AP State Syllabus AP Board 6th Class Maths Solutions Chapter 3 HCF and LCM Ex 3.7 Textbook Questions and Answers.

## AP State Syllabus 6th Class Maths Solutions 3rd Lesson HCF and LCM Ex 3.7

Question 1.

Find the LCM and HCF of the following numbers and check their relationship.

i) 15, 24

ii) 8, 25

iii) 12, 48

iv) 30, 48

Answer:

i) 15,24

The given numbers are 15 and 24.

LCM of 15 and 24 = 2 × 2 × 2 × 3 × 5 = 120

HCF of 15 and 24 = 3

Now, LCM × HCF = 120 × 3 = 360

Product of two numbers = 15 × 24 = 360,

∴ LCM × HCF = Product of the numbers.

ii) 8,25

The given numbers are 8 and 25.

LCM of 8 and 25 = 2 × 2 × 2 × 5 × 5 = 200

HCF of 8 and 25 = 1

Now LCM × HCF = 200 × 1 = 200

Product of two numbers = 8 × 25 = 200

∴ LCM × HCF = Product of the numbers

iii) 12, 48

The given numbers are 12 and 48.

LCM of 12 and 48 = 2 × 2 × 2 × 2 × 3 = 48

HCF of 12 and 48 = 12

Now LCM × HCF = 48 × 12 = 576

Product of two numbers = 12 × 48 = 576

∴ LCM × HCF = Product of the numbers

iv) 30, 48

Given numbers are 30 and 48.

LCM of 30 and 48 = 2 × 2 × 2 × 2 × 3 × 5 = 240

HCF of 30 and 48 = 6

LCM × HCF = 240 × 6 = 1440

Product of the numbers = 30 × 48 = 1440

∴ LCM × HCF = Product of the numbers.

Question 2.

If the LCM of two numbers is 290 and their product is 7250, what will be its HCF?

Answer:

Given LCM of two numbers = 290

Product of two numbers = 7250

HCF of two numbers = ?

We know that, LCM × HCF = Product of two numbers

290 × HCF = 7250

HCF = \(\frac{7250}{290}\) = 25

∴ HCF of two numbers = 25

Question 3.

The product of two numbers is 3276. If their HCF is 6, find their LCM.

Answer:

Given HCF of two numbers = 6

Product of two numbers = 3276

LCM of two numbers = ?

We know that, LCM × HCF = Product of two numbers

LCM × 6 = 3276

LCM = \(\frac{3276}{6}\) = 546

∴ LCM of two numbers = 546

Question 4.

The HCF of two numbers is 6 and their LCM is 36. If one of the numbers is 12, find the other.

Answer:

Given LCM of two numbers = 36

HCF of two numbers = 6

One of the numbers a = 12

Let second number b = ?

We know that, product of two numbers = LCM × HCF

a × b = LCM × HCF

12 × b = 36 × 6

b = \(\frac{36 × 6}{12}\) = 18

∴ Second number b = 18

Question 5.

Can two numbers have 16 as their HCF and 384 as their LCM? Give reason.

Answer:

Here, if we divide 384 by 16

\(\frac{384}{16}\) = 24 (No remainder)

If HCF is the factor LCM of two numbers, then it is possible.

Question 6.

Can two numbers have 14 as their HCF and 204 as their LCM? Give reasons in support of your answer.

Answer:

Here, if we divide 204 by 14

\(\frac{204}{14}\) = 14 and the remainder is 8.

HCF is not the factor of LCM to two numbers. So, it is not possible.