---
title: "NCERT Solutions Class 7 Maths Chapter 11 Finding Common Ground"
url: https://www.swavid.com/maths/class/7/chapter/finding-common-ground/ncert-solutions
dateModified: 2026-10-07T15:11:15+00:00
---

# NCERT Solutions Class 7 Maths Chapter 11 Finding Common Ground

This chapter's questions cover concepts related to common factors, common multiples, prime factorisation, HCF, LCM, and their properties and applications. Students practice finding HCF and LCM through various methods and solving word problems based on them.

Free PDF (27 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-7/swavid-ncert-solutions-class-7-maths-chapter-11-finding-common-ground-339f8770e6.pdf

## Try This

### Question 1

*3 marks · Short answer*

How many tiles of this size should she purchase?
What if Sameeksha did not insist on the length of the tile to be a whole number of feet and the length could be a fractional number of feet? Would the answer change?

**Part (i)**

1. The breadth of the room is $12\text{ ft}$ and the length is $16\text{ ft}$.
2. The size of the square tile is the HCF of 12 and 16, which is $4\text{ ft}$.
3. Area of the room $= 16\text{ ft} \times 12\text{ ft} = 192\text{ sq ft}$, and area of one tile $= 4\text{ ft} \times 4\text{ ft} = 16\text{ sq ft}$.
4. Number of tiles required $= \frac{192}{16} = 12$ tiles.

Answer (i): 12 tiles

**Part (ii)**

1. If the tile length can be fractional, we look for common factors that are fractions.
2. The largest common measure of 12 and 16 is still 12 if we consider common divisors, but since tiles must fit both dimensions, the greatest common divisor of 12 and 16 is 4 when restricted to whole numbers.
3. Without the whole number restriction, we could use a tile of side $12\text{ ft}$, but it would not fit the length of $16\text{ ft}$ exactly unless we cut it. If only whole number of feet is relaxed for the tile size, the answer would change.

Answer (ii): Yes, the answer would change as larger fractional sizes or other common divisors become possible.

**Answer:** Sameeksha needs 12 tiles of size $4\text{ ft} \times 4\text{ ft}$. If fractional tile sizes were allowed, the tile size could be larger (up to 12 ft), so the number of tiles would change.

> Common mistake: Dividing the room perimeter instead of the area, or forgetting to compute the total number of tiles.

## Try This

### Question 1

*3 marks · Short answer*

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 $42$
Is the longest jump size for the numbers the same as their HCF? Explain why it is so.

**Part (a)**

1. Factors of $14$ are $1, 2, 7, 14$ and factors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$.
2. The common factors are $1$ and $2$.
3. The HCF is $2$, so the longest jump size is $2$.

Answer (a): $2$

**Part (b)**

1. $7$ and $11$ are prime numbers, so their only common factor is $1$.
2. The HCF is $1$, so the longest jump size is $1$.

Answer (b): $1$

**Part (c)**

1. Factors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$ and factors of $50$ are $1, 2, 5, 10, 25, 50$.
2. The common factors are $1, 2, 5,$ and $10$.
3. The HCF is $10$, so the longest jump size is $10$.

Answer (c): $10$

**Part (d)**

1. Factors of $28$ are $1, 2, 4, 7, 14, 28$ and factors of $42$ are $1, 2, 3, 6, 7, 14, 21, 42$.
2. The common factors are $1, 2, 7,$ and $14$.
3. The HCF is $14$, so the longest jump size is $14$.

Answer (d): $14$

**Answer:** The longest jump size is the HCF of the two numbers in each case.

> Common mistake: Confusing common factors with common multiples when finding jump sizes.

## Figure it Out

### Question 1

*3 marks · Short answer*

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)

**Part (a)**

1. Prime factorisation of $90 = 2 \times 3 \times 3 \times 5$.
2. Systematically combining prime factors gives subparts and factors.
3. The factors of $90$ are $1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45,$ and $90$.

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

**Part (b)**

1. Prime factorisation of $105 = 3 \times 5 \times 7$.
2. Systematically combining prime factors gives subparts and factors.
3. The factors of $105$ are $1, 3, 5, 7, 15, 21, 35,$ and $105$.

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

**Part (c)**

1. Prime factorisation of $132 = 2 \times 2 \times 3 \times 11$.
2. Systematically combining prime factors gives subparts and factors.
3. The factors of $132$ are $1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66,$ and $132$.

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

**Part (d)**

1. Prime factorisation of $360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5$.
2. Systematically combining prime factors gives all 24 subparts.
3. 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 $360$.

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

**Part (e)**

1. Prime factorisation of $840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7$.
2. Systematically combining prime factors gives all 32 subparts.
3. 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 $840$.

Answer (e): 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, 840

**Answer:** Factors listed for all five numbers using prime factorisation subparts.

> Common mistake: Missing some factors while combining prime factors.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Part (a)**

1. Prime factorisation of $50 = 2 \times 5 \times 5$ and $60 = 2 \times 2 \times 3 \times 5$.
2. The common prime factors are $2$ and $5$.
3. Taking the minimum power of common primes, $\text{HCF} = 2 \times 5 = 10$, and the common factors are $1, 2, 5, 10$.

