Class 6 Exam > Class 6 Notes > Mathematics (Maths) Class 6 > RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.5)
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Page 1
Exercise 4.5 page: 4.23
1. Without drawing a diagram, find
(i) 10
th
square number
(ii) 6
th
triangular number
Solution:
(i) 10
th
square number
The square number can be remembered using the following rule
Nth square number = n × n
So the 10
th
square number = 10 × 10 = 100
(ii) 6
th
triangular number
The triangular number can be remembered using the following rule
Nth triangular number = n × (n + 1)/ 2
So the 6
th
triangular number = 6 × (6 + 1)/ 2 = 21
2. (i) Can a rectangular number also be a square number?
(ii) Can a triangular number also be a square number?
Solution:
(i) Yes. A rectangular number can also be a square number.
Example – 16 is a rectangular number which can also be a square number.
(ii) Yes. A triangular number can also be a square number.
Example – 1 is a triangular number which can also be a square number.
3. Write the first four products of two numbers with difference 4 starting from in the following order:
1, 2, 3, 4, 5, 6, ………
Identify the pattern in the products and write the next three products.
Solution:
We know that
1 × 5 = 5
2 × 6 = 12
3 × 7 = 21
4 × 8 = 32
So the first four products of two numbers with difference 4
5 – 1 = 4
6 – 2 = 4
7 – 3 = 4
8 – 4 = 4
4. Observe the pattern in the following and fill in the blanks:
9 × 9 + 7 = 88
Page 2
Exercise 4.5 page: 4.23
1. Without drawing a diagram, find
(i) 10
th
square number
(ii) 6
th
triangular number
Solution:
(i) 10
th
square number
The square number can be remembered using the following rule
Nth square number = n × n
So the 10
th
square number = 10 × 10 = 100
(ii) 6
th
triangular number
The triangular number can be remembered using the following rule
Nth triangular number = n × (n + 1)/ 2
So the 6
th
triangular number = 6 × (6 + 1)/ 2 = 21
2. (i) Can a rectangular number also be a square number?
(ii) Can a triangular number also be a square number?
Solution:
(i) Yes. A rectangular number can also be a square number.
Example – 16 is a rectangular number which can also be a square number.
(ii) Yes. A triangular number can also be a square number.
Example – 1 is a triangular number which can also be a square number.
3. Write the first four products of two numbers with difference 4 starting from in the following order:
1, 2, 3, 4, 5, 6, ………
Identify the pattern in the products and write the next three products.
Solution:
We know that
1 × 5 = 5
2 × 6 = 12
3 × 7 = 21
4 × 8 = 32
So the first four products of two numbers with difference 4
5 – 1 = 4
6 – 2 = 4
7 – 3 = 4
8 – 4 = 4
4. Observe the pattern in the following and fill in the blanks:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = ………
98765 × 9 + 3 = ………
987654 × 9 + 2 = ……….
9876543 × 9 + 1 = ……….
Solution:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = 88888
98765 × 9 + 3 = 888888
987654 × 9 + 2 = 8888888
9876543 × 9 + 1 = 88888888
5. Observe the following pattern and extend it to three more steps:
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
….. × ……. - ….. = …….
….. × ……. - ….. = …….
….. × ……. - ….. = …….
Solution:
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
10 × 6 - 45 = 15
11 × 7 - 60 = 17
12 × 8 - 77 = 19
6. Study the following pattern:
1 + 3 = 2 × 2
1 + 3 + 5 = 3 × 3
1 + 3 + 5 + 7 = 4 × 4
Page 3
Exercise 4.5 page: 4.23
1. Without drawing a diagram, find
(i) 10
th
square number
(ii) 6
th
triangular number
Solution:
(i) 10
th
square number
The square number can be remembered using the following rule
Nth square number = n × n
So the 10
th
square number = 10 × 10 = 100
(ii) 6
th
triangular number
The triangular number can be remembered using the following rule
Nth triangular number = n × (n + 1)/ 2
So the 6
th
triangular number = 6 × (6 + 1)/ 2 = 21
2. (i) Can a rectangular number also be a square number?
(ii) Can a triangular number also be a square number?
