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

Answer:

(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?

Answer:
(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.


Answer:
⒜ 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.