Answer (a): Common factors: $1, 2, 5, 10$; $\text{HCF} = 10$

**Part (b)**

1. Prime factorisation of $140 = 2 \times 2 \times 5 \times 7$ and $275 = 5 \times 5 \times 11$.
2. The only common prime factor is $5$.
3. Taking the minimum power of the common prime, $\text{HCF} = 5$, and the common factors are $1, 5$.

Answer (b): Common factors: $1, 5$; $\text{HCF} = 5$

**Part (c)**

1. Prime factorisation of $77 = 7 \times 11$ and $725 = 5 \times 5 \times 29$.
2. There are no common prime factors between $77$ and $725$.
3. Therefore, the only common factor is $1$, and $\text{HCF} = 1$.

Answer (c): Common factor: $1$; $\text{HCF} = 1$

**Part (d)**

1. Prime factorisation of $370 = 2 \times 5 \times 37$ and $592 = 2 \times 2 \times 2 \times 2 \times 37$.
2. The common prime factors are $2$ and $37$.
3. Taking the minimum power of common primes, $\text{HCF} = 2 \times 37 = 74$, and the common factors are $1, 2, 37, 74$.

Answer (d): Common factors: $1, 2, 37, 74$; $\text{HCF} = 74$

**Part (e)**

1. Prime factorisation of $81 = 3 \times 3 \times 3 \times 3$ and $243 = 3 \times 3 \times 3 \times 3 \times 3$.
2. The common prime factor is $3$, occurring at least 4 times in both.
3. Taking the minimum power, $\text{HCF} = 3 \times 3 \times 3 \times 3 = 81$, and the common factors are powers of $3$ from $3^0$ to $3^4$.

Answer (e): Common factors: $1, 3, 9, 27, 81$; $\text{HCF} = 81$

**Answer:** See sub-parts for the common factors and HCF of each pair.

> Common mistake: Listing only the HCF and forgetting to list all the common factors as asked in the question.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Part (a)**

1. Prime factorisation of $24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$.
2. Prime factorisation of $180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$.
3. Taking the minimum power of common prime factors, $\text{HCF} = 2^2 \times 3 = 12$.

Answer (a): 12

**Part (b)**

1. Prime factorisation of $42 = 2 \times 3 \times 7$.
2. Prime factorisation of $75 = 3 \times 5 \times 5 = 3 \times 5^2$.
3. Prime factorisation of $24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3$.
4. Taking the common prime factor with the minimum power, $\text{HCF} = 3$.

Answer (b): 3

**Part (c)**

1. Prime factorisation of $240 = 2^4 \times 3 \times 5$.
2. Prime factorisation of $378 = 2 \times 3^3 \times 7$.
3. Taking the minimum power of common prime factors, $\text{HCF} = 2 \times 3 = 6$.

Answer (c): 6

**Part (d)**

1. Prime factorisation of $400 = 2^4 \times 5^2$.
2. Prime factorisation of $2500 = 2^2 \times 5^4$.
3. Taking the minimum power of common prime factors, $\text{HCF} = 2^2 \times 5^2 = 100$.

Answer (d): 100

**Part (e)**

1. Prime factorisation of $300 = 2^2 \times 3 \times 5^2$.
2. Prime factorisation of $800 = 2^5 \times 5^2$.
3. Taking the minimum power of common prime factors, $\text{HCF} = 2^2 \times 5^2 = 100$.

Answer (e): 100

**Answer:** HCF values are: (a) $12$, (b) $3$, (c) $6$, (d) $400$, (e) $100$

> Common mistake: Including prime factors that are not common to all the given numbers.

### Question 2

*3 marks · Short answer*

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

**Solution**

1. No, one cannot say that they have no common factor other than $1$.
2. A factorisation into composite numbers does not show all the prime factors of the numbers.
3. Both $6 \times 12$ and $8 \times 18$ can be further factorised into primes, revealing common prime factors like $2$ and $3$, so their HCF is actually $36$.

**Answer:** No, because composite factorisations do not reveal all prime factors; $72$ and $144$ share many common factors.

> Common mistake: Assuming that using composite factors instead of prime factors gives complete information about divisibility and common factors.

## Try This

### Question 1

*3 marks · Short answer*

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.)

**Solution**

1. The sweet shop distributes free gajak every Monday, which means the days when free gajak is available are multiples of $7$ ($7, 14, 21, 28, 35, 42, 49, 56, 63, 70, \dots$).
2. Kabamai visits the sweet shop once every $10$ days, so her visit days are multiples of $10$ ($10, 20, 30, 40, 50, 60, 70, \dots$).
3. To find when she will get free gajak again, we need to find the lowest common multiple (LCM) of $7$ and $10$, which is $70$ days.

**Answer:** 70 days

> Common mistake: Adding the numbers $7$ and $10$ instead of finding their least common multiple.

## Try This

### Question 1

*3 marks · Short answer*

Do you remember the ‘Idli-Vada’ game from Grade 6 (see chapter ‘Prime Time’)? Two numbers are chosen and whenever players come to their multiples, ‘idli’ or ‘vada’ should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then ‘idli-vada’ should be called out. In each problem below, the two numbers corresponding to ‘idli’ and ‘vada’ are given. Find 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 $55$
Is the answer always the LCM of the two numbers? Explain.