Solution:
(i) Yes. A rectangular number can also be a square number.
Example – 16 is a rectangular number which can also be a square number.
(ii) Yes. A triangular number can also be a square number.
Example – 1 is a triangular number which can also be a square number.
3. Write the first four products of two numbers with difference 4 starting from in the following order:
1, 2, 3, 4, 5, 6, ………
Identify the pattern in the products and write the next three products.
Solution:
We know that
1 × 5 = 5
2 × 6 = 12
3 × 7 = 21
4 × 8 = 32
So the first four products of two numbers with difference 4
5 – 1 = 4
6 – 2 = 4
7 – 3 = 4
8 – 4 = 4
4. Observe the pattern in the following and fill in the blanks:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = ………
98765 × 9 + 3 = ………
987654 × 9 + 2 = ……….
9876543 × 9 + 1 = ……….
Solution:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = 88888
98765 × 9 + 3 = 888888
987654 × 9 + 2 = 8888888
9876543 × 9 + 1 = 88888888
5. Observe the following pattern and extend it to three more steps:
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
….. × ……. - ….. = …….
….. × ……. - ….. = …….
….. × ……. - ….. = …….
Solution:
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
10 × 6 - 45 = 15
11 × 7 - 60 = 17
12 × 8 - 77 = 19
6. Study the following pattern:
1 + 3 = 2 × 2
1 + 3 + 5 = 3 × 3
1 + 3 + 5 + 7 = 4 × 4
1 + 3 + 5 + 7 + 9 = 5 × 5
By observing the above pattern, find
(i) 1 + 3 + 5 + 7 + 9 + 11
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(iii) 21 + 23 + 25 + ….. + 51
Solution:
(i) 1 + 3 + 5 + 7 + 9 + 11
By using the pattern
1 + 3 + 5 + 7 + 9 + 11 = 6 × 6
= 36
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
By using the pattern
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8
= 64
(iii) 21 + 23 + 25 + ….. + 51
We know that
21 + 23 + 25 + ….. + 51 can be written as (1 + 3 + 5 + 7 + ….. + 49 + 51) – (1 + 3 + 5 + …… + 17 + 19)
By using the pattern
(1 + 3 + 5 + 7 + ….. + 49 + 51) = 26 × 26 = 676
(1 + 3 + 5 + …… + 17 + 19) = 10 × 10 = 100
So we get
21 + 23 + 25 + ….. + 51 = 676 – 100 = 576
7. Study the following pattern:
1 × 1 + 2 × 2 = (2 × 3 × 5)/ 6
1 × 1 + 2 × 2 + 3 × 3 = (3 × 4 × 7) / 6
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = (4 × 5 × 9)/ 6
By observing the above pattern, write next two steps.
Solution:
By using the pattern
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5
On further calculation
= (5 × 6 × 11)/6
So we get
= 55
By using the pattern
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6
On further calculation
= (6 × 7 × 13)/6
So we get
= 91
8. Study the following pattern:
1 = (1 × 2)/ 2
1 + 2 = (2 × 3)/ 2
Page 4
Exercise 4.5 page: 4.23
1. Without drawing a diagram, find
(i) 10
th
square number
(ii) 6
th
triangular number
Solution:
(i) 10
th
square number
The square number can be remembered using the following rule
Nth square number = n × n
So the 10
th
square number = 10 × 10 = 100
(ii) 6
th
triangular number
The triangular number can be remembered using the following rule
Nth triangular number = n × (n + 1)/ 2
So the 6
th
triangular number = 6 × (6 + 1)/ 2 = 21
2. (i) Can a rectangular number also be a square number?
(ii) Can a triangular number also be a square number?
Solution:
(i) Yes. A rectangular number can also be a square number.
Example – 16 is a rectangular number which can also be a square number.
(ii) Yes. A triangular number can also be a square number.
Example – 1 is a triangular number which can also be a square number.
3. Write the first four products of two numbers with difference 4 starting from in the following order:
1, 2, 3, 4, 5, 6, ………
Identify the pattern in the products and write the next three products.
