happens: the floor area is fixed at 192 sq ft. A tile of side s covers s × s sq ft,
so the number of tiles is 192 ÷ s². Making s as large as possible makes s² as large as
possible, and so the count as small as possible.

In-text Questions — Page 48
Section 3.1 The Greatest of All — HCF and Lekhana's rice bags

Q1 How many tiles of this size should she purchase?

She should purchase 12 tiles.

Along the breadth: 12 ÷ 4 = 3 tiles

Along the length: 16 ÷ 4 = 4 tiles

Total = 3 × 4 = 12 tiles

Checking by area: 192 sq ft ÷ 16 sq ft per tile = 12 tiles ✓

Tip: counting rows × columns and counting by area must agree. If they do not, one
of the two is wrong — a quick way to catch a slip.

Page 3 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
What if Sameeksha did not insist on the length of the tile to be a whole number of
e
Q2

com
feet and the length could be a fractional number of feet? Would the answer
g l as
. a
em
change?

a s
a gl

m
No, the answer would not change. Even with fractional sides allowed, 4 ft is still the largest tile
. co ag
m
that fits, so 12 tiles is still the smallest number.

as e
g l
Many fractional sides do fit. For example a side of 4⁄3 ft works:
a
12 ÷ 4⁄3 = 12 × 3⁄4 = 9 tiles across the breadth ✓
co m
em.
m l as
.co g
16 ÷ 4⁄3 = 16 × 3⁄4 = 12 tiles along the length ✓

em of tiles = 9 × 12 = 108 — far more than 12 a
a s
gl
Number
a
But no side bigger than 4 ft can fit. Here is the reason.
m a s
m .co agl
l a se
g
Let the side be s ft, with 12 ÷ s = a tiles and 16 ÷ s = b tiles, a and b whole numbers

So 12 = a × s and 16 = b × s
a

co m
Subtracting: 16 − 12 = (b − a) × s
m .
m as e
.co
4 = (b − a) × s
a g l
se m
l a
Since 16 > 12, (b − a) is a whole number that is at least 1
ag So s = 4⁄(b − a) ≤ 4

se m
com g l a
m . a
ase
agl
Why it happens: the strip of floor left over when a 12 ft length is laid against a 16 ft
length is exactly 4 ft wide, and that strip must itself hold a whole number of tiles. So
the tile can never be wider than 4 ft — fractions or no fractions. Dropping the whole-

co m
.
number condition only adds smaller tiles such as 2 ft, 4⁄3 ft, 1 ft, 4⁄5 ft …, and every
m
m as e
.co l
one of them needs more tiles.
a g
se m
g l a
a Tip: notice what the argument really used — that 4 = 16 − 12 must also be a whole
c
m .
e
number of tiles. Taking differences like this is a very old and very quick way to hunt
m a s
co agl
for the HCF.
m .
as e
a g l

co m
m .
m ase
.co


a g l Page 4 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q3 Lekhana purchases rice from two farms and sells it in the market. She bought 84 kg
of rice from one farm and 108 kg from the other farm. She wants the rice to be
packed in bags, so each bag has rice from only one farm and all bags have the same
weight that is a whole number of kg. If she wants to use as few bags as possible,
what should the weight of each bag be?

Each bag should hold 12 kg.

Factors of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

Factors of 108 = 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108

Common factors = 1, 2, 3, 4, 6, 12

HCF = 12

With 12 kg bags:

From the first farm: 84 ÷ 12 = 7 bags

From the second farm: 108 ÷ 12 = 9 bags
Total = 16 bags

Why it happens: the bag weight must divide 84 exactly (no rice left loose) and must
divide 108 exactly as well, since the same size bag is used for both farms. So it has to
be a common factor. Heavier bags hold more, so the fewest bags come from the
heaviest allowed bag — the HCF.

Q4 She can use any of these weights to pack rice from both farms in bags of equal
weight. But, she wants to minimise the number of bags. Which weight should she
choose to minimise the number of bags?

She should choose the largest common factor, 12 kg.

Page 5 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

BAG WEIGHT BAGS FOR 84 KG BAGS FOR 108 KG TOTAL BAGS

1 kg 84 108 192

2 kg 42 54 96

3 kg 28 36 64

4 kg 21 27 48

6 kg 14 18 32

12 kg 7 9 16

Why it happens: the total rice is 84 + 108 = 192 kg whatever she does. The number
of bags is 192 ÷ (bag weight), so the heaviest bag gives the fewest bags. The
heaviest bag allowed is the HCF, 12 kg.

Try This — Page 48
Section 3.1 The Greatest of All — the 'Jump Jackpot' game from Grade 6

TRY THIS

Q1 Do you remember the ‘Jump Jackpot’ game from Grade 6 (see the chapter ‘Prime
Time’)? Grumpy places a treasure on a number and Jumpy chooses a jump size and
tries to collect the treasure. In each case below, the two numbers upon which
treasures are kept are given. Find the longest jump size (starting from 0) using
which Jumpy can land on both the numbers having the treasure. (a) 14 and 30 (b) 7
and 11 (c) 30 and 50 (d) 28 and 42

Jumping from 0 in steps of size j, Jumpy lands on 0, j, 2j, 3j, … — the multiples of j. To land on a
treasure number, j must be a factor of it.

NUMBERS COMMON FACTORS LONGEST JUMP

(a) 14 and 30 1, 2 2

(b) 7 and 11 1 1

(c) 30 and 50 1, 2, 5, 10 10

(d) 28 and 42 1, 2, 7, 14 14

Page 6 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

(a) 14 = 2 × 7, 30 = 2 × 3 × 5 → only 2 is shared → jump 2

(b) 7 and 11 are both prime and different → only 1 is shared → jump 1

(c) 30 = 2 × 3 × 5, 50 = 2 × 5 × 5 → 2 and 5 shared → jump 2 × 5 = 10
(d) 28 = 2 × 2 × 7, 42 = 2 × 3 × 7 → 2 and 7 shared → jump 2 × 7 = 14

Check it yourself: for (c), jumps of 10 land on 0, 10, 20, 30, 40, 50 — both treasures
collected. A jump of 15 lands on 30 but skips over 50.

Q2 Is the longest jump size for the numbers the same as their HCF? Explain why it is so.

Yes. The longest jump size is exactly the HCF of the two numbers.

(a) HCF(14, 30) = 2 = longest jump ✓

(b) HCF(7, 11) = 1 = longest jump ✓

(c) HCF(30, 50) = 10 = longest jump ✓

(d) HCF(28, 42) = 14 = longest jump ✓

Why it happens: starting at 0 with jump size j, Jumpy visits exactly the multiples of j.
He lands on a number only when that number is a multiple of j — that is, only when j
is a factor of it. To land on both treasures, j must be a factor of both numbers, so j is
a common factor. The question asks for the longest such jump, which is the greatest
common factor — the HCF. The two questions are word-for-word the same question.

In-text Questions — Page 49

Page 7 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Section 3.1 Primes and Prime Factorisation

Q1 So far, we have been listing all the factors to find the HCF. This can become
cumbersome for numbers with many factors, as you would have observed for the
numbers 30 and 50, and 28 and 42. Sometimes, we may also miss some factors
which can lead to errors. Can this process be simplified? Can it be made more
reliable?

Yes — prime factorisation simplifies it and makes it reliable.
Listing factors has two weaknesses. It is long for numbers such as 360 or 840, and it is easy to
miss one.

Listing method: 30 → 1, 2, 3, 5, 6, 10, 15, 30

50 → 1, 2, 5, 10, 25, 50

Compare the two lists, pick the largest match = 10

Prime method: 30 = 2 × 3 × 5, 50 = 2 × 5 × 5

Common primes: one 2 and one 5

HCF = 2 × 5 = 10

Why it happens: a number has many factors but only one prime factorisation.
Working with the primes means working with a short, fixed list instead of a long one
you must build yourself — so there is nothing to miss.

Q2 The number 90 could also have been factorised as 3 × 30 or 2 × 45 or in a few other
different ways. Will these all lead to the same prime factors?

Yes. The prime factors always come out the same, only the order may differ.

Page 8 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
e m.
90 = 9 × 10 = (3 × 3) × (2 × 5) = 2 × 3 × 3 × 5
m l as
.co
90 = 3 × 30 = 3 × (2 × 15) = 3 × 2 × 3 × 5
m a g
l a se
g
90 = 2 × 45 = 2 × (9 × 5) = 2 × 3 × 3 × 5
a90 = 6 × 15 = (2 × 3) × (3 × 5) = 2 × 3 × 3 × 5

co m
Every route ends with one 2, two 3s and onem5.. ag
l a se
ag
Why it happens: however you split the number, you keep breaking composite pieces
until nothing composite is left. The pieces you finish with are the primes hidden
co m
inside 90 from the start — splitting differently only changes the order in which you
em.
m l as
.co g
dig them out, not which ones are there.
m a
l a se
a g Did you know? This fact — that every number has exactly one prime factorisation
apart from order — is so important that it is called the Fundamental Theorem of
m a s
Arithmetic.
m .co agl
l a se
a g

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

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

co m
m .
m ase
.co


a g l Page 9 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q3 Procedure for Prime Factorisation. Can you see what is happening below?

3 105 2 30

5 35 3 15

7 5

The two figures from the book: a circled number with another number written to its
left, a circled number below it in the same way, and a plain number at the foot.

At each step the number is divided by a prime, and the quotient is written below.

105 ÷ 3 = 35, so write 3 outside and 35 below

35 ÷ 5 = 7, so write 5 outside and 7 below

7 is prime — stop

30 ÷ 2 = 15, so write 2 outside and 15 below

15 ÷ 3 = 5, so write 3 outside and 5 below

5 is prime — stop

Read the numbers outside from top to bottom, then the last number at the bottom:

105 = 3 × 5 × 7

30 = 2 × 3 × 5

Page 10 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Why it happens: each circled number is the product of the number to its left and
the number below it — 105 = 3 × 35 and 35 = 5 × 7. Putting these together, 105 = 3 ×
(5 × 7). Every step peels off one prime, so when the peeling stops, the primes
collected are the prime factorisation.

