{"id":22472,"date":"2020-02-05T12:23:28","date_gmt":"2020-02-05T06:53:28","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=22472"},"modified":"2021-07-02T15:25:49","modified_gmt":"2021-07-02T09:55:49","slug":"1st-puc-basic-maths-question-bank-chapter-12","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-question-bank-chapter-12\/","title":{"rendered":"1st PUC Basic Maths Question Bank Chapter 12 Linear Functions"},"content":{"rendered":"

Students can Download Basic Maths Chapter 12 Linear Functions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Basic Maths Question Bank Chapter 12 Linear Functions<\/h2>\n

Question 1.
\nA manufacturer produces and sells pens ’10 per unit. His fixed costs are ‘ 600 and variable cost per pen is ‘ 3.50. Calculate
\n(i) Revenue function
\n(ii) Cost function
\n(iii) Profit function
\n(iv) Breakeven point.
\nAnswer:
\n(i) Revenue function R(x) = p.x. = lQx
\n(ii) Cost function C(x) = ax + b
\n(iii) Profit function p(x) = Rx – C(x) = 8x – [3.50x + 6500]
\n(iv) Breakeven point at BEP \u21d2 TR = TC \u21d2 R(x) = R(x) \u21d2 P(x) = 0
\n\u2234 4.50x – 6500 = 0 \u21d2 x = \\(\\frac{6500}{4.50}\\) = 1445. units
\nBreakeven point revenue is RS = 1445 \u00d7 10 = Rs. 14450<\/p>\n

\"1st<\/p>\n

Question 2.
\nThe daily cost of production ‘C’ in R for ‘x unit of an assembly is C(x) = 12.5x + 6400. If each unit is sold for ’25\/- then determine the minimum number of units that should be produced and sold to ensure no loss. If the selling price is reduced by Rs. 2.5 unit, What would be the break even point.
\nAnswer:
\nBy date C(x) = 12.5x + 6400, S(x) = 25x
\nBreakeven point \u21d2 C(x) = S(x)
\n\u21d2 12.5x + 6400 = 25x \u21d2 6400 = 12.5x
\n\u21d2 x = \\(\\frac{6400}{12.5}\\) = 512
\nSelling price is reduced by 2.5 unit
\nS(x) = (25 – 2.5) = (22.5)x
\nBreakeven point \u21d2 C(x) = S(x)
\n12.5x + 640 = 22.5x
\n\u21d2 6400 = 10x = x = \\(\\frac{6400}{10}\\) = 640<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Chapter 12 Linear Functions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Basic Maths Question Bank Chapter 12 Linear Functions Question 1. A manufacturer produces and …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22472"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=22472"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22472\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=22472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=22472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=22472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}