NCERT Solutions for Class 7 Maths Chapter 3 - Finding Common Ground

3.1 The Greatest of All

Figure It Out (Page 51)

Question 1.
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)

Answer:

(a) 90

Prime factorisation:

90 = 2 × 3 × 3 × 5

$\begin{array}{r|l}2&90\\\hline3&45\\\hline3&15\\\hline5&5\\\hline&1\end{array}$

Therefore, the factors of 90 are:

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90

Total number of factors = 12

(b) 105

Prime factorisation:

105 = 3 × 5 × 7

$\begin{array}{r|l}3&105\\\hline5&35\\\hline7&7\\\hline&1\end{array}$

Therefore, the factors of 105 are:

1, 3, 5, 7, 15, 21, 35 and 105

Total number of factors = 8

(c) 132

Prime factorisation:

132 = 2 × 2 × 3 × 11

$\begin{array}{r|l}2&132\\\hline2&66\\\hline3&33\\\hline11&11\\\hline&1\end{array}$

Therefore, the factors of 132 are:

1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66 and 132

Total number of factors = 12

(d) 360

Prime factorisation:

360 = 2 × 2 × 2 × 3 × 3 × 5

$\begin{array}{r|l}2&360\\\hline2&180\\\hline2&90\\\hline3&45\\\hline3&15\\\hline5&5\\\hline&1\end{array}$

Therefore, the factors of 360 are:

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

Total number of factors = 24

(e) 840

Prime factorisation:

840 = 2 × 2 × 2 × 3 × 5 × 7

$\begin{array}{r|l}2&840\\\hline2&420\\\hline2&210\\\hline3&105\\\hline5&35\\\hline7&7\\\hline&1\end{array}$

Therefore, the factors of 840 are:

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

Total number of factors = 32

Figure It Out (Page 53)

Question 1.
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
How do we directly find the HCF without listing all the factors?

Answer:

We can find the HCF directly by prime factorisation and multiplying the common prime factors raised to their lowest powers.

(a) 50, 60

$\begin{gathered}\begin{array}{r|l}2&50\\\hline5&25\\\hline&5\end{array}\\[8pt]\therefore\ 50=\boxed{2}\times5\times\boxed{5}\\[12pt]\text{and }60\\[6pt]\begin{array}{r|l}2&60\\\hline2&30\\\hline3&15\\\hline5&5\\\hline&1\end{array}\\[8pt]\therefore\ 60=\boxed{2}\times2\times3\times\boxed{5}\end{gathered}$

50 = 2 × 5 × 5

60 = 2 × 2 × 3 × 5

Common factors = 1, 2, 5 and 10

HCF = 2 × 5 = 10

(b) 140, 275

$\begin{array}{r|l}2&140\\\hline2&70\\\hline5&35\\\hline7&7\\\hline&1\end{array}\qquad\begin{array}{r|l}5&275\\\hline5&55\\\hline11&11\\\hline&1\end{array}$

140 = 2 × 2 × 5 × 7

275 = 5 × 5 × 11

The only common prime factor is 5.

Therefore,

HCF (140, 275) = 5

(c) 77, 725

$\begin{array}{r|l}7&77\\\hline11&11\\\hline&1\end{array}\qquad\text{and}\qquad\begin{array}{r|l}5&725\\\hline5&145\\\hline29&29\\\hline&1\end{array}$

77 = 7 × 11

725 = 5 × 5 × 29

There is no common prime factor.

Therefore,

HCF (77, 725) = 1

(d) 370, 592

$\begin{array}{r|l}2&370\\\hline5&185\\\hline37&37\\\hline&1\end{array}\qquad\text{and}\qquad\begin{array}{r|l}2&592\\\hline2&296\\\hline2&148\\\hline2&74\\\hline37&37\\\hline&1\end{array}$

370 = 2 × 5 × 37

592 = 2 × 2 × 2 × 2 × 37

The common prime factors are 2 and 37.

