Class 5

RBSE Solutions for Class 5 Maths Chapter 1 Numbers Ex 1.1

June 2, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 5 Maths Chapter 1 Numbers Ex 1.1 is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 1 Numbers Exercise 1.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 5
Subject	Maths
Chapter	Chapter 1
Chapter Name	Numbers
Exercise	Ex 1.1
Number of Questions	7
Category	RBSE Solutions

Rajasthan Board RBSE Class 5 Maths Chapter 1 Numbers Ex 1.1

1. Write the following numbers in wrods-
(i) 24056
(ii) 40009
(iii) 99999
(iv) 80511
(v) 67725
Solution.
(i) 24056 Twenty four thousand fifty six
(ii) 40009 Forty thousand nine
(iii) 99999 Ninety nine(RBSESolutions.com)thousand nine hundred ninety nine
(iv) 80511 Eighty thousand five hundred eleven
(v) 67725 Sixty seven thousand seven hundred twenty five

Question 2.
Write the expanded form of following numbers –
(i) 12372
(ii) 23434
(iii) 45302
(iv) 75004
(v) 68877
Solution.
(i) 12372 = 10000 + 2000 + 300 + 70 + 2
(ii) 23434 = 20000 + 3000 + 400 + 30 + 4
(iii) 45302 = 40000 + 5000 + 300 + 00 + 2
(iv) 75004 = 70000 + 5000 + 000 + 00 + 4
(v) 68877 = 60000 + 8000 + 800 + 70 + 7

Question 3.
Write the equivalent numbers of following expanded forms –
(i) 40000 + 5000 + 700 + 70 + 2
(ii) 60000 + 0000 + 000 + 20 + 6
(iii) 30000 + 9000 + 900 + 00 + 8
(iv) 50000 + 2000 + 800 + 10 + 1
(v) 80000 + 0000 + 000 + 00 + 8
Solution.
(i) 40000 + 5000 + 700 + 70 + 2 = 45772
(ii) 60000 + 0000 + 000 + 20 + 6 = 60026
(iii) 30000 + 9000 + 900 + 00 + 8 = 39908
(iv) 50000 + 2000 + 800 + 10 + 1 = 52811
(v) 80000 + 0000 + 000 + 00 + 8 = 80008

For larger numbers, it is easier to use exponents for the place values. Example B: … The 4 is in the thousandths place, so multiplying by 10-3 (1/1000).

Question 4.
Find out the place(RBSESolutions.com)value of 6 and 2 in the following numbers –
(i) 28506
(ii) 36265
(iii) 52266
(iv) 69242
(v) 82563
Solution.
(i) 28506 in
Place value of 6 = 6
Place value of 2 = 20000

(ii) 36265 in
Place value of 6 = 6000
Again Place value of 6 = 60
Place value of 2 = 200

(iii) 52266 in
Place value of 6 = 60
Again Place value of 6 = 6
Place value of2 = 2000
Again Place value of = 200

(iv) 69242 in
Place value of 6 = 60000
Place value of 2 = 200
AgainPlance value of 2 = 2

(v) 82563 in
Place value of 6 = 60
Place value of 2 = 2000

Question 5.
Mark the right symbol (<, >, =) between the given numbers
(i) 2979 ………….. 2932
(ii) 5423 ………….. 5432
(iii) 8952 ………….. 8952
(iv) 6850 ………….. 6852
(v) 3675 ………….. 3675
(vi) 9821 ………….. 9799
Solution.
(i) 2979 > 2932
(ii) 5423 < 5432
(iii) 8952 = 8952
(iv) 6850 < 6852
(v) 3675 = 3675
(vi) 9821 > 9799

Question 6.
Write the following(RBSESolutions.com)numbers in ascending order –
(i) 26886, 37725, 30840, 25975, 40021
(ii) 59307, 53907, 59703, 57039, 57903
(iii) 74443, 74434, 74344, 77444, 77555
Solution.
(i) 25975, 26886, 30840, 37725, 40021
(ii) 53907, 57039, 57903, 59307, 59703
(iii) 74344, 74434, 74443, 77444, 77555

Question 7.
Write the following numbers in descending order-
(i) 41525, 51425, 34152, 42325, 50925
(ii) 86067, 81316, 85032, 82511, 81154
(iii) 76543, 73456, 74356, 76435, 74653
Solution.
(i) 51425, 50925, 42325, 41525, 34152
(ii) 86067, 85032, 82511, 81316, 81154
(iii) 76543, 76435, 74653, 74356, 73456