In-text Questions — Page 50
Section 3.1 The Division Method · Factors of a Number Using Prime Factorisation

Q1 Can you write the prime factorisation of 105 and 30 using these two figures?

3 105 2 30

5 35 3 15

7 5

The two figures from the book: a circled number with another number written to its
left, a circled number below it in the same way, and a plain number at the foot.

Collect the primes written on the left, then the prime left at the bottom.

Page 11 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

3 | 105

5 | 35

7
105 = 3 × 5 × 7

2 | 30

3 | 15

5

30 = 2 × 3 × 5

Check: 3 × 5 × 7 = 105 ✓ and 2 × 3 × 5 = 30 ✓

Tip: the last number at the bottom is a prime, so it counts too. Forgetting it is the
commonest slip in this method.

Q2 Try finding the prime factorisation of 1200 using the method above.

Divide by the smallest prime that works, again and again.

2 | 1200
2 | 600

2 | 300

2 | 150

3 | 75

5 | 25

5

1200 = 2 × 2 × 2 × 2 × 3 × 5 × 5

Check: 2 × 2 × 2 × 2 = 16, and 16 × 3 = 48, and 48 × 5 × 5 = 48 × 25 = 1200 ✓

Page 12 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Tip: keep dividing by 2 while the number is even, then by 3, then by 5, then by 7.
Working up the primes in order means you never have to guess.

Q3 If we had used the earlier method, our calculation would have been as follows: 1200
= 40 × 30 = 5 × 8 × 5 × 6 = … Which calculation is easier to carry out?

The division method is easier.

EARLIER METHOD DIVISION METHOD

You must spot a pair of factors of 1200 yourself You only test 2, 3, 5, 7 … in order

Several unfinished pieces to keep track of (8 and 6 still One number at a time, written in a neat
composite) column

Easy to stop too early and leave a composite factor You stop only when a prime is reached

Finishing the earlier calculation: 1200 = 40 × 30 = 5 × 8 × 5 × 6 = 5 × (2 × 2 × 2) × 5 × (2 × 3) = 2 × 2
× 2 × 2 × 3 × 5 × 5 — the same answer, but with more bookkeeping.

Why it happens: the division method turns factorising into a fixed routine — divide,
write, repeat — so there is nothing to invent at each step. The earlier method needs
a fresh idea every time you meet a new composite piece.

Q4 Consider the number 840 and its prime factorisation 2 × 2 × 2 × 3 × 5 × 7. Is 2 × 2 × 7 =
28 a factor of 840?

Yes, 28 is a factor of 840.
Reorder the prime factors so that 2 × 2 × 7 sit together (reordering does not change a product):

840 = 2 × 2 × 2 × 3 × 5 × 7

= (2 × 2 × 7) × (2 × 3 × 5)

= 28 × 30

So 840 ÷ 28 = 30, a whole number. Hence 28 is a factor.

Page 13 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: 2 × 2 × 7 uses two 2s and one 7, and 840 has three 2s and one 7 to

m l a se
spare. Since every prime asked for is available in 840's factorisation, the group can
o g prime
c and what remains is the other factor. A subpart of athe
be pulled.out
se m
g l a
factorisation is always a factor.
a
o m
If yes, what should it be multipliedm
e
. c ag
s
by to get 840?
a
Q5

agl

co m
m.
It should be multiplied by 30.

as e
. co×m2 × 7) × (2 × 3 × 5) a g l
em
840 = (2
a s
a gl The primes left over are 2, 3 and 5
s
2 × 3 × 5 = 30
m a
m .co agl
se
Check: 28 × 30 = 840 ✓

g l a
a
Tip: the partner of a factor is simply the primes that are left behind. You never need

co m
.
to divide.
e m
m l as
.co a g
a s em
a l
gIn-text Questions — Page 51
Section 3.1 Factors of a Number Using Prime Factorisation
se m
com g l a
m . a
ase
agl
Q1 Similarly, is 2 × 7 = 14 a factor of 840? Why or why not?

m

. co
m
Yes, 14 is a factor of 840.

m as e
.co a g l
s e m 840 = 2 × 2 × 2 × 3 × 5 × 7
agla
.c
= (2 × 7) × (2 × 2 × 3 × 5)

s e m
m a
co agl
= 14 × 60

m .
Check: 14 × 60 = 840 ✓
as e
a g l

co m
m .
m ase
.co


a g l Page 14 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Why it happens: 14 asks for one 2 and one 7. The factorisation of 840 has three 2s
and one 7, so both are available. The group can be lifted out, and 2 × 2 × 3 × 5 = 60 is
what remains.

Q2 Is 2 × 2 × 2 a factor of 840? Why or why not?

Yes, 2 × 2 × 2 = 8 is a factor of 840.

840 = (2 × 2 × 2) × (3 × 5 × 7)

= 8 × 105

Check: 8 × 105 = 840 ✓

Why it happens: 840 contains exactly three 2s, and 8 asks for exactly three 2s. There
are just enough, so 8 fits.

Check it yourself: is 16 = 2 × 2 × 2 × 2 a factor of 840? No — that would need four 2s
and 840 has only three. Indeed 840 ÷ 16 = 52.5.

Q3 Is 3 × 3 × 3 a factor of 840? Why or why not?

No, 3 × 3 × 3 = 27 is not a factor of 840.

840 = 2 × 2 × 2 × 3 × 5 × 7

Number of 3s in 840 = only one

27 needs three 3s

Check: 840 ÷ 27 = 31.11… — not a whole number ✗

Why it happens: a factor cannot ask for more copies of a prime than the number
actually has. 840 has just one 3, so 3 is a factor and 9 = 3 × 3 is not, let alone 27.

Page 15 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q4 Can we use this idea to list down all the possible factors of a number using just its
prime factors?

Yes. Every factor is a subpart of the prime factorisation, so listing all the subparts lists all the
factors.
Take 30 = 2 × 3 × 5.

No prime at all: 1

One prime at a time: 2, 3, 5

Two primes together: 2 × 3 = 6, 2 × 5 = 10, 3 × 5 = 15

All three: 2 × 3 × 5 = 30

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

Why it happens: a factor of 30 can only be built out of the primes inside 30.
Choosing which of those primes to keep and which to drop produces every factor
exactly once — and choosing none of them leaves 1, which is why 1 is always a
factor.

Q5 Find the factors of 225 using prime factorisation.

First factorise 225 by the division method.

5 | 225

5 | 45

3| 9

3| 3

1

225 = 3 × 3 × 5 × 5

Now form every subpart, systematically:

Page 16 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

No prime: 1

One prime: 3, 5

Two primes: 3 × 3 = 9, 5 × 5 = 25, 3 × 5 = 15

Three primes: 3 × 3 × 5 = 45, 3 × 5 × 5 = 75

Four primes: 3 × 3 × 5 × 5 = 225

Factors of 225 = 1, 3, 5, 9, 15, 25, 45, 75, 225 — nine factors in all.

Tip: 225 = 3² × 5². You may take 0, 1 or 2 threes (3 ways) and 0, 1 or 2 fives (3 ways),
so there are 3 × 3 = 9 factors. That count matches the list.

Q6 Check that all the factors of 225 occur in this list.

Test every number from 1 to 225 that could divide it — or simply pair the factors up.

PAIR PRODUCT

1 × 225 225 ✓
3 × 75 225 ✓
5 × 45 225 ✓
9 × 25 225 ✓
15 × 15 225 ✓

The nine factors fall into four pairs plus the middle one, 15. Nothing is missing, and nothing
extra has crept in.

Why it happens: factors come in pairs that multiply to the number, and the two
members of a pair sit on either side of √225 = 15. So it is enough to test 1 to 15: only
1, 3, 5, 9 and 15 divide 225, and each brings its partner. That gives exactly nine
factors.

Figure it Out — Page 51

Page 17 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Section 3.1 Factors of a Number Using Prime Factorisation

Q1 List all the factors of the following numbers: (a) 90 (b) 105 (c) 132 (d) 360 (this
number has 24 factors) (e) 840 (this number has 32 factors)

Factorise each number into primes, then build every subpart.

(a) 90 = 2 × 3 × 3 × 5

(b) 105 = 3 × 5 × 7

(c) 132 = 2 × 2 × 3 × 11

(d) 360 = 2 × 2 × 2 × 3 × 3 × 5

(e) 840 = 2 × 2 × 2 × 3 × 5 × 7

NUMBER ALL FACTORS HOW
MANY

(a) 90 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 12

(b) 105 1, 3, 5, 7, 15, 21, 35, 105 8

(c) 132 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132 12

(d) 360 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 24
360

(e) 840 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 32
105, 120, 140, 168, 210, 280, 420, 840

Tip — count before you list. Write the factorisation with powers and add 1 to each
power, then multiply.
90 = 2¹ × 3² × 5¹ → 2 × 3 × 2 = 12 factors
105 = 3¹ × 5¹ × 7¹ → 2 × 2 × 2 = 8 factors
132 = 2² × 3¹ × 11¹ → 3 × 2 × 2 = 12 factors
360 = 2³ × 3² × 5¹ → 4 × 3 × 2 = 24 factors
840 = 2³ × 3¹ × 5¹ × 7¹ → 4 × 2 × 2 × 2 = 32 factors
The counts for (d) and (e) match what the book says, so nothing has been missed.