HCF = 2 × 37

= 74

(e) 81, 243

$\begin{array}{r|l}3&81\\\hline3&27\\\hline3&9\\\hline3&3\\\hline&1\end{array}\qquad\text{and}\qquad\begin{array}{r|l}3&243\\\hline3&81\\\hline3&27\\\hline3&9\\\hline3&3\\\hline&1\end{array}$

81 = 3 × 3 × 3 × 3

243 = 3 × 3 × 3 × 3 × 3

HCF = 3 × 3 × 3 × 3

= 81

Figure It Out (Page 54)

Question 1.
Find the HCF of the following numbers:
(a) 24, 180
(b) 42, 75, 24
(c) 240, 378
(d) 400, 2500
(e) 300, 800

Answer:

(a) 24, 180

24 = 2 × 2 × 2 × 3

180 = 2 × 2 × 3 × 3 × 5

The common prime factors are 2 × 2 × 3.

Therefore,

HCF (24, 180) = 12

(b) 42, 75, 24

42 = 2 × 7 × 3

75 = 5 × 5 × 3 

24 = 2 × 2 × 3 × 2

The only prime factor common to all three numbers is 3.

Therefore,

HCF (42, 75, 24) = 3

(c) 240, 378

$\begin{gathered}\begin{array}{r|l}2&240\\\hline2&120\\\hline2&60\\\hline2&30\\\hline3&15\\\hline5&5\\\hline&1\end{array}\qquad\begin{array}{r|l}2&378\\\hline3&189\\\hline3&63\\\hline3&21\\\hline7&7\\\hline&1\end{array}\\[10pt]\therefore\quad240=\boxed{2}\times2\times2\times2\times\boxed{3}\times5\\[6pt]\text{and}\quad378=\boxed{2}\times\boxed{3}\times3\times3\times7\end{gathered}$

The common prime factors are 2 and 3.

Therefore,

HCF (240, 378) = 2 × 3 = 6

(d) 400, 2500

$\begin{gathered}\begin{array}{r|l}2&400\\\hline2&200\\\hline2&100\\\hline2&50\\\hline5&25\\\hline5&5\\\hline&1\end{array}\qquad\begin{array}{r|l}2&2500\\\hline2&1250\\\hline5&625\\\hline5&125\\\hline5&25\\\hline5&5\\\hline&1\end{array}\\[10pt]\therefore\quad400=\boxed{2\times2}\times2\times2\times\boxed{5\times5}\\[6pt]\text{and}\quad2500=\boxed{2\times2}\times\boxed{5\times5}\times5\times5\end{gathered}$

Common prime factors = 2 × 2 × 5 × 5

Therefore,

HCF (400, 2500) = 100

(e) 300, 800

$\begin{gathered}\begin{array}{r|l}2&300\\\hline2&150\\\hline3&75\\\hline5&25\\\hline5&5\\\hline&1\end{array}\qquad\text{and}\qquad\begin{array}{r|l}2&800\\\hline2&400\\\hline2&200\\\hline2&100\\\hline2&50\\\hline5&25\\\hline5&5\\\hline&1\end{array}\\[10pt]\therefore\quad300=\boxed{2\times2}\times3\times\boxed{5\times5}\\[6pt]800=\boxed{2\times2}\times2\times2\times2\times\boxed{5\times5}\end{gathered}$

Common prime factors = 2 × 2 × 5 × 5

Therefore,

HCF (300, 800) = 100

Question 2.
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?

Answer:

No. We cannot conclude that 72 and 144 have no common factor other than 1 because 6, 12, 8 and 18 are composite numbers, not prime factors.

Prime factorisation of 72:

72 = 2 × 2 × 2 × 3 × 3

Prime factorisation of 144:

144 = 2 × 2 × 2 × 2 × 3 × 3

The common prime factors are:

2 × 2 × 2 × 3 × 3 = 72

Therefore,

HCF (72, 144) = 72

So, 72 and 144 have several common factors, and their HCF is 72. 

3.2 Least, but not Last!

Figure It Out (Page 58)

Question 1.
Find the LCM of the following numbers:
(a) 30, 72
(b) 36, 54
(c) 105, 195, 65
(d) 222, 370

Answer:

(a) 30, 72

30 = 2 × 3 × 5

72 = 2 × 2 × 2 × 3 × 3

For the LCM, take each prime factor with its highest power.

