Maharashtra Board Class 5 Math Chapter 8 Multiples and Factors Solution

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:

Maharashtra Board Class 5 Math Solution Chapter 8 – Multiples and Factors

PROBLEM SET 32

PROBLEM SET 33

PROBLEM SET 33

PROBLEM SET 35

For More Solutions, Click Below:

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