Page 18 of 74

ase
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: to build a factor of 360 = 2³ × 3² × 5 you decide how many 2s to

m l a se
keep (0, 1, 2 or 3 — four choices), how many 3s (0, 1 or 2 — three choices) and how
o
.c or 1 — two choices). Each different set of choices gives
a ga different factor,
m
many 5s (0
so a
l se are 4 × 3 × 2 = 24 of them.
ag
there

. c om ag
In-text Questions — Pages 51–53 em
g l asPrime Factorisation
a
Section 3.1 Finding the HCF of Numbers Using

. c om
After observing a few prime factorisations, Anshu claims “The larger a number is,
em claim?
Q1

m a s
the longer its prime factorisation will be”. What do you think of Anshu’s

. co a gl
as em
g l

a Anshu's claim is false. One counterexample is enough to settle it.
a s
. com agl
em
96 = 2 × 2 × 2 × 2 × 2 × 3 → 6 prime factors
a s
agl
121 = 11 × 11 → 2 prime factors

121 is the larger number, but its factorisation is shorter
com
m .
m one. as e
cojust l
More counterexamples: 128 = 2⁷ has seven prime factors, while the much larger 1009 is prime
. a g
em
and has

a s
agl Why it happens: the length of a prime factorisation depends on how small the

se m
com A number that is a product of two large a
primes are, not on how big the number is. A number built from many 2s grows
. a g l
emfactors in its list.
slowly and collects a long factorisation.

a s
agl
primes is huge but has only two

co
Did you know? A claim made without proof is called a conjecture. Anshu's claim is a
m
m .
e
conjecture, and we disproved it by finding a counterexample — a single case where
m l as
.co g
it fails. To prove a conjecture true you must cover every case; to prove it false one

em a
s
example is enough.
l a
ag c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 19 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q2 Example 1: Find the common factors, and the HCF of 45 and 75. [So the common
factors should be subparts of both the factorisations. Can you write them down?]

The common factors are 1, 3, 5 and 15, and the HCF is 15.

45 = 3 × 3 × 5

75 = 3 × 5 × 5

A common factor must be a subpart of both lists. Compare them:

SUBPART INSIDE 3 × 3 × 5? INSIDE 3 × 5 × 5? COMMON FACTOR?

3 yes yes yes

5 yes yes yes

3 × 5 = 15 yes yes yes

3×3=9 yes no (only one 3) no

5 × 5 = 25 no (only one 5) yes no

Adding 1, the common factors are 1, 3, 5, 15. The highest is 15.

HCF(45, 75) = 15

Why it happens: 45 has two 3s and one 5; 75 has one 3 and two 5s. A shared
subpart can use at most one 3 (that is all 75 has) and at most one 5 (that is all 45
has). So the biggest shared subpart is 3 × 5 = 15.

Q3 Example 2: Find the common factors, and the HCF of 112 and 84.

The common factors are 1, 2, 4, 7, 14 and 28, and the HCF is 28.

Page 20 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

112 = 2 × 2 × 2 × 2 × 7

84 = 2 × 2 × 3 × 7

Common primes: 2 (112 has four, 84 has two → at most two) and 7 (one each).

COMMON SUBPART VALUE

2 2

7 7

2×2 4

2×7 14

2×2×7 28

Highest common subpart = 2 × 2 × 7 = 28

Check: 112 ÷ 28 = 4 ✓ and 84 ÷ 28 = 3 ✓

Why it happens: 84 has only two 2s, so no common factor may use more than two.
84 has one 3 but 112 has none, so 3 can never appear. Both have exactly one 7.
Taking the most that is allowed of each shared prime — two 2s and one 7 — gives
the highest common factor.

Q4 Example 3: Find the common factors and the HCF of 96 and 275.

The only common factor is 1, so the HCF is 1.

96 = 2 × 2 × 2 × 2 × 2 × 3

275 = 5 × 5 × 11

Primes in 96: 2, 3

Primes in 275: 5, 11

Shared primes: none

Page 21 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

With no prime in common, no subpart other than the empty one is shared. So 1 is the only
common factor, and it is also the HCF.

Why it happens: any common factor bigger than 1 would have to contain at least
one prime, and that prime would have to sit in both factorisations. Here the two lists
of primes do not overlap at all, so nothing bigger than 1 can be common.

Did you know? Two numbers whose HCF is 1 are called co-prime. Neither number
need be prime — 96 and 275 are both composite, yet they are co-prime.

Figure it Out — Page 53
Section 3.1 Finding the HCF of Numbers Using Prime Factorisation

Q1 Find the common factors and the HCF of the following numbers: (a) 50, 60 (b) 140,
275 (c) 77, 725 (d) 370, 592 (e) 81, 243

Factorise both numbers, keep the shared primes, then list every subpart built from them.

(a) 50 = 2 × 5 × 5, 60 = 2 × 2 × 3 × 5 → shared: one 2, one 5 → HCF = 2 × 5 = 10

(b) 140 = 2 × 2 × 5 × 7, 275 = 5 × 5 × 11 → shared: one 5 → HCF = 5

(c) 77 = 7 × 11, 725 = 5 × 5 × 29 → shared: none → HCF = 1

(d) 370 = 2 × 5 × 37, 592 = 2 × 2 × 2 × 2 × 37 → shared: one 2, one 37 → HCF = 2 × 37 =

74

(e) 81 = 3 × 3 × 3 × 3, 243 = 3 × 3 × 3 × 3 × 3 → shared: four 3s → HCF = 3 × 3 × 3 × 3 =

81

Page 22 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

NUMBERS COMMON FACTORS HCF

(a) 50, 60 1, 2, 5, 10 10

(b) 140, 275 1, 5 5

(c) 77, 725 1 1

(d) 370, 592 1, 2, 37, 74 74

(e) 81, 243 1, 3, 9, 27, 81 81

Why it happens: the common factors are exactly the factors of the HCF. That is why
each row's list is just the factor list of the number in the last column — 1, 2, 5, 10 are
the factors of 10; 1, 3, 9, 27, 81 are the factors of 81. Once you know the HCF, you
know all the common factors for free.

Tip: in (e), 81 is itself a factor of 243 (243 = 81 × 3). Whenever one number divides the
other, the smaller one is the HCF.

In-text Questions — Pages 53–54
Section 3.1 Finding the HCF Directly from the Prime Factorisations

Q1 How do we directly find the HCF without listing all the factors?

Build the HCF prime by prime: keep only the primes that occur in every number, and take each
one the fewest number of times it occurs.

Step 1 — write the prime factorisation of each number

Step 2 — mark the primes that appear in all of them

Step 3 — for each such prime, count how many times it occurs in each number and take the

minimum

Step 4 — multiply what you have collected

Page 23 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: the HCF is the largest subpart common to both factorisations. A

m l a
prime missing from one number can never appear in a common subpart. And a
o se
.coccurs twice in one number but three times in the other
a g can be used at
m
prime that

l a setwice — the smaller count is the limit. Taking the most that is allowed of each
g
most
ashared prime gives the largest common subpart.

com
e m . ag
g l as
a
Q2 Example 4: Find the HCF of 30 and 72. [How many 2s will it contain? How many 3s
will it contain?]

co m
em.
m l as
.co a g
HCF(30, 72) = 6.

se m
g l a
a 30 = 2 × 3 × 5

om a s
agl
72 = 2 × 2 × 2 × 3 × 3
. c
s e mso it is out)
Primes in both: 2 and 3 (5 is only in 30,
a
agl
How many 2s? 30 has one 2, 72 has three 2s. The minimum is one.

co m
.
How many 3s? 30 has one 3, 72 has two 3s. The minimum is one.

e m
co=m2 × 3 = 6 g l as
m . a
ase
HCF

agl Check: 30 ÷ 6 = 5 ✓ and 72 ÷ 6 = 12 ✓

se m
om not ask for more 2s than 30 can supply, and
cmay g l a
. a
emfor the 3s. So one 2 and one 3 is the most any
Why it happens: a common factor
a s
l 2 × 3 = 6 is the highest.
agand
30 supplies only one. The same
common factor can hold,

co m
m .
mExample 5: Find the HCF of 225 and 750. as e
.co a g l
m
Q3

l a se
ag
.c

s e m
m a
HCF(225, 750) = 75.

e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 24 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

225 = 3 × 3 × 5 × 5

750 = 2 × 3 × 5 × 5 × 5

Primes in both: 3 and 5 (2 is only in 750)

How many 3s? 225 has two, 750 has one. The minimum is one.
How many 5s? 225 has two, 750 has three. The minimum is two.

HCF = 3 × 5 × 5 = 75

Check: 225 ÷ 75 = 3 ✓ and 750 ÷ 75 = 10 ✓

Tip: the same recipe works for three or more numbers — take the primes common
to all of them, each the minimum number of times across all the factorisations.

Figure it Out — Page 54
Section 3.1 Finding the HCF of Numbers Using Prime Factorisation

Q1 Find the HCF of the following numbers: (a) 24, 180 (b) 42, 75, 24 (c) 240, 378 (d) 400,
2500 (e) 300, 800

Factorise, mark the shared primes, take the minimum count of each.

Page 25 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

PRIME FACTORISATIONS SHARED PRIMES (MINIMUM COUNT) HCF

(a) 24 = 2 × 2 × 2 × 3 two 2s, one 3 2 × 2 × 3 = 12
180 = 2 × 2 × 3 × 3 × 5

(b) 42 = 2 × 3 × 7 one 3 only (75 has no 2) 3
75 = 3 × 5 × 5
24 = 2 × 2 × 2 × 3

(c) 240 = 2 × 2 × 2 × 2 × 3 × 5 one 2, one 3 2×3=6
378 = 2 × 3 × 3 × 3 × 7

(d) 400 = 2 × 2 × 2 × 2 × 5 × 5 two 2s, two 5s 2 × 2 × 5 × 5 = 100
2500 = 2 × 2 × 5 × 5 × 5 × 5

(e) 300 = 2 × 2 × 3 × 5 × 5 two 2s, two 5s 2 × 2 × 5 × 5 = 100
800 = 2 × 2 × 2 × 2 × 2 × 5 × 5

Checks:

(a) 24 ÷ 12 = 2, 180 ÷ 12 = 15 ✓

(b) 42 ÷ 3 = 14, 75 ÷ 3 = 25, 24 ÷ 3 = 8 ✓

(c) 240 ÷ 6 = 40, 378 ÷ 6 = 63 ✓

(d) 400 ÷ 100 = 4, 2500 ÷ 100 = 25 ✓

(e) 300 ÷ 100 = 3, 800 ÷ 100 = 8 ✓

Tip: in (b), 75 is odd, so 2 cannot be in the HCF however many 2s the other two
numbers have. Scanning for a number that lacks a prime is the fastest way to throw
that prime out.