LCM = 2 × 2 × 2 × 3 × 3 × 5

= 8 × 9 × 5

= 360

(b) 36, 54

36 = 2 × 2 × 3 × 3

54 = 2 × 3 × 3 × 3

LCM = 2 × 2 × 3 × 3 × 3

= 108

(c) 105, 195, 65

105 = 3 × 5 × 7

195 = 3 × 5 × 13

65 = 5 × 13

LCM = 3 × 5 × 7 × 13

= 1365

(d) 222, 370

$\begin{array}{r|l}2&222\\\hline3&111\\\hline37&37\\\hline&1\end{array}\qquad\begin{array}{r|l}2&370\\\hline5&185\\\hline37&37\\\hline&1\end{array}$

222 = 2 × 3 × 37

370 = 2 × 5 × 37

LCM = 2 × 3 × 5 × 37

= 1110

3.3 Patterns, Properties, and a Pretty Procedure!

Figure It Out (Page 59)

Question 1.
Make a general statement about the HCF for the following pairs of numbers. Consider examples before making 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.

Answer:

(a) Two consecutive even numbers

Examples:

HCF (2, 4) = 2

HCF (6, 8) = 2

HCF (10, 12) = 2

General statement: The HCF of two consecutive even numbers is 2.

Both numbers are divisible by 2, and since consecutive even numbers differ by 2, they cannot have a common factor greater than 2.

(b) Two consecutive odd numbers

Examples:

HCF (3, 5) = 1

HCF (7, 9) = 1

HCF (11, 13) = 1

General statement: The HCF of two consecutive odd numbers is 1.

Consecutive odd numbers have no common factor other than 1.

(c) Two even numbers

Examples:

HCF (4, 10) = 2

HCF (8, 12) = 4

HCF (14, 20) = 2

General statement: The HCF of two even numbers is always even.

Since every even number is divisible by 2, their HCF must contain 2 as a factor.

(d) Two consecutive numbers

Examples:

HCF (7, 8) = 1

HCF (14, 15) = 1

HCF (20, 21) = 1

General statement: The HCF of any two consecutive numbers is 1.

Two consecutive numbers cannot have a common factor greater than 1.

(e) Two co-prime numbers

Examples:

HCF (4, 9) = 1

HCF (5, 8) = 1

HCF (7, 10) = 1

General statement: The HCF of two coprime numbers is always 1 because coprime numbers have no common factor other than 1.

Question 2.
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.

Answer:

(a) Some examples are:

• LCM (2, 4) = 4

• LCM (5, 10) = 10

• LCM (6, 12) = 12

• LCM (7, 49) = 49

• LCM (10, 100) = 100

In each pair, the smaller number divides the larger number.

(b) Let the two positive integers be a and b, where a < b.

If b is a multiple of a, then:

b = k × a

where k is a positive integer.

In such a case:

LCM (a, b) = b

Therefore, the LCM of two numbers equals the larger number whenever the smaller number divides the larger number.

Question 3.
Make a general statement about the LCM for the following pairs of numbers. Consider examples before writing your 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

Answer:

(a) Two multiples of 3

Examples:

LCM (6, 9) = 18

LCM (9, 12) = 36

LCM (12, 18) = 36

General statement: The LCM of two multiples of 3 is always a multiple of 3.

Since both numbers contain 3 as a factor, their LCM will also contain 3 as a factor.

(b) Two consecutive even numbers

Examples:

LCM (2, 4) = 4

LCM (6, 8) = 24

LCM (10, 12) = 60

General statement: The LCM of two consecutive even numbers is half their product.

If the numbers are 2n and 2n + 2:

LCM (2n, 2n + 2) = [2n × (2n + 2)] ÷ 2

= n(2n + 2)

= 2n² + 2n

(c) Two consecutive numbers

Examples:

LCM (7, 8) = 56

LCM (9, 10) = 90

LCM (10, 11) = 110

General statement: The LCM of two consecutive numbers equals their product.

This happens because consecutive numbers have no common factor other than 1.

(d) Two coprime numbers

Examples:

LCM (4, 9) = 36

LCM (5, 8) = 40

LCM (7, 10) = 70

General statement: The LCM of two coprime numbers equals their product because their only common factor is 1. 

Figure It Out (Pages 63-64)

Question 1.
In the two rows below, colours repeat as shown. When will the black stars meet next?

Answer:

In the first row, the black stars occur at positions:

4, 10, 16, 22, 28, ...

The positions increase by 6 each time.

In the second row, the black stars occur at:

4, 8, 12, 16, 20, ...

The positions increase by 4 each time.

The next common position in both patterns is 16.

Therefore, the black stars will meet again at the 16th position.

Question 2.
(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?

Answer:

(a) No.

