Question 1(i)
Write 140 as a product of prime factors.
Steps only
140 = 2 × 2 × 5 × 7
Answer
140 = 2 × 2 × 5 × 7
CLASS 10 · MATHEMATICS · CHAPTER 1
Real Numbers
Work through prime factorisation, HCF, LCM and proofs of irrationality. Short working comes first, with explanations of each method available beside the answer.
21 worked answersAll 2 exercise sets
21 worked answers · short steps first, explanations when you need them
Exercise 1.1
Textbook pages 5–6
Question 1(ii)
Write 156 as a product of prime factors.
Steps only
156 = 2 × 2 × 3 × 13
Answer
156 = 2 × 2 × 3 × 13
Question 1(iii)
Write 3825 as a product of prime factors.
Steps only
3825 = 3 × 3 × 5 × 5 × 17
Answer
3825 = 3 × 3 × 5 × 5 × 17
Question 1(iv)
Write 5005 as a product of prime factors.
Steps only
5005 = 5 × 7 × 11 × 13
Answer
5005 = 5 × 7 × 11 × 13
Question 1(v)
Write 7429 as a product of prime factors.
Steps only
7429 = 17 × 19 × 23
Answer
7429 = 17 × 19 × 23
Question 2(i)
Find the HCF and LCM of 26 and 91, then check their product.
Steps only
26 = 2 × 13
91 = 7 × 13
HCF = 13; LCM = 182
13 × 182 = 26 × 91 = 2366
Answer
HCF = 13; LCM = 182.
Question 2(ii)
Find the HCF and LCM of 510 and 92, then check their product.
Steps only
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2; LCM = 23460
2 × 23460 = 510 × 92 = 46920
Answer
HCF = 2; LCM = 23460.
Question 2(iii)
Find the HCF and LCM of 336 and 54, then check their product.
Steps only
336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 6; LCM = 3024
6 × 3024 = 336 × 54 = 18144
Answer
HCF = 6; LCM = 3024.
Question 3(i)
Find the HCF and LCM of 12, 15, 21 by prime factorisation.
Steps only
12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3; LCM = 420
Answer
HCF = 3; LCM = 420.
Question 3(ii)
Find the HCF and LCM of 17, 23, 29 by prime factorisation.
Steps only
17 = 17
23 = 23
29 = 29
HCF = 1; LCM = 11339
Answer
HCF = 1; LCM = 11339.
Question 3(iii)
Find the HCF and LCM of 8, 9, 25 by prime factorisation.
Steps only
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
HCF = 1; LCM = 1800
Answer
HCF = 1; LCM = 1800.
Question 4
The HCF of 306 and 657 is 9. Find their LCM.
Steps only
LCM = (306 × 657)/9
= 201,042/9
= 22,338
Answer
22,338
Question 5
Can 6ⁿ end in 0 for a natural number n?
Steps only
6ⁿ = 2ⁿ × 3ⁿ
Ending in 0 requires a factor 5, which is absent.
Answer
No.
Question 6 · first expression
Explain why 7 × 11 × 13 + 13 is composite.
Steps only
7 × 11 × 13 + 13
= 13(7 × 11 + 1)
= 13 × 78
Answer
Composite: 13 × 78.
Question 6 · second expression
Explain why 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is composite.
Steps only
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × 1,009
Answer
Composite: 5 × 1,009.
Question 7
Two people complete laps in 18 minutes and 12 minutes. When do they next meet at their starting point?
Steps only
18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM(18, 12) = 2 × 2 × 3 × 3 = 36
Answer
After 36 minutes.
Exercise 1.2
Textbook pages 9
Question 1
Prove that √5 is irrational.
Steps only
Assume √5 = a/b in lowest terms, b ≠ 0.
a² = 5b² ⇒ 5 divides a
a = 5k ⇒ b² = 5k² ⇒ 5 divides b
This contradicts a and b being coprime.
Answer
√5 is irrational.
Question 2
Prove that 3 + 2√5 is irrational.
Steps only
Assume 3 + 2√5 = r, where r is rational.
√5 = (r − 3)/2
The right side is rational, contradicting the irrationality of √5.
Answer
3 + 2√5 is irrational.
Question 3(i)
Prove that 1/√2 is irrational.
Steps only
Assume 1/√2 = r, where r is rational.
√2 = 1/r
The right side is rational, contradicting the irrationality of √2.
Answer
1/√2 is irrational.
Question 3(ii)
Prove that 7√5 is irrational.
Steps only
Assume 7√5 = r, where r is rational.
√5 = r/7
The right side is rational, contradicting the irrationality of √5.
Answer
7√5 is irrational.
Question 3(iii)
Prove that 6 + √2 is irrational.
Steps only
Assume 6 + √2 = r, where r is rational.
√2 = r − 6
The right side is rational, contradicting the irrationality of √2.
Answer
6 + √2 is irrational.