NCERT-SOLUTIONS-CLASS-10 MATH CH-1 REAL-NUMBERS-EX-1.2 has detailed explanations of concepts as well as answers just for the excellence of the students. We, at cbseinsights.com aspire to increase the intellectual levels of the students for various competitive exams……
Question 1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:
Ex 1.2 Class 10 Maths Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers:
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
Expressing 26 and 91 as product of its prime factors, we get,
26 = 2 × 13 × 1
91 = 7 × 13 × 1
Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182
(Product of the greatest power of each prime factor, involved in the numbers)
and HCF (26, 91) = 13
(Product of the smallest power of each common prime factor in the numbers)
Verification
Now, product of 26 and 91 = 26 × 91 = 2366
And Product of LCM and HCF = 182 × 13 = 2366
Hence, LCM × HCF = product of the 26 and 91.
(ii) 510 and 92
Expressing 510 and 92 as product of its prime factors, we get,
510 = 2 × 3 × 17 × 5 × 1
92 = 2 × 2 × 23 × 1= 22 × 23 × 1
Therefore, LCM (510, 92) = 22 × 3 × 5 × 17 × 23 = 23460
(Product of the greatest power of each prime factor, involved in the numbers)
And HCF (510, 92) = 2
(Product of the smallest power of each common prime factor in the numbers)
Verification
Now, product of 510 and 92 = 510 × 92 = 46920
And Product of LCM and HCF = 23460 × 2 = 46920
Hence, LCM × HCF = product of the 510 and 92.
(iii) 336 and 54
Expressing 336 and 54 as product of its prime factors, we get,
336 = 2 × 2 × 2 × 2 × 7 × 3 × 1= 24 × 7 × 3 × 1
54 = 2 × 3 × 3 × 3 × 1= 2 x 33 x 1
Therefore, LCM(336, 54) = 24 x 33 x 7 = 3024
(Product of the greatest power of each prime factor, involved in the numbers)
And HCF(336, 54) = 2×3 = 6
(Product of the smallest power of each common prime factor in the numbers)
Verification
Now, product of 336 and 54 = 336 × 54 = 18,144
And Product of LCM and HCF = 3024 × 6 = 18,144
Hence, LCM × HCF = product of the 336 and 54.
Ex 1.2 Class 10 Maths 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
Solution:
(i) 12, 15 and 21
Writing the product of prime factors for all the three numbers, we get,
12 = 2 × 2 × 3 = 22 ×3
15 = 5 × 3
21 = 7 × 3
Therefore,
HCF (12, 15, 21) = 3
(Product of the smallest power of each common prime factor in the numbers)
LCM (12, 15, 21) = 22 × 3 × 5 × 7 = 420
(Product of the greatest power of each prime factor, involved in the numbers)
(ii) 17, 23 and 29
Writing the product of prime factors for all the three numbers, we get,
17=17 × 1
23 = 23 × 1
29 = 29 × 1
Therefore,
HCF (17, 23, 29) = 1
(Product of the smallest power of each common prime factor in the numbers)
LCM (17, 23, 29) = 17 × 23 × 29 = 11339
(Product of the greatest power of each prime factor, involved in the numbers)
(iii) 8, 9 and 25
Writing the product of prime factors for all the three numbers, we get,
8=2 × 2 × 2 × 1= 23 x 1
9 = 3 × 3 × 1= 32 x 1
25 = 5 × 5 × 1= 52 x 1
Therefore,
HCF ( 8, 9, 25 ) = 1
(Product of the smallest power of each common prime factor in the numbers)
LCM (8, 9, 25) = 23 × 32 × 52 = 1800
(Product of the greatest power of each prime factor, involved in the numbers)
Ex 1.2 Class 10 Maths Question 4:
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
As we know that,
HCF × LCM = Product of the two given numbers
Therefore,
9 × LCM = 306 × 657
LCM = (306 × 657) / 9 = 22338
Hence, LCM (306, 657) = 22338
Ex 1.2 Class 10 Maths Question 5.
Check whether 6n can end with the digit 0 for any natural number n.
Solution:
If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.
Prime factorization of 6n = (2 × 3)n
Therefore, the prime factorization of 6n doesn’t contain the prime number 5.
Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.
Ex 1.2 Class 10 Maths Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution: By the definition of composite number, we know, if a number is composite, then it means it has factors other than 1 and itself.
Therefore, for the given expression;
7 × 11 × 13 + 13
Taking 13 as common factor, we get,
=13 (7 × 11 × 1 + 1) = 13 (77 + 1) = 13 × 78 = 13 × 3 × 2 × 13
The no. (13 x 78) has factors
(13 x 78), (13), (78), (1), (2), (26), (39), (169). which are more than 2 factors.
Therefore, the no. (13 x 78) is composite number.
Now let’s take the other number,
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 as a common factor, we get,
=5 (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 (1008 + 1) = 5 × 1009
The no. (5 × 1009) has factors
(5 × 1009), (5), (1009), (1) which are more than 2 factors.
Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.
NCERT-SOLUTIONS-CLASS-10 MATH CH-1 REAL-NUMBERS-EX-1.2
Ex 1.2 Class 10 Maths 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?
Solution:
Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time, when they will be meeting again at the starting point, is LCM of 18 and 12.
Expressing 18 and 12 as product of its prime factors, we get,
18= 2 x 32
12= 22 x 3
Therefore, LCM (18, 12) = 22 x 32 = 36
(Product of the greatest power of each prime factor, involved in the numbers)
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.