{"id":4594,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-6-1\/"},"modified":"2016-05-17T10:52:43","modified_gmt":"2016-05-17T05:22:43","slug":"ncert-solutions-class-11-maths-exercise-6-1","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-6-1\/","title":{"rendered":"NCERT Solutions class-11 Maths Exercise 6.1"},"content":{"rendered":"

Exercise 6.1<\/strong><\/p>\n

1. Solve <\/strong> when:<\/strong><\/p>\n

(i) is a natural number <\/strong><\/p>\n

(ii) is an integer<\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

Dividing both sides by 24, <\/p>\n

<\/p>\n

(i)<\/strong> When is a natural number then values of that make statement true are 1, 2, 3, 4. The solution set of inequality is {1, 2, 3, 4}.<\/p>\n

(ii)<\/strong> When is an integer then values of that make statement true are The solution set of inequality is <\/p>\n

\n

2. Solve <\/strong> when:<\/strong><\/p>\n

(i) is a natural number <\/strong><\/p>\n

(ii) is an integer<\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

Dividing both sides by <\/p>\n

<\/p>\n

(i)<\/strong> When is a natural number then values of that make statement true are none.<\/p>\n

(ii)<\/strong> When is an integer then values of that make statement true are The solution set of inequality is <\/p>\n

\n

3. Solve <\/strong> when<\/strong><\/p>\n

(i) is an integer <\/strong><\/p>\n

(ii) is a real number<\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

(i)<\/strong> When is an integer then values of that make statement true are The solution set of inequality is <\/p>\n

(ii)<\/strong> When is a real number then solution set of inequality is <\/p>\n

\n

4. Solve <\/strong> when:<\/strong><\/p>\n

(i) is an integer <\/strong><\/p>\n

(ii) is a real number<\/strong><\/p>\n

Ans. <\/strong>Given: <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

(i)<\/strong> When is an integer then values of that make statement true are The solution set of inequality is <\/p>\n

(ii)<\/strong> When is a real number then solution set of inequality is <\/p>\n

\n

Solve the inequalities in Exercises 5 to 16 for real <\/strong><\/p>\n

5. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

6. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

7. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

8. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

9. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

10. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

11. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

12. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

13. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

14. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

15. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

16. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, the solution set is <\/p>\n

\n

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line:<\/strong><\/p>\n

17. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

The solution set is <\/p>\n

<\/p>\n

\n

18. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

The solution set is <\/p>\n

<\/p>\n

\n

19. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

The solution set is <\/p>\n

<\/p>\n

\n

20. <\/strong><\/p>\n

Ans. <\/strong>Here <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

The solution set is <\/p>\n

<\/p>\n

\n

21. Ravi obtained 70 and 75 marks in first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.<\/strong><\/p>\n

Ans. <\/strong>Let the marks obtained by Ravi in third test be <\/p>\n

According to question,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, minimum marks needed to be obtained by Ravi is 35.<\/p>\n

\n

22. To receive Grade \u2018A\u2019 in a course, one must obtain an average 90 marks or more in five examinations (each of 100 marks). If Sunita\u2019s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get Grade \u2018A\u2019 in the course.<\/strong><\/p>\n

Ans. <\/strong>Let the marks obtained by Sunita in fifth examination be <\/p>\n

According to question,<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

Therefore, minimum marks needed to be obtained by Sunita is 82.<\/p>\n

\n

23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.<\/strong><\/p>\n

Ans. <\/strong>Let and be two consecutive odd positive integers.<\/p>\n

and <\/p>\n

and <\/p>\n

and <\/p>\n

<\/p>\n

and 7<\/p>\n

Therefore, the required pairs of odd positive integers are (5, 7) and (7, 9).<\/p>\n

\n

24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.<\/strong><\/p>\n

Ans. <\/strong>Let and be two consecutive even positive integers.<\/p>\n

and <\/p>\n

and <\/p>\n

and <\/p>\n

<\/p>\n

and 10<\/p>\n

Therefore, the required pairs of even positive integers are (6,8), (8 10) and (10, 12).<\/p>\n

\n

25. The longest side of a triangle s 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm. Find the minimum length of the shortest side.<\/strong><\/p>\n

Ans. <\/strong>Let the length of the shortest side be cm.<\/p>\n

Length of longest side = cm and length of third side = cm<\/p>\n

Perimeter of triangle = = cm<\/p>\n

Now, <\/p>\n

<\/p>\n

<\/p>\n

Therefore, the minimum length of shortest side is 9 cm.<\/p>\n

\n

26. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?<\/strong><\/p>\n

Ans. <\/strong>Let the length of the shortest side be cm.<\/p>\n

Length of the second board = cm and length of third side = cm<\/p>\n

Now, and <\/p>\n

and <\/p>\n

and <\/p>\n

and <\/p>\n

Therefore, minimum length of shortest board is 8 cm and maximum length is 22 cm.<\/p>\n","protected":false},"excerpt":{"rendered":"

Exercise 6.1 1. Solve when: (i) is a natural number (ii) is an integer Ans. Given: Dividing both sides by 24, (i) When is a natural number then values of that make statement true are 1, 2, 3, 4. The solution set of inequality is {1, 2, 3, 4}. (ii) When is an integer then … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47,281],"tags":[],"class_list":["post-4594","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Exercise 6.1 | myCBSEguide<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-exercise-6-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions class-11 Maths Exercise 6.1 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"Exercise 6.1 1. Solve when: (i) is a natural number (ii) is an integer Ans. Given: Dividing both sides by 24, (i) When is a natural number then values of that make statement true are 1, 2, 3, 4. The solution set of inequality is {1, 2, 3, 4}. (ii) When is an integer then ... 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