We hope the RBSE Solutions for Class 5 Maths Chapter 1 Numbers Ex 1.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 1 Numbers Exercise 1.1, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 5 Maths Chapter 2 Addition and Subtraction In Text Exercise

June 2, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 5 Maths Chapter 2 Addition and Subtraction In Text Exercise is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 2 Addition and Subtraction In Text Exercise.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 5
Subject	Maths
Chapter	Chapter 2
Chapter Name	Addition and Subtraction
Exercise	In Text Exercise
Number of Questions	14
Category	RBSE Solutions

Rajasthan Board RBSE Class 5 Maths Chapter 2 Addition and Subtraction In Text Exercise

Page No. 7

Solve :

Question 6.
580 + 27 + 306 = …………
Solution.
9133

1/32 to Decimals are a set of numbers lying between integers on a number line.

Question 7.
473 – 296 = …………
Solution.
177

Question 8.
364+67 – 199 = 431 – 199 = …………
Solution.
232

Question 9.
37 – 15 + 10 = 47 – 15 = …………
Solution.
32

Question 10.
185 boys and 162 girls are studying in Government(RBSESolutions.com)Higher Secondary School, Udliyas. What is the total number of students studying ¡n this school?
Solution.

Question 11.
In a garden 255 roses, 156 marigold, 178 Jasmine(RBSESolutions.com)plants are there. What is the total number of flowering plants ?
Solution.

Question 12.
Gita’s father is at the mart to sell pots. This(RBSESolutions.com)time her father built 523 pots, of which 484 were sold out. Then how many pots are left unsold ?
Solution.

Question 13.
Do addition and(RBSESolutions.com)subtraction with Indian numbers.

Solution:

Page No. 8

Practice

Solution.

We hope the RBSE Solutions for Class 5 Maths Chapter 2 Addition and Subtraction In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 2 Addition and Subtraction In Text Exercise , drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area Ex 14.1

June 2, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area Ex 14.1 is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 14 Perimeter and Area Exercise 14.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 5
Subject	Maths
Chapter	Chapter 14
Chapter Name	Perimeter and Area
Exercise	Ex 14.1
Number of Questions	10
Category	RBSE Solutions

Rajasthan Board RBSE Class 5 Maths Chapter 14 Perimeter and Area Ex 14.1

Question 1.
Find out the perimeter of given shapes.

Solution.
(a) Perimeter of the shape = 10 + 7 + 10 + 8 + 7 + 7 + 9 + 7 = 65 cm.
(b) Perimeter of the shape = 10 + 5 + 4 + 5 + 5 + 4 + 5 = 38 cm.
(c) Perimeter of the shape = 4 × side = 4 × 7 = 28 cm.

Question 2.
Find out the perimeter of(RBSESolutions.com)rectangular shapes with given measurements :
(a) Length = 30 cm. , Width = 48 cm.
(b) Length=20 cm. , Width = 34 cm.
(c) Length = 60 cm. , Width = 20 cm.
(d) Length = 30 cm. , Width = 12 cm.
Solution.
(a) Perimeter = 2 × (Length + Width)
= 2 × (30 + 48)
= 2 × 78
= 156 cm.

(b) Perimeter = 2 × (Length + Width)
= 2 × (20 + 34)
= 2 × 54
= 108 cm.

(d) Perimeter = 2 × (Length + Width)
= 2 × (30 + 12)
= 2 × 42
= 84 cm.

Question 3.
Find out the perimeter of(RBSESolutions.com)following regular shapes with the help of formula.

Solution.
(a) Perimeter of triangle = 3 × side = 3 × 14 = 42 cm.
(b) Perimeter of square = 4 × side = 4 × 6 = 24 cm.
(c) Perimeter of Hexagon = 6 × side = 6 × 6 = 36 cm.
(d) Perimeter of Pentagon = 5 × side = 5 × 9 = 45 cm.

Question 4.
Vijay has made a rectangle. Find its perimeter and area?

Can you extend its length and width in such a way, so that it has equal perimeter and area. (There is no need to increase both length and width)
OR
A rectangle is shown in the picture. Find the (RBSESolutions.com)and the area of this rectangle. Can you increase or decrease its length and width in this way so that its perimeter and area become the same. If yes then write their measurements.

