Finding Common Ground Class 7 Solutions Maths Ganita Prakash Part 2 Chapter 3
Figure it Out (Page 51)
Q. 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)
Solution:
(a) 90
1 × 90 = 90
2 × 45 = 90
3 × 30 = 90
5 × 18 = 90
6 × 15 = 90
9 × 10 = 90
Therefore, factors of 90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.
(b) 105
1 × 105 = 105
3 × 35 = 105
5 × 21 = 105
7 × 15 = 105
Therefore, factors of 105 = 1, 3, 5, 7, 15, 21, 35, 105.
(c) 132
1 × 132 = 132
2 × 66 = 132
3 × 44 = 132
4 × 33 = 132
6 × 22 = 132
11 × 12 = 132
Therefore, factors of 132 = 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
(d) 360
1 × 360 = 360
2 × 180 = 360
3 × 120 = 360
4 × 90 = 360
5 × 72 = 360
6 × 60 = 360
8 × 45 = 360
9 × 40 = 360
10 × 36 = 360
12 × 30 = 360
15 × 24 = 360
18 × 20 = 360
Therefore, factors of 360 =
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360.
Figure it Out (Page 53)
Q. 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
Solution:
(a) 50, 60
50 = 2 × 5 × 5
60 = 2 × 2 × 3 × 5
Common factors = 1, 2, 5, 2 × 5 = 1, 2, 5, 10.
HCF = 10.
(b) 140, 275
140 = 2 × 2 × 5 × 7
275 = 5 × 5 × 11
Common factors = 5
HCF = 5.
(c) 77, 725
77 = 7 × 11
725 = 5 × 5 × 29
Common factors = 1.
HCF = 1.
(d) 370, 592
370 = 2 × 5 × 37
592 = 2 × 2 × 2 × 2 × 37
Common factors = 1, 2, 37, 2 × 37 = 1, 2, 37, 74.
HCF = 74.
(e) 81, 243
81 = 3 × 3 × 3 × 3
243 = 3 × 3 × 3 × 3 × 3
Common factors = 1, 3, 3 × 3, 3 × 3 × 3, 3 × 3 × 3 × 3 = 1, 3, 9, 27, 81.
HCF = 81.
Figure it Out (Page 54)
1. Find the HCF of the following numbers:
(a) 24, 180
(b) 42, 75, 24
(c) 240, 378
(d) 400, 2500
(e) 300, 800
Solution:
(a) 24, 180
24 = 2 × 2 × 2 × 3
180 = 2 × 2 × 3 × 3 × 5
HCF = 2 × 2 × 3 = 12.
(b) 42, 75, 24
42 = 2 × 3 × 7
75 = 3 × 5 × 5
24 = 2 × 2 × 2 × 3
HCF = 3.
(c) 240, 378
240 = 2 × 2 × 2 × 2 × 3 × 5
378 = 2 × 3 × 3 × 3 × 7
HCF = 2 × 3 = 6.
(d) 400, 2500
400 = 2 × 2 × 2 × 2 × 5 × 5
2500 = 2 × 2 × 5 × 5 × 5 × 5
HCF = 2 × 2 × 5 × 5 = 100.
(e) 300, 800
300 = 2 × 2 × 3 × 5 × 5
800 = 2 × 2 × 2 × 2 × 2 × 5 × 5
HCF = 2 × 2 × 5 × 5 = 100.
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?
Solution:
No, we cannot say that 72 and 144 have no common factor other than 1.
This is because we cannot use composite factorisation (like 6 × 12 or 8 × 18) to decide common factors. Only prime factorization correctly shows common factors.
Figure it Out (Page 58)
Q. Find the LCM of the following numbers:
(a) 30, 72
(b) 36, 54
(c) 105, 195, 65
(d) 222, 370
Solution:
(a) 30, 72
30 = 2 × 3 × 5
72 = 2 × 2 × 2 × 3 × 3
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 = 4 × 27 = 108.
(c) 105, 195, 65
105 = 3 × 5 × 7
195 = 3 × 5 × 13
65 = 5 × 13
LCM = 3 × 5 × 7 × 13 = 15 × 91 = 1365.
(d) 222, 370
222 = 2 × 3 × 37
370 = 2 × 5 × 37
LCM = 2 × 3 × 5 × 37 = 6 × 185 = 1110.
Figure it Out (Page 59)
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
Solution:
(a) Two Consecutive Even Numbers:
General Statement: The HCF of any two consecutive even numbers is 2.
Reason: Both numbers are divisible by 2 and have no other common factor.
Example:
Numbers: 8 and 10
HCF = 2
(b) Two consecutive odd numbers:
General Statement: The HCF of any two consecutive odd numbers is 1.
Reason: Consecutive odd numbers have no common factor.
Example:
Numbers: 11 and 13
HCF = 1
(c) Two even numbers:
General Statement: The HCF of any two even numbers is at least 2.
Reason: Every even number has 2 as a factor.
Example:
Numbers: 12 and 18
HCF = 6 (at least 2)
(d) Two consecutive numbers:
General Statement: The HCF of any two consecutive natural numbers is 1.
Explanation: Consecutive numbers have no common factor except 1.