**Part (a)**

1. Find the LCM of $4$ and $6$.
2. Multiples of $4$: $4, 8, 12, 16, \dots$
3. Multiples of $6$: $6, 12, 18, \dots$
4. The lowest common multiple is $12$.

Answer (a): $12$

**Part (b)**

1. Find the LCM of $7$ and $11$.
2. Since $7$ and $11$ are prime numbers, their LCM is their product.
3. $7 \times 11 = 77$

Answer (b): $77$

**Part (c)**

1. Find the LCM of $14$ and $30$.
2. $14 = 2 \times 7$ and $30 = 2 \times 3 \times 5$.
3. LCM $= 2 \times 3 \times 5 \times 7 = 210$.

Answer (c): $210$

**Part (d)**

1. Find the LCM of $15$ and $55$.
2. $15 = 3 \times 5$ and $55 = 5 \times 11$.
3. LCM $= 3 \times 5 \times 11 = 165$.

Answer (d): $165$

**Answer:** The first number for which 'idli-vada' is called out is the LCM of the two numbers. For (a) it is $12$, for (b) it is $77$, for (c) it is $210$, and for (d) it is $165$. Yes, the answer is always the LCM because the first common multiple where both names are called together is the least common multiple.

> Common mistake: Listing only multiples of one number instead of finding the common multiples.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Part (a)**

1. Prime factorisation of $30 = 2 \times 3 \times 5$.
2. Prime factorisation of $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$.
3. Taking the highest power of each prime factor, $\text{LCM} = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$.

Answer (a): $360$

**Part (b)**

1. Prime factorisation of $36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^3$.
2. Prime factorisation of $54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3$.
3. Taking the highest power of each prime factor, $\text{LCM} = 2^2 \times 3^3 = 4 \times 27 = 108$.

Answer (b): $108$

**Part (c)**

1. Prime factorisations are $105 = 3 \times 5 \times 7$, $195 = 3 \times 5 \times 13$, and $65 = 5 \times 13$.
2. Taking the highest power of each prime factor, $\text{LCM} = 3 \times 5 \times 7 \times 13$.
3. Multiplying the factors, $\text{LCM} = 15 \times 91 = 1365$.

Answer (c): $1365$

**Part (d)**

1. Prime factorisations are $222 = 2 \times 3 \times 37$ and $370 = 2 \times 5 \times 37$.
2. Taking the highest power of each prime factor, $\text{LCM} = 2 \times 3 \times 5 \times 37$.
3. Multiplying the factors, $\text{LCM} = 30 \times 37 = 1110$.

Answer (d): $1110$

**Answer:** The LCM values are (a) 360, (b) 108, (c) 1365, and (d) 1110.

> Common mistake: Taking the lowest powers instead of the highest powers of prime factors when finding the LCM.

## Try This

### Question 1

*3 marks · Short answer*

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?

**Part (a)**

1. We are given that $m$ is a number and one number is a multiple of the other such that $m$ is their HCF.
2. This means the other number must be a multiple of $m$.
3. Therefore, the other number could be $nm$ where $n$ is a positive integer.

Answer (a): $nm$ for any positive integer $n$

**Part (b)**

1. We are given that $7k$ is a number and it is the HCF of the pair.
2. The other number must be a multiple of $7k$.
3. Therefore, the other number could be $7kn$ where $n$ is a positive integer.

Answer (b): $7kn$ for any positive integer $n$

**Answer:** (a) $nm$ for any integer $n$, (b) $7kn$ for any integer $n$

> Common mistake: Confusing multiples with factors when finding the other number.

## Figure it Out

### Question 1

*3 marks · Short answer*

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.

**Part (a)**

1. Examples: HCF of 2 and 4 is 2; HCF of 6 and 8 is 2.
2. The HCF of any two consecutive even numbers is always 2.

Answer (a): 2

**Part (b)**

1. Examples: HCF of 3 and 5 is 1; HCF of 7 and 9 is 1.
2. The HCF of any two consecutive odd numbers is always 1.

Answer (b): 1

**Part (c)**

1. Examples: HCF of 4 and 10 is 2; HCF of 12 and 18 is 6.
2. The HCF of two even numbers depends on their common factors, so it is always an even number greater than or equal to 2.

Answer (c): An even number (at least 2)

**Part (d)**

1. Examples: HCF of 4 and 5 is 1; HCF of 12 and 13 is 1.
2. The HCF of any two consecutive numbers is always 1.

Answer (d): 1

**Part (e)**

1. Definition: Co-prime numbers have no common factors other than 1.
2. The HCF of any two co-prime numbers is always 1.

Answer (e): 1

**Answer:** General statements for the HCF of various number pairs.

> Common mistake: Confusing HCF with LCM for consecutive numbers.

### Question 2

*3 marks · Short answer*

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.

**Part (a)**

1. Consider pairs where one number is a multiple of the other, such as 4 and 12, or 5 and 25.
2. The LCM of 4 and 12 is 12, and the LCM of 5 and 25 is 25.

