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)

Accepted Answer

(a) 90
Here
∴ 90 = 2 × 3 × 3 × 5
Hence, factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Total factors = 12
(b) 105
Here
∴ 105 = 3 × 5 × 7
Hence, factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105.
Total factors = 8
(c) 132
Here
∴ 132 = 2 × 2 × 3 × 11
Hence, factors of 132 are 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
Total factors = 12
(d) 360
Here
∴ 360 = 2 × 2 × 2 × 3 × 3 × 5
Hence, 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, 360.
Total factors = 24
(e) 840
Here
∴ 840 = 2 × 2 × 2 × 3 × 5 × 7
Hence, 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, 840.
Total factors = 32
Figure It Out (Page 53)

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?

Accepted Answer

No, one cannot say that 72 and 144 have no common factor other than 1 because their factorisations have composite numbers.
Here 72 = 6 × 12
144 = 8 × 18
These are not prime factorisations.
Both 6 and 12 are composite numbers, as are 8 and 18.
Prime factorisation of 72 = 2 × 2 × 2 × 3 × 3
Prime factorisation of 144 = 2 × 2 × 2 × 2 × 3 × 3
∴ HCF(72, 144) = 2 × 2 × 2 × 3 × 3 = 8 × 9 = 72
Since the HCF is 72, which is greater than 1, the numbers have common factors other than 1.
3.2 Least, but not Last!
Figure It Out (Page 58)

Question 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

Accepted Answer

(a) Two Multiples of 3
Examples:
(i) (6, 9)
LCM (6, 9) = 18
(ii) (9, 12)
LCM (9, 12) = 36
(iii) (12, 18)
LCM (12, 18) = 36
Observation: The LCM of two multiples of 3 is also a multiple of 3.
Reason: Since both numbers are divisible by 3, their common multiples will also be divisible by 3.
Hence, the LCM must include 3 as a factor.
General Statement: The LCM of two multiples of 3 is always a multiple of 3.
(b) Two Consecutive Even Numbers
Examples:
(i) (2, 4)
LCM (2, 4) = 4
(ii) (6, 8)
LCM (6, 8) = 24
(iii) (10, 12)
LCM (10, 12) = 60
Observation: The LCM of two consecutive even numbers is half of their product.
Reason: Consecutive even numbers always share a common factor of 2, but not more.
Therefore, when finding the LCM, one factor of 2 overlaps, so the LCM becomes smaller than their product.
General Statement: The LCM of two consecutive even numbers 2n and 2n + 2 is always equal to half of their product.
or LCM (2n, 2n + 2) = $\frac{2 n \times(2 n+2)}{2}$
= n(2n + 2)
= 2n
2
+ 2n
(c) Two Consecutive Numbers
Examples:
(i) (7, 8)
LCM (7, 8) = 56
(ii) (9, 10)
LCM (9, 10) = 90
(iii) (10, 11)
LCM (10, 11) = 110
Observation: The LCM of two consecutive numbers is equal to their product.
Reason: Consecutive numbers have no common factors other than 1, so their product is the smallest number divisible by both.
General Statement: The LCM of two consecutive numbers is their product.
(d) Two Co-prime Numbers
Examples:
(i) (4, 9)
LCM (4, 9) = 36
(ii) (5, 8)
LCM (5, 8) = 40
(iii) (7, 10)
LCM (7, 10) = 70
Observation: The LCM of two co-prime numbers is equal to their product.
Reason: Co-prime numbers do not share any common factors except 1, so the smallest number that contains both is simply their product.
General Statement: The LCM of two co-prime numbers is equal to their product.
Note: Co-prime numbers are any two natural numbers that have no common factor other than 1.
Figure It Out (Pages 63-64)

Question 4

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

Accepted Answer

For two numbers a and b, the product of the numbers is equal to the product of their LCM and HCF.
∴ a × b = HCF(a, b) × LCM(a, b)
Here HCF = 1, LCM = 66
⇒ a × b = 1 × 66 = 66
∴ Pairs of factors of 66 are (1, 66), (2, 33), (3, 22), (6, 11)
Any of the following pairs of numbers will satisfy the conditions: (1, 66), (2, 33), (3, 22), (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 again 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 the folklore mathematics from Karnataka).