For one number to be a multiple of another, it must contain all the required prime factors.

The number:

5 × 7 × 11 × 11

does not contain the prime factor 2, which is present in:

5 × 7 × 7 × 11 × 2

Therefore,

5 × 7 × 11 × 11 is not a multiple of 5 × 7 × 7 × 11 × 2.

(b) No.

For the first number to be a factor of the second, the second number must contain all the first number's prime factors with at least the same powers.

The first number has two factors of 11, whereas the second has only one.

Therefore,

5 × 7 × 11 × 11 is not a factor of 5 × 7 × 7 × 11 × 2.

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

Answer:

(a)

First number:

3 × 3 × 5 × 7 × 7

Second number:

12 × 7 × 11 = 2 × 2 × 3 × 7 × 11

Common prime factors are 3 and 7.

Therefore,

HCF = 3 × 7

For the LCM, take all prime factors with their highest powers:

LCM = 2 × 2 × 3 × 3 × 5 × 7 × 7 × 11

(b)

45 = 3 × 3 × 5

36 = 2 × 2 × 3 × 3

Therefore,

HCF = 3 × 3

and

LCM = 2 × 2 × 3 × 3 × 5

Question 4.
Find two numbers whose HCF is 1, and LCM is 66.

Answer:

For two numbers a and b:

a × b = HCF × LCM

Given:

HCF = 1

LCM = 66

Therefore:

a × b = 1 × 66 = 66

Possible factor pairs of 66 are:

(1, 66), (2, 33), (3, 22), (6, 11)

Each pair consists of coprime numbers.

Therefore, possible answers are:

(1, 66), (2, 33), (3, 22) or (6, 11)

Question 5.
A cowherd took all his cows to graze in the fields. The cows can go to a crossing with 3 gates. An equal number of cows passed through each gate. Later, at another crossing with 5 gates, an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had fewer than 200 cows, how many cows did he have? (Based on Karnataka folklore mathematics).

Answer:

The number of cows must be divisible by 3, 5 and 7.

Therefore, find their LCM:

LCM (3, 5, 7)

= 3 × 5 × 7

= 105

The multiples of 105 are:

105, 210, 315, ...

Since the cowherd had fewer than 200 cows, the only possible number is 105.

Therefore, the cowherd had 105 cows.

Question 6.
The length, width, and height of a box are 12 cm, 18 cm, and 36 cm, respectively. Which of the following cube sizes can be packed in this box without leaving gaps?
(a) 9 cm
(b) 6 cm
(c) 4 cm
(d) 3 cm
(e) 2 cm

Answer:

For cubes to fit without gaps, the side length of each cube must divide 12 cm, 18 cm and 36 cm exactly.

Checking the options:

(a) 9 cm – does not divide 12 exactly. ✗

(b) 6 cm – divides 12, 18 and 36 exactly. ✓

(c) 4 cm – does not divide 18 exactly. ✗

(d) 3 cm – divides all three dimensions exactly. ✓

(e) 2 cm – divides all three dimensions exactly. ✓

Therefore, the cube sizes that can be packed without leaving gaps are:

6 cm, 3 cm and 2 cm — options (b), (d) and (e).

Question 7.
Among the numbers below, which is the largest number that perfectly divides both 306 and 36?
(a) 36
(b) 612
(c) 18
(d) 3
(e) 2
(f) 360

Answer:

We need the largest number that divides both 306 and 36, so we find their HCF.

306 = 2 × 3 × 3 × 17

36 = 2 × 2 × 3 × 3

Common prime factors:

2 × 3 × 3 = 18

$\begin{gathered}\text{Now}\quad\begin{array}{r|l}2&306\\\hline3&153\\\hline3&51\\\hline17&17\\\hline&1\end{array}\\[10pt]\therefore\quad306=\boxed{2}\times\boxed{3\times3}\times17\\[6pt]36=\boxed{2}\times2\times\boxed{3\times3}\end{gathered}$

Therefore,

HCF (306, 36) = 18

Hence, option (c) 18 is correct.

Question 8.
Find the smallest number divisible by 3, 4, 5, and 7, but leaving a remainder of 10 when divided by 11.

Answer:

First, find the LCM of 3, 4, 5 and 7:

LCM (3, 4, 5, 7)

= 3 × 4 × 5 × 7

= 420

So, the required number must be a multiple of 420.