Answer (a): Examples include (4, 12) and (5, 25).

**Part (b)**

1. When one number is a multiple of the other, the larger number is a multiple of the smaller number.
2. Using algebra, if $n$ is a number and $mn$ is its multiple (where $m$ is a positive integer), the LCM of $n$ and $mn$ is $mn$.

Answer (b): The LCM of $n$ and $mn$ is $mn$.

**Answer:** Number pairs and their general algebraic description where LCM is one of the numbers.

> Common mistake: Stating that the numbers must be co-prime instead of one being a multiple of the other.

### Question 3

*3 marks · Short answer*

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

**Part (a)**

1. Examples: LCM of 3 and 6 is 6; LCM of 6 and 15 is 30.
2. The LCM of two multiples of 3 is a multiple of 3.

Answer (a): A multiple of 3

**Part (b)**

1. Examples: LCM of 2 and 4 is 4; LCM of 6 and 8 is 24.
2. The LCM of two consecutive even numbers is half of their product.

Answer (b): Half of their product

**Part (c)**

1. Examples: LCM of 2 and 3 is 6; LCM of 4 and 5 is 20.
2. Since consecutive numbers are always co-prime, their LCM is equal to their product.

Answer (c): Their product

**Part (d)**

1. Co-prime numbers have no common factors other than 1.
2. The LCM of any two co-prime numbers is always equal to their product.

Answer (d): Their product

**Answer:** General statements for the LCM of various number pairs.

> Common mistake: Forgetting that consecutive numbers are always co-prime.

## Try This

### Question 1

*3 marks · Short answer*

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?

**Solution**

1. When both numbers are doubled, each number gets an extra factor of $2$ in its prime factorisation.
2. This additional factor of $2$ becomes common to both numbers and is included in the largest common subpart.
3. Consequently, the Highest Common Factor (HCF) of the two numbers also doubles.

**Answer:** The HCF of two numbers also doubles when both numbers are doubled because an extra factor of $2$ is added to the common prime factors.

> Common mistake: Thinking that doubling both numbers increases the HCF by a factor of four instead of two.

## Try This

### Question 1

*3 marks · Short answer*

Consider the following two multiples of $14$ — $14 \times 6, 14 \times 9$. What is their HCF?
Clearly, $14$ is a common factor. Is it also the highest common factor? To see it, let us calculate the prime factorisations.

**Solution**

1. Given numbers are $14 \times 6$ and $14 \times 9$.
2. Express each number in terms of its prime factorisation: $14 \times 6 = (2 \times 7) \times (2 \times 3)$ and $14 \times 9 = (2 \times 7) \times (3 \times 3)$.
3. Identify the common factors in both prime factorisations, which are $2 \times 7$ (or $14$) and $3$.
4. Multiply the common prime factors to find the Highest Common Factor: $\text{HCF} = 14 \times 3 = 42$.

**Answer:** $42$

> Common mistake: Only considering the common multiplier $14$ as the HCF without checking for common factors in the remaining multipliers $6$ and $9$.

## Try This

### Question 1

*3 marks · Short answer*

Here are some more numbers where both numbers are multiples of the same number. Find their HCF:
(a) $18 \times 10, 18 \times 15$
(b) $10 \times 38, 10 \times 21$
(c) $5 \times 13, 5 \times 20$
(d) $12 \times 16, 12 \times 20$
In which of these cases is the HCF the same as the common multiplier, like problem (b) where the HCF is $10$? Explore a few more examples of this type to understand when this happens.

**Part (a)**

1. $18 \times 10 = 2 \times 3^2 \times 2 \times 5$ and $18 \times 15 = 2 \times 3^2 \times 3 \times 5$.
2. The common prime factors are $2, 3^2, \text{ and } 5$.
3. The HCF is $18 \times 5 = 90$.

Answer (a): 90

**Part (b)**

1. $10 \times 38 = 2 \times 5 \times 2 \times 19$ and $10 \times 21 = 2 \times 5 \times 3 \times 7$.
2. The common prime factors are $2 \text{ and } 5$.
3. The HCF is $10 \times 1 = 10$.

Answer (b): 10

**Part (c)**

1. $5 \times 13 = 5 \times 13$ and $5 \times 20 = 5 \times 2^2 \times 5$.
2. The only common prime factor is $5$.
3. The HCF is $5 \times 13 = 65$.

Answer (c): 65

**Part (d)**

1. $12 \times 16 = 2^2 \times 3 \times 2^4$ and $12 \times 20 = 2^2 \times 3 \times 2^2 \times 5$.
2. The common prime factors are $2^2 \times 3$ and the remaining factors are $4 \text{ and } 5$.
3. The HCF is $12 \times 4 = 48$.

Answer (d): 48

**Answer:** The HCFs are (a) 90, (b) 10, (c) 65, (d) 48. The HCF is equal to the common multiplier when the remaining factors are co-prime.

> Common mistake: Assuming the HCF of two multiples is always equal to the common multiplier without checking the remaining factors.