Example:
Numbers: 14 and 15
HCF = 1
(e) Two co-prime numbers:
General Statement: The HCF of any two co-prime numbers is 1.
Explanation: Co-prime numbers have no common prime factor.
Example:
Numbers: 9 and 20
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.
Solution:
(a) LCM of 4 and 20 is 20
LCM of 6 and 18 is 18
LCM of 8 and 32 is 32
LCM of 5 and 15 is 15
(b) General statement:
If one number is a factor of the other, then the LCM of the two numbers is the larger number.
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
Solution:
(a) Two multiples of 3:
General Statement: The LCM of two multiples of 3 is also a multiple of 3.
Reason: Both numbers contain 3 as a factor.
Example:
Numbers: 9 and 15
LCM = 45, which is a multiple of 3.
(b) Two consecutive even numbers:
General Statement: The LCM of two consecutive even numbers is a multiple of the larger number.
Reason: The numbers differ by 2 and have no common factor other than 2.
Example:
Numbers: 10 and 12
LCM = 60, which is a multiple of the larger number 12.
(c) Two consecutive numbers:
General Statement: The LCM of two consecutive numbers is their product.
Reason: The consecutive numbers are co-prime.
Example:
Numbers: 7 and 8
LCM = 56, which is their product.
(d) Two co-prime numbers:
General Statement: The LCM of two co-prime numbers is their product.
Reason: The numbers have no common factor.
Example:
Numbers: 8 and 15
LCM = 120, which is their product.
Figure it Out (Page 63)
1. In the two rows below, colours repeat as shown. When will the blue stars meet next?
Solution:
Blue star positions in top row: 4, 10, 16, 22, … (adding +6 each time).
Blue star positions in bottom row: 4, 8, 12, 16, 20, … (adding +4 each time).
The next common position after 4 is 16.
So, they will meet at the 16th star.
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?
Solution:
(a) To be a multiple, the first number must contain all the factors of the second number.
But the second number has an extra 7 and 2, which the first number does not have.
So, it is NOT a multiple.
(b) For the first number to be a factor, the second number must contain all parts of the first number.
But the first number here has an extra 11.
So, it is not a factor.
3. Find the HCF and LCM of the following (state your answers in the form of prime factorisations):
(a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11
(b) 45 and 36
Solution:
(a) First number = 3 × 3 × 5 × 7 × 7
Second number = 12 × 7 × 11 = 2 × 2 × 3 × 7 × 11
HCF = 3 × 7 = 21.
LCM = 2 × 2 × 3 × 3 × 5 × 7 × 7 × 11 = 97,020.
(b) 45 = 3 × 3 × 5
36 = 2 × 2 × 3 × 3
HCF = 3 × 3 = 9.
LCM = 2 × 2 × 3 × 3 × 5 = 180.
4. Find two numbers whose HCF is 1 and LCM is 66.
Solution:
Prime factorisation of 66 = 2 × 3 × 11.
Take factors with no common prime:
6 = 2 × 3, 11 = 11.
So, HCF(6, 11) = 1 and LCM(6, 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.)
Solution:
The number of cows must be divisible by 3, 5, and 7.
LCM of 3, 5, 7 = 3 × 5 × 7 = 105.
Since 105 is less than 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
Solution:
The side of the cube must divide each of the dimensions 12 cm, 18 cm, and 36 cm.
HCF of 12, 18, 36 = 6.
Since factors of 6 = 1, 2, 3 and 6.
So, cubes of side 2 cm, 3 cm or 6 cm can be packed in the box without leaving gaps.
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
Solution:
To find the largest number that divides both 306 and 36, we find their HCF.
306 = 2 × 3 × 3 × 17.
36 = 2 × 2 × 3 × 3.
HCF = 2 × 3 × 3 = 18.
So, 18 is the largest number that divides 306 and 36.
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:
Since the required number is divisible by 3, 4, 5, and 7, we first find their LCM.
LCM of 3, 4, 5, and 7 = 3 × 4 × 5 × 7 = 420.
So, the required number must be a multiple of 420.
Thus, 2100 leaves a remainder of 10 when divided by 11.
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?
Solution:
When 6 and 9 were called, no one got out.
So, the number of children must be divisible by 6 and 9.
LCM of 6 and 9 = 2 × 3 × 3 = 18
Therefore, the possible number of children could be:
18, 36, 54, …
When 10 was called, some children got out, so the number of children is not divisible by 10.
The smallest number that satisfies all conditions is 18.
Since 18 is not among the given options,
(f) None of these
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
Solution:
Since prime numbers have no common factor other than 1,
LCM of m and n = m × n.
Therefore,
(c) Greater than both numbers is the correct statement.
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:
Head start of rabbit = 150 feet
Dog’s jump = 9 feet
Rabbit’s jump = 7 feet
In one leap, the dog gains = 9 − 7 = 2 feet.
The number of leaps required = 150 ÷ 2 = 75.
Thus, the dog catches 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?
Solution:
LCM of 1, 2, 3, 4, 5, 6, 8, 9, 10 = 2 × 2 × 2 × 3 × 3 × 5 = 360.
Therefore, 360 is the required smallest number.
13. 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:
LCM of 15, 20, 36, 63, and 21 = 2 × 2 × 3 × 3 × 5 × 7 = 1260.
Adding,
.