by Tiwari Academy Team
NCERT Solutions for Class 7 Maths Ganita Prakash Part 2 Chapter 3 Finding Common Ground – Question Answers with explanation for session 2026-27. Chapter 3 of NCERT Class 7 Maths Part 2, “Finding Common Ground”, uses situations like tiling a floor and packing rice into bags to build up HCF and LCM. Students learn to find both efficiently through prime factorisation and a faster “ladder” division method and discover the rule connecting HCF, LCM and the product of two numbers. These NCERT solutions offer clear, step-by-step answers to every “Figure it Out” question in the chapter.
Class 7 Maths all Chapters
NCERT Class 7 Maths Ganita Prakash Part 2 Chapter 3 Solutions
Page 51 – Figure it Out
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) Factors of 90
Prime factorisation: 90 = 2 ร 3ยฒ ร 5
Factors: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
(b) Factors of 105
Prime factorisation: 105 = 3 ร 5 ร 7
Factors: 1, 3, 5, 7, 15, 21, 35, 105
(c) Factors of 132
Prime factorisation: 132 = 2ยฒ ร 3 ร 11
Factors: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132
(d) Factors of 360
Prime factorisation: 360 = 2ยณ ร 3ยฒ ร 5
Factors (24 factors):
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360
(e) Factors of 840
Prime factorisation: 840 = 2ยณ ร 3 ร 5 ร 7
Factors (32 factors):
1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210, 280, 420, 840
Page 53 – Figure it Out
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:
(a) 50 and 60
Factors of 50: 1, 2, 5, 10, 25, 50
Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factors: 1, 2, 5, 10
HCF = 10
(b) 140 and 275
Factors of 140: 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140
Factors of 275: 1, 5, 11, 25, 55, 275
Common factors: 1, 5
HCF = 5
(c) 77 and 725
Factors of 77: 1, 7, 11, 77
Factors of 725: 1, 5, 25, 29, 145, 725
Common factors: 1
HCF = 1
(d) 370 and 592
Factors of 370: 1, 2, 5, 10, 37, 74, 185, 370
Factors of 592: 1, 2, 4, 8, 16, 37, 74, 148, 296, 592
Common factors: 1, 2, 37, 74
HCF = 74
(e) 81 and 243
Factors of 81: 1, 3, 9, 27, 81
Factors of 243: 1, 3, 9, 27, 81, 243
Common factors: 1, 3, 9, 27, 81
HCF = 81
We can find the HCF directly by using prime factorisation:
Steps:
- Write the prime factorisation of each number.
- Identify the common prime factors.
- Take the smallest power of each common prime factor.
- Multiply them to get the HCF.
Example:
For 50 and 60
50 = 2 ร 5ยฒ
60 = 2ยฒ ร 3 ร 5
Common primes: 2 and 5
Smallest powers: 2ยน and 5ยน
HCF = 2 ร 5 = 10
This method is faster and easier than listing all factors.
Page 54 – Figure it Out
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 and 180
Prime factorisation:
24 = 2ยณ ร 3
180 = 2ยฒ ร 3ยฒ ร 5
Common prime factors: 2ยฒ and 3
HCF = 2ยฒ ร 3 = 12
(b) 42, 75 and 24
Prime factorisation:
42 = 2 ร 3 ร 7
75 = 3 ร 5ยฒ
24 = 2ยณ ร 3
Common prime factor in all three numbers: 3
HCF = 3
(c) 240 and 378
Prime factorisation:
240 = 2โด ร 3 ร 5
378 = 2 ร 3ยณ ร 7
Common prime factors: 2ยน and 3ยน
HCF = 2 ร 3 = 6
(d) 400 and 2500
Prime factorisation:
400 = 2โด ร 5ยฒ
2500 = 2ยฒ ร 5โด
Common prime factors: 2ยฒ and 5ยฒ
HCF = 2ยฒ ร 5ยฒ = 4 ร 25 = 100
(e) 300 and 800
Prime factorisation:
300 = 2ยฒ ร 3 ร 5ยฒ
800 = 2โต ร 5ยฒ
Common prime factors: 2ยฒ and 5ยฒ
HCF = 4 ร 25 = 100
2. Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72 = 6 ร 12 and 144 = 8 ร 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?
