.jpg)
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution 1:
(i)\(140 = 2 × 2 × 5 × 7 = 2^2 × 5 × 7\)
(ii)\(156 = 2 × 2 × 3 × 13 = 2^2 × 3 × 13\)
(iii)\(3825 = 3 × 3 × 5 × 5 × 17 = 3^2 × 5^2 × 17\)
(iv)\(5005 = 5 ×7× 11 × 13\)
(v)\(7429 = 17 × 19 × 23\)
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 \(91\)
\(26 = 2 × 13\)
\(91 = 7 × 13\)
\(\text{HCF} = 13\)
\(\text{LCM} = 2 × 7 × 13 = 182\)
\(\text{Product of the two numbers} = 26 × 91 = 2366\)
\(\text{HCF × LCM} = 13 × 182 = 2366\)
\(\text{Hence, product of two numbers = HCF × LCM}\)
(ii)\(510\) and \(92\)
\(510 = 2 × 3 × 5 × 17\)
\(92 = 2 × 2 × 23\)
\(\text{HCF} = 2\)
\(\text{LCM} = 2 × 2 × 3 × 5 × 17 × 23 = 23460\)
\(\text{Product of the two numbers }= 510 × 92 = 46920\)
\(\text{HCF × LCM }= 2 × 23460= 46920\)
\(\text{Hence, product of two numbers = HCF × LCM}\)
(iii)\(336\) and \(54\)
\(336 = 2 × 2 × 2 × 2 × 3 × 7\)
\(336 = 2^4× 3× 7\)
\(54 = 2 × 3 × 3 × 3\)
\(54 = 2 × 3^3\)
\(\text{HCF} = 2 × 3 = 6\)
\(\text{LCM} = 2^4 × 3^3 × 7 = 3024\)
\(\text{Product of the two numbers }= 336 × 54 = 18144\)
\(\text{HCF × LCM} = 6 × 3024 = 18144\)
\(\text{Hence, product of two numbers = HCF × LCM}\).
Question 3.
Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
\(12 = 2\times 2× 3\)
\(15 = 3 × 5\)
\(21 = 3 × 7\)
\(\text{HCF} = 3\)
\(\text{LCM }= 2\times2× 3 × 5 × 7 = 420\)
\(12 = 2\times 2× 3\)
\(15 = 3 × 5\)
\(21 = 3 × 7\)
\(\text{HCF} = 3\)
\(\text{LCM }= 2\times2× 3 × 5 × 7 = 420\)
(ii) 17, 23 and 29
\(17 = 1 × 17\)
\(23 = 1 × 23\)
\(29 = 1 × 29\)
\(\text{HCF} = 1\)
\(\text{LCM} = 17 × 23 × 29 = 11339\)
(iii) 8, 9 and 25
\(8 = 2 × 2 × 2\)
\(9 = 3 × 3\)
\(25 = 5 × 5\)
\(\text{HCF} = 1\)
\(\text{LCM } = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800\)
Solution:
\(\text{HCF} (306, 657) = 9\)
\(\text{Product of two numbers = HCF × LCM}\).
\(\therefore \text{HCF × LCM}=306 × 657 \)
\(\text{LCM}= \dfrac{306 × 657 }{ \text{HCF}} = \dfrac{306 × 657 }{9}\)
\(\text{LCM}= {22338}\)
Check whether 6n can end with the digit 0 for any natural number n.
Solution 5:
Prime factorization of \(6^n = (2 ×3)^n\)
It can be observed that \(5\) is not in the prime factorization of \(6^n\).
Hence, for any value of \(n, 6^n\) will not be divisible by \(5\).
Therefore, \(6^n\) cannot end with the digit \(0\) for any natural number \( n\).
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution 6:
\(7 × 11 × 13 + 13 \)
\(= 13 × (7 × 11 + 1) \)
\(= 13 × (77 + 1) \)
\(= 13 × 78\)
\(= 13 ×13 × 6\)
The given expression has \(6\) and \(13\) as its factors.
\(= 13 ×13 × 6\)
The given expression has \(6\) and \(13\) as its factors.
Therefore, it is a composite number.
\(7 × 6 × 5 × 4 × 3 × 2 × 1 + 5\)
\(= 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)\)
\(= 5 × (1008 + 1)\)
\(= 5 ×1009\)
\(1009\) cannot be factorized further.
Therefore, the given expression has \(5\) and \(1009\) as its factors.
Hence, it 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?
Solution 7:
LCM of \(18\) minutes and \(12\) minutes.
\(18 = 2 ×3 ×3\)
And, \(12 = 2 ×2 ×3\)
LCM of \(12\) and \(18 = 2 × 2 × 3 × 3 = 36\)
Therefore, Ravi and Sonia will meet together at the starting point after \(36\) minutes.
.jpg&description=Class 10 Real Numbers Ex 1.1){kind=link}
0 Comments