RBSE Solutions for Class 5 Maths Chapter 8 Patterns Additional Questions

RBSE Solutions for Class 5 Maths Chapter 8 Patterns Additional Questions is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 8 Patterns Additional Questions.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 5
Subject	Maths
Chapter	Chapter 8
Chapter Name	Patterns
Exercise	Additional Questions
Number of Questions	32
Category	RBSE Solutions

Rajasthan Board RBSE Class 5 Maths Chapter 8 Patterns Additional Questions

Multiple choice Questions

Question 1.
Extend the patterns given below

Question 2.
Extend the following pattern-

Question 3.
Next term of 2, 5, 8, 11, ………..
(a) 12
(b) 13
(c) 14
(d) 3

Question 4.
Sixth term of 1, 7, 13, 19, 25 is-
(a) 31
(b) 26
(c) 27
(d) 28

Question 5.
Next term of 4, 7, 11, 16 is-
(a) 18
(b) 20
(c) 24
(d) 22

Question 6.
Wrong term in 3, 8, 13, 17, 23 is-
(a) 8
(b) 17
(c) 13
(d) 23

Question 7.

Complete the above pattern

Question 8.

Complete the above pattern

Question 9.
Wrong term in 1, 6, 11, 17, 21 is-
(a) 17
(b) 6
(c) 11
(d) 21

Question 10.
Next term of 3, 5, 1,9, 11 is-
(a) 12
(b) 13
(c) 14
(d) 15

Question 11.
Next term of 729, 243, 81 is-
(a) 18
(b) 9
(c) 27
(d) 36

Question 12.
Next term of 125, 115, 105, 95 is-
(a) 75
(b) 65
(c) 90
(d) 85

Question 13.

Next figure of the above given figure-

Answers
1. (c)
2. (a)
3. (c)
4. (a)
5. (d)
6. (b)
7. (d)
8. (c)
9. (a)
10. (b)
11. (c)
12. (d)
13. (c)

Fill in the blanks in following

1. Pattern of shapes has ……………….
2. Patterns of ………………. can be(RBSESolutions.com)formed.
3. Sometimes shapes of pattern have a ………………. of patterns.

Ans.
1. Sequence
2. Numbers
3. Sequence

Very Short Answer Type Questions

Question 1.
Pattern ofis formed complete it-

Solution.

Question 2.
Now, using → in place of make a pattern like this-

Solution.

Question 3.
Using in place of make the same patern below.

Solution.

Question 4.
Complete the following

Solution.

Question 5.
See the different patterns of(RBSESolutions.com)numbers given below and proceed further-
9 × 9 = 81
99 × 99 = 9801
999 × 999 = 998001
………………. ……………….
Solution.
9 × 9 = 81
99 × 99 = 9801
999 × 999 = 998001
9999 × 9999 =99980001
99999 × 99999 = 9999800001

Question 6.
Give the shortest method to multiply any number by 5.
Solution.
To multiply any number by 5. Put 0 (zero) before unit(RBSESolutions.com)digit of the number and divide the obtained number by 2. The number obtained is the resultant product of 5 with the given number.
For Example- 217 × 5 = 2170 ÷ 2 = 1085

Question 7.
Find the next term of 4, 5, 10, 13, 52.
Solution.
In series 4, 5, 10, 13, 52, …
4 + 1 = 5, 5 × 2 = 10,10 + 3 = 13, 13 × 4 = 52
Therefore next term is 52 + 5 = 57.

Question 8.
Find the next term of 3, 6, 9, 15, 24.
Solution.
In 3, 6, 9, 15, 24, ….
Succeeding term is(RBSESolutions.com)obtained by adding preceding two terms.
Here 3 + 6 = 9, 6 + 9 = 15, 9 + 15 = 24
Therefore, next term is 15 + 24 = 39.

Question 9.
Forward the pattern given below by 2 steps.
(1)1 6, 10, 14, 18, …… , ……
(2) 35,29,23,17, …… , ……
Solution.
(1) 6, 10, 14, 18, 22, 26
(2) 35, 29, 23, 17, 11, 5

Short Answer Type Question

Question 1.
Identify the pattern and proceed further-

Solution.

Question 2.
Identify the pattern and proceed further.


Solution:

Question 3.
Fill in the blanks by numbers, in 8, 6, 11,11,14,16,?,?
Solution.
There is no continuity in(RBSESolutions.com)the terms of series. 8,6, 11, 11, 14, 16, ….
Therefore, sequence can be written as follows-
(i) 8 11 14 = ?
(ii) 6 11 16 =?
Here, we see that number pattern is formed by combining two Arithmetic progression. The common difference of these Arithmetic pro-gressions are 3 and 5 respectively, next term of first sequence is 14 + 3 = 17 and next term of second sequence is 16 + 5 = 21.

Question 4.
Proceed further-

Solution.
.

Question 5.
Find next term of 3, 7, 16, 35, 74, .
Solution.
In number series 3, 7, 16, 35, 74,… the difference between terms(RBSESolutions.com)increasing continuously. Therefore, there is a possibility of getting next term by multiplying the terms of series by any number.
Here 3 × 2 + 1 = 7, 7 × 2 + 2 = 16, 16 × 2 + 3 = 35, 35 × 2 + 4 = 74
Therefore, next term is 74 × 2 + 5 = 153.

Question 6.
See the given number pattern and proceed further-
7 × 7 = 49
67 × 67 = 4489
667 × 667 = 444889
……………….. ………………..
……………….. ………………..
……………….. ………………..
Solution.
7 × 7 = 49
67 × 67 = 4489
667 × 667 = 444889
6667 × 6667 = 44448889
66667 × 66667 = 4444488889
666667 × 666667 = 444444888889

Question 7.
Find next term of 4, 6, 9, 13, 18,
Solution.
Terms of sequence 4,6,9, 13, 18, ………….. are continuously increasing.
Therefore, next term may be obtained by(RBSESolutions.com)adding some number to its preceding term. Next terms are obtained by adding 2, 3,4, 5, ………… respectively in their preceding terms.

Therefore next term is 18 + 6 = 24

Question 8.
See by doing this

Solution:

Question 9.
Stand the following patterns and then fill in the blanks.?

Solution.

We hope the RBSE Solutions for Class 5 Maths Chapter 8 Patterns Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 8 Patterns Additional Questions, drop a comment below and we will get back to you at the earliest.