Answer:
Although 72 is written as 6 ร 12 and 144 as 8 ร 18, the numbers 6, 12, 8 and 18 are composite numbers, not prime numbers.
If we break these composite numbers further into prime factors, we get:
72 = 6 ร 12
= (2 ร 3) ร (2ยฒ ร 3)
= 2ยณ ร 3ยฒ
144 = 8 ร 18
= (2ยณ) ร (2 ร 3ยฒ)
= 2โด ร 3ยฒ
Now we can clearly see that both numbers have common prime factors 2 and 3.
So, 72 and 144 have many common factors, such as 2, 3, 4, 6, 12, 18, 36 and 72.
Page 58 – Figure it Out
Find the LCM of the following numbers:
(a) 30, 72
(b) 36, 54
(c) 105, 195, 65
(d) 222, 370
Answer:
(a) 30 and 72
Prime factorisation:
30 = 2 ร 3 ร 5
72 = 2ยณ ร 3ยฒ
Take the highest powers of each prime: 2ยณ, 3ยฒ, 5
LCM = 2ยณ ร 3ยฒ ร 5
LCM = 8 ร 9 ร 5 = 360
(b) 36 and 54
Prime factorisation:
36 = 2ยฒ ร 3ยฒ
54 = 2 ร 3ยณ
Highest powers: 2ยฒ, 3ยณ
LCM = 2ยฒ ร 3ยณ = 4 ร 27 = 108
(c) 105, 195 and 65
Prime factorisation:
105 = 3 ร 5 ร 7
195 = 3 ร 5 ร 13
65 = 5 ร 13
Highest powers:
3ยน, 5ยน, 7ยน, 13ยน
LCM = 3 ร 5 ร 7 ร 13 = 1365
(d) 222 and 370
Prime factorisation:
222 = 2 ร 3 ร 37
370 = 2 ร 5 ร 37
Highest powers: 2ยน, 3ยน, 5ยน, 37ยน
LCM = 2 ร 3 ร 5 ร 37
LCM = 1110
Page 59 – Figure it Out
1. 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.
Answer:
(a)
Statement: The HCF of two consecutive even numbers is 2.
Example:
8 and 10
HCF = 2
Reason: Both numbers are divisible by 2 and have no bigger common factor.
(b)
Statement: The HCF of two consecutive odd numbers is 1.
Example:
9 and 11
HCF = 1
Reason: They have no common factor except 1.
(c)
Statement: The HCF of two even numbers is at least 2. It can be more than 2.
Example:
12 and 18
HCF = 6
Reason: All even numbers are divisible by 2.
(d)
Statement: The HCF of two consecutive numbers is always 1.
Example:
7 and 8
HCF = 1
Reason: Consecutive numbers have no common factor except 1.
(e)
Statement: The HCF of two co-prime numbers is 1
Example:
4 and 9
HCF = 1
Reason: Co-prime numbers have only 1 as a common factor.
Observation:
- Consecutive numbers always have HCF = 1.
- Even numbers always have 2 as a common factor.
- Co-prime numbers have HCF = 1.
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:
Given:
The LCM of 3 and 24 is 24.
Here, 24 is one of the given numbers.
(a)
In these pairs, one number is a multiple of the other, so the LCM is the bigger number.
Examples:
2 and 10 โ LCM = 10
4 and 20 โ LCM = 20
6 and 18 โ LCM = 18
5 and 25 โ LCM = 25
8 and 32 โ LCM = 32
(b)
General Statement:
If one number is a multiple of the other number, then the LCM of the two numbers is the bigger number.
Algebraic description:
Let the smaller number be a.
Let the bigger number be ka, where k is a whole number.
So the two numbers are: a and ka
Since ka is already a multiple of a, the LCM of a and ka = ka.
3. 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
Answer:
a)
Statement: The LCM of two multiples of 3 is also a multiple of 3.
Example:
6 and 15
LCM = 30
Reason:
Both numbers have 3 as a common factor, so their LCM must include 3.
(b)
Statement: The LCM of two consecutive even numbers is their product.
Example:
8 and 10
LCM = 40
Reason:
Consecutive even numbers have only 2 as a common factor.
