NCERT Solutions for Class 7 maths Chapter 11: Finding Common Ground

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.

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.
Solution:
(a) Here are more number pairs where the LCM of the two numbers is one of the given numbers:
(i) LCM of 2 and 4: The LCM is 4.
Prime factorization of 2: 2
Prime factorization of 4: 2 × 2
LCM (2, 4) = 2 × 2 = 4

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
Solution:
(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.

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

Solution:
The positions of the black star in the first row are 4 and 10.
The difference in position = 10 – 4 = 6.
So, the next positions of the black star in the first row are: 4, 10, 16, 22, 28, …
The positions of the black star in the second row are 4 and 8.
The difference in position = 8 – 4 = 4.
So, the next positions of the black star in the second row are: 4, 8, 12, 16, 20, …
The 16th position is common in both rows.
Hence, the black stars meet again at the 16th position.

(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?
Solution:
(a) Here let a = 5 × 7 × 11 × 11 and b = 5 × 7 × 7 × 11 × 2
For a number a to be a multiple of number b, all prime factors of b must be present in a with at least the same power.
Number a does not have the prime factor 2.
Number a has 71, while number b has 7.
Hence, a is not a multiple of b.

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
Solution:
(a) Here 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11 = 2 × 2 × 3 × 7 × 11
∴ HCF = 3 × 7 = 21
∴ LCM = 2 × 2 × 3 × 3 × 5 × 7 × 7 × 11.

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

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).
Solution:
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.

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
Solution:
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.

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
Solution:
Longest number that perfectly divides (306, 36) = HCF (306, 36)

HCF (306, 36) = 2 × 3 × 3 = 18.
∴ (c) is the correct option.

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

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
Solution:
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.

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
Solution:
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.

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:
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 =
=
= 75 leaps

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 = 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.

Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together and . What do you get? How can we find this sum efficiently?
Solution:
Here
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