Q2 Consider the numbers 72 and 144. Suppose they are factorised into composite
numbers as: 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two
numbers have no common factor other than 1? Why not?

No, one cannot say that. The splittings 6 × 12 and 8 × 18 share no visible factor, but the
numbers themselves share plenty.

Page 26 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

72 = 6 × 12 = (2 × 3) × (2 × 2 × 3) = 2 × 2 × 2 × 3 × 3

144 = 8 × 18 = (2 × 2 × 2) × (2 × 3 × 3) = 2 × 2 × 2 × 2 × 3 × 3

Shared: three 2s and two 3s

HCF = 2 × 2 × 2 × 3 × 3 = 72

In fact 144 = 72 × 2, so 72 divides 144 and is itself the HCF. The common factors are all twelve
factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.

Why it happens: 6, 12, 8 and 18 are composite, so each still hides primes inside it. 6
and 8 look unrelated, but both contain a 2. Only when every piece has been broken
down to primes can you trust what you see. Comparing half-finished factorisations is
exactly the mistake this chapter's method is designed to prevent.

Check it yourself: 6 and 8 already share the factor 2, and 12 and 18 share 6. Even
the given splittings are not really factor-free.

In-text Questions — Page 55
Section 3.2 Least, but not Last!

Q1 Anshu and Guna make torans out of strips of cloth. Multiple strips are placed one
next to another to make a toran. Anshu uses strips of length 6 cm and Guna uses
strips of 8 cm length. If both have to make torans of the same length, what is the
smallest possible length, the torans could be? What is the length of the shortest
toran that they can both make?

The shortest common length is 24 cm.

Anshu's torans (multiples of 6): 6, 12, 18, 24, 30, 36, 42, 48, 54, …

Guna's torans (multiples of 8): 8, 16, 24, 32, 40, 48, 56, 64, 72, …

Common multiples: 24, 48, 72, …

Lowest common multiple = 24 cm

Anshu uses 24 ÷ 6 = 4 strips; Guna uses 24 ÷ 8 = 3 strips.

Page 27 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Why it happens: Anshu can only build lengths that are whole numbers of 6 cm
strips, so his toran is always a multiple of 6. Guna's is always a multiple of 8. For the
two torans to be the same length, that length must be a multiple of both — a
common multiple. The shortest one is the lowest common multiple.

Tip: 24 is the Lowest Common Multiple (LCM) of 6 and 8.

Q2 What about the largest common multiple? Does such a number exist?

No. There is no largest common multiple.

Common multiples of 6 and 8: 24, 48, 72, 96, 120, 144, …

Whatever common multiple you name, add 24 to it

The new number is still a multiple of 6 and still a multiple of 8

So it is a bigger common multiple

Why it happens: the multiples of a number go on forever, and so do the common
multiples — they are exactly the multiples of the LCM. A never-ending increasing list
has a smallest member but no largest one. That is why we speak of the lowest
common multiple and the highest common factor, and never the other way round:
the list of common factors is finite (nothing bigger than the numbers themselves), so
it has a greatest member.

Q3 A sweet shop gives out free gajak to school children on Mondays. Today is a Monday
and Kabamai enjoyed eating the gajak. But she visits the sweet shop once every 10
days. When is the next time she would be able to get free gajak from Sweet shop?
(Answer in number of days.)

After 70 days.

Page 28 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
e m.
Free gajak days (Mondays, multiples of 7): 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, …
m l as
.co
Kabamai's visits (multiples of 10): 10, 20, 30, 40, 50, 60, 70, …
m a g
l a se
g
First day common to both lists = 70
aLCM(7, 10) = 7 × 10 = 70

. c om ag
70 days is 10 weeks, and also Kabamai's 7thm
s e visit.

a gla
Why it happens: the shop gives gajak only on days that are multiples of 7 from
today, and Kabamai comes only on days that are multiples of 10. The two must line
co m
up, so the day has to be a common multiple. The next such day is the lowest
em.
m l as
.co g
common multiple. Since 7 and 10 have no common prime factor, their LCM is simply
m a
l a se
their product.

a g
m a s
In-text Questions — Page 56
m .co agl
l a se
g
Section 3.2 Least, but not Last! — the 'Idli-Vada' game from Grade 6
a
Q1 Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two
co m
m .
se
numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’

o m l a
ag be called out. In each
should be called out depending on whose multiple the number is. If the number
.chappens to be a common multiple, then ‘idli-vada’ should
m problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find
l a se
ag the first number for which ‘idli-vada’ will be called out: (a) 4 and 6 (b) 7 and 11 (c) 14
and 30 (d) 15 and 55
se m
com g l a
m . a
ase
agl

'Idli-vada' is called out at the first number that is a multiple of both — the LCM.

co m
NUMBERS PRIME FACTORISATIONS
.
FIRST 'IDLI-VADA'
m
m 4 and 6 l a se12
.co ag
(a) 4 = 2 × 2, 6 = 2 × 3

a s em (b)
gl
7 and 11 7, 11 (both prime)
a
77

.c
(c) 14 and 30 14 = 2 × 7, 30 = 2 × 3 × 5 210
s e m
m a
15 and 55
e m . co
15 = 3 × 5, 55 = 5 × 11 agl
as
(d) 165

a g l

co m
m .
m ase
.co


a g l Page 29 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

(a) LCM = 2 × 2 × 3 = 12 (multiples of 4: 4, 8, 12…; of 6: 6, 12…)

(b) LCM = 7 × 11 = 77

(c) LCM = 2 × 3 × 5 × 7 = 210
(d) LCM = 3 × 5 × 11 = 165

Tip: for (c) and (d), listing multiples takes a long time — 210 is the 15th multiple of
14. Prime factorisation gets there in one line.

Q2 Is the answer always the LCM of the two numbers? Explain.

Yes, always.

Why it happens: 'idli' is called at every multiple of the first number and 'vada' at
every multiple of the second. 'Idli-vada' is called only when a number is on both lists,
that is, at a common multiple. Counting starts at 1 and goes up, so the first such
number reached is the smallest common multiple — the LCM, by definition.

(a) LCM(4, 6) = 12 ✓ (8 is only 'idli', 6 is only 'vada')

(b) LCM(7, 11) = 77 ✓

(c) LCM(14, 30) = 210 ✓

(d) LCM(15, 55) = 165 ✓

Check it yourself: every later 'idli-vada' is a multiple of the first one — 12, 24, 36, …
for (a). The common multiples of two numbers are exactly the multiples of their LCM.

Q3 How do we find the LCM of two numbers using their prime factors?

Collect every prime that appears in either number, and take each one the greatest number of
times it occurs in either factorisation.

Page 30 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Step 1 — write both prime factorisations

Step 2 — list every prime that appears anywhere

Step 3 — for each prime, take the maximum number of occurrences

Step 4 — multiply

Example: 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3.

2s: 12 has two, 18 has one → take two
3s: 12 has one, 18 has two → take two

LCM = 2 × 2 × 3 × 3 = 36

Why it happens: a multiple of 12 must contain 2 × 2 × 3 inside it, and a multiple of
18 must contain 2 × 3 × 3. To hold both, the number needs at least two 2s and at
least two 3s. Taking exactly that many — and no spare primes — gives the smallest
such number. Notice the contrast with the HCF: there we took the minimum count,
here the maximum.

In-text Questions — Pages 57–58
Section 3.2 Finding LCM through Prime Factorisation

Q1 We get, 36 = 2 × 2 × 3 × 3, 648 = 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3). What do you
observe? We can see that the prime factors of the multiple contain the prime
factors of the number along with some more prime factors. Will this happen with
every multiple?

Yes, with every multiple.

36 = 2 × 2 × 3 × 3

648 = 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3) = 2 × 2 × 2 × 3 × 3 × 3 × 3

The block 2 × 2 × 3 × 3 is still sitting inside 648

Try two more multiples of 36:

Page 31 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

36 × 5 = 180 = (2 × 2 × 3 × 3) × 5 ✓

36 × 14 = 504 = (2 × 2 × 3 × 3) × (2 × 7) ✓

Why it happens: a multiple of 36 is 36 × k for some whole number k. Factorising it
means factorising 36 and factorising k and putting the two lists side by side. So the
primes of 36 are always there, joined by whatever primes k brings. In short: every
multiple of a number contains that number's whole prime factorisation.

Q2 Can this be used to find the LCM?

Yes. It tells us exactly what a common multiple must contain.

A common multiple of a and b must contain

• the whole prime factorisation of a, and

• the whole prime factorisation of b

The smallest such number carries nothing extra

So for each prime, keep just enough copies to cover whichever number needs more of it — the
maximum of the two counts.

Why it happens: if a common multiple held fewer copies of some prime than one of
the numbers needs, that number's factorisation would not fit inside it, and it would
not be a multiple. If it held more copies than needed, or an extra prime, it would still
be a common multiple but a larger one. So the lowest common multiple takes each
prime the maximum number of times and nothing beyond.

Q3 Example 6: Find the LCM of 14 and 35. [What is the lowest among all the common
multiples of 14 and 35?]

LCM(14, 35) = 70.

Page 32 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

14 = 2 × 7

35 = 5 × 7

Primes appearing: 2, 5, 7

2s: one (in 14) → take one

5s: one (in 35) → take one

7s: one in each → take one

LCM = 2 × 5 × 7 = 70

Check: 70 ÷ 14 = 5 ✓ and 70 ÷ 35 = 2 ✓

Why it happens: 2 × 5 × 7 contains 2 × 7 = 14 as a subpart and 5 × 7 = 35 as a
subpart, so it is a common multiple. Remove any one of the three primes and it
stops working — drop the 2 and 14 no longer fits; drop the 5 and 35 no longer fits;
drop the 7 and neither fits. Nothing smaller can do the job, so 70 is the lowest.

