Std 7 Mathematics Chapter :  3 HCF and LCM

Title : Std 7 Mathematics Chapter :  3 HCF and LCM

Maharashtra state board solutions for class 7 SSC board maths  STD 7-  HCF and LCM

 Know everything about HCF and LCM Maharashtra state board class 7 SSC .Get detailed Questions and answers for chapter3 maths.

Important question and answers

Introduction :

 Co-prime numbers :

• Two numbers which have only 1 as a common factor are said to be co-prime or relatively prime or mutually prime numbers.
• For example, 10 and 21 are co-primes, because the divisors of 10 are 1, 2, 5, 10 while the divisors of 21 are 1, 3, 7, 21 and the only factor common to both 10 and 21 is 1.
• Some other co-prime numbers are (3, 8) ; (4, 9); (21, 22) ; (22, 23) ; (23, 24).

1. Twin Prime Numbers

 If the difference between two co-prime numbers is 2, the numbers are said to be twin prime numbers.

For example : (3, 5) ; (5, 7) .

Q.1 Which number is neither a prime number nor a composite number?

 Solution : 1 is neither a primer number nor a composite number.

2. Factorising a number into its Prime Factors A simple but important rule given by Euclid is often used to find the GCD or HCF and LCM of numbers.

 The rule says that any composite number can be written as the product of prime numbers. To write a given number as a product of its prime factors is to factorise it into primes

 Q.1 Find Prime Factors of : 20

Solution : 20= 2 x 2 x 5

Q.2 Find Prime factors of : 40

 Solution : 2 x 2 x 2 x 5 3. Greatest Common Divisor (GCD) or Highest Common Factor (HCF) HCF or the GCD of given numbers is their greatest common divisor or factor. If one of the given numbers is a divisor of all the others, then it is the HCF of the given numbers. If no prime number is a common divisor of all the given numbers, then 1 is their HCF because it is the only common divisor. 2 is the HCF of any two consecutive even numbers and 1 is the HCF of any two consecutive odd numbers.

 **Q.1 Find the HCF of 24 and 32 by the prime factors method

. Solution: 24 = 4 × 6 = 2 × 2 × 2 × 3 32 = 8 × 4 = 2 × 2 × 2 × 2 × 2 The common factor 2 occurs thrice in each number. Therefore, the HCF = 2×2×2 = 8

 **Q.2 Find the HCF of 195, 312, 546.

 Solution : 195= 5 x 39 =5 x 3 x 13 312= 4 x 78 = 2 x 2 x 2 x 39 = 2 x 2 x 2 x 3 x 13 546= 2 x 373 = 2 x 3 x 91 = 2 x 3 x 7 x 13 The common factors 3 and 13 each occur once in all the numbers. ∴ HCF = 3 × 13 = 39 Example Find the HCF of 195, 312, 546.

The Division Method for Finding the HCF Steps :

 Divide the bigger number by the smaller one. Divide the previous divisor by the remainder in this division.

 Divide the divisor of step 2 by the remainder obtained in the division in step 2.

3. Continue like this till the remainder becomes zero. The divisor in the division in which the remainder is zero is the HCF of the given number

*Q.1 Find the HCF of 144 and 252.

1. Divide the bigger number by the smaller one.

2. Divide the previous divisor by the remainder in this division.

 3. Divide the divisor of step 2 by the remainder obtained in the division in step 2.

4. Continue like this till the remainder becomes zero

. The divisor in the division in which the remainder is zero is the HCF of the given numbers. ∴ The HCF of 144 and 252 = 36