Checking its multiples:

420 ÷ 11 leaves a remainder of 2.

840 ÷ 11 leaves a remainder of 4.

1260 ÷ 11 leaves a remainder of 6.

1680 ÷ 11 leaves a remainder of 8.

2100 ÷ 11 leaves a remainder of 10.

Therefore, the smallest required number is 2100.

Question 9.
Children are playing ‘Fire in the Mountain.’ When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?
(a) 72
(b) 90
(c) 45
(d) 3
(e) 36
(f) None of these

Answer:

If no one gets out when 6 or 9 is called, the number of children must be divisible by both 6 and 9.

LCM (6, 9) = 18

Therefore, the number must be a multiple of 18.

It must also not be divisible by 10, because some children got out when 10 was called.

Checking the options:

72 → divisible by 18 but not by 10 ✓

90 → divisible by 18 and also by 10 ✗

45 → not divisible by 18 ✗

3 → not divisible by 18 ✗

36 → divisible by 18 but not by 10 ✓

Therefore, the possible numbers of children are 36 and 72.

Question 10.
Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be:
(a) Less than both numbers
(b) In between the two numbers
(c) Greater than both numbers
(d) Less than m × n
(e) Greater than m × n

Answer:

Two different prime numbers have no common factor other than 1.

Therefore:

LCM (m, n) = m × n

Since their product is greater than either prime number, the LCM is greater than both.

Therefore, the correct statement is:

(c) Greater than both numbers

Question 11.
A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?

Answer:

Distance covered by the dog in one leap = 9 feet

Distance covered by the rabbit in one leap = 7 feet

Distance gained by the dog in each leap:

9 - 7 = 2 feet

The rabbit has a 150-foot head start.

Number of leaps required:

150 ÷ 2 = 75

Therefore, the dog catches up with the rabbit after 75 leaps.

Question 12.
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?

Answer:

We need to find:

LCM (1, 2, 3, 4, 5, 6, 8, 9, 10)

Using prime factors:

4 = 2 × 2

8 = 2 × 2 × 2

9 = 3 × 3

10 = 2 × 5

Take the highest powers of the required prime factors:

LCM = 2 × 2 × 2 × 3 × 3 × 5

= 8 × 9 × 5

= 360

Therefore, the smallest number that is a multiple of all the given numbers is 360.

Yes, this is the same result students may remember from Grade 6, Chapter 5, Prime Time.

Question 13.
Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together 815,120,736,1163 and 121. What do you get? How can we find this sum efficiently?

Answer:
The denominators of the given fractions are 15, 20, 36, 63 and 21. To add the fractions efficiently, first find the LCM of these denominators.

15 = 3 × 5
20 = 2 × 2 × 5
36 = 2 × 2 × 3 × 3
63 = 3 × 3 × 7
21 = 3 × 7

Therefore,

LCM = 2 × 2 × 3 × 3 × 5 × 7
= 1260

$\begin{aligned}\text{Now}\quad\dfrac{8}{15}+\dfrac{1}{20}+\dfrac{7}{36}+\dfrac{11}{63}+\dfrac{1}{21}\\[6pt]=\dfrac{8\times84}{1260}+\dfrac{1\times63}{1260}+\dfrac{7\times35}{1260}\\[6pt]=\dfrac{11}{1260}\times20+\dfrac{1\times60}{1260}\\[6pt]=\dfrac{1260}{1260}=1\end{aligned}$

Strengthen Your Understanding of Finding Common Ground

The NCERT Solutions for Class 7 Maths Chapter 3 help students understand factors, multiples, HCF, and LCM through simple, step-by-step explanations.

• Understand key concepts: Learn HCF, LCM, prime factorisation, and number patterns.

• Practise word problems: Apply these concepts to real-life situations and textbook questions.

• Check your answers: Use the NCERT Solutions for Maths Class 7 Chapter 3 to identify mistakes and improve accuracy.

• Revise effectively: Refer to the Class 7 Maths Chapter 3 solutions PDF to practise difficult questions before exams.

Vedantu's CBSE NCERT Solutions for Class 7 Maths Chapter 3 make it easier to review important concepts and build confidence in problem-solving.

Important Study Material Links for Chapter 3: Finding Common Ground

NCERT Solutions for Class 7 Maths Ganita Prakash I and II (New Syllabus)

Important Related Links for NCERT Class 7 Maths