## Try This

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Write both numbers 84 and 180 side by side and divide by their common prime factors.
2. Divide by 2 to get the quotients 42 and 90, then divide by 2 again to get 21 and 45.
3. Divide by 3 to get 7 and 15, which have no further common prime factors.
4. Multiply the common prime factors: $\text{HCF} = 2 \times 2 \times 3 = 12$.

**Answer:** 12

> Common mistake: Dividing further when the remaining numbers no longer share a common prime factor.

## Try This

### Question 1

*3 marks · Short answer*

Find the HCF in the following cases.
[Two worked-out problems on page 15 are shown with diagrams]

**Part (i)**

1. Divide both numbers 300 and 150 by the common prime factor 2 to get 150 and 75.
2. Divide both by 5 to get 30 and 15, then again by 5 to get 6 and 3.
3. Divide both by 3 to get 2 and 1, which have no common factors.
4. Multiply the common prime divisors: $2 \times 5 \times 5 \times 3 = 150$.

Answer (i): 150

**Part (ii)**

1. Divide both numbers 630 and 770 by the common prime factor 2 to get 315 and 385.
2. Divide both by 5 to get 63 and 77, then divide by 7 to get 9 and 11.
3. Since 9 and 11 have no common prime factors, the division stops.
4. Multiply the common prime divisors: $2 \times 5 \times 7 = 70$.

Answer (ii): 70

**Answer:** The HCF for the first case is 150 and for the second case is 70.

> Common mistake: Stopping the division before dividing by all common prime factors.

## Try This

### Question 1

*3 marks · Short answer*

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?]

**Solution**

1. The product of the factors marked in the division method includes all common prime factors along with the remaining prime factors of the numbers.
2. Thus, this product contains the prime factorisations of both $300$ and $150$ as subparts, making it a common multiple.
3. Since we only multiply the necessary unique prime factors with their highest occurrences, it is the smallest such number, which is the LCM.

**Answer:** The product gives the LCM because it is the smallest number that contains the prime factorisations of both numbers.

> Common mistake: Thinking that any common multiple is the LCM without checking if it is the smallest possible one.

## Try This

### Question 1

*3 marks · Short answer*

You can try this method for these pairs of numbers.
(a) $90$ and $150$
(b) $84$ and $132$

**Part (a)**

1. Divide both numbers by their common factor 30 to get quotients 3 and 5.
2. The numbers 3 and 5 have no common prime factor.
3. HCF = $30$ and $\text{LCM} = 30 \times 3 \times 5 = 450$.

Answer (a): HCF = 30, LCM = 450

**Part (b)**

1. Divide both numbers by their common factor 12 to get quotients 7 and 11.
2. The numbers 7 and 11 have no common prime factor.
3. HCF = $12$ and $\text{LCM} = 12 \times 7 \times 11 = 924$.

Answer (b): HCF = 12, LCM = 924

**Answer:** For (a) HCF = 30, LCM = 450; for (b) HCF = 12, LCM = 924

> Common mistake: Stopping division before fully dividing out all common factors or incorrectly multiplying the final quotients for the LCM.

## Try This

### Question 1

*3 marks · Short answer*

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

**Solution**

1. The product of any two numbers is always a common multiple of those two numbers.
2. By definition, the Lowest Common Multiple (LCM) is the smallest of all common multiples of the numbers.
3. Since the LCM is the lowest among all common multiples, it can never be greater than the product, which is also a common multiple.

**Answer:** The product of two numbers is always greater than or equal to their LCM because the product itself is a common multiple, and the LCM is the lowest common multiple.

> Common mistake: Confusing LCM with HCF or assuming that the product is always strictly greater without considering numbers where one is a factor of the other.

## Try This

### Question 1

*3 marks · Short answer*

Consider the numbers $105$ and $95$. Find their LCM.
Factorising them into their primes:
$105 = 3 \times 5 \times 7$
$95 = 5 \times 19$
LCM = $3 \times 5 \times 7 \times 19$.
Let us consider the product in the factorised form:
$105 \times 95 = 3 \times 5 \times 5 \times 7 \times 19$
Is the LCM a factor of the product? If yes, what should it be multiplied with to get the product? It can be seen that
$105 \times 95 = \text{LCM} \times 5$.

**Solution**

1. The prime factorisations are given as $105 = 3 \times 5 \times 7$ and $95 = 5 \times 19$.
2. The LCM of $105$ and $95$ is $3 \times 5 \times 7 \times 19 = 9975$.
3. Yes, the LCM is a factor of the product, and it should be multiplied by $5$ (which is the HCF of $105$ and $95$) to get the product.

**Answer:** Yes, the LCM is a factor of the product and it should be multiplied by $5$.

> Common mistake: Forgetting that the number multiplying the LCM to get the product is always the HCF of the two numbers.

## Try This

### Question 1

*3 marks · Short answer*

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) $222, 370$

**Part (a)**

1. For $45$ and $105$, prime factorisations are $45 = 3 \times 3 \times 5$ and $105 = 3 \times 5 \times 7$.
2. The HCF is $15$ and the LCM is $315$.
3. The product $45 \times 105 = 4725$, which when divided by the LCM $315$ gives $15$, which is the HCF.