Tip: the single 7 is shared. Writing 14 × 35 = 490 would also give a common multiple,
but it counts the 7 twice, so it is 7 times bigger than it needs to be.

Q4 Common multiples should contain each prime factor as a subpart: 2 × 7 as a subpart
and 5 × 7 as a subpart. For example, 2 × 7 × 5 × 7 × 3 is a common multiple of 14 and
35. 2 × 2 × 5 × 7 × 7 × 11 is another common multiple. Is 2 × 3 × 5 × 7 also a common
multiple?

Yes. 2 × 3 × 5 × 7 = 210 is a common multiple of 14 and 35.

2 × 3 × 5 × 7 = 210

Contains 2 × 7 = 14 as a subpart ✓ → 210 = 14 × 15

Contains 5 × 7 = 35 as a subpart ✓ → 210 = 35 × 6

210 ÷ 14 = 15 ✓ and 210 ÷ 35 = 6 ✓

But it is not the lowest: 210 = 70 × 3, and the extra 3 is not needed by either number.

Page 33 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
Why it happens: any number that holds one 2, one 5 and one 7 is a common

m l a se
multiple of 14 and 35, whatever else it holds. Extra primes such as the 3 here only
o
c
make the.number a gleaves 2 × 5 × 7 =
m
bigger. Stripping away everything not required
se LCM.
70,athe
l
ag
o m
Example 7: Find the LCM of 96 and m
e
. c ag
s
360. [How many 2s should the LCM contain? How
a How many 5s should the LCM contain?]
Q5

agl
many 3s should the LCM contain?

co m
em.
as
LCM(96, 360) = 1440.
m l
.co a g
a s e=m2 × 2 × 2 × 2 × 2 × 3
gl
96
a 360 = 2 × 2 × 2 × 3 × 3 × 5

m a s
.co agl
Primes appearing anywhere: 2, 3, 5

a s em
a gl three. Take the maximum, five. Five 2s contain three 2s, so
How many 2s? 96 has five, 360 has
both are covered.

co m
.
How many 3s? 96 has one, 360 has two. Take two.

e m
m as
How many 5s? 96 has none, 360 has one. Take one.

.co a g l
s m
eLCM
gl a =2×2×2×2×2×3×3×5
a = 32 × 9 × 5
se m
com g l a
= 32 × 45
m. a
= 1440
gl ase
a
Check: 1440 ÷ 96 = 15 ✓ and 1440 ÷ 360 = 4 ✓

co m
m .
o m l a se
.cmaximum count of all the primes. Here HCF(96, 360) = a2g× 2 × 2 × 3 = 24, and 24 ×
Tip: HCF asks for the minimum count of the shared primes; LCM asks for the

se m
g l a
a
1440 = 34560 = 96 × 360 ✓

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

co m
m .
m ase
.co


a g l Page 34 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q6 Choosing more than five occurrences of 2s will give a common multiple; but it will
not be the lowest. Are you able to see why?

Because every extra 2 doubles the number without being needed by 96 or by 360.

Five 2s: 2⁵ × 3 × 3 × 5 = 1440

Six 2s: 2⁶ × 3 × 3 × 5 = 2880 = 1440 × 2 — still a common multiple, but twice as big

Seven 2s: 5760 = 1440 × 4 — bigger still

Why it happens: 96 needs five 2s and 360 needs three. Five copies already cover
both demands. A sixth 2 satisfies no new demand, so the only thing it changes is the
size. Since the LCM is defined as the smallest common multiple, we stop at exactly
what is required — not one prime more.

Figure it Out — Page 58
Section 3.2 Finding LCM through Prime Factorisation

Q1 Find the LCM of the following numbers: (a) 30, 72 (b) 36, 54 (c) 105, 195, 65 (d) 222,
370

Take every prime that appears, each the maximum number of times.

Page 35 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

PRIME MAXIMUM COUNT OF EACH LCM
FACTORISATIONS PRIME

(a) 30 = 2 × 3 × 5 three 2s, two 3s, one 5 2×2×2×3×3×5=
72 = 2 × 2 × 2 × 3 × 3 360

(b) 36 = 2 × 2 × 3 × 3 two 2s, three 3s 2 × 2 × 3 × 3 × 3 = 108
54 = 2 × 3 × 3 × 3

(c) 105 = 3 × 5 × 7 one 3, one 5, one 7, one 13 3 × 5 × 7 × 13 = 1365
195 = 3 × 5 × 13
65 = 5 × 13

(d) 222 = 2 × 3 × 37 one 2, one 3, one 5, one 37 2 × 3 × 5 × 37 = 1110
370 = 2 × 5 × 37

Checks:

(a) 360 ÷ 30 = 12, 360 ÷ 72 = 5 ✓

(b) 108 ÷ 36 = 3, 108 ÷ 54 = 2 ✓

(c) 1365 ÷ 105 = 13, 1365 ÷ 195 = 7, 1365 ÷ 65 = 21 ✓

(d) 1110 ÷ 222 = 5, 1110 ÷ 370 = 3 ✓

Tip: in (d) the LCM 1110 is much smaller than the product 222 × 370 = 82,140 —
because the two numbers share 2 and 37. In fact HCF(222, 370) = 74, and 74 × 1110 =
82,140.

In-text Questions — Pages 58–59
Section 3.3 Patterns, Properties, and a Pretty Procedure!

Q1 The HCF of 6 and 18 is 6, which is one of the two given numbers. Find more such
number pairs where the HCF is one of the two numbers. How can we describe such
pairs of numbers?

This happens exactly when one number is a factor of the other.

Page 36 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

PAIR RELATION HCF

6, 18 18 = 6 × 3 6

5, 40 40 = 5 × 8 5

12, 60 60 = 12 × 5 12

9, 9 9=9×1 9

14, 98 98 = 14 × 7 14

The general statement: if one number is a factor of the other, their HCF is the smaller
number — and the other number is a multiple of it.

Why it happens: the HCF of two numbers can never be bigger than the smaller
number, since it has to divide it. So the best possible case is that the smaller number
divides the larger one too. When it does, it is a common factor and it is as large as a
common factor can be — so it is the HCF.

Did you know? A statement that holds in every possible case, like the one above, is
called a general statement, and the process of arriving at it is generalisation.

Q2 For number pairs satisfying this property (i.e., one of the numbers is the HCF), (a) if
m is a number, what could be the other number? (b) if 7k is a number, what could be
the other number?

The other number must be a multiple of the first.
(a) If one number is m, the other could be 2m, 3m, 4m, … — in general am, where a is any
positive integer.

HCF of m and am = m

Example: m = 15 and a = 4 → 15 and 60, HCF = 15 ✓

(b) If one number is 7k, the other could be any multiple of 7k, that is 7ka for a positive integer a
— for instance 14k, 21k, 28k.

Page 37 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

HCF of 7k and 14k = 7k

Example: k = 3 → 21 and 42, HCF = 21 = 7 × 3 ✓

Why it happens: in the general form n and an, the prime factorisation of an
contains every prime of n plus whatever a adds. So the largest subpart shared by the
two is the whole of n, and HCF(n, an) = n. Writing it with letters shows that the
pattern does not depend on which particular numbers you picked.

Tip: the other number could also be 7k itself (taking a = 1), since HCF(7k, 7k) = 7k. A
number is always a factor of itself.

Figure it Out — Page 59
Section 3.3 Patterns, Properties, and a Pretty Procedure!

MATH TALK

Q1 Make a general statement about the HCF for the following pairs of numbers. You
could consider examples before coming up with general statements. Look for
possible explanations of why they hold. (a) Two consecutive even numbers (b) Two
consecutive odd numbers (c) Two even numbers (d) Two consecutive numbers (e)
Two co-prime numbers. Share your observations with the class.

Try examples first, then state the rule.

Page 38 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
EXAMPLES HCF GENERAL STATEMENT

m6, 8 → 2 as e
(a) consecutive
.co 14, 16 → 2
always 2
a g l
The HCF of two consecutive even numbers is 2.
even
se m
g l a 30, 32 → 2
a
(b) consecutive 7, 9 → 1 always 1 Two consecutive odd numbers are co-prime; their

m
.co ag
odd 15, 17 → 1 HCF is 1.

sem
21, 23 → 1

(c) two even 6, 8 → 2 a gla
an even The HCF of two even numbers is always even (at least
12, 18 → 6 number 2), but it can be anything even.
20, 100 →
co m
em.
20

m l as
m .co
(d) consecutive 8, 9 → 1 always 1
a g
Two consecutive numbers are always co-prime.

l a se 20, 21 → 1

a g 99, 100 → 1

a s
com
4, 9 → 1 The HCF of two co-prime numbers is 1, by definition.

agl
(e) co-prime always 1
25, 42 → 1
m .
ase
Why they hold: agl
(d) Write them as n and n + 1. A common factor must divide the difference, which
co m
m .
is 1. The only number that divides 1 is 1 itself.
m as e
.co a
(a) Write them as 2n and 2n + 2 = 2(n + 1). Both contain a 2, so 2 is common.
g l
se m Beyond that, the HCF of n and n + 1 is 1 by (d), so nothing more can be shared.
g l a
a HCF = 2 × 1 = 2.
(b) Their difference is 2, so any common factor divides 2 — it is 1 or 2. But both
se m
com g l a
. a
numbers are odd, so 2 cannot divide them. Only 1 is left.
m
ase
(c) Both are multiples of 2, so 2 is a common factor and the HCF is at least 2. How

agl
much more they share depends on the numbers, so no fixed value can be
promised.

co m
.
(e) "Co-prime" means "no common prime factor". With no shared prime, the

e m
as
largest common subpart is empty, which is 1.
m l
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 ase
.co


a g l Page 39 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q2 The LCM of 3 and 24 is 24 (it is one of the two given numbers). (a) Find more such
number pairs where the LCM is one of the two numbers. (b) Make a general
statement about such numbers. Describe such number pairs using algebra.

