{"id":32628,"date":"2022-03-30T10:30:21","date_gmt":"2022-03-30T05:00:21","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=32628"},"modified":"2022-03-30T10:53:29","modified_gmt":"2022-03-30T05:23:29","slug":"ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nA dietician has to develop a special diet using two foods P and Q. Each packet (containing 30g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and.atmost 300 units of cholesterol. How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maxi\u00acmum amount of vitamin A in the diet?
\nSolution:
\nLet x and y be the number of packets of food P and Q. We have the data as
\n\"NCERT
\nCalcium constraint: 12x + 3y \u2265 240
\ni.e., 4x + y \u2265 80
\nIron constraint: 4x + 20y \u2265 460
\ni.e., x + 5y \u2265 115
\nCholesterol constraint: 6x + 4y \u2264 300
\ni.e., 3x + 2y \u2264 150
\nQuantity of vitamin A in the diet Z = 6x + 3y
\nThe LPP is Minimise Z = 6x + 3y
\nsubject to the constraints
\n4x + y \u2265 80
\nx + 5y \u2265 115
\n3x + 2y \u2264 150
\nx \u2265 0, y \u2265 0
\n\"NCERT
\nThe feasible region is shaded in the figure. We use corner point method to find the minimum value of Z.
\n\"NCERT
\nMinimum value of Z is at B (15,20). Hence, the amount of vitamin A under constraints given in the problem will be minimum, if 15 packets of food P and 20 packets of food Q are used in the special diet. The minimum quantity of vitamin A will be 15 0 units.
\n\u2234 Maximum value of Z is 285 at C(40, 15).
\nHence 40 packets of food P and 15 packets of food Q should be used to maximise the amount of vitamin A. The maximum amount of vitamin A in the diet is 285 units.<\/p>\n

Question 2.
\nA farmer mixes two brands P and Q of cattle feed. Brand P, costing \u20b9 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing \u20b9 200 per bag, contains 1.5 units of nutritional element A, 1.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
\nSolution:
\nLet x and y denote the number of bags of brand P and brand Q respectively.
\n\"NCERT
\nElement A constraint: 3x +1, 5y \u2265 18
\nElement B constraint: 2.5x + 11.25y \u2265 45
\nElement C constraint: 2x + 3y \u2265 24
\nCost function : Z = 250x + 200y
\nThe L.P.P is minimise Z = 250x + 200y
\nsubject to the constraints 3x + 1.5y \u2265 18, 2.5x + 11.25y \u2265 45, 2x + 3y \u2265 24, x, y \u2265 0.
\n\"NCERT
\nThe feasible region is shaded in the figure.
\n\"NCERT
\nFrom the table the minimum value of Z is 1950. Since the region is unbounded 1950 may or may not be the minimum value of Z. Consider the inequality 250x + 200y < 1950. This half plane has no point common with the feasible region The minimum value of Z is 1950 at C(3, 6). To produce a mixture having minimum cost, the farmer should mix 3 bags’ of brand P and 6 bags of brand Q.
\nThe minimum cost of the mixture is \u20b9 1950<\/p>\n

Question 3.
\nA dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains atleast 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:
\n\"NCERT
\nOne kg of food X costs Rs. 16 and one kg of food Y costs Rs.20. Find the least cost of the mixture which will produce the required diet.
\nSolution:
\nLet x kg of food X and y kg of food Y be mixed in the diet. We have the data as
\n\"NCERT
\nVitamin A constraint: x + 2y \u2265 10
\nVitamin B constraint: 2x + 2y \u2265 12
\nVitamin C constraint: 3x + y \u2265 18
\nCost function Z = 16x + 20y
\nThe L.P.P. is
\nMinimise Z = 16x + 20y
\nsubject to the constraints
\nx + 2y \u2265 10, 2x + 2y \u2265 12, 3x + y \u2265 8, x, y \u2265 0
\n\"NCERT
\nFrom the table, the minimum value of Z is 112. Since the region is unbounded 112 may or may not be the minimum value of Z. Consider the inequality 16x + 20y =112. This half plane has no point common with the feasible region.
\n\u2234 Minimum value of Z is 112 at B(2, 4). Hence 2kg of food X and 4kg of food Y should be mixed. The least cost of the mix-ture is \u20b9 112<\/p>\n