(c)
Statement: The LCM of two consecutive numbers is their product.
Example:
7 and 8
LCM = 56
Reason:
Consecutive numbers have no common factor except 1.
(d)
Statement: The LCM of two co-prime numbers is their product.
Example:
5 and 12
LCM = 60
Reason:
Co-prime numbers have no common factor except 1.
Page 63 – Figure it Out
1. In the two rows below, colours repeat as shown. When will the blue stars meet next?
Answer:
Top row pattern:
Yellow โ Green โ Orange โ Blue โ Pink โ Grey
This pattern repeats after 6 stars.
So, in the top row, blue appears every 6th position.
Bottom row pattern:
Green โ Orange โ Yellow โ Blue
This pattern repeats after 4 stars.
So, in the bottom row, blue appears every 4th position.
The blue stars will meet when a position number is a common multiple of 6 and 4.
So, we find the least common multiple (LCM) of 6 and 4.
Multiples of 6: 6, 12, 18, โฆ
Multiples of 4: 4, 8, 12, โฆ
The smallest common multiple is 12.
The blue stars will meet next at the 12th star, because the colour patterns repeat every 6 stars in the top row and every 4 stars in the bottom row and 12 is the first position common to both patterns.
2. (a) Is 5 ร 7 ร 11 ร 11 a multiple of 5 ร 7 ร 7 ร 11 ร 2? (b) Is 5 ร 7 ร 11 ร 11 a factor of 5 ร 7 ร 7 ร 11 ร 2?
Answer:
(a) To check whether 5 ร 7 ร 11 ร 11 is a multiple of 5 ร 7 ร 7 ร 11 ร 2, we compare the factors.
Both numbers have common factors 5, 7 and 11.
The first number has one extra 11.
The second number has one extra 7 and a factor 2.
Since the first number does not contain the factors 7 (twice) and 2, it cannot be a multiple.
No, 5 ร 7 ร 11 ร 11 is not a multiple of 5 ร 7 ร 7 ร 11 ร 2.
(b) To check whether 5 ร 7 ร 11 ร 11 is a factor of 5 ร 7 ร 7 ร 11 ร 2, all the factors of the first number must be present in the second number.
The first number contains two 11s.
The second number contains only one 11.
So the first number cannot divide the second number completely.
No, 5 ร 7 ร 11 ร 11 is not a factor of 5 ร 7 ร 7 ร 11 ร 2.
3. Find the HCF and LCM of the following (state your answers in the form of prime factorisations):
(a) 3 ร 3 ร 5 ร 7 ร 7 and 12 ร 7 ร 11
(b) 45 and 36
Answer:
(a)
First number = 3 ร 3 ร 5 ร 7 ร 7
Second number = 12 ร 7 ร 11
= 2 ร 2 ร 3 ร 7 ร 11
HCF = Common prime factors with the smallest powers:
= 3 ร 7
LCM = All prime factors with the greatest powers:
= 2 ร 2 ร 3 ร 3 ร 5 ร 7 ร 7 ร 11
(b)
45 = 3 ร 3 ร 5
36 = 2 ร 2 ร 3 ร 3
HCF = Common prime factors with the smallest powers:
= 3 ร 3
LCM = All prime factors with the greatest powers:
= 2 ร 2 ร 3 ร 3 ร 5
4. Find two numbers whose HCF is 1 and LCM is 66.
Answer:
HCF ร LCM = product of the two numbers
HCF = 1
LCM = 66
So, Product of the two numbers = 1 ร 66 = 66
Now we need two numbers whose product is 66 and HCF is 1 (that is, they are co-prime).
66 = 6 ร 11
6 and 11 have no common factor except 1.
HCF of 6 and 11 = 1
LCM of 6 and 11 = 66
5. A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have?
(Based on the folklore mathematics from Karnataka.)
Answer:
We are looking for a number of cows less than 200 that is divisible by 3, 5 and 7 (so that an equal number of cows pass through each gate at the crossings).
LCM of 3, 5 and 7
LCM = 3 ร 5 ร 7 = 105
Checking if it fits the condition:
105 รท 3 = 35 cows per gate at the first crossing
105 รท 5 = 21 cows per gate at the second crossing
105 รท 7 = 15 cows per gate at the third crossing
All conditions are satisfied and 105 < 200.