Accepted Answer

No. of cows is divisible by 3, 5, 7.
This means the number of cows is a multiple of the LCM(3, 5, 7).
The total number of cows is also less than 200.
LCM (3, 5, 7) = 3 × 5 × 7 = 105
No. of cows must be a multiple of 105.
Multiples of 105 are 105, 210, 315.
The problem states that the cowherd had fewer than 200 cows.
Only multiples of 105 that are less than 200 are 105.
Hence 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-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

Accepted Answer

Here, dimensions of box: Length = 12 cm, Width = 18 cm, Height = 36 cm
The size of the largest cube that can exactly fit (without gaps)
That means the side of the cube must exactly divide all three dimensions of the box.
So, we need to find the HCF (Highest Common Factor) of 12, 18, and 36.
Prime factorisation 12 = 2 × 2 × 3, 18 = 2 × 3 × 3, and 36 = 2 × 2 × 3 × 3
Common factors = 2 × 3 = 6
HCF = 6 cm
The cube must have a side length equal to a factor of the HCF.
From the options:
(a) 9 cm ✗ (9 doesn’t divide 12 evenly)
(b) 6 cm ✓ (divides 12, 18, and 36 exactly)
(c) 4 cm ✗ (doesn’t divide 18 evenly)
(d) 3 cm ✓ (also divides all)
(e) 2 cm ✓ (also divides all)
Hence, the largest possible cube that fits without gaps is (b) 6 cm.

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

Accepted Answer

Longest number that perfectly divides (306, 36) = HCF (306, 36)
HCF (306, 36) = 2 × 3 × 3 = 18.
∴ (c) is the correct option.

Source: This question is from Finding Common Ground, Mathematics &mdash; Class 7, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Finding Common Ground: Numbers, Largest, Number, Perfectly, Divides, Finding Common Ground. These are fundamental topics in Mathematics that students are expected to master as part of the CBSE Class 7 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Mathematics papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberAlways show your working steps clearly.
Verify your answer by substituting values back into the equation.
Practice similar problems from the textbook exercises.
Memorise important formulae and their conditions of applicability.
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Question 8

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

Accepted Answer

LCM(3, 4, 5,7) = 3 × 4 × 5 × 7 = 420
The number must be a multiple of 420, so the number can be written as 420k, where k is a whole number.
N = 420 × 1 = 420
When divided by 11 leaves a remainder of 2
11 × 38 + 2 = 420
But we require 10 as a remainder
∴ 2k = 10
⇒ k = 5
Hence number = 420 × 5 = 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

Accepted Answer

Interpretation is that when a number k is called, the children are grouped into rows of size k, and “no one got out” means the children formed complete rows with no remainder.
So: “No one got out” when 6 was called ⇒ the total number N is divisible by 6.
“No one got out” when 9 was called ⇒ N is divisible by 9.
“Some people got out” when 10 was called ⇒ N is not divisible by 10.
If N is divisible by both 6 and 9, then it must be divisible by their LCM:
LCM (6, 9) = 18.
So N is a multiple of 18, but not a multiple of 10.
Now check the options:
(a) 72 = 18 × 4 – divisible by 18 and not by 10 → possible.
(b) 90 = 18 × 5 – divisible by 18 but is divisible by 10 → not possible.
(c) 45 – not divisible by 18 → not possible.
(d) 3 – not divisible by 18 → not possible.
(e) 36 = 18 × 2 – divisible by 18 and not by 10 → possible.
So the numbers that could have been playing 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

Accepted Answer

LCM of two different prime numbers, m and n, is their product of m × n.
Since both m and n are prime numbers, their only common factor is 1.
Hence, LCM is always greater than both individual numbers.
Hence, option (c) is correct.

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?

Accepted Answer

Here, distance gained per leap = Dog’s leap distance – Rabbit’s leap distance
= 9 – 7
= 2
The dog needs to close a 150-foot head start.
No. of leaps = $\frac{\text { Head Start Distance }}{\text { Distance Gained Per Leap }}$
= $\frac {150}{2}$
= 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?

Accepted Answer

1 = 1
2 = 2
3 = 3
4 = 2 × 2
5 = 5
6 = 2 × 3
7 = 7
8 = 2 × 2 × 2
9 = 3 × 3
10 = 2 × 5
LCM(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 2 × 2 × 2 × 3 × 3 × 5
= 8 × 9 × 5
= 360
Yes, this is the same answer.
You might 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 $\frac{8}{15}, \frac{1}{20}, \frac{7}{36}, \frac{11}{63}$ and $\frac {1}{21}$. What do you get? How can we find this sum efficiently?

Accepted Answer

Here $\frac{8}{15}, \frac{1}{20}, \frac{7}{36}, \frac{11}{63}, \frac{1}{21}$
Now 15 = 3 × 5
20 = 2 × 2 × 5
36 = 2 × 2 × 3 × 3
63 = 3 × 3 × 7
21 = 3 × 7
LCM of denominators = 2 × 2 × 3 × 3 × 5 × 7
= 4 × 9 × 5 × 7
= 1260