Question 4.
\nA manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given be-low:
\n\"NCERT
\nEach machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is \u20b9 7.50 and that on each toy of type B is \u20b9 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
\nSolution:
\nLet x be the no. of toys of type A and y be the no. of toys of type B.
\nWe have the data as
\n\"NCERT
\nMachine I constraint: 12x + 6y \u2264 360
\nMachine II constraint: 18x \u2264 360
\nMachine II constraint: 6x + 9y \u2264 360
\nProfit function : Z = 7.5 x + 5y
\nThe LPP is maximise Z = 7.5 x + 5y
\nsubject to the constraints 12x + 6y \u2264 360, 18x + 0y \u2264 360, 6x + 9y \u2264 360, x, y \u2265 0.
\n\"NCERT
\n\u2234 Maximum value of Z is 262.50 at C(15, 30).
\nHence 15 toys of type A and 30 toys of type B should be manufactured in a day to get maximum profit.<\/p>\n

Question 5.
\nAn aeroplane can carry a maximum of 200 passengers. A profit of \u20b9 1000 is made on each executive class ticket and a profit of \u20b9 600 is made on each economy class ticket. The airline reserves atleast 20 seats for executive class. However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?
\nSolution:
\nLet x tickets of executive class and y tickets of economy class be sold.
\nWe have the data as
\n\"NCERT
\nTotal ticket constraint: x + y \u2264 200
\nExecutive ticket constraint: x \u2265 20
\nEconomy ticket constraint: y \u2265 Ax
\nProfit function : Z = 1000x + 600y
\nThe L.P.P. is maximise Z = 1000x + 600y
\nsubject to the constraints x + y \u2264 200, x \u2265 20, y \u2265 4x, x, y \u2265 0.
\n\"NCERT
\nThe feasible region is shaded in the figure.
\n\"NCERT
\nHence 40 tickets of executive class and 160 tickets of economy class are to be sold to get the maximum profit. The maximum profit is \u20b9 136000<\/p>\n

Question 6.
\nTwo godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60,50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:
\n\"NCERT
\nHow should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?
\nSolution:
\nLet x quintals of grains are transported from godown Ato ration shop D andy quintals be transported from godown Ato ration shop E.
\nWe have the data as
\n\"NCERT
\nThe transported quantity of rice is non negative x \u2265 0, y \u2265 0, 100 – (x + y) \u2265 0, 60 – x \u2265 0
\n50 – y \u2265 0 , x + y – 60 \u2265 0
\ni.e., x + y \u2264 100, x + y \u2265 60, x \u2264 60, y \u2264 50, x \u2265 0, y \u2265 0
\nThe total transportation cost
\nZ = 6x + 3y + 2.5 (100 – (x + y) + 4(60 – x) + 2(50 – y) + 3 (x + y – 60)
\nZ = 2.5x + 1.5y + 410
\nThe L.P.P. is Minimise Z = 2.5x + 1.5y + 410
\nsubject to the constraints x \u2264 60, y \u2264 50, x + y \u2265 60, x + y \u2264 100, x, y \u2265 0
\n\"NCERT
\nThe minimum value of Z is 510 at D( 10,50)
\nHence the supply from godown A are 10,50,40 quintals and from godown B are 50,0,0 quintals to shops D, E and F respectively and the minimum cost is \u20b9 510<\/p>\n

Question 7.
\nAn oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500 L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:
\n\"NCERT
\nAssuming that the transportation cost of 10 litres of oil is \u20b9 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?
\nSolution:
\nLet x litre of oil be supplied from depot A to pump D and y litre of oil from depot B to pumb E. Then 7000 – (x + y) litre of oil will be transported to pump F.
\nThe oil supplied from B to D, E and F are 4500 – x, 3000 – y, x + y – 3500
\n\"NCERT
\nThe transportation cost of 10 litres of oil for 1 km is \u20b9 1.
\n\u2234 Transportation cost
\n\"NCERT
\nZ = 0.3 x + 0.1 y + 3950
\nThe transported quantity of oil is non negative.
\ni.e., x \u2265 0, y \u2265 0, 7000 – (x + y) \u2265 0
\n4500 – x \u2265 0, 3000 – y \u22650, x + y – 3500 \u2265 0
\ni.e., x, y \u2265 0
\nx + y \u2264 7000
\nx + y \u2265 3500
\nx \u2264 4500
\ny \u2264 3000
\nHence the L.P.P. is
\nMinimise Z = 0.3x + 0. 1 y + 3950 subject to the constraints x \u2264 4500, y \u2264 3000, x + y \u2264 7000, x + y \u2265 3500, x, y \u2265 0
\n\"NCERT
\nThe feasible region is shaded in the figure. We use comer point method to find the minimum value of Z.
\n\"NCERT
\nThe minimum of Z is 4400 at E (500, 3000)
\nWhen x = 500 and y = 3000 500 l, 3000 l, 3500 l of oil should be trans\u00acported from depot A to pumps D, E and F and 4000 l, 0 l, 0 l are transported from B to pumps D, E and F<\/p>\n