Solution:
We know that
1 × 5 = 5
2 × 6 = 12
3 × 7 = 21
4 × 8 = 32
So the first four products of two numbers with difference 4
5 – 1 = 4
6 – 2 = 4
7 – 3 = 4
8 – 4 = 4
4. Observe the pattern in the following and fill in the blanks:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = ………
98765 × 9 + 3 = ………
987654 × 9 + 2 = ……….
9876543 × 9 + 1 = ……….
Solution:
9 × 9 + 7 = 88
98 × 9 + 6 = 888
987 × 9 + 5 = 8888
9876 × 9 + 4 = 88888
98765 × 9 + 3 = 888888
987654 × 9 + 2 = 8888888
9876543 × 9 + 1 = 88888888
5. Observe the following pattern and extend it to three more steps:
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
….. × ……. - ….. = …….
….. × ……. - ….. = …….
….. × ……. - ….. = …….
Solution:
6 × 2 – 5 = 7
7 × 3 – 12 = 9
8 × 4 – 21 = 11
9 × 5 – 32 = 13
10 × 6 - 45 = 15
11 × 7 - 60 = 17
12 × 8 - 77 = 19
6. Study the following pattern:
1 + 3 = 2 × 2
1 + 3 + 5 = 3 × 3
1 + 3 + 5 + 7 = 4 × 4
1 + 3 + 5 + 7 + 9 = 5 × 5
By observing the above pattern, find
(i) 1 + 3 + 5 + 7 + 9 + 11
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(iii) 21 + 23 + 25 + ….. + 51
Solution:
(i) 1 + 3 + 5 + 7 + 9 + 11
By using the pattern
1 + 3 + 5 + 7 + 9 + 11 = 6 × 6
= 36
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
By using the pattern
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8
= 64
(iii) 21 + 23 + 25 + ….. + 51
We know that
21 + 23 + 25 + ….. + 51 can be written as (1 + 3 + 5 + 7 + ….. + 49 + 51) – (1 + 3 + 5 + …… + 17 + 19)
By using the pattern
(1 + 3 + 5 + 7 + ….. + 49 + 51) = 26 × 26 = 676
(1 + 3 + 5 + …… + 17 + 19) = 10 × 10 = 100
So we get
21 + 23 + 25 + ….. + 51 = 676 – 100 = 576
7. Study the following pattern:
1 × 1 + 2 × 2 = (2 × 3 × 5)/ 6
1 × 1 + 2 × 2 + 3 × 3 = (3 × 4 × 7) / 6
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = (4 × 5 × 9)/ 6
By observing the above pattern, write next two steps.
Solution:
By using the pattern
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5
On further calculation
= (5 × 6 × 11)/6
So we get
= 55
By using the pattern
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6
On further calculation
= (6 × 7 × 13)/6
So we get
= 91
8. Study the following pattern:
1 = (1 × 2)/ 2
1 + 2 = (2 × 3)/ 2
1 + 2 + 3 = (3 × 4)/ 2
1 + 2 + 3 + 4 = (4 × 5)/ 2
By observing the above pattern, find
(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
(ii) 50 + 51 + 52 + ……. + 100
(iii) 2 + 4 + 6 + 8 + 10 + …….. + 100
Solution:
(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
We get
= (10 × 11)/2
On further calculation
= 55
(ii) 50 + 51 + 52 + ……. + 100
We can write it as
(1 + 2 + 3 + …… + 99 + 100) – (1 + 2 + 3 + 4 + ….. + 47 + 49)
So we get
(1 + 2 + 3 + …… + 99 + 100) = (100 × 101)/2
(1 + 2 + 3 + 4 + ….. + 47 + 49) = (49 × 50)/2
By substituting the values
50 + 51 + 52 + ……. + 100 = (100 × 101)/2 + (49 × 50)/2
On further calculation
= 5050 – 1225
We get
= 3825
(iii) 2 + 4 + 6 + 8 + 10 + …….. + 100
We can write it as
2 (1 + 2 + 3 + 4 + …… + 49 + 50)
So we get
= 2 × (50 × 51)/2
On further calculation
= 2 × (1275)
We get
= 2550
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Document Description: RD Sharma Solutions: Operations on Whole Numbers (Exercise 4.5) for Class 6 2024 is part of Mathematics (Maths) Class 6 preparation.
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