{"id":4612,"date":"2016-05-17T09:49:00","date_gmt":"2016-05-17T04:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-part-2\/"},"modified":"2016-05-17T11:14:15","modified_gmt":"2016-05-17T05:44:15","slug":"ncert-solutions-class-11-maths-miscellaneous-part-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-part-2\/","title":{"rendered":"NCERT Solutions class-11 Maths Miscellaneous Part-2"},"content":{"rendered":"

17. If <\/strong>\u00a0are in G.P., prove that <\/strong>\u00a0are in G.P.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0are in G.P.<\/p>\n

To prove: \u00a0are in G.P.<\/p>\n

\u00a0\u00a0<\/p>\n

Let <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

And <\/p>\n

\u00a0\u00a0<\/p>\n

Also <\/p>\n

\u00a0\u00a0<\/p>\n

Now, <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Therefore, \u00a0are in G.P.<\/p>\n

\n

18. If <\/strong>\u00a0and <\/strong>\u00a0are the roots <\/strong>\u00a0and <\/strong>\u00a0are roots of <\/strong>\u00a0where <\/strong>\u00a0form a G.P. Prove that <\/strong>\u00a0<\/strong><\/p>\n

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

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

And <\/p>\n

\u00a0\u00a0<\/p>\n

Also <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0\u00a0and \u00a0are the roots <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

And <\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Also \u00a0are roots of <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(iii)<\/p>\n

And <\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0\u00a0\u2026\u2026\u2026.(iv)<\/p>\n

Dividing eq. (iii) by eq. (i),\u00a0 <\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Now <\/p>\n

= <\/p>\n

Therefore, <\/p>\n

\n

19. The ratio of the A.M. and G.M. of two positive numbers <\/strong>\u00a0and <\/strong>\u00a0is <\/strong>\u00a0Show that <\/strong>\u00a0= <\/strong>\u00a0<\/strong><\/p>\n

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

\u00a0\u00a0<\/p>\n

By component and dividend,<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Again by component and dividend,<\/p>\n

<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Therefore, \u00a0= <\/p>\n

\n

20. If <\/strong>\u00a0are in A.P.; <\/strong>\u00a0are in G.P. and <\/strong>\u00a0are in A.P., prove that <\/strong>\u00a0are in G.P.<\/strong><\/p>\n

Ans. <\/strong>Since, \u00a0are in A.P.<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

Since, \u00a0are in G.P. \u00a0\u00a0<\/p>\n

\u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

\u00a0Also \u00a0are in A.P.<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Putting values of \u00a0and \u00a0in eq. (i),<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0which shows that \u00a0are in G.P.<\/p>\n

\n

21. Find the sum of the following series up to <\/strong>\u00a0terms:<\/strong><\/p>\n

(i) 5 + 55 + 555 + \u2026\u2026\u2026.<\/strong><\/p>\n

(ii) .6 + .66 + .666 + \u2026\u2026\u2026\u2026.<\/strong><\/p>\n

Ans. (i)<\/strong> \u00a0= 5 + 55 + 555 + \u2026\u2026\u2026. up to \u00a0terms<\/p>\n

= 5 [1 + 11 + 111 + \u2026\u2026\u2026. up to \u00a0terms]\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

(ii)<\/strong> \u00a0= .6 + .66 + .666 + \u2026\u2026\u2026\u2026. up to \u00a0terms<\/p>\n

= 6 [.1 + .11 + .111 + \u2026\u2026\u2026. up to \u00a0terms]\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

\n

22. Find the 20th<\/sup> term of the series <\/strong>\u00a0terms.<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0terms<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0\u00a0= <\/p>\n

\u00a0\u00a0<\/p>\n

\n

23. Find the sum of the first <\/strong>\u00a0terms of the series: 3 + 7 + 13 + 21 + 21 + \u2026\u2026\u2026.<\/strong><\/p>\n

Ans. <\/strong>Given: \u2026\u2026\u2026.(i)<\/p>\n

Also \u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

Subtracting eq. (ii) from eq. (i), <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0\u00a0= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

\n

24. If <\/strong>are the sum of first <\/strong>\u00a0natural numbers, their squares and their cubes respectively, show that <\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Here , \u00a0and <\/p>\n

Now, <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0\u00a0 Proved.<\/p>\n