Answer (a): LCM is $315$, which is a factor of the product and is multiplied by the HCF ($15$) to get the product.

**Part (b)**

1. For $275$ and $352$, prime factorisations are $275 = 5 \times 5 \times 11$ and $352 = 2 \times 2 \times 2 \times 2 \times 2 \times 11$.
2. The HCF is $11$ and the LCM is $8800$.
3. The product $275 \times 352 = 96800$, which when divided by the LCM $8800$ gives $11$, which is the HCF.

Answer (b): LCM is $8800$, which is a factor of the product and is multiplied by the HCF ($11$) to get the product.

**Part (c)**

1. For $222$ and $370$, prime factorisations are $222 = 2 \times 3 \times 37$ and $370 = 2 \times 5 \times 37$.
2. The HCF is $74$ and the LCM is $1110$.
3. The product $222 \times 370 = 82140$, which when divided by the LCM $1110$ gives $74$, which is the HCF.

Answer (c): LCM is $1110$, which is a factor of the product and is multiplied by the HCF ($74$) to get the product.

**Answer:** In each case, the LCM is a factor of the product, and it is multiplied by the HCF of the numbers to get the product.

> Common mistake: Forgetting that the number by which LCM must be multiplied to obtain the product is always the HCF of the two numbers.

## Try This

### Question 1

*3 marks · Short answer*

Do you see that, in each case, the number by which the LCM is multiplied to get the product is actually the HCF?
Thus, our observations seem to suggest the following:
$\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$.
Why does this happen? Can you give an explanation or proof?
[Hint: Consider the prime factorisation of the given numbers. Among their prime factors, some are common to both factorisations, and the rest occur in only one of them. Between the HCF and the LCM, see how the common and non-common prime factors get distributed. In the product, observe how these two kinds of prime factors occur. Compare them.]

**Solution**

1. Let the prime factorisations of two numbers be written, separating the common prime factors and the non-common prime factors.
2. The HCF is the product of all the common prime factors, while the LCM is the product of all the common prime factors and the remaining non-common prime factors from both numbers.
3. When we multiply the HCF and the LCM, every common prime factor and non-common prime factor appears exactly the same number of times as it appears in the product of the two numbers.

**Answer:** The product of HCF and LCM equals the product of the two numbers because both sides contain the exact same set of prime factors.

> Common mistake: Confusing how common and non-common factors are distributed between the HCF and the LCM.

## Try This

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Let us test the property for three numbers, say $4, 6,$ and $8$.
2. The prime factorisations are $4 = 2^2$, $6 = 2 \times 3$, and $8 = 2^3$.
3. Their HCF is $2$ and their LCM is $2^3 \times 3 = 24$.
4. The product of the three numbers is $4 \times 6 \times 8 = 192$, while $\text{HCF} \times \text{LCM} = 2 \times 24 = 48$.
5. Since $192 \neq 48$, the property $\text{HCF} \times \text{LCM} = \text{Product of numbers}$ does not hold when $3$ numbers are considered.

**Answer:** The property does not hold for three numbers.

> Common mistake: Assuming that the property $\text{HCF} \times \text{LCM} = \text{Product}$ extends directly from two numbers to three numbers.

## Figure it Out

### Question 1

*3 marks · Short answer*

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

**Solution**

1. Observe the repeating pattern lengths for the star colors in the two rows from the figure in the textbook (Fig. 3.x on page 63).
2. Let the repetition periods of the two rows be $a$ and $b$ units respectively.
3. The blue stars will meet next at a position corresponding to the Lowest Common Multiple (LCM) of their repetition intervals.

**Answer:** The blue stars will meet next at the LCM of the repetition intervals of the two rows.

> Common mistake: Adding the repetition periods instead of finding their LCM.

### Question 2

*3 marks · Short answer*

2. (a) Is $5 \times 7 \times 11 \times 11$ a multiple of $5 \times 7 \times 7 \times 11 \times 2$?
(b) Is $5 \times 7 \times 11 \times 11$ a factor of $5 \times 7 \times 7 \times 11 \times 2$?

**Part (a)**

1. A number is a multiple of another if its prime factorisation contains all the prime factors of the second number with at least the same counts.
2. Here, $5 \times 7 \times 11 \times 11$ does not contain the prime factor $2$ and has fewer $7$s compared to $5 \times 7 \times 7 \times 11 \times 2$.

Answer (a): No, it is not a multiple.

**Part (b)**

1. A number is a factor of another if its prime factorisation is contained as a subpart in the second number.
2. Here, $5 \times 7 \times 11 \times 11$ has two $11$s, whereas $5 \times 7 \times 7 \times 11 \times 2$ has only one $11$, so it cannot be a factor.

Answer (b): No, it is not a factor.

**Answer:** Part (a) is No, and Part (b) is No.

> Common mistake: Confusing the definitions of factors and multiples in prime factorisation form.

### Question 3

*3 marks · Short answer*

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

**Part (a)**

1. Write the prime factorisations as given: $3 \times 3 \times 5 \times 7 \times 7$ and $12 \times 7 \times 11 = 2^2 \times 3 \times 7 \times 11$.
2. The common prime factors with minimum powers give the HCF: $7$.
3. The highest powers of all prime factors give the LCM: $2^2 \times 3^2 \times 5 \times 7^2 \times 11$.

