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

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

Question 1. The L. C. M. and H. C. F. of the two numbers are 252 and 6 respectively. What is the product of these two numbers?

Solution:

Given:

The L. C. M. and H. C. F. of the two numbers are 252 and 6 respectively.

 We know that, the product of two numbers = L. C. M. x H. C. F.

 = 252 × 6 

= 1512.

∴ The required product = 1512.

WBBSE Class 6 HCF and LCM Simplification Notes

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

Question 2. The H. C. F. and L. C. M. of the two numbers are 8 and 280 respectively; if one of them is 56, then what is the other number?

Solution:

Given:

The H. C. F. and L. C. M. of the two numbers are 8 and 280 respectively; if one of them is 56

We know that, the product of two numbers = H. C. F. x L. C. M.

Here the product of two numbers = 8 × 280.

One number = 56. 

∴ The other number = \(\frac{8 x 280}{56}

= 40

∴ The required other number = is 40.

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

Short Questions on HCF and LCM Problems

Class 6 Math Solution WBBSE

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Question 3. The circumference of the front wheel of a rail engine is 140 cm and that of the rear wheel is 350 cm. Find the least distance to be covered so that both wheels complete their full rotation simultaneously.

Solution:

Given:

The circumference of the front wheel of a rail engine is 140 cm and that of the rear wheel is 350 cm.

The L. C. M. of 140 cm and 350 cm will be the required distance.

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 17 Q 1

∴ 140 = 2 x 2 x 5 x 7

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 17 Q 2

∴ 350 = 2 x 5 x 5 x 7

The required L. C. M. of 140 and 350 = 2 x 2 x 5 x 7 x 5

= 700

So the required distance = 700 cm

= 70 dm.

Class 6 Math Solution WBBSE

Question 4.

1. With the help of H. C. F. of 45 and 60, find their L. C. M.

Solution:

Given:

H. C. F. of 45 and 60

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 18 Q 1

 

∴ The H.C.F. of 45, 60 = 15

Now, the product of numbers = H.C.F x L.C.M

∴ 45 x 60 = 15 x L.C.M

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 18 Q 2

 

 2. With the help of L. C. M. of 105 and 225, find their H. C. F.

Solution:

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 18 Q 3

 

∴ 105 = 3 x 5 x 7

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 18 Q 4

Common Questions About Finding HCF and LCM

∴ 225 = 3 x 3 x 5 x 5

L.C. M. of 105 and 225 = 3 x 5 x 7 x 3 x 5 = 1575 

Now, the product of the numbers = H. C. F. x L. C. M.

or, 105 x 225 = H. C. F. x 1575

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 18 Q 5

 

∴ The required H.C.F of 150 and 225 = 15.

Question 5. Find the H. C. F. of 24, 33, and 130.

Solution:

Given:

The H. C. F. of 24, 33, and 130

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 19 Q 1

∴ 24 = 1 x 2 x 2 x 2 x 3

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 19 Q 2

 

∴ 33 = 3 x 11

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 19 Q 3

 

∴ 130 = 1 x 2 x 5 x 13

Here 33 and 130 are prime to each other.

The given numbers have no common factor except 1.

Hence the required H. C. F. is 1.

Practice Problems on HCF and LCM of Integers

Question 6. Examine whether 278 and 365 are prime to each other.

Solution:

Given:

278 and 365

Here we shall find the H. C. F. of the given numbers by the method of division

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 20WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 20

 

∴ The H. C. F. of 278 and 365 is 1. Hence 278 and 365 are prime to each other.

Question 7. Find the H. C. F of 906, 1057, and 1510 by the division method.

Solution:

Given:

906, 1057, and 1510

First, we find the H. C. F. of 906 and 1057.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 21 Q 1

 

∴ The H. C. F. of 906 and 1057 is 151.

Now we shall find H. C. F. of 151 and 1510.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 21 Q 2

 

∴ The H. C. F. of 151 and 1510 = 151.

Hence the required H. C. F. of 906, 1057, and 1510 = 151.

Question 8. Find the H. C. F. of 84, 112, 140.

Solution:

Given:

84, 112, 140.

Simplification For Class 6

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 22

 

Here we divide all the given numbers by their common prime factor 2 in the 1st row.

The quotients obtained are written in the 2nd row and divided into all the numbers in the 2nd row by their common prime factor 2 and the quotients are written in the 3rd row.

They are also divided by their common factor 7 and the quotients are written in the 4th row.