The cowherd had 105 cows.
6. The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes 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:
Box dimensions: Length = 12 cm, Width = 18 cm, Height = 36 cm
Cube sides: 9 cm, 6 cm, 4 cm, 3 cm, 2 cm
We need cubes that fit perfectly without leaving gaps.
This means the side of the cube must exactly divide the length, width and height of the box.
Check each cube size
(a) 9 cm cube
12 รท 9 = 1.33
18 รท 9 = 2
36 รท 9 = 4
Cannot pack perfectly
(b) 6 cm cube
12 รท 6 = 2
18 รท 6 = 3
36 รท 6 = 6
Can pack perfectly.
(c) 4 cm cube
12 รท 4 = 3
18 รท 4 = 4.5
Cannot pack perfectly.
(d) 3 cm cube
12 รท 3 = 4
18 รท 3 = 6
36 รท 3 = 12
Can pack perfectly.
(e) 2 cm cube
12 รท 2 = 6
18 รท 2 = 9
36 รท 2 = 18
Can pack perfectly.
Hence, the cubes that can be packed without leaving gaps are: 6 cm, 3 cm, and 2 cm.
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:
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 306: 1, 2, 3, 6, 9, 17, 18, 34, 51, 102, 153, 306
Common factors of 36 and 306: 1, 2, 3, 6, 9, 18
The largest number that divides both = 18
8. Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.
Answer:
LCM of 3, 4, 5 and 7:
3 = 3
4 = 2 ร 2
5 = 5
7 = 7
LCM = 2ยฒ ร 3 ร 5 ร 7 = 420
So, the number must be a multiple of 420, i.e., 420, 840, 1260, 1680, 2100, โฆ
Now, the number leaves remainder 10 when divided by 11. To check, divide each multiple of 420 by 11:
420 รท 11 = remainder 2
840 รท 11 = remainder 4
1260 รท 11 = remainder 6
1680 รท 11 = remainder 8
2100 รท 11 = remainder 10
So, the smallest number is 2100.
9. Children are playing โFire in the Mountainโ. 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:
Children are playing Fire in the Mountain. In this game, children get out only when the called number is not divisible by a hidden number.
From the problem:
When 6 was called, no one got out โ the hidden number divides 6
When 9 was called, no one got out โ the hidden number divides 9
When 10 was called, some children got out โ the hidden number does not divide 10
So, the hidden number must divide both 6 and 9.
The HCF of 6 and 9 is 3, so the hidden number is 3.
This works because 3 does not divide 10, so some children get out when 10 is called.
The total number of children must be a multiple of the hidden number 3, since they are arranged in a circle according to the game.
Checking the given options, the numbers divisible by 3 are: 72, 90, 45, 3, 36
The hidden number is 3 and the number of children could be 72, 90, 45, 3 or 36
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 ร n
(e) Greater than m ร n
Answer:
The LCM of two different prime numbers (m) and (n) is the smallest number divisible by both.
Since prime numbers have no common factors other than 1, the LCM of two different primes is simply their product, m ร n.
This product is always greater than both numbers.
Therefore, the LCM cannot be less than either number, cannot lie in between them, cannot be less than (m ร n), and cannot be greater than (m ร n).
The only correct statement is that the LCM is greater than both numbers.
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:
The rabbit has a head start of 150 feet.
Every time the rabbit jumps 7 feet, the dog jumps 9 feet.
So, in one leap of each, the dog gains: 9 – 7 = 2 feet on the rabbit
This means the dog closes the gap by 2 feet per leap.
The rabbit is initially 150 feet ahead, so the number of leaps the dog needs to catch the rabbit is:
Number of leaps = Head start/Distance gained per leap = 150/2 = 75
The dog will catch up with the rabbit in 75 leaps
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:
LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10
Prime Factorization:
First, write each number as a product of prime factors:
1 = 1
2 = 2
3 = 3
4 = 2ยฒ
5 = 5
6 = 2 ร 3
8 = 2ยณ
9 = 3ยฒ
10 = 2 ร 5
For LCM: Take the highest power of each prime factor
Highest power of 2 = 2ยณ (from 8)
Highest power of 3 = 3ยฒ (from 9)
Highest power of 5 = 5ยน (from 5, 10)
LCM = 2ยณ ร 3ยฒ ร 5 = 8 ร 9 ร 5 = 360
13. Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together 8/15, 1/20, 7/36, 11/63 and 1/21. What do you get?
