NCERT Solutions For Class 10 Maths Chapter 1 Real Numbers
EXERCISE 1.1
2. Show that any positive odd integer is of form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
SolutionLet `a` be the positive integer and `b` = 6
Applying Euclid's division lemma
`a = 6q + r, 0 ≤ r < 6`
Case I If `r = 0` then `a = 6q`
Case II If `r = 1` then `a = 6q +1`
Case III If `r = 2` then `a = 6q +2`
Case IV If `r = 3` then `a = 6q + 3`
Case V If `r = 4` then `a = 6q + 4`
Case VI If `r = 5` then `a = 6q + 5`
`therefore` Case II `a = 6q +1`, Case IV `a = 6q + 3` , Case VI `a = 6q + 5` are odd integers.
4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form `3m` or `3m + 1` for some integer `m.`
Solution
Let `a` be the positive integer and `b = 3`
Applying Euclid's division lemma
`a = 3q + r , 0 ≤ r < 3`
Case I If `r = 0` then `a = 3q`
`a^2 = (3q)^2 = 9q^2 = 3(3q^2) = 3m` where `m = 3q^2`
Case II If `r = 1` then `a = 3q +1`
`a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3q(3q +2) + 1 = 3m + 1`, where `m = q(3q + 2)`
Case III If `r = 2` then `a = 3q +2`
`a^2 = (3q + 2)^2 = 9q^2 + 12q + 4`
`= 9q^2 + 12q + 3 + 1`
`= 3(3q^2 + 4q + 1) + 1 =3m + 1`, where `m = 3q^2 + 4q + 1`
Hence, the square of any positive integer is either of the form `3m` or `3m + 1` for some integer `m.`
EXERCISE 1.2
1. Express each number as a product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Solutions:
(i) 140
By prime factorisation, we get
`140 = 2^2 × 5 × 7`
(ii) 156
By prime factorisation, we get
`156 = 2^2 × 3 × 13`
(iii) 3825
By prime factorisation, we get
`3825 = 3^2 × 5^2 × 17`
(iv) 5005
By prime factorisation, we get
`5005 = 5 × 7 × 11 × 13`
(v) 7429
By prime factorisation, we get
`7429 = 17 × 19 × 23`
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Solutions:
(i) 26 and 91
By prime factorisation, we get
`26 = 2 × 13`
`91 = 7 × 13`
`therefore LCM (26, 91) = 2 × 7 × 13 = 182`
`HCF (26, 91) = 13`
`HCF(a, b) × LCM (a, b) = a × b`
`HCF(26, 91) × LCM (26, 91) = 26 × 91`
`13 × 182 = 26 × 91 = 2366`
Hence, verified.
(ii) 510 and 92
By prime factorisation, we get
`510 = 2 × 3 × 5 × 17`
`92 = 2^2 × 23`
`LCM (92, 510) = 2^2 × 3 × 5 × 17 × 23 = 23460`
`HCF (92, 510) = 2`
`HCF(a, b) × LCM (a, b) = a × b`
`HCF(92, 510) × LCM (92, 510) = 92 × 510`
`2 × 23460 = 92 × 510 =46920`
Hence, verified.
(iii) 336 and 54
By prime factorisation, we get
`336 = 2^4 × 3 × 7`
`54 = 2 × 3^3`
`LCM (54, 336) = 2^4 × 36^3 × 7 = 3024`
`HCF (54, 336) = 2× 3 = 6`
`HCF(a, b) × LCM (a, b) = a × b`
`HCF(54, 336) × LCM (54, 336) = 6 × 3024`
`6 × 3024 = 54 × 336 = 18144`
Hence, verified.
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Solutions:
(i) 12, 15 and 21
`12 = 2^2 × 3`
`15 = 3 × 5`
`21 = 3 × 7`
`therefore LCM (12, 15, 21) = 2^2 × 3 × 5 × 7 = 420`
`HCF (12, 15, 21) = 3`
Work in progress
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