Class 10 1. Real Numbers Exercise 1.1 : NCERT Book Solutions
Class 10 chapter 1. Real Numbers Important key points for quick revision for board exams, ssc and upsc exams preparaion.
NCERT Solutions
All chapters of ncert books Mathematics 1. Real Numbers Exercise 1.1 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 Important key points for quick revision for board exams, ssc and upsc exams preparaion. - 1. Real Numbers - Exercise 1.1 : NCERT Book Solutions for class 10th. All solutions and extra or additional solved questions for 1. Real Numbers : Exercise 1.1 Mathematics class 10th:English Medium NCERT Book Solutions. Class 10 chapter 1. Real Numbers Important key points for quick revision for board exams, ssc and upsc exams preparaion.
1. Real Numbers : Exercise 1.1 Mathematics class 10th:English Medium NCERT Book Solutions
Class 10 chapter 1. Real Numbers Important key points for quick revision for board exams, ssc and upsc exams preparaion. - 1. Real Numbers - Exercise 1.1 : NCERT Book Solutions for class 10th. All solutions and extra or additional solved questions for 1. Real Numbers : Exercise 1.1 Mathematics class 10th:English Medium NCERT Book Solutions.
Class 10 1. Real Numbers Exercise 1.1 : NCERT Book Solutions
NCERT Books Subjects for class 10th Hindi Medium
1. Real Numbers
Class 10 chapter 1. Real Numbers Important key points for quick revision for board exams, ssc and upsc exams preparaion.
Exercise 1.1
Exercise 1.1
1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Sol:
(1) 135 and 225
a = 225, b = 135 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
225 = 135 ×1 + 90
135 = 90 ×1 + 45
90 = 45 × 2 + 0 {when we get r=0, our computing get stopped}
b = 45 {b is HCF}
Hence: HCF = 45
Sol:
(ii) 196 and 38220
a = 38220, b = 196 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
Sol:
(iii) 867 and 255
a = 867, b = 255 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
38220= 196 ×195 + 0 {when we get r=0, our computing get stopped}
b = 196 {b is HCF}
Hence: HCF = 196
2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Sol:
Let a is the positive odd integer
Where b = 6,
When we divide a by 6 we get reminder 0, 1, 2, 3, 4 and 5, {r < b}
Here a is odd number then reminder will be also odd one.
We get reminders 1, 3, 5
Using Euclid’s division algorithm
So we get
a = 6q + 1, 6q+3 and 6q+5
3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Sol:
Maximum number of columns = HCF (616, 32)
a = 616, b = 32 {Greatest number is ‘a’ and smallest number is ‘b’}
Using Euclid’s division algorithm
a = bq + r (then)
616 = 32 ×19 + 8 {when we get r=0, our computing get stopped}
32 = 8 × 4 + 0
b = 8 {b is HCF}
HCF = 8
Hence: Maximum number of columns = 8
4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]
Solution :
To Show :
a2 = 3m or 3m + 1
a = bq + r
Let a be any positive integer, where b = 3 and r = 0, 1, 2 because 0 ≤ r < 3
Then a = 3q + r for some integer q ≥ 0
Therefore, a = 3q + 0 or 3q + 1 or 3q + 2
Now we have;
⇒ a2 = (3q + 0)2 or (3q + 1)2 or (3q +2)2
⇒ a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 4
⇒ a2 = 9q2 or 9q2 + 6q + 1 or 9q2 + 12q + 3 + 1
⇒ a2 = 3(3q2) or 3(3q2 + 2q) + 1 or 3(3q2 + 4q + 1) + 1
Let m = (3q2) or (3q2 + 2q) or (3q2 + 4q + 1)
Then we get;
a2 = 3m or 3m + 1 or 3m + 1
5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let , a is any positive integer
By using Euclid’s division lemma;
a = bq + r where; 0 ≤ r < b
Putting b = 9
a = 9q + r where; 0 ≤ r < 9
when r = 0
a = 9q + 0 = 9q
a3 = (9q)3 = 9(81q3) or 9m where m = 81q3
when r = 1
a = 9q + 1
a3 = (9q + 1)3 = 9(81q3 + 27q2 + 3q) + 1
= 9m + 1 where m = 81q3 + 27q2 + 3q
when r = 2
a = 9q + 2
a3 = (9q + 2)3 = 9(81q3 + 54q2 + 12q) + 8
= 9m + 2 where m = 81q3 + 54q2 + 12q
⇒ The End
Class 10 chapter 1. Real Numbers Important key points for quick revision for board exams, ssc and upsc exams preparaion.
See other sub-topics of this chapter:
Class 10 chapter 1. Real Numbers Important key points for quick revision for board exams, ssc and upsc exams preparaion. - 1. Real Numbers - Exercise 1.1 : NCERT Book Solutions for class 10th. All solutions and extra or additional solved questions for 1. Real Numbers : Exercise 1.1 Mathematics class 10th:English Medium NCERT Book Solutions. Class 10 chapter 1. Real Numbers Important key points for quick revision for board exams, ssc and upsc exams preparaion.
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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