## Tuesday, 2 May 2017

### CBSE Class 6 Maths - Playing With Numbers - NCERT Exercise 3.6 (#cbsenotes)

Playing With Numbers

## EXERCISE 3.6

CBSE Class 6 Maths

Q1: Find the HCF of the following numbers:
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75

(a) Factors of 18 = 2 × 3 × 3
Factors of 48 = 2 × 2 × 2 × 2 × 3
HCF (18, 48) = 2 × 3 = 6

(b) Factors of 30 = 2 × 3 × 5
Factors of 42 = 2 × 3 × 7
HCF (30, 42) = 2 × 3 = 6

(c) Factors of 18 = 2 × 3 × 3
Factors of 60 = 2 × 2 × 3 × 5
HCF (18, 60) = 2 × 3 = 6

(d) Factors of 27 = 3 × 3 × 3
Factors of 63 = 3 × 3 × 7
HCF (27, 63) = 3 × 3 = 9

(e) Factors of 36 = 2 × 2 × 3 × 3
Factors of 84 = 2 × 2 × 3 × 7
HCF (36, 84) = 2 × 2 × 3 = 12

(f) Factors of 34 = 2 × 17
Factors of 102 = 2 × 3 × 17
HCF (34, 102) = 2 × 17 = 34

Q2: What is the HCF of two consecutive
(a) numbers?
(b) even numbers?
(c) odd numbers?

(a) HCF of two consecutive numbers be 1.
(b) HCF of two consecutive even numbers be 2.
(c) HCF of two consecutive odd numbers be 1.

Q3: HCF of co-prime numbers 4 and 15 was found as follows by factorisation :
4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Answer: It is incorrect. Since there is no common factor, the correct HCF is 1.

Other Questions on HCF

Q4: State True (T) or FALSE (F)
The Highest Common Factor of two or more numbers is greater than their Lowest Common Multiple.

HCF of two or more numbers may be one of the numbers.

Any two consecutive numbers are coprime.

The HCF of two numbers is smaller than the smaller of the numbers.

False
True
True
False

Q5: Find the largest number that divides 92 and 74 leaving 2 as remainder.

Answer: Number divides 92 and leaves remainder 2.
∴ The number divides 90 ( = 92 - 2) completely.
The number also divides 74 leaving 2 as remainder
∴ The number divides 72 ( = 74 - 2) completely.

Factors of 90 = 3 × 3 × 5 × 2
Factors of 72 = 3 × 3 × 2 × 2 × 2
HCF of (90, 72) is = 3 × 3 × 2 = 18

∴ The required number is 18.  