But the numbers in the 4th row have no common factor except 1 and the process is now completed.

∴ The required H. C. F. = product of the common prime factors

= 2 × 2 × 7 = 28.

The required H. C. F = 28.

Examples of Real-Life Applications of HCF and LCM

Question 9. Find the L. C. M. of 28, 35, 63, 84, 96.

Solution:

Given:

28, 35, 63, 84, 96

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 23

 

∴ The required L.C.M = 2 x 2 x 3 x 7 x 5 x 3 x 8

= 10080.

The required L.C.M = 10080.

Question 10

1. Find the least number which is exactly divisible by 18, 24, and 42.

Solution:

Given:

18, 24, and 42

Find the least number which is exactly divisible by 18, 24, and 42.

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 24 Q 1

 

 The L. C. M. of 18, 24, 42 = 2 x 3 x 3 x 4 x7 = 504.

∴ The required least number = 504.

2. Find the least number which is exactly divisible by 18, 39, 56, and 64.

Solution:

Given:

18, 39, 56, and 64

The required least number will be the L. C. M. of 18, 39, 56, and 64. 

Now,

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 24 Q 2

Simplification For Class 6

L.C.M. of 18, 39, 56 and 64 = 2 x 2 x 2 x 3 x 3 x 13 x 7 x 8 = 52416.

∴ The required least number = 52416.

Question 11.

The honorable Education Minister of the West Bengal Government has sent some books on Mathematics, Physical Science, and Life Science for class VI of Raghunath Vidyamandir. These books can be arranged in the school library in 20, 24, and 30 rows so that each row contains an equal number of books. How many least number of books has the minister sent?

Solution:

Given:

The honorable Education Minister of the West Bengal Government has sent some books on Mathematics, Physical Science, and Life Science for class VI of Raghunath Vidyamandir. These books can be arranged in the school library in 20, 24, and 30 rows so that each row contains an equal number of books.

Since each row contains an equal number of books the required number of books will be the L. C. M. of 20, 24, 30.

Now,

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 25
∴ L.C.M. of 20, 24, 30 = 2 x 2 x 5 x 3 x 2
= 120.
∴ The honorable Education minister has sent at least 120 books.
Conceptual Questions on Using LCM in Problem Solving

Question 12.

Find the H. C. F. and L. C. M. of 35, 45, and 50. Examine whether the product of the numbers and the product of their H. C. F. and L. C. M. are equal or not.

Solution:

Given:

35, 45, and 50

 

WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Question 26

 

∴ H. C. E. of 35, 45, 50 = 5 and their L. C. M. = 5 x 7 x 9 x 10 = 3150. Product of H. C. F. and L. C. M. = 5 x 3150 = 15750.

Again the product of the numbers 35 x 45 x 50 = 78750

∴ Product of the numbers ≠ product of H.C.F. and L.C.M.

Question 13.

The sum of the two numbers is 150 and their H. C. F. is 15. Find the numbers.

Solution:

Since the H. C. F. of two numbers is 15, the numbers must be multiples of 15.

Let the numbers be 15x and 15 y where x and y are primes to each other.

∴ Their sum = 15x + 15y= 15 (x + y)

∴ By the given condition, we get, 15 (x + y) = 150

Dividing both sides by 15, we get, x + y = [latex]\frac{150}{15}\) = 10.

Now,

10 = 1+9=2+8=3+7=4+6=5+5.

Among them, 1, and 9 are prime to each other, and 3, and 7 are also prime to each other. 

So the numbers are either 15 x 1 = 15; 15 x 9 = 135

or, 15 x 3 = 45; 15 x 7 = 105 . 

∴ The required numbers are either 15, 135, or 45, 105.

Real-Life Scenarios Involving Equal Distribution Using HCF

Question 14.

If the H. C. F. and L. C. M. of two numbers be 5 and 75 respectively, then find the numbers.

Solution:

Since the H. C. F. of two numbers is 5, let the numbers be 5x and 5y, where x and y are primes to each other.

Then L. C. M. of 5x and 5y = 5xy. But the L. C. M. is 75.

∴ 5xy = 75.

Now dividing both sides by 5, we get, xy= 15.

But 151 x 15 = 3 x 5. Both the sets of numbers 1, 15, and 3, 5 are prime to each other.

∴ The numbers are either 5 x 1, 5 x 15 = 5, 75.

or, 5 x 3, 5 x 5 = 15, 25.

∴ The required numbers are either 5, 75

or 15, 25.

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