\n

25. Find the sum of the following series up to <\/strong>\u00a0terms:\u00a0 <\/strong>\u00a0<\/strong><\/p>\n

Ans. <\/strong>Given: \u00a0up to \u00a0terms<\/p>\n

\u00a0\u00a0<\/p>\n

= <\/p>\n

= <\/p>\n

\u00a0\u00a0\u00a0= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

\n

26. Show that <\/strong>\u00a0<\/strong><\/p>\n

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

= \u00a0= <\/p>\n

= <\/p>\n

= <\/p>\n

= <\/p>\n

= \u00a0= \u00a0= <\/p>\n

\n

27. A farmer buys a used tractor for Rs. 12000. He pays Rs. 6000 cash and agrees to pay the balance in annual installments of Rs. 500 plus 12% interest on the unpaid amount. How much will the tractor cost him?<\/strong><\/p>\n

Ans. <\/strong>Total cost of the tractor = Rs. 12000, Cash amount paid = Rs. 6000<\/p>\n

Balance amount = 12000 \u2013 6000 = Rs. 6000<\/p>\n

Interest of 1st<\/sup> installment = \u00a0= Rs. 720<\/p>\n

Amount of 1st<\/sup> installment = 500 + 720 = Rs. 1220<\/p>\n

Interest of 2nd<\/sup> installment = \u00a0= Rs. 660<\/p>\n

Amount of 2nd<\/sup> installment = 500 + 660 = Rs. 1160<\/p>\n

Interest of 3rd<\/sup> installment = \u00a0= Rs. 600<\/p>\n

Amount of 3rd<\/sup> installment = 500 + 600 = Rs. 1100<\/p>\n

\u00a0\u00a0Sequence of installments is 1220, 1160, 1100, \u2026\u2026\u2026.<\/p>\n

Here, \u00a0and <\/p>\n

\u00a0\u00a0<\/p>\n

= <\/p>\n

\u00a0= 6 [2440 \u2013 660] = Rs. 10680<\/p>\n

Therefore, the total cost of tractor is (10680 + 6000) = Rs. 16680.<\/p>\n

\n

28. Shams had Ali buys a scooter for Rs. 22000. He pays Rs. 4000 cash and agrees to pay the balance in annual installment of Rs. 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?<\/strong><\/p>\n

Ans. <\/strong>Total cost of the scooter = Rs. 22000, Cash amount paid = Rs. 4000<\/p>\n

Balance amount = 22000 \u2013 4000 = Rs. 18000<\/p>\n

Interest of 1st<\/sup> installment = \u00a0= Rs. 1800<\/p>\n

Amount of 1st<\/sup> installment = 1000 + 1800 = Rs. 2800<\/p>\n

Interest of 2nd<\/sup> installment = \u00a0= Rs. 1700<\/p>\n

Amount of 2nd<\/sup> installment = 1000 + 1700 = Rs. 2700<\/p>\n

Interest of 3rd<\/sup> installment = \u00a0= Rs. 1600<\/p>\n

Amount of 3rd<\/sup> installment = 1000 + 1600 = Rs. 2600<\/p>\n

\u00a0\u00a0Sequence of installments is 2800, 2700, 2600, \u2026\u2026\u2026.<\/p>\n

Here, \u00a0and <\/p>\n

\u00a0\u00a0\u00a0= <\/p>\n

\u00a0= 9 [5600 \u2013 1700] = Rs. 35100<\/p>\n

Therefore, the total cost of tractor is (35100 + 4000) = Rs. 39100.<\/p>\n

\n

29. A person writes a letter to four of his friends. He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly. Assuming that the chain is not broken and that it costs 50 paise to mail one letter. Find the amount spent on the postage when 8th<\/sup> set of letter is mailed.<\/strong><\/p>\n

Ans. <\/strong>Total letters in the first set = 4, Total letters in the second set = \u00a0= 16<\/p>\n

Total letters in the third set = \u00a0= 64<\/p>\n

\u00a0\u00a0Sequence of letters is 4, 16, 64, \u2026\u2026\u2026. In G.P.<\/p>\n

Here \u00a0and <\/p>\n

\u00a0\u00a0<\/p>\n

= <\/p>\n

= <\/p>\n

= \u00a0= 87380<\/p>\n

Since, the amount of postage on each letter is 50 paise.<\/p>\n

Therefore, amount spent on postage = = Rs. 43690.<\/p>\n

\n

30. A man deposited Rs. 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15th<\/sup> year since he deposited the amount and also calculate the total amount after 20 years.<\/strong><\/p>\n

Ans. <\/strong>Total amount deposited = Rs. 10000, Rate of interest = 5% per annum<\/p>\n

