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- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.3 Real Number
- NCERT Solutions For Class 10 Maths – Chapter 1 Exercise 1.4 Real Numbe
CHAPTER 1 – REAL NUMBER – EXERCISE 1.2
Q1. Express each number as a product of its prime factors:
(i) 140
Solution: 140
140 =2X2X5X7
140 =2²X5X7
Therefore, prime factors of 140 =2²X5X7 .
(ii) 156
Solution: 156
156 =2X2X3X13
156 =2²X3X13
Therefore, prime factors of 156 =2²X3X13 .
(iii) 3825
Solution: 3825
3825 =3X3X5X5X17
3825 =3²X5²X17
Therefore, prime factors of 3825 =3²X5²X17 .
(iv) 5005
Solution: 5005
5005 =5X7X11X13
Therefore, prime factors of 5005 =5X7X11X13.
(v) 7429
Solution: 7429
7429 =17X19X23
Therefore, prime factors of 7429 =17X19X23.
Q2. 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
Solution: 26 and 91
Prime factors of 26 and 91 are
26 =2X13
91 =7X13
Therefore, LCM(26,91) =2X7X13 =182
And HCF(26,91) = 13
Verification:-
product of 26 and 91 =26X91 =2366 and product of LCM and HCF =182X13 =2366 Hence, LCM X HCF = product of 26 and 91.
(ii) 510 and 92
Solution: 510 and 92
Prime factors of 510 and 92 are
510 =2X3X5X17
92 = 2X2X33
Therefore, LCM(510,92) =2X2X3X5X17X23 =23460 and HCF (510,92) = 2
Verification:- product of 510 and 92 =510X92 =46920 and product of LCM and HCF =23460X2 =46920 Hence, LCM X HCF = product of 510 and 92.
(iii) 336 and 54
Solution:
336 and 54 Prime factors of 336 and 54 are
336 =2X2X2X2X3X7
54 =2X3X3X3
Therefore, LCM(336,54) =2^4X3³X7 =3024 and HCF(336,54) =2X3 =6.
Verification:- product of 336 and 54 =336X54 =18144 and product of LCM and HCF =336X54 =18144
Hence, LCM X HCF = product of 336 and 54 .
Q3) Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution:
12, 15 and 21 prime factor of three number are
12 =2X2X3
15 =3X5
21 =3X7
Therefore, LCM(12,15,21) =2X2X3X5X7 =420 HCF(12,15,21) =3 Hence, LCM and HCF(12,15,21) is 420 and 3 respectively.
(ii) 17, 23 and 29
Solution:
17, 23 and 29 prime factor of three number are
17 =17X1
23 =23X1
29 =29X1
Therefore, LCM(17,23,29) =17X23X29 =11339
HCF(17,23,29) =1
Hence, LCM and HCF(17,23,29) is 11339 and 1 respectively.
(iii) 8, 9 and 25
Solution:
8, 9 and 25 prime factor of three numberare
8 =2X2X2X1
9 =3X3X1
25 =5X5X1
Therefore, LCM(8,9,25) =2X2X2X3X3X5X5 =1800
HCF(8,9,25) =1
Hence, LCM and HCF(8,9,25) is 1800 and 1 respectively.
Q4) Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
LCM X HCF =Product of two numbers
Therefore, LCM X 9 =306X657
LCM = 306X657/9
LCM = 22338
Hence, LCM (306, 657) is 22338.
Q5) Check whether 6^n can end with the digit 0 for any natural number n.
Solution:
If the number 6^n, for any n, were to end with the digit zero, then it would be divisible by 5. As we know that any number with unit place as 0 or 5 divisible by 5.
Prime factorization of 6^n =(2X3)^n Therefore, the Prime factorization of 6^n doesn’t contain prime number 5. Hence, it is clear that there is no natural number n for which 6^n ends with the digit zero.
Q6) Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
composite numbers:-
Composite numbers are those number which have more than two factors other than one and the number itself.
In the expression we have, 7X11X13+13 13(7X11+1) (Taking 13 as common)
13X78 =1014 =13X13X2X3
Hence, 7X11X13+13 number have more than two factor so it is a composite numbers.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 5(7X6X1X4X3X2X1+1) (By taking 5 as a common.)
5X1009 = 5045 =5X1009 Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 number have more than two factor so it is a composite number.
Q7) 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, Sonia and Ravi are moving in same direction and at the same time. To find the time of meeting again at starting point, we will take LCM of 18 and 12 Therefore, LCM(18,12) 18 =2X3X3 12 =2X2X3 LCM(18,12) =2X3X2X3 = 36 Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.
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