Solution.
Perimeter of a rectangle made by Vijay = 2 × (Length + Width)
= 2 × (5 + 3)
= 2 × 8
= 16 cm.
Area of rectangle = Length × Width = 5 × 3 = 15 square cm.
If length of this rectangle is decreased by 1 cm. and breadth increased by 1 cm.
then its measurements become.
Then perimeter and(RBSESolutions.com)area of this square become equal.
Perimeter of square = 4 × side = 4 × 4 = 16 cm.
Area of square = (side)²= (4)² = 16 sq.cm.

Question 5.
Length and width of a rectangular field are 25 meter and 30 meter respectively. Find its area?
Solution.
Area of rectangular field = Length × Width
= 25 × 30
= 750 square meter.

Question 6.
Length and width of a rectangular towel are 125 cm and 60 cm respectively. What will be the perimeter of the towel?
Solution.
Perimeter of(RBSESolutions.com)rectangular towel = 2 × (Length + Width)
= 2 × (125 + 60)
= 2 × 185
= 370 cm.

Question 7.
A square field needs 260 meter long barbed wire for throughout fencing. Find its one side?
Solution.
Perimeter of square = 4 × side

Question 8.
The length and width of a room floor is 8 meter and 7 meter respectively. In this room one(RBSESolutions.com)mat covers the entire floor. Find out the area of this mat.
Solution.
Length of floor = 8 meter
Width of floor = 7 meter
Area of floor = Area of mat
= Length × Width
= 8 × 7
= 56 square meter
Therefore Area of mat = 56 square meter.

Question 9.
Find out the perimeter of a square stool, whose side is 60 centimeter.
Solution.
Perimeter of square stool = 4 × side
= 4 × 60
= 240 centimeter
Therefore perimeter of stool = 240 centimeter.

Meter to Mile Calculator is an online tool that converts the value of a distance or length from meters to miles.

Question 10.
To take around of a square field, Dev had to walk 40 meters. Find the side of the square field.
Solution.
Distance walked in two round = 40 meter
∴ Distance walked (RBSESolutions.com)one round = \(\frac { 40 }{ 2 }\) = 20 meter
Distance walked in one round is called perimeter
Therefore perimeter of square field = 20 meter
∵ 4 × side = Perimeter

Therefore side of square field = 5 meter.

We hope the RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area Ex 14.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 14 Perimeter and Area Exercise 14.1, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 5 Maths Chapter 5 Multiples and Factors Ex 5.1

June 1, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 5 Maths Chapter 5 Multiples and Factors Ex 5.1 is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Exercise 5.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 5
Subject	Maths
Chapter	Chapter 5
Chapter Name	Multiples and Factors
Exercise	Ex 5.1
Number of Questions	10
Category	RBSE Solutions

Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Ex 5.1

Question 1.
Write down 4 multiples of given numbers-
(i) 4 – ………………….
(ii) 7 – ………………….
(iii) 14 – ………………….
(iv) 19 – ………………….
Solution.
(i) Multiples of 4 = 8, 12, 16, 20
(ii) Multiples of 7 = 14, 21, 28, 35
(iii) Multiples of 14 = 28, 42, 56, 70
(iv) Multiples of 19 = 38, 57, 76, 95

Question 2.
In circle only the(RBSESolutions.com)multiples of given number.

Solution.

Question 3.
Circle those numbers, which are common multiple of 3 and 4 –
6, 12, 15, 18, 24, 30
Solution.
In numbers 6, 12, 15, 18, 24, 30 the common multiples of 3 and 4 both are

Question 4.
Write the multiples of 7 which lie between 10 and 30.
Solution.
Multiples of 7 are 14, 21, 28.

Question 5.
Write three(RBSESolutions.com)multiples of 4, which are greater than 25.
Solution.
Multiples of 4 greater than 25 = 28, 32, 36.

Question 6.
Find out the least common multiple (LCM) of 2 and 5.
Solution.

Question 7.
Find out the least common multiple (LCM) of 8 and 12.
Solution.

Common multiples of 8 and 12 = 24,48,……..
Therefore least(RBSESolutions.com)common multiples of 8 and 12 = 24

To sum up, the lcm of 12 and 16 is 48.

Question 8.
Find out the least common multiple (LCM) of 6, 9 and 15.
Solution.
Least common multiple of 6, 9, 15 is –

Page No. 28

Question 1.
Factors of 6 – 1, 2, 3, 6
Factors of 8 – ……………..
Factors of 15 – ……………..
Solution.
Factors of 8 = 1, 2, 4, 8
Factors of 15 = 1, 3, 5, 15

Question 2.
Find out the factors of 9 and 27. Write their(RBSESolutions.com)common factors too and identify highest common factor (HCF).
Solution.

