Maharashtra Board Class 5 Math Solution Chapter 8 – Multiples and Factors
Balbharati Maharashtra Board Class 5 Math Solution Chapter 8: Multiples and Factors. Marathi or English Medium Students of Class 5 get here Multiples and Factors full Exercise Solution.
Std |
Maharashtra Class 5 |
Subject |
Math Solution |
Chapter |
Multiples and Factors |
PROBLEM SET 32
Write the factors of the following numbers.
(1) factors of 8 are – 1,2,4,8
(2) factors of 5 are – 1,5
(3) factors of 14 are – 1,2,7,14
(4) factors of 10 are – 1,2,5,10
(5) factors of 7 are – 1,7
(6) factors of 22 are – 1,2,11,22
(7) factors of 25 are – 1,5,25
(8) factors of 32 are – 1,2,4,8,16,32
(9) factors of 33 are – 1,3,11,33
PROBLEM SET 33
(1) Write five three-digit numbers that are multiples of 2.
If there is 0, 2, 4, 6 or 8 in the units place, the number is a multiple of 2, or is exactly divisible by 2.
So, the five three-digit numbers that are multiples of 2 are – 100, 122, 154, 186, 208
(2) Write five three-digit numbers that are multiples of 5.
Any number with 5 or 0 in the units place is a multiple of 5 or, is divisible by 5.
So, the five three-digit numbers that are multiples of 5 are – 100, 125, 150, 205, 250
(3) Write five three-digit numbers that are multiples of 10.
Any number that has 0 in the units place is a multiple of 10.
So, the five three-digit numbers that are multiples of 10 are – 100, 200, 250, 360, 430
2.) Write 5 numbers that are multiples of 2 as well as of 3.
6, 12, 18, 66, 24 are multiples of 2 as well as of 3.
3.) A ribbon is 3 metres long. Can we cut it into 50 cm pieces and have nothing left over ?
Write the reason why or why not.
The ribbon is 3 meters long.
1 meter = 100 cm
So, 3 meter = 3 × 100 = 300 cm
We know any number with 5 or 0 in the units place is a multiple of 5 or, is divisible by 5.
As 300 cm also has a 0 in the unit place it is divisible by 5.
So we can cut it and have nothing left over.
4.) A ribbon is 3 metres long. I need 8 pieces of ribbon each 40 cm long. How many centimetres shorter is the ribbon than the length I need ?
The ribbon is 3 meters long.
1 meter = 100 cm
So, 3 meter = 3 × 100 = 300 cm
we need 8 pieces of ribbon 40 cm each, so total length we need = 8 × 40cm = 320 cm
subtracting the ribbon length we have from the the ribbon length we need = ( 320 – 300 ) = 20 cm
answer = the ribbon is 20 cm shorter than the length we need.
5.) If the number given in the table is divisible by the given divisor, put ✓ in the box. If it is not divisible by the divisor, put × in the box.
Divisor
Number |
2 | 5 | 10 |
15 | × | ✓ | × |
30 | ✓ | ✓ | |
34 | ✓ | × | × |
46 | ✓ | × | × |
55 | × | ✓ | × |
63 | × | × | × |
70 | ✓ | ✓ | ✓ |
84 | ✓ | × | × |
PROBLEM SET 33
- ) Write all the prime numbers between 1 and 20.
A number which has only two factors, 1 and the number itself, is called a prime number.
So the prime numbers between 1 and 20 are – 2, 3, 5, 7, 11, 13, 17, 19
2.) Write all the composite numbers between 21 and 50.
A number which has more than two factors is called a composite number.
So, the composite numbers between 21 and 50 are – 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50
3.) Circle the prime numbers in the list given below.
1, 37, 43, 53, 91, 57, 59, 79, 97
4.) Which of the prime numbers are even numbers ?
The only even prime number is 2.
PROBLEM SET 35
Determine whether the pairs of numbers given below are co-prime numbers.
(1) 22,24
Factors of 22 – 1,2,11,22
Factors of 24 – 1,2,3,4,6,8,12,24
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1,2 as common factors, so they are not co-prime numbers.
(2) 14,21
Factors of 14 – 1,2,7,14
Factors of 21 – 1,3,7,21
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1,7 as common factors, so they are not co-prime numbers.
(3) 10,33
Factors of 10 – 1, 2, 5, 10
Factors of 33 – 1, 3, 11, 33
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1 as common factor, so they are co-prime numbers.
(4) 11,30
Factors of 11 – 1, 11
Factors of 30 – 1, 2, 5, 10, 15, 30
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1 as common factor , so they are co-prime numbers.
(5) 5,7
Factors of 5 – 1, 5
Factors of 7 – 1, 7
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1 as common factor, so they are co-prime numbers.
(6) 15,16
Factors of 15 – 1, 3, 5, 15
Factors of 16 – 1,2,4,8,16
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1 as common factor, so they are co-prime numbers.
(7) 50,52
Factors of 50 – 1, 2, 5, 10, 25, 50
Factors of 52 – 1, 2, 4, 13, 26, 52
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1,2 as common factors, so they are not co-prime numbers.
(8) 17,18
Factors of 17 – 1, 17
Factors of 18 – 1, 2, 3, 6, 9, 18
Numbers which have only 1 as a common factor are called co-prime numbers,
So, we can see they have 1 as common factor, so they are co-prime numbers.
For More Solutions, Click Below:
Part One:
- Chapter 1 Roman Numerals
- Chapter 2 Number Work
- Chapter 3 Addition and Subtraction
- Chapter 4 Multiplication and Division
- Chapter 5 Fractions
- Chapter 6 Angles
- Chapter 7 Circles
Part Two:
- Chapter 9 Decimal Fractions
- Chapter 10 Measuring Time
- Chapter 11 Problems and Measurement
- Chapter 12 Perimeter and Area
- Chapter 15 Patterns
- Chapter 16 Preparations for Algebra