Students can Download Class 10 Maths Chapter 8 Real Numbers Ex 8.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 10 Maths Chapter 8 Real Numbers Ex 8.2
I. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Answer:
i) 140
140 = 2 × 2 × 5 × 7
140 = 22 × 5 × 7
ii) 156
156 = 2 × 2 × 3 × 13
156 = 22 × 3 × 13
iii) 3825
3825 = 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17
iv) 5005
5005 = 5 × 7 × 11 × 13
v) 7429
7429= 17 × 19 × 23
Question 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
Answer:
i) 26 and 91
∴ H.C.F = 13
L.C.M = 2 × 7 × 13
= 14 × 13 = 182
∴ H.C.F × L.C.M = a × b
13 × 182 = 26 × 96
2366 = 2366
ii) 510 and 92
510 = 2 × 5 × 3 × 17
92 = 2 × 2 × 23 = 22 × 23
LCM of (510, 92)
= 22 × 3 × 5 × 17 × 23 = 23460
HCF of (510, 92) = 2
Verification.
LCM (510, 92) × HCF (510, 92)
= 23460 × 2 = 46920
Product of two numbers
= 510 × 92 = 46920
∴ LCM × HCF = Product of two numbers,
iii) 336 and 54
336 = 2 × 2 × 2 × 2 × 3 × 7 = 24 × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 33
LCM of (336, 54) = 24 × 33 × 7
= 16 × 27 × 7 = 3024
HCF of (336, 54) = 2 × 3 = 6
Verification
LCM (336, 54) × H C F (336, 54)
= 3024 × 6 = 18144
Product of two numbers
= 336 × 54 = 18144
∴ L C M × H C F = Product of two numbers
Question 3.
Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Answer:
i) 12,15 and 21
12 = 2 × 2 × 3 = 22 × 3
15 = 3 × 5
21 = 3 × 7
LCM of (12, 15, 21)
= 22 × 3 × 5 × 7
= 4 × 3 × 5 × 7 = 420
∴ LCM of (12, 15, 21) = 420
HCF of (12, 15, 21) = 3
ii) 17, 23 and 29
17, 23 and 29 are prime numbers
LCM of (17, 23, 29) = 17 × 23 × 29
= 391 × 29
= 11339
HCF of (17, 23, 29) = 1
(∵ 17, 23 and 29 have no common Factor)
iii) 8, 9 and 25
(iii) 8, 9 and 25
8 = 2 × 2 × 2= 23
9 = 3 × 3 = 32
24 = 5 × 5 = 52
∴ H.C.F. = 1
L.C.M. = 23 × 32 × 52 = 8 × 9 × 25 = 1800
∴ HCF × LCM = Product of three numbers
1 × 1800= 8 × 9 × 25
1800 = 1800
Question 4.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:
Question 5.
Check whether 6n can end with the digit 0 for any natural number n.
Answer:
We have to check whether 6n can end with the digit 0 for any natural number ‘n1.
6n = 10 x q
(2 × 3)n = 2 × 5 × q
2n × 3n = 2 × 5 × q
5 is the prime factor of 2n × 3n
Only 2 and 3
∴ This is not possible.
Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
i) 7 × 11 × 13 + 13 = 13 × [7 × 11 + 1]
= 13 × 78
This is composite number.
ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 × {7 × 6 × 5 × 4 × 3 × 2 + 1}
= 5 × 1009.
This is a composite number.
Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?
Answer:
LCM of 18 and 12
18 = 2 × 3 × 3 = 2 × 32
12 = 2 × 2 × 3 = 22 × 3
LCM of (18, 12) = 22 × 32 = 4 × 9 = 36
∴ Sonia and Ravi meet again at the starting point after 36 minutes.