Interest of first year = \u00a0= Rs. 500<\/p>\n

Here <\/p>\n

\u00a0\u00a0Amount in 15th<\/sup> year = \u00a0= 10000 + 7000 = Rs. 17000<\/p>\n

Total amount after 20 years = <\/p>\n

\u00a0 = 10 (20000 + 9500) = Rs. 295000<\/p>\n

\n

31. A manufacturer reckons that the value of a machine, which cost him Rs. 15625 will depreciate each year by 20%. Find the estimated value at the end of 5 years.<\/strong><\/p>\n

Ans. <\/strong>After 1 year value of machine = \u00a0= 15625 \u2013 3125 = Rs. 12500<\/p>\n

After 2 year value of machine = \u00a0= 12500 \u2013 2500 = Rs. 10000<\/p>\n

After 3 year value of machine = \u00a0= 10000 \u2013 2000 = Rs. 8000<\/p>\n

\u00a0\u00a0Sequence of values of machine is 12500, 10000, 8000, \u2026\u2026. in G.P.<\/p>\n

Here \u00a0\u00a0and <\/p>\n

\u00a0\u00a0\u00a0= \u00a0= Rs. 5120<\/p>\n

Therefore, the value of machine at the end of 5 years is Rs. 5120.<\/p>\n

\n

32. 150 workers were engaged to finish a job in a certain number of boys. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work find the number of days in which the work was completed.<\/strong><\/p>\n

Ans. <\/strong>Number of workers on the first day = 150<\/p>\n

Number of workers on the second day = 150 \u2013 4 = 146<\/p>\n

Number of workers on the third day = 146 \u2013 4 = 142<\/p>\n

\u00a0\u00a0Sequence of values of workers is 150, 146, 142, \u2026\u2026\u2026 in A.P.<\/p>\n

Here <\/p>\n

\u00a0Total number of workers required to finish the work in \u00a0days<\/p>\n

= <\/p>\n

= \u00a0\u00a0\u2026\u2026\u2026.(i)<\/p>\n

If no worker had dropped out, then<\/p>\n

Total number of workers required to finish the work \u00a0days = \u00a0\u00a0\u2026\u2026\u2026.(ii)<\/p>\n

From eq. (i) and (ii), <\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0\u00a0and \u00a0which is not possible<\/p>\n

Therefore, the work is completed in 25 days.<\/p>\n","protected":false},"excerpt":{"rendered":"

17. If \u00a0are in G.P., prove that \u00a0are in G.P. Ans. Given: \u00a0are in G.P. To prove: \u00a0are in G.P. \u00a0\u00a0 Let \u00a0\u00a0 \u00a0\u00a0 And \u00a0\u00a0 Also \u00a0\u00a0 Now, \u00a0 \u00a0 \u00a0 \u00a0 Therefore, \u00a0are in G.P. 18. If \u00a0and \u00a0are the roots \u00a0and \u00a0are roots of \u00a0where \u00a0form a G.P. Prove that \u00a0 Ans. … 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-4612","post","type-post","status-publish","format-standard","hentry","category-cbse-class-11","category-ncert-solutions"],"yoast_head":"\nNCERT Solutions class-11 Maths Miscellaneous Part-2 | 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-miscellaneous-part-2\/\" \/>\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 Miscellaneous Part-2 | myCBSEguide\" \/>\n<meta property=\"og:description\" content=\"17. If \u00a0are in G.P., prove that \u00a0are in G.P. Ans. Given: \u00a0are in G.P. To prove: \u00a0are in G.P. \u00a0\u00a0 Let \u00a0\u00a0 \u00a0\u00a0 And \u00a0\u00a0 Also \u00a0\u00a0 Now, \u00a0 \u00a0 \u00a0 \u00a0 Therefore, \u00a0are in G.P. 18. If \u00a0and \u00a0are the roots \u00a0and \u00a0are roots of \u00a0where \u00a0form a G.P. Prove that \u00a0 Ans. ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-class-11-maths-miscellaneous-part-2\/\" \/>\n<meta property=\"og:site_name\" content=\"myCBSEguide\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/mycbseguide\/\" \/>\n<meta property=\"article:published_time\" content=\"2016-05-17T04:19:00+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2016-05-17T05:44:15+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/11\/maths\/ch09\/Misc\/image235.png\" \/>\n<meta name=\"author\" content=\"myCBSEguide\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:site\" content=\"@mycbseguide\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"myCBSEguide\" \/>\n\t<meta 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