We hope the RBSE Solutions for Class 5 Maths Chapter 5 Multiples and Factors Ex 5.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Exercise 5.1, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 5 Maths Chapter 5 Multiples and Factors Ex 5.2

May 31, 2022 by Fazal Leave a Comment

RBSE Solutions for Class 5 Maths Chapter 5 Multiples and Factors Ex 5.2 is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Exercise 5.2.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 5
Subject	Maths
Chapter	Chapter 5
Chapter Name	Multiples and Factors
Exercise	Ex 5.2
Number of Questions	7
Category	RBSE Solutions

Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Ex 5.2

Find the factors of 18, 32, 28, 42, 45, 56 through division.

Question 1.
Find out the factors of given numbers.
(i) 7
(ii) 9
(iii) 16
(iv) 25
(v) 48
(vi) 63
Solution.
(i) Factors of 7 = 1, 7
(ii) Factors of 9= 1,3,9
(iii) Factors of 16 = 1, 2, 4, 8, 16
(iv) Factors of 25= 1, 5,25
(v) Factors of 48 = 1,2, 3, 4, 6, 8, 12, 16, 24, 48
(vi) Factors of 63 = 1, 3, 7, 9, 21, 63

Question 2.
Find out the factors of given numbers,
(i) 8 and 12
(ii) 10 and 20
(iii) 7 and 16
(iv) 18 and 32
Solution.
(i) 8 and 12
Factors of 8 = 1,2, 4, 8
Factors of 12 = 1,2, 3, 4, 6, 12
Therefore, common(RBSESolutions.com)factors of 8 and 12 = 1, 2, 4

(ii) 10 and 20
Factors of 10 = 1,2, 5, 10
Factors of 20 = 1, 2, 4, 5, 10, 20
Therefore, common factors of 10 and 20 = 1, 2, 5, 10

(iii) 7 and 16
Factors of 7 = 1 , 7
Factors of 16 = 1,2, 4, 8, 16
Therefore, common factors of 7 and 16 = 1

(iv) 18 and 32
Factors of 18 =1,2, 3, 6, 9, 18
Factors of 32 = 1,2, 4, 8, 16, 32
Therefore, common factors of 18 and 32 = 1, 2.

Question 3.
Find out the highest(RBSESolutions.com)common factors of 21 and 28.
Solution.

Therefore highest common factors of 21 and 28 = 7

Question 4.
Find out the highest number which can exactly divides 45 and 75.
Solution.
Factors of 45 = 1, 3, 5, 9, 15, 45.
Factors of 75 = 1, 3, 5, 15, 25, 75
Common factors of 45 and 75 = 1, 3, 5, 15
Highest common factor in factors = 15
Therefore 15 is the highest number which can exactly divides 45 and 75.

Question 5.
Find out the highest common common factors of 12,18 and 24.
Solution.
Factors of 12 = 1, 2, 3, 4, 6, 12
Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Common factors of 12, 18, 24 = 1, 2, 3, 6
Therefore highest(RBSESolutions.com)common factor of 12, 18 and 24 = 6

Question 6.
Find out the highest common factors of 15, 27 and 36.
Solution.
Factors of 15 = 1, 3, 5, 15
Factors of 27 = 1, 3, 9, 27
Facotors of 36 = 1, 2, 3, 4, 6, 12, 18, 36
Common factors of 15, 27 and 36 = 1,3
Therefore, highest common factor of 15, 27 and 36 = 3

Question 7.
Two milk containers filled with 20 litre and 30 litre of milk respectively. What will be the greatest(RBSESolutions.com)measurement of a container that can measure total milk of both container exactly ?
Solution.
Factors of 20 = 1, 2, 4, 5, 10, 20
Similarly factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
Highest common factor = 10
Therefore, greatest measurement of a container that can measure total milk of both container exactly is 10 litre.

We hope the RBSE Solutions for Class 5 Maths Chapter 5 Multiples and Factors Ex 5.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Exercise 5.2, drop a comment below and we will get back to you at the earliest.

Class 5

Rajasthan Board RBSE Class 5 Maths Chapter 1 Numbers Ex 1.1

Rajasthan Board RBSE Class 5 Maths Chapter 2 Addition and Subtraction In Text Exercise

Rajasthan Board RBSE Class 5 Maths Chapter 14 Perimeter and Area Ex 14.1

Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Ex 5.1

Rajasthan Board RBSE Class 5 Maths Chapter 5 Multiples and Factors Ex 5.2

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