(a) More such pairs:

PAIR RELATION LCM

3, 24 24 = 3 × 8 24

5, 20 20 = 5 × 4 20

7, 21 21 = 7 × 3 21

6, 36 36 = 6 × 6 36

11, 11 11 = 11 × 1 11

(b) General statement: the LCM is one of the two numbers exactly when one number is a
factor of the other — and then the LCM is the larger number.

In algebra: the pair is n and an, where a is a positive integer

LCM(n, an) = an

HCF(n, an) = n

Why it happens: an is already a multiple of n and a multiple of itself, so it is a
common multiple. No common multiple can be smaller than the larger number,
because a multiple of an is at least an. So an is the lowest — the LCM.

Tip: this is the same family of pairs as in the HCF question on page 58. For such a
pair, the smaller number is the HCF and the larger is the LCM — and indeed n × an =
HCF × LCM ✓

Page 40 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q3 Make a general statement about the LCM for the following pairs of numbers. You
could consider examples before coming up with these general statements. Look for
possible explanations of why they hold. (a) Two multiples of 3 (b) Two consecutive
even numbers (c) Two consecutive numbers (d) Two co-prime numbers

EXAMPLES GENERAL STATEMENT

(a) two multiples of 3 6, 9 → 18 The LCM is always a multiple of 3.
12, 18 → 36
15, 21 → 105

(b) consecutive even 6, 8 → 24 The LCM is half the product of the two numbers.
10, 12 → 60
14, 16 → 112

(c) consecutive 8, 9 → 72 The LCM is the product of the two numbers.
11, 12 → 132
20, 21 → 420

(d) co-prime 4, 9 → 36 The LCM is the product of the two numbers.
7, 11 → 77
25, 42 → 1050

Why they hold:

(a) The LCM is a multiple of each number, and each number is a multiple of 3. So
the LCM contains a 3 in its prime factorisation.
(c) and (d) Co-prime numbers share no prime. So the LCM must take every prime
of the first and every prime of the second, with nothing overlapping — which is
exactly their product. Consecutive numbers are always co-prime, so (c) is a special
case of (d).
(b) Two consecutive even numbers are 2n and 2(n + 1), with HCF 2. Their product
is 4n(n + 1), while the LCM only needs one of the two 2s, giving 2n(n + 1) — half
the product. Check with 6 and 8: product 48, LCM 24 ✓

Tip: all four statements are the same fact seen from different sides — HCF × LCM =
product. When the HCF is 1, LCM = product; when the HCF is 2, LCM = half the
product.

Page 41 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

In-text Questions — Pages 59–60
Section 3.3 Doubling both numbers · Multiples of the same number

Q1 What happens to the HCF of two numbers if both numbers are doubled? Take some
pairs of numbers and explore. Are you able to see why the HCF will also double?

The HCF also doubles.

PAIR HCF DOUBLED PAIR NEW HCF

270, 50 10 540, 100 20

12, 18 6 24, 36 12

30, 72 6 60, 144 12

7, 11 1 14, 22 2

270 = 2 × 3 × 3 × 3 × 5, 50 = 2 × 5 × 5 → HCF = 2 × 5 = 10
540 = 2 × 2 × 3 × 3 × 3 × 5, 100 = 2 × 2 × 5 × 5 → HCF = 2 × 2 × 5 = 20

Why it happens: doubling a number puts one extra 2 into its prime factorisation. Do
it to both numbers and each gains one 2, so the count of 2s that they share goes up
by exactly one. Every other prime is untouched, so the largest common subpart is
the old HCF with one more 2 in it — that is, twice as big.

Check it yourself: the same reasoning works for tripling (HCF triples) or multiplying
both by 10 (HCF becomes ten times). In general, HCF(ka, kb) = k × HCF(a, b).

Q2 Consider the following two multiples of 14 — 14 × 6, 14 × 9. What is their HCF?

HCF = 42, not 14.

Page 42 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

14 × 6 = 84 = 2 × 7 × 2 × 3

14 × 9 = 126 = 2 × 7 × 3 × 3

Shared primes: one 2, one 3, one 7

HCF = 2 × 3 × 7 = 14 × 3 = 42

Check: 84 ÷ 42 = 2 ✓ and 126 ÷ 42 = 3 ✓

Why it happens: 14 is certainly a common factor, but it is not the highest one. The
multipliers 6 and 9 are themselves not co-prime — both contain a 3. That extra
shared 3 joins the 14, giving 14 × 3 = 42. In general, HCF(14a, 14b) = 14 × HCF(a, b),
and here HCF(6, 9) = 3.

Q3 Here are some more numbers where both numbers are multiples of the same
number. Find their HCF: (a) 18 × 10, 18 × 15 (b) 10 × 38, 10 × 21 (c) 5 × 13, 5 × 20 (d) 12 ×
16, 12 × 20

Use HCF(ka, kb) = k × HCF(a, b) — the common multiplier comes out, and whatever the
multipliers still share joins it.

NUMBERS MULTIPLIER K HCF OF THE MULTIPLIERS HCF

(a) 180, 270 18 HCF(10, 15) = 5 18 × 5 = 90

(b) 380, 210 10 HCF(38, 21) = 1 10 × 1 = 10

(c) 65, 100 5 HCF(13, 20) = 1 5×1=5

(d) 192, 240 12 HCF(16, 20) = 4 12 × 4 = 48

Checks by prime factorisation:

(a) 180 = 2 × 2 × 3 × 3 × 5, 270 = 2 × 3 × 3 × 3 × 5 → 2 × 3 × 3 × 5 = 90 ✓
(b) 380 = 2 × 2 × 5 × 19, 210 = 2 × 3 × 5 × 7 → 2 × 5 = 10 ✓

(c) 65 = 5 × 13, 100 = 2 × 2 × 5 × 5 → 5 ✓

(d) 192 = 2⁶ × 3, 240 = 2⁴ × 3 × 5 → 2 × 2 × 2 × 2 × 3 = 48 ✓

Page 43 of 74

ase
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
In which of these cases is the HCF the same as the common multiplier, like problem
e
Q4

comhappens.
(b) where the HCF is 10? Explore a few more examples of this type to understand
when.this g l as
em a
a s
a gl

m
In (b) and (c) only.
. co ag
em
MULTIPLIERS
l as
ARE THEY CO-PRIME?
g
HCF = MULTIPLIER?

(a) 10 and 15 (share 5)
a
no no — HCF 90, multiplier 18

co m
m.
(b) 38 and 21 yes yes — both 10

m as e
.co a g l
m
(c) 13 and 20 yes yes — both 5

l a se
a g (d) 16 and 20 (share 4) no no — HCF 48, multiplier 12

m a s
agl
The rule: HCF(ka, kb) equals k exactly when a and b are co-prime.

m .co
l a se
More examples:
a g
9 × 4, 9 × 25 → 4 and 25 co-prime → HCF = 9 ✓

co m
9 × 4, 9 × 26 → 4 and 26 share 2 → HCF = 9 × 2 = 18 ✗
m .
m as e
.co
7 × 6, 7 × 35 → 6 and 35 co-prime → HCF = 7 ✓
a g l
se m
g l a
a Why it happens: HCF(ka, kb) = k × HCF(a, b). The answer is k itself only when the
se m
com that extra piece rides along and makes the
second piece, HCF(a, b), equals 1 — that is, when the multipliers share no prime
g l a
. a
em
factor at all. If they do share something,
a s
agl
HCF bigger than k.

co m
m .
In-text Questions — Pages 60–61
m as e
.co a g l
m
Section 3.3 Efficient Procedures for HCF and LCM

l a se
ag See the procedure on the right. Can you explain how it has been carried out? [2 | 84,
.c
m
Q1
180 · 2 | 42, 90 · 3 | 21, 45 · 7, 15]
m a s e
e m . co agl
g l as
a
Both numbers are divided by a common prime at each step, and the two quotients are written
in the next row.

co m
m .
m ase
.co


a g l Page 44 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

2 | 84, 180 → 84 ÷ 2 = 42 and 180 ÷ 2 = 90

2 | 42, 90 → 42 ÷ 2 = 21 and 90 ÷ 2 = 45

3 | 21, 45 → 21 ÷ 3 = 7 and 45 ÷ 3 = 15

7, 15 → 7 and 15 have no common prime factor, so we stop

This is the division method of prime factorisation, run on two numbers at once.

Why it happens: at every step the divisor must divide both numbers, so it is a
common factor. We stop when the two numbers left share nothing — 7 is prime and
does not divide 15. Reading the work back gives 84 = 2 × 2 × 3 × 7 and 180 = 2 × 2 × 3
× 15.

Q2 How do we use this to find the HCF of 84 and 180? Explore. [Hint: Observe that 84 = 2
× 2 × 3 × 7, and 180 = 2 × 2 × 3 × 15 similar to prime factorisation]

Multiply the numbers in the left column.

Left column: 2, 2, 3

HCF(84, 180) = 2 × 2 × 3 = 12

Check: 84 ÷ 12 = 7 ✓ and 180 ÷ 12 = 15 ✓

Why it happens: the hint spells it out — the ladder rewrites the numbers as 84 = (2 ×
2 × 3) × 7 and 180 = (2 × 2 × 3) × 15. The bracket, which is the left column, is a
common factor of both. And it is the highest one, because we only stopped when the
leftovers 7 and 15 had nothing more in common. Anything still shared would have
given one more row.

Tip: this is much quicker than factorising each number separately and comparing —
one ladder does both numbers at once.

Page 45 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q3 Find the HCF in the following cases. [2 | 300, 150 · 5 | 150, 75 · 5 | 30, 15 · 3 | 6, 3 · 2, 1
— HCF = 2 × 5 × 5 × 3] and [2 | 630, 770 · 5 | 315, 385 · 7 | 63, 77 · 9, 11 — HCF = 2 × 5 ×
7]

Multiply out the left column of each ladder.