How can we find the sum efficiently?
Answer:
LCM of the denominators
Prime factorization:
15 = 3 ร 5
20 = 2ยฒ ร 5
36 = 2ยฒ ร 3ยฒ
63 = 3ยฒ ร 7
21 = 3 ร 7
LCM = 2ยฒ ร 3ยฒ ร 5 ร 7
LCM = 4 ร 9 ร 5 ร 7 = 1260
Convert each fraction to denominator 1260
8/15 = (8 ร 84)/(15 ร 84) = 672/1260
1/20 = (1 ร 63)/(20 ร 63) = 63/1260
7/36 = (7 ร 35)/(36 ร 35) = 245/1260
11/63 = (11 ร 20)/(63 ร 20) = 220/1260
1/21 = (1 ร 60)/(21 ร 60) = 60/1260
Add the numerators
672 + 63 + 245 + 220 + 60 = 1260
1260/1260 = 1
FAQs – Class 7 Maths Ganita Prakash Part 2 Chapter 3
Is Class 7 Maths Part 2 Chapter 3 difficult?
Chapter 3 starts out very approachable – the idea of HCF grows naturally out of Sameeksha choosing tile sizes and LCM grows out of matching up strip lengths for a toran, so the definitions never feel arbitrary. The prime-factorisation method and the faster “ladder” division shortcut are also genuinely easy to pick up with practice. Where it gets trickier is the final “Figure it Out” set, which hides HCF and LCM inside word problems – a cowherd’s gates, a game called Fire in the Mountain, a dog chasing a rabbit – where the first challenge is recognising which concept even applies before you can start calculating.
How to solve Class 7 Maths Ganita Prakash Part 2, Chapter 3 in one day?
To get through Chapter 3 in one sitting, work through it in order rather than jumping to the exercises. Start with Section 3.1 to see why HCF is needed and practise finding it first by listing factors, then by prime factorisation. Move to Section 3.2 for LCM the same way, then read Section 3.3, which is where the chapter’s real payoff sits – the “ladder” division method that finds HCF and LCM together and the rule HCF ร LCM = the product of the two numbers. Save the final thirteen-part “Figure it Out” set for last, since it draws on everything before it, including a historical problem from the mathematician Mahaviracharya.
How many questions are there in Class 7 Maths Part 2 Chapter 3?
Chapter 3 has six separate “Figure it Out” sections woven through its three parts – several short ones for practising factor-listing, prime factorisation, HCF and LCM individually, a set built around forming general statements and patterns and a substantial final exercise with thirteen questions that range from a page-number colour puzzle to a classic cows-and-gates folklore problem and a centuries-old fraction sum from the mathematician Mahaviracharya. Working through them roughly in order makes the most sense, since later ones assume you’re comfortable with prime factorisation from the earlier ones.
Is Class 7 Maths Ganita Prakash Part 2, Chapter 3 important for the exam?
While exact exam weightage varies by school and year, HCF and LCM are used throughout later Maths – simplifying fractions, comparing ratios and solving time-and-work or pipe-and-cistern style problems all lean on being able to find them quickly and correctly. The chapter’s own word problems (tiling a room, packing rice into bags, cows passing through gates, a dog chasing a rabbit) are exactly the style examiners like, since they test whether you can spot HCF versus LCM from the situation, not just calculate once you’re told which one to use. That recognition skill is worth practising specifically.
How can NCERT Solutions help while solving Class 7 Maths Ganita Prakash Chapter 3?
A good number of this chapter’s questions ask you to reason about a pattern or prove a general statement rather than just compute an answer – for example, explaining why HCF ร LCM always equals the product of two numbers or working out what happens to the HCF when both numbers are doubled. NCERT Solutions walk through this reasoning using prime factorisation, showing exactly which prime factors are “common” versus “extra” and how that explains the pattern, so students can check whether their own explanation actually proves the rule or just happens to work for one example.
