Class 10 1. Real Numbers Exercise 1.2 : NCERT Book Solutions
Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams.
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All chapters of ncert books Mathematics 1. Real Numbers Exercise 1.2 is solved by exercise and chapterwise for class 10 with questions answers also with chapter review sections which helps the students who preparing for UPSC and other competitive exams and entrance exams.
Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams. - 1. Real Numbers - Exercise 1.2 : NCERT Book Solutions for class 10th. All solutions and extra or additional solved questions for 1. Real Numbers : Exercise 1.2 Mathematics class 10th:English Medium NCERT Book Solutions. Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams.
1. Real Numbers : Exercise 1.2 Mathematics class 10th:English Medium NCERT Book Solutions
Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams. - 1. Real Numbers - Exercise 1.2 : NCERT Book Solutions for class 10th. All solutions and extra or additional solved questions for 1. Real Numbers : Exercise 1.2 Mathematics class 10th:English Medium NCERT Book Solutions.
Class 10 1. Real Numbers Exercise 1.2 : NCERT Book Solutions
NCERT Books Subjects for class 10th Hindi Medium
1. Real Numbers
Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams.
Exercise 1.2
EXERCISE 1.2
Q1. Express each number as a product of its prime factors:
(i) 140
Solution:
= 22 × 5 × 7
(ii) 156
Solution:
= 22 × 3 × 13
(iii) 3825
Solution:
= 32 × 52 × 17
(iv) 5005
Solution:
= 5 × 7 × 11 × 13
(v) 7429
Solution:
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 = 2 × 13
91 = 7 × 13
Common factors = 13
∴ HCF = 13
LCM = 2 × 7 × 13 = 182
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
26 × 91 = 13 × 182
2366 = 2366
Hence Varified,
(ii) 510 and 92
Solution:
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
Common factors = 2
∴ HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
510 × 92 = 2 × 23460
46920 = 46920
Hence Varified,
(iii) 336 and 54
Solution:
336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
Common factors = 2 × 3
∴ HCF = 6
LCM = 2 × 2 × 2× 2 × 3 × 3 × 3 × 7 = 3024
Now varification,
product of the two numbers = LCM × HCF
N1 × N2 = LCM × HCF
336 × 54 = 6 × 3024
18144 = 18144
Hence Varified,
Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution:
12 = 2 × 2 × 3
15 = 5 × 3
21 = 7 × 3
Common Factors = 3
HCF = 3
LCM = 3 × 2 × 2 × 5 × 7 = 420
(ii) 17, 23 and 29
Solution:
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
Solution:
8 = 2 × 2 × 2
9 = 3 × 3
25 = 5 × 5
There is no common factor except 1.
∴ HCF = 1
LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5
= 8 × 9 × 25
= 1800
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9
LCM × HCF = N1 × N2
LCM = 22338
Q5. Check whether 6n can end with the digit 0 for any natural number n.
Solution:
Prime factorisation of 6n = (2 × 3 )n
While, Any natural number which end with digit 0 has
the prime factorisation as form of (2 × 5 )n
Therefore, 6n will not end with digit 0.
Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Let A = 7 × 11 × 13 + 13
= 13 (7 × 11 + 1)
= 13 (77 + 1)
= 13 × 78
Hence this is composite number because It has at least one positive divisor other than one.
Similarily,
Let B = 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
Hence this is also a composite number because It has at least one positive divisor other than one.
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:
Sonia takes 18 minutes in one round.
Ravi takes 12 minutes in one round
they will meet again at the starting point after LCM(18, 12) minutes
18 = 2 × 3 × 3
12 = 2 × 2 × 3
HCF = 2 × 3 = 6
= 36 minutes
Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams.
See other sub-topics of this chapter:
Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams. - 1. Real Numbers - Exercise 1.2 : NCERT Book Solutions for class 10th. All solutions and extra or additional solved questions for 1. Real Numbers : Exercise 1.2 Mathematics class 10th:English Medium NCERT Book Solutions. Class 10 chapter 1. Real Numbers ncert exercise questions and textual questions with solution for board exams and term 1 and term 2 exams.
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Mathematics Chapter List
1. Real Numbers
2. Polynomials
3. Pair of Linear Equations in Two Variables
4. Quadratic Equations
5. Arithmetic Progressions
6. Triangles
7. Coordinate Geometry
8. Introduction to Trigonometry
9. Some Applications of Trigonometry
10. Circles
11. Constructions
12. Areas Related to Circles
13. Surface Areas and Volumes
14. Statistics
15. Probability