Question 8.
\nA fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs atleast 240 kg of phosphoric acid, atleast 270 kg of potash and atmost 310 kg of chlorine.
\nIf the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?
\n\"NCERT<\/p>\n

Solution:
\nLet x bags of brand P and y bags of brand Q be used in the garden.
\nWe have the data as
\n\"NCERT
\nPhosphoric acid constraint: x + 2y \u2265 240
\nPotash constraint: 3x + 1.5y \u2265 270
\nChlorine constraint: 1.5x + 2y \u2264 310
\nQuantity of nitrogen : Z = 3x + 3.5y
\nThe L.P.P. is Minimise Z = 3x + 3.5y
\nsubject to the constraints x + 2y \u2265 240, 3x + 1.5y \u2265 270, 1.5x + 2y \u2264 310, x, y \u2265 0.
\n\"NCERT
\nThe feasible region is shaded in the figure.
\n\"NCERT
\n\u2234 Minimum value of Z is 470 at C(40, 100)
\nHence 40 bags of brand P and 100 bags of brand Q are used.
\nThe minimum amount of nitrogen is 470.<\/p>\n

Question 9.
\nA fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs atleast 240 kg of phos-phoric acid, atleast 270 kg of potash and atmost 310 kg of chlorine. If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?
\nSolution:
\nLet x bags of brand P and y bags of brand Q be used in the garden.
\nWe have the data as
\n\"NCERT
\nPhosphoric acid constraint: x + 2y \u2265 240
\nPotash constraint: 3x + 1.5y \u2265 270
\nChlorine constraint: 1.5x + 2y \u2264 310
\nQuantity of nitrogen : Z = 3x + 3.5y
\nThe L.P.P. is Minimise Z = 3x + 3.5y
\nsubject to the constraints x + 2y \u2265 240, 3x + 1.5y \u2265 270, 1.5x + 2y \u2264 310, x, y \u2265 0.
\n\"NCERT
\nThe feasible region is shaded in the figure.
\n\"NCERT
\nThe maximum value of Z is 595 obtained at A(140,50)
\nHence 140 bags of brand P and 50 bags of brand Q are used.
\nThe maximum amount of nitrogen is 595.<\/p>\n

Question 10.
\nA toy company manufactures two types of dolls, A and B. Market tests and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is utmost half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by utmost 600 units. If the company makes profit of \u20b9 12 and \u20b9 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?
\nSolution:
\nLet the number of dolls of type A be x and that of type B be y.
\nWe have the data as
\n\"NCERT
\nTotal dolls constraint: x + y \u2264 1200
\nDoll A constraint: x \u2264 3y + 600 or x – 3y \u2264 600
\nProfit function : Z = 12x +16y
\nThe L.P.P is maximise Z = 12x + 16y
\nsubject to the constraints
\nx + y \u2264 1200, x – 3y \u2264 600, y \u2264 \\(\\frac { x }{ 2 }\\), x \u2265 0, y \u2265 0
\n\"NCERT
\nThe maximum value of Z is 16000 at C(800,400) Hence 800 dolls of type A and 400 dolls of type B should be manufactured to get maximum profit.
\nThe maximum profit is \u20b9 16000.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise Question 1. A dietician has to develop a special diet using two foods P and Q. Each packet …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-12-miscellaneous-exercise\/ NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Miscellaneous Exercise Question 1. 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A dietician has to develop a special diet using two foods P and Q. 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