300 and 150

2 | 300, 150 → 150, 75

5 | 150, 75 → 30, 15

5 | 30, 15 → 6, 3

3 | 6, 3 → 2, 1

2 and 1 have no common factor — stop

HCF = 2 × 5 × 5 × 3 = 150

630 and 770

2 | 630, 770 → 315, 385

5 | 315, 385 → 63, 77

7 | 63, 77 → 9, 11

9 and 11 have no common factor — stop

HCF = 2 × 5 × 7 = 70

Checks: 300 ÷ 150 = 2 and 150 ÷ 150 = 1 ✓; 630 ÷ 70 = 9 and 770 ÷ 70 = 11 ✓

Tip: for 300 and 150 the HCF is 150, one of the two numbers — as expected, since
150 is a factor of 300.

Q4 This procedure not only gives the HCF but can also be used to find the LCM! Can you
see how?

Multiply the left column together with the last row — the whole L-shape.

Page 46 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

300 and 150: left column 2, 5, 5, 3 and last row 2, 1

LCM = 2 × 5 × 5 × 3 × 2 × 1 = 300

630 and 770: left column 2, 5, 7 and last row 9, 11

LCM = 2 × 5 × 7 × 9 × 11 = 70 × 99 = 6930

Note that for 630 and 770 the book carries the ladder one step further, dividing 9 by 3 to get 3
— which is why it writes LCM = 2 × 5 × 7 × 3 × 3 × 11. That is the same number: 3 × 3 = 9.

Why it happens: the ladder splits each number into (common part) × (leftover). Here
630 = 70 × 9 and 770 = 70 × 11. A common multiple must contain the common part
70, the leftover 9 and the leftover 11 — and once the leftovers share nothing, that is
all it needs.

Q5 Why are these the LCMs? [Hint: Will the product of the factors marked as the LCM of
300 and 150 contain the prime factorisations of both 300 and 150? Is this the
smallest such number?]

Because the L-shaped product contains both numbers, and nothing in it can be dropped.

Does it contain both?

2 × 5 × 5 × 3 × 2 × 1 = 300

Reading 300 = (2 × 5 × 5 × 3) × 2 — the left column with the 2 from the last row ✓

Reading 150 = (2 × 5 × 5 × 3) × 1 — the left column with the 1 from the last row ✓

So 300 is a common multiple of 300 and 150

Is it the smallest?

Left column = HCF = 150, leftovers 2 and 1

Any common multiple must hold the HCF 150, the leftover 2 and the leftover 1

The leftovers share nothing, so none of them can be spared

Smallest = 150 × 2 × 1 = 300

Page 47 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Same reasoning for 630 and 770: HCF 70, leftovers 9 and 11, which are co-prime. LCM = 70 × 9 ×
11 = 6930.

Why it happens: write the two numbers as h × p and h × q, where h is the HCF and
p, q are the leftovers. The number h × p × q is a multiple of both (it is h × p times q,
and h × q times p). It cannot be made smaller: dropping anything from p would stop
it being a multiple of h × p, and dropping anything from q would stop it being a
multiple of h × q. So h × p × q is the LCM. Notice too that HCF × LCM = h × (h × p × q)
= (h × p) × (h × q) = the product of the two numbers.

Try This — Page 62
Section 3.3 Efficient Procedures — Guna's and Anshu's shortcut

TRY THIS

Q1 Guna says “I found a better way to factorise to find HCF/LCM. This is faster than
what was taught in class! For the numbers 300 and 150, I can first directly divide
both numbers by 50. The HCF will be 50 × 3. The LCM will be 50 × 3 × 2 × 1”. Anshu
also tried to remove the bigger common factors. “For 630 and 770, I will divide both
numbers by 10 first. Now, I can divide them by 7. The HCF will be 10 × 7 = 70. The
LCM will be 10 × 7 × 9 × 11 = 6930”. Can you see why this works?

It works because a step of the ladder never has to be a prime — any common factor will do.

Guna, on 300 and 150

50 | 300, 150 → 6, 3

3 | 6, 3 → 2, 1

HCF = 50 × 3 = 150

LCM = 50 × 3 × 2 × 1 = 300

Page 48 of 74

ase
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
e m.
Anshu, on 630 and 770
m l as
.co
10 | 630, 770 → 63, 77
m a g
l a se
g
7 | 63, 77 → 9, 11
aHCF = 10 × 7 = 70

co m
. ag
LCM = 10 × 7 × 9 × 11 = 6930

se m
l a
agslower prime-by-prime ladders on page 61.
Both answers agree exactly with the

. com
Why it happens: dividing both numbers by 50 in one go is the same as dividing by 2,

m a s
then by 5, then by 5 again — the three rows are merged into one. The emleft column
. co to the same thing, because 50 = 2 × 5 × 5. Nothing
a gl is skipped; the
em are only collected in bigger bundles. The one condition is that the divisor
still multiplies

a s
gl
primes
a must be a common factor of both numbers, and that you keep going until the

s
leftovers share nothing.
m a
m .co agl
l a se
g
Tip: the shortcut is only safe if you finish the job. If Guna had stopped after dividing
a
by 50, he would have got 50 as the HCF instead of 150 — because 6 and 3 still share

m
a 3.
. co
e m
m l as
.co a g
a s em You can try this method for these pairs of numbers. (a) 90 and 150 (b) 84 and 132
gl
Q2

a
se m
com a

. a g l
m
(a) 90 and 150 — spot the common factor 30 straight away.

l a se
30 | 90, 150 → 3, 5 ag

co m
.
3 and 5 share nothing — stop

em
m l as
.co
HCF = 30
a g
a s em LCM = 30 × 3 × 5 = 450

agl
.c
Check with primes: 90 = 2 × 3 × 3 × 5 and 150 = 2 × 3 × 5 × 5. HCF = 2 × 3 × 5 = 30 ✓, LCM = 2 × 3 ×
s e m
3 × 5 × 5 = 450 ✓
. c om a g la
s e m
a
(b) 84 and 132 — both end in an even digit and both have digit sums divisible by 3, so 12 is

agl
worth trying.

co m
m .
m ase
.co


a g l Page 49 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

12 | 84, 132 → 7, 11

7 and 11 share nothing — stop

HCF = 12
LCM = 12 × 7 × 11 = 924

Check with primes: 84 = 2 × 2 × 3 × 7 and 132 = 2 × 2 × 3 × 11. HCF = 2 × 2 × 3 = 12 ✓, LCM = 2 × 2
× 3 × 7 × 11 = 924 ✓

Check it yourself: HCF × LCM should equal the product. (a) 30 × 450 = 13,500 = 90 ×
150 ✓ (b) 12 × 924 = 11,088 = 84 × 132 ✓

In-text Questions — Pages 62–63
Section 3.3 Property Involving both the HCF and the LCM

Q1 Which is greater — the LCM of two numbers or their product?

The product is greater, or the two are equal. The LCM is never bigger than the product.

NUMBERS PRODUCT LCM WHICH IS GREATER?

6, 8 48 24 product

12, 18 216 36 product

7, 11 77 77 equal

4, 9 36 36 equal

15, 25 375 75 product

They are equal exactly when the two numbers are co-prime.

Page 50 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Q2 You could analyse the above statement using examples. Then try to reason or
prove, why the LCM is never greater than the product of the numbers. [Hint: Is the
product also a common multiple of the two numbers?]

Because the product is itself a common multiple, and the LCM is the smallest common multiple.

Take the two numbers a and b

a × b = a × b, so a × b is a multiple of a ✓

a × b = b × a, so a × b is a multiple of b ✓

So a × b is a common multiple of a and b

The LCM is the smallest common multiple

Therefore LCM ≤ a × b

Why it happens: the LCM has to be less than or equal to every common multiple,
since it is the least of them. The product is one of the common multiples on that list.
So the LCM cannot overtake it. The two are equal only when the product is itself the
smallest — that is, when a and b share no prime factor and nothing can be trimmed
away.

Tip: in prime terms, the product counts every shared prime twice while the LCM
counts it once. The extra copies are exactly the HCF — which is why LCM = (a × b) ÷
HCF.

Q3 Consider the numbers 105 and 95. Find their LCM. [Is the LCM a factor of the
product? If yes, what should it be multiplied with to get the product?]

LCM(105, 95) = 1995, and it is a factor of the product — multiply it by 5 to get the product.

Page 51 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

105 = 3 × 5 × 7

95 = 5 × 19

LCM = 3 × 5 × 7 × 19 = 1995

Product in factorised form: 105 × 95 = 3 × 5 × 5 × 7 × 19 = 9975

Comparing: 3 × 5 × 5 × 7 × 19 = (3 × 5 × 7 × 19) × 5

So 105 × 95 = LCM × 5

Check: 1995 × 5 = 9975 ✓

And 5 is exactly HCF(105, 95).

Why it happens: both numbers carry a 5. When you multiply them, that 5 is counted
twice; the LCM keeps only one copy. The one spare 5 is what is left over — and one
spare copy of each shared prime is precisely the HCF.

Q4 Explore whether the LCM is a factor of the product in the following cases. If yes,
identify the number that the LCM should be multiplied by to get the product. Do
you see any pattern? Use these numbers: (a) 45, 105 (b) 275, 352 (c) 222, 370

In every case the LCM is a factor of the product, and the multiplier is the HCF.

PRIME FACTORISATIONS LCM PRODUCT MULTIPLIER

(a) 45 = 3 × 3 × 5 3 × 3 × 5 × 7 = 315 4725 15 = HCF
105 = 3 × 5 × 7

(b) 275 = 5 × 5 × 11 2⁵ × 5 × 5 × 11 = 8800 96800 11 = HCF
352 = 2 × 2 × 2 × 2 × 2 × 11

(c) 222 = 2 × 3 × 37 2 × 3 × 5 × 37 = 1110 82140 74 = HCF
370 = 2 × 5 × 37

Page 52 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

(a) 4725 ÷ 315 = 15, and HCF(45, 105) = 3 × 5 = 15 ✓

(b) 96800 ÷ 8800 = 11, and HCF(275, 352) = 11 ✓

(c) 82140 ÷ 1110 = 74, and HCF(222, 370) = 2 × 37 = 74 ✓

The pattern: product ÷ LCM = HCF, that is, HCF × LCM = product.

