WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Highest Common Factor And Lowest Common Multiple Of Integers

Chapter 1 Simplification Highest Common Factor And Lowest Common Multiple Of Integers

Question 1. 

1. Write down 2 numbers of two digits which are multiples of 4

Solution:

Two multiples of 4 of two digits are 12 and 16, because 4 × 3 = 12 and 4 × 4 = 16.

2. Write 6 multiples of 5 except 0;

Solution:

6 multiples of 5 except 0 are 5 × 15, 5 × 2 = 10, 5 × 3 = 15, 5 × 4 = 20, 5 × 5, 25 and 5 × 6 = 30.

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

3. Write two numbers whose L. C. M. is 12 and whose sum is 10;

Solution:

Two numbers are 4 and 6; because of the L. C. M. of 4 and 6 = 12 and whose sum 4+6= 10.

(Here 4 2 × 2 and 6 = 2 x 3. .. L. C. M.2 × 2 × 3 = 12).

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Highest Common Factor And Lowest Common Multiple Of Integers

WBBSE Class 6 HCF and LCM Simplification Notes

4. Write three numbers each of which has a factor of 4

Solution:

The required 3 numbers are 8, 12, and 16 because each of these numbers has a factor of 4.

5. Write three multiples of 7 greater than 50.

Solution:

Three multiples of 7 greater than 50 are 56, 63, and 70 because 56 > 50; 63 50; 70 > 50

56 = 7 × 8,

63 = 7 × 9, 

70 = 7 × 10.

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Question 2.

1. Which is the smallest even prime number?

Solution:

The smallest even prime number is 2.

2. Which is the smallest odd prime number?

Solution:

The smallest odd prime number is 3.

3. What are the prime factors of 14?

Solution:

We know that 14= 1 × 2 × 7.

The factors of 14 are 1, 2, 7, and 14.

The prime factors of 14 are 2 and 7.

Short Questions on HCF and LCM

Question 3. 42 is the multiple of which of the following numbers? 

1. 5

2. 6

3. 7

4. 13.

Solution:

42 is divisible by both the numbers 6 and 7 but 42 is not divisible by 5 or 13.

So, 42 is the multiple of 6 and 7.

42 is not a multiple of 5. and 13.

Question 4. 11 is a factor of which of the following numbers?

1. 101

2. 111

3. 121

4. 112.

Solution: We know that 121 = 11 × 11.

∴ 11 is a factor of 121. 

Again the numbers 101, 111, and 112 are not divisible by 11.

So 11 is not a factor of 101, 111, or 112.

Question 5.

Find the H.C.F. by the resolution into prime factors of the numbers in each of the following cases:

1. 22, 44

Solution:

22 = 1 × 2 × 11

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 1

44 = 1 × 2 × 2 × 11

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 2

The factors of 22 are 1, 2, 11, 22 and that of 44 are 1, 2, 4, 11, 22, 44

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 3

∴ The common factors of 22 and 44 are 1, 2, 11, and 22.

Among these factors, the highest factor is 22

∴ The required H. C. F. = 22.

2. 54, 72.

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 1

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 2

∴ So the factors of 54 are 1, 2, 3, 6,9, 18, 27, and 54.

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 3

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 5 Q 4

∴ So the factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

∴ The common factors of 54 and 72 are 1, 2, 3, 6, 9, and 18.

Among these factors, the highest factor is 18.

∴ The required H. C. F. = 18.

Common Questions About Simplifying Fractions Using HCF

Question 6. Find the H. C. F. of 36 and 48 by resolution into factors.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 6 Q 1

∴ 36 = 2 × 2 × 3 × 3

∴ 48 = 2 × 2 × 2 × 2 × 3

Now

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 6 Q 2

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 6 Q 3

∴ The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

The common factors of 36 and 48 are 1, 2, 3, 4, 6, and 12.

Among these factors, the highest factor is 12.

∴ The required H. C. F. = 12.

Alternative Method:

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 6 Q 4

v

Here

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 6 Q 6 markings are common to both 36 and 48.

∴ The required H.C.F = 1 × 2 × 2× 3 =12.

Practice Problems on HCF and LCM

Question 7. Find the H. C. F. of 75 and 105 by factorization.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 7 Q 1

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 7 Q 2

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 7 Q 3

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 7 Q 4

Here

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question Q 7 Q 5 marking is common to both 75 and 105.

∴ the required H.C.F = 1 × 3 × 5 = 15.

Question 8. Write two numbers whose H. C. F. is 7.

 Solution : 7 × 2 = 14; 7 × 3 = 21

2 and 3 are prime to each other, and the H. C. F. of 14 and 21 is 7.

The required two numbers are 14 and 21.

N.B. We can get an infinite number of such two numbers.

Question 9. By division method find the H. C. F. in the following cases:

1. 28, 35

Solution: 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 9 Q 1

 

∴ the required H.C.F. = 7

2. 54, 72

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 9 Q 2

 

∴ The required H.C.f. = 18.

Examples of Real-Life Applications of HCF and LCM

Question 10. By division method, find the H. C. F. of 90 and 144.

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 10

 

∴ The required H.C.F. = 18.

Question 11. What will be the greatest number by which 45 and 60 will be exactly divisible so that there will be no remainder in each case?

Solution:

The required greatest number will be the H. C. F. of 45 and 60.

Now,

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 11

 

∴ The H. C. F. of 45 and 60 = 15. So the required greatest number is 15.

Question 12.

1. The H. C. F. of two numbers is 1. How many such two numbers may exist? Find one pair of such numbers. What conclusion can you draw about such two numbers?

Solution:

There are infinite numbers of two numbers that have the H. C. F. 1. For example 2, 3; 5, 7; 9, 11, etc.

The required numbers are 9 and 11.

The numbers which have the H. C. F. 1 are prime to each other.

Conceptual Questions on Using HCF in Fraction Simplification

2. How many greatest number of persons can be distributed equally two kinds of sweets: 48 sweets of one kind and 64 sweets of another kind, without breaking the sweets?

Solution:

The required number of persons will be the H. C. F. of 48 and 64.

Now,

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 12

H.C. F. of 48 and 64 = 16

So the required greatest number of people is 16.

Question 13. Find the L. C. M. by factorization in each of the following cases:

 1. 25, 80

Solution:

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 13 Q 1

∴ 25 = 1 x 5 x 5

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 13 Q 2

∴ The required L.C.M = 1 x 5 x 5 x 2 x 2 x 2 x 2

= 400.

2. 36, 39

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 13 Q 3

∴ 36 = 2 x 2 x 3 x 3

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 13 Q 4

∴ 39 = 3 x 13

∴ The required L. C. M. = 2 x 2 x 3 x 3 x 13

= 468

Question 14. Find the L. C. M. by prime factors in each of the following cases:

1. 33, 132

Solution:

∴ 33 = 3 x 11

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 14 Q 2

∴ 132 = 2 x 2 x 3 x 11

∴ The required L. C. M. = 3 x 11 x 2 x 2 = 132.

Real-Life Scenarios Involving Equal Distribution Using HCF

2. 90, 144

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 14 Q 3

∴ 90 = 2 x 3 x 3 x 5

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 14 Q 4

∴ 144 = 2 x 2 x 2 x 2 x 3

 The required L. C. M. = 2 x 3 x 3 x 5 x 2 x 2 x 2 = 720.

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