Answer (a): HCF = $7$, LCM = $2^2 \times 3^2 \times 5 \times 7^2 \times 11$

**Part (b)**

1. Find the prime factorisations: $45 = 3^2 \times 5$ and $36 = 2^2 \times 3^2$.
2. Identify the common prime factors with minimum powers to find the HCF: $3^2 = 9$.
3. Identify the highest powers of all prime factors to find the LCM: $2^2 \times 3^2 \times 5 = 180$.

Answer (b): HCF = $9$, LCM = $180$

**Answer:** Part (a): HCF = $7$, LCM = $2^2 \times 3^2 \times 5 \times 7^2 \times 11$. Part (b): HCF = $9$, LCM = $180$.

> Common mistake: Confusing minimum powers for HCF with maximum powers for LCM.

### Question 4

*3 marks · Short answer*

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

**Solution**

1. We know that the product of two numbers equals the product of their HCF and LCM.
2. Product of the two numbers = $1 \times 66 = 66$.
3. We need to find two numbers whose HCF is $1$ (co-prime) and product is $66$, such as $2$ and $33$, or $1$ and $66$, or $3$ and $22$, or $6$ and $11$.

**Answer:** The numbers can be $6$ and $11$ (or $2$ and $33$).

> Common mistake: Not checking if the HCF of the chosen numbers is actually 1.

### Question 5

*3 marks · Short answer*

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.)

**Solution**

1. Since an equal number of cows passed through each gate at each crossing, the total number of cows must be a multiple of $3$, $5$, and $7$.
2. Find the LCM of $3$, $5$, and $7$. Since they are prime numbers, their LCM = $3 \times 5 \times 7 = 105$.
3. The total number of cows must be a multiple of $105$ and less than $200$, so the number is $105 \times 1 = 105$.

**Answer:** 105 cows

> Common mistake: Finding the HCF instead of the LCM.

### Question 6

*1 mark · MCQ*

6. The length, width, and height of a box are $12\text{ cm}, 18\text{ cm},$ and $36\text{ cm}$ respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

- $9\text{ cm}$
- $6\text{ cm}$
- $4\text{ cm}$
- $3\text{ cm}$
- $2\text{ cm}$

**Solution**

1. The dimensions of the box are $12\text{ cm}$, $18\text{ cm}$, and $36\text{ cm}$.
2. For cubes to pack the box without leaving gaps, the side length of the cube must be a common factor of $12$, $18$, and $36$.
3. The factors of $12$ are $1, 2, 3, 4, 6, 12$, and among the given options, $6\text{ cm}$ is the largest common factor that divides all three dimensions.

**Answer:** (b) $6\text{ cm}$

> Common mistake: Choosing a number that divides only two of the dimensions.

### Question 7

*1 mark · MCQ*

7. Among the numbers below, which is the largest number that perfectly divides both $306$ and $36$?

- $36$
- $612$
- $18$
- $3$
- $2$
- $360$

**Solution**

1. The largest number that perfectly divides both numbers is their Highest Common Factor (HCF).
2. Prime factorisation gives $306 = 2 \times 3 \times 3 \times 17$ and $36 = 2 \times 2 \times 3 \times 3$, so their HCF is $2 \times 3 \times 3 = 18$.

**Answer:** (c) $18$

> Common mistake: Confusing HCF with LCM.

### Question 8

*3 marks · Short answer*

8. Find the smallest number that is divisible by $3, 4, 5$ and $7$, but leaves a remainder of $10$ when divided by $11$.

**Solution**

1. First, find the LCM of $3, 4, 5,$ and $7$.
2. The prime factorisations are $3 = 3$, $4 = 2^2$, $5 = 5$, and $7 = 7$, so their LCM is $2^2 \times 3 \times 5 \times 7 = 420$.
3. Any number divisible by $3, 4, 5,$ and $7$ must be a multiple of $420$, which can be written in the form $420k$.
4. Test multiples of $420$ plus $10$ to find the smallest number that leaves a remainder of $10$ when divided by $11$: for $k = 5$, $420 \times 5 + 10 = 2100 + 10 = 2110$.
5. Dividing $2110$ by $11$ gives a quotient of $191$ and a remainder of $10$.

**Answer:** $2110$

> Common mistake: Forgetting to add the remainder after finding the LCM.

### Question 9

*1 mark · MCQ*

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?

- $72$
- $90$
- $45$
- $3$
- $36$
- None of these

**Solution**

1. The number of children must be a multiple of both $6$ and $9$ so that no one gets out when these numbers are called.
2. The LCM of $6$ and $9$ is $18$, so possible numbers of children are multiples of $18$ ($18, 36, 54, 72, 90, \dots$).
3. When $10$ is called out, some people get out, meaning the total number is not a multiple of $10$. Checking the options, $90$ is a multiple of $10$, while $72, 45, 36$ are not multiples of $10$ and fit the conditions.
4. Since $90$ is a multiple of $10$, it is incorrect, making 'None of these' the correct choice as $72$ or $36$ are valid possibilities among the given options.