Q5 Do you see that, in each case, the number by which the LCM is multiplied to get the
product is actually the HCF?

Yes. In every case the multiplier turned out to be the HCF.

NUMBERS HCF LCM HCF × LCM PRODUCT

105, 95 5 1995 9975 9975 ✓

45, 105 15 315 4725 4725 ✓

275, 352 11 8800 96800 96800 ✓

222, 370 74 1110 82140 82140 ✓

HCF × LCM = Product of the two numbers

Tip: this gives a quick way to find one from the other. If you know HCF(84, 132) = 12,
then LCM = (84 × 132) ÷ 12 = 11088 ÷ 12 = 924 — no factorising needed.

Try This — Page 63

Page 53 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

Section 3.3 Property Involving both the HCF and the LCM
co m
e m.
m as
TRY THIS

.co a g l
a s em our observations seem to suggest the following: HCF × LCM = Product of the
gl
Thus,
a
Q1
two numbers. Why does this happen? Can you give an explanation or proof? [Hint:
Consider the prime factorisation of the given numbers. Among their prime factors,

. com
some are common to both factorisations, and the rest occur in only one of them.
ag
Between the HCF and the LCM, see
a s emhow the common and non-common prime
a gl product, observe how these two kinds of prime
factors get distributed. In the
factors occur. Compare them.]

co m
e m.
as

m l
.co
Follow one prime at a time. That is enough to prove the whole statement.
m a g
l a se
Take any prime p. Suppose it occurs i times in the first number and j times in the second.
a g
m a s
agl
In the HCF, p occurs the smaller of i and j times
. c o
s e m j times
In the LCM, p occurs the larger of i and

So in HCF × LCM, p occurs a gla + (larger) = i + j times
(smaller)

co m
m .
as e
In the product of the two numbers, p occurs i + j times as well
m l
m .co
(the two lists of factors are simply written side by side)
a g
l a se
ag Every prime occurs the same number of times on both sides, so the two numbers are equal.

se m
com g l a
m . a
ase
HCF × LCM = product of the two numbers

a gl
A worked case, 60 and 90:

co m
m .
e
PRIME IN 60 = IN 90 = IN HCF IN LCM MIN +

m l a s(MAX)
.co ag
2×2×3×5 2×3×3×5 (MIN) MAX

a s em 2
gl
2 1 1 2 3

a c
m .
e
3 1 2 1 2 3

m a s
5 1
e
1
m . co 1 1 2
agl
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

HCF = 2 × 3 × 5 = 30, LCM = 2 × 2 × 3 × 3 × 5 = 180

HCF × LCM = 30 × 180 = 5400

60 × 90 = 5400 ✓

Why it happens: the key is that "smallest + largest = first + second" for any two
numbers. Whichever of i and j is bigger, the pair (min, max) is just the pair (i, j)
rearranged. So the HCF and the LCM between them use up each prime exactly as
many times as the plain product does — the HCF takes the shared copies, the LCM
takes everything else, and nothing is lost or double-counted.

Q2 Explore whether this property holds when 3 numbers are considered.

It does not hold in general for three numbers.

NUMBERS HCF LCM HCF × LCM PRODUCT EQUAL?

2, 3, 5 1 30 30 30 yes

2, 4, 8 2 8 16 64 no

4, 6, 9 1 36 36 216 no

6, 10, 15 1 30 30 900 no

It happens to work when the three numbers are pairwise co-prime, as in 2, 3, 5 — but that is a
special case, not a rule about three numbers.

Why it fails: run the same prime-counting argument with three numbers. For a
prime p occurring i, j and k times, the HCF takes the smallest and the LCM takes the
largest, so together they use (min + max) copies. But the product uses i + j + k copies.
With two numbers those two counts always match; with three they match only when
the middle count is 0 — which is why 4, 6, 9 fails (the prime 2 occurs 2, 1 and 0 times:
min + max = 0 + 2 = 2, while the product needs 3).

Page 55 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Tip: for three numbers the correct statement is different. For 2, 4 and 8, HCF = 2 and
LCM = 8, and 2 × 8 = 16, nowhere near 64. Do not carry the two-number rule across
without checking.

Figure it Out — Pages 63–64
End-of-chapter question set

TRY THIS MATH TALK

Q1 In the two rows below, colours repeat as shown. When will the blue stars meet
next? [Row 1: yellow, green, orange, blue, pink, grey, repeating. Row 2: green,
orange, yellow, blue, repeating.]

The blue stars meet next at the 16th star — that is, 12 places after they last met.

Row 1 repeats every 6 stars, and blue is the 4th

So row-1 blues are at 4, 10, 16, 22, 28, …

Row 2 repeats every 4 stars, and blue is the 4th

So row-2 blues are at 4, 8, 12, 16, 20, …

They already meet at position 4

After that they meet every LCM(6, 4) = 12 stars

Next meeting = 4 + 12 = position 16

Row 1

Row 2

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

The blue stars meet at the 4th star and again at the 16th

Page 56 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Row 1 repeats every 6 stars, row 2 every 4 stars, and blue is the 4th star in both. They line up again 12
places later — at the 16th star.

Why it happens: counting from the first meeting at position 4, a row-1 blue appears
after every 6 more stars and a row-2 blue after every 4 more stars. For both to
appear together, the number of extra stars must be a multiple of 6 and a multiple of
4 — a common multiple. The first such number is the LCM, 12.

Check it yourself: LCM(6, 4) = 12 because 6 = 2 × 3 and 4 = 2 × 2, so the LCM needs
two 2s and one 3: 2 × 2 × 3 = 12. Not 24 — the shared 2 must not be counted twice.

Q2 (a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2? (b) Is 5 × 7 × 11 × 11 a factor of 5 ×
7 × 7 × 11 × 2?

(a) No. (b) No.

PRIME IN 5 × 7 × 11 × 11 IN 5 × 7 × 7 × 11 × 2

2 0 times 1 time

5 1 time 1 time

7 1 time 2 times

11 2 times 1 time

(a) For the first number to be a multiple of the second, it must contain the second's whole
factorisation. But the second has a 2 and two 7s, and the first has no 2 and only one 7. So no.
(b) For the first to be a factor of the second, the second must contain the first. But the first has
two 11s and the second has only one. So no.

5 × 7 × 11 × 11 = 4235

5 × 7 × 7 × 11 × 2 = 5390

4235 ÷ 5390 is not a whole number, and 5390 ÷ 4235 is not either ✓

Page 57 of 74

Class 7 Maths Chapter 11 Finding Common Ground AglaSem · NCERT Solutions

Why it happens: "A is a multiple of B" and "B is a factor of A" say the same thing, so
(a) and (b) are asking about the two possible directions. Neither factorisation sits
inside the other: each has a prime the other is short of. When two numbers each
hold something the other lacks, neither can divide the other.

Q3 Find the HCF and LCM of the following (state your answers in the form of prime
factorisations): (a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11 (b) 45 and 36

First write both numbers fully in primes, then take minimums for the HCF and maximums for
the LCM.
(a) 3 × 3 × 5 × 7 × 7 is already prime. 12 × 7 × 11 = (2 × 2 × 3) × 7 × 11.

PRIME 3×3×5×7×7 2 × 2 × 3 × 7 × 11 HCF (MIN) LCM (MAX)

2 0 2 0 2

3 2 1 1 2

5 1 0 0 1

7 2 1 1 2

11 0 1 0 1

HCF = 3 × 7 = 21

LCM = 2 × 2 × 3 × 3 × 5 × 7 × 7 × 11 = 97020

(b) 45 = 3 × 3 × 5 and 36 = 2 × 2 × 3 × 3.

HCF = 3 × 3 = 9

LCM = 2 × 2 × 3 × 3 × 5 = 180

Check it yourself: (a) 2205 × 924 = 2,037,420 and 21 × 97020 = 2,037,420 ✓ (b) 45 ×
36 = 1620 and 9 × 180 = 1620 ✓

Page 58 of 74

as e
Class 7 Maths Chapter 11 Finding Common Ground
a g l AglaSem · NCERT Solutions

co m
m.
Find two numbers whose HCF is 1 and LCM is 66.
e
Q4

m l as
.co a g
a
s em
agl 33 is one such pair. So are 6 and 11, 3 and 22, and 1 and 66.
2 and

co m
. ag
HCF = 1 means the two numbers are co-prime
e m
For co-prime numbers, LCM = product
g l as
So the product must be 66
a

co m
m.
66 = 2 × 3 × 11

mprimes into two groups so that no prime is shared: glas e
. c o a
e m
Split the three

g l as
a SPLIT PAIR HCF LCM

m a s
.co agl
(none) and 2 × 3 × 11 1, 66 1 66 ✓
se m
2 and 3 × 11
g l a 2, 33 1 66 ✓
3 and 2 × 11
a 3, 22 1 66 ✓

. com66 ✓
e m
11 and 2 × 3 11, 6 1

m l as
.co
m it happens: HCF × LCM = product, so 1 × 66 = 66 tells us the product at once. a g
a s eWhy
a gl Each of the three primes 2, 3 and 11 must go entirely to one number or the other —
if a prime were split between them it would be shared, and the HCF would not be 1.
se m
com g l a
.
Three primes give 2 × 2 × 2 = 8 ways to hand them out, which is the 4 pairs above
m a
ase
counted twice.

agl

. com
A cowherd took all his cows to graze in the fields. The cows came to a crossing with
emLater at another
Q5

m a s
3 gates. An equal number of cows passed through each gate.
l
. co crossing with 5 gates again an equal number of cows
a gpassed
em same happened at the third crossing with 7 gates. If the cowherd had less than 200
through each gate. The

a s
agl cows, how many cows did he have? (Based on the folklore mathematics from
c
m .
e
Karnataka.)
m a s
e m . co agl
as

He had 105 cows.
a g l

co m
m .
m ase
.co


a g l Page 59 of 74