**Answer:** (f) None of these

> Common mistake: Selecting a number that is a multiple of 10 because of misreading the elimination condition.

### Question 10

*1 mark · MCQ*

10. Tick the correct statement(s). The LCM of two different prime numbers ($m, n$) can be:

- Less than both numbers
- In between the two numbers
- Greater than both numbers
- Less than $m \times n$
- Greater than $m \times n$

**Solution**

1. The LCM of two different prime numbers $m$ and $n$ is their product $m \times n$.
2. Therefore, the LCM is greater than both numbers and equal to $m \times n$, making the statement 'Greater than both numbers' correct.

**Answer:** (c) Greater than both numbers

> Common mistake: Assuming LCM of primes is their sum.

### Question 11

*3 marks · Short answer*

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?

**Solution**

1. Given: The rabbit has a head start of $150$ feet, the dog jumps $9$ feet per leap, and the rabbit jumps $7$ feet per leap.
2. Difference covered by the dog in each leap relative to the rabbit is $9 - 7 = 2$ feet.
3. Number of leaps required to cover the $150$ feet head start is $\frac{150}{2} = 75$ leaps.

**Answer:** $75$ leaps

> Common mistake: Dividing the head start by the dog's jump length instead of the relative difference.

### Question 12

*3 marks · Short answer*

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?

**Solution**

1. We need to find the smallest number that is a multiple of $1, 2, 3, 4, 5, 6, 8, 9, 10$, which is the LCM of these numbers.
2. Prime factorisations are: $2 = 2$, $3 = 3$, $4 = 2^2$, $5 = 5$, $6 = 2 \times 3$, $8 = 2^3$, $9 = 3^2$, $10 = 2 \times 5$.
3. Taking the highest power of each prime factor, we get $2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$.

**Answer:** $360$

> Common mistake: Including numbers like $6$ and $10$ prime factors redundantly instead of taking maximum powers.

### Question 13

*3 marks · Short answer*

13. Here is a problem posed by the ancient Indian Mathematician Mahaviracharya ($850$ C.E.). Add together $\frac{8}{15}, \frac{1}{20}, \frac{7}{36}, \frac{11}{63}$ and $\frac{1}{21}$. What do you get? How can we find this sum efficiently?

**Solution**

1. Find the LCM of the denominators $15, 20, 36, 63,$ and $21$ to use as the common denominator.
2. The prime factorisations are $15 = 3 \times 5$, $20 = 2^2 \times 5$, $36 = 2^2 \times 3^2$, $63 = 3^2 \times 7$, and $21 = 3 \times 7$, giving an LCM of $1260$.
3. Convert each fraction to an equivalent fraction with the denominator $1260$: $\frac{8}{15} = \frac{672}{1260}$, $\frac{1}{20} = \frac{63}{1260}$, $\frac{7}{36} = \frac{245}{1260}$, $\frac{11}{63} = \frac{220}{1260}$, and $\frac{1}{21} = \frac{60}{1260}$.
4. Add the numerators together: $672 + 63 + 245 + 220 + 60 = 1260$.
5. Divide the sum of the numerators by the common denominator: $\frac{1260}{1260} = 1$.

**Answer:** $1$

> Common mistake: Making calculation errors when finding equivalent fractions for large denominators.

## Frequently asked questions

### How many questions are there in Class 7 Maths Chapter 11 Finding Common Ground?

This chapter for the 2026-27 session follows the new NCERT book based on the NCF 2023 guidelines. It includes multiple sections with specific question counts across Try This and Figure it Out exercises, such as 13 questions in the final Figure it Out set covering topics like LCM and HCF. You can find all these questions solved step by step in SwaVid's free PDF available on this page only.

### What topics are covered in Class 7 Maths Chapter 11 Finding Common Ground?

The chapter covers essential concepts like HCF and its application to room tiling, jump sizes, prime factorisation, and the division method. It also explores Lowest Common Multiple (LCM), the relation between HCF, LCM and the product of numbers, and properties for three numbers. SwaVid's free PDF on this page provides clear explanations for each of these topics.

### Which are the most challenging question types in this chapter and how should we approach them?

Word problems involving applications of LCM and HCF, such as cube packing and divisibility with remainders, are often found challenging by students. To approach them, first identify whether you need a common factor or a common multiple based on the context. SwaVid's step-by-step solutions on this page break down these difficult problems into simple steps.

### How should I write answers to get full marks in Class 7 Maths exams?

To secure full marks, you should clearly state the given information, show the complete prime factorisation or division method steps, and state the final answer with proper units. Writing down each logical step ensures the examiner can follow your method easily. You can refer to SwaVid's free PDF on this page to see model answers structured for maximum scores.

### Is a free PDF available for NCERT Solutions of Class 7 Maths Chapter 11?

Yes, comprehensive solutions for this chapter based on the new NCERT book for the 2026-27 session are available for students. These materials help you practice concepts ranging from basic factorisation to advanced HCF and LCM relations. You can access SwaVid's free PDF and detailed step-by-step solutions directly on this page only.

## Related pages

- [Class 7 Maths chapters](https://www.swavid.com/maths/class/7)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student