Page 1 CBSE TEST PAPER-01 CLASS - XII MATHEMATICS (Linear Programming) Topic: - Linear Programming [ANSWERS] Ans 01. Let food P consist of x Kg and food Q consists of Y Kg. Z = 60x + 80y 3x + 4y = 8 5x + 2y = 11 x = 0 y = 0 Hence, Cost is minimum = Rs 160 When x = 2, y = 1 2 Ans 02. Let x be number of cakes of first kind and y the number of cakes of other kind. Z = x + y 200x + 100y = 5000 ? 2x + y = 50 25x + 50y = 1000 ? x + 2y = 40 x = 0, y = 0 Maximum number of cakes Z = 30 When x = 20, y = 10. Ans 03. Let the number of cricket and the number of cricket bats to be made in a day be x and y respectively. Z = x + y and also P = 20x + 10y 3 3 42 2 2 28 3 24 0, 0 x y x y x y x y + = ? + = + = = = (i) Maximum Z = 16 at x = 4 y = 12 (ii) P = 20×4 +10×12 = 200 Page 2 CBSE TEST PAPER-01 CLASS - XII MATHEMATICS (Linear Programming) Topic: - Linear Programming [ANSWERS] Ans 01. Let food P consist of x Kg and food Q consists of Y Kg. Z = 60x + 80y 3x + 4y = 8 5x + 2y = 11 x = 0 y = 0 Hence, Cost is minimum = Rs 160 When x = 2, y = 1 2 Ans 02. Let x be number of cakes of first kind and y the number of cakes of other kind. Z = x + y 200x + 100y = 5000 ? 2x + y = 50 25x + 50y = 1000 ? x + 2y = 40 x = 0, y = 0 Maximum number of cakes Z = 30 When x = 20, y = 10. Ans 03. Let the number of cricket and the number of cricket bats to be made in a day be x and y respectively. Z = x + y and also P = 20x + 10y 3 3 42 2 2 28 3 24 0, 0 x y x y x y x y + = ? + = + = = = (i) Maximum Z = 16 at x = 4 y = 12 (ii) P = 20×4 +10×12 = 200 Ans 04. Let the manufacture produce x nuts and y bolts Z = 17.50x + 7y x + 3y = 12 3x + y = 12 X 1 y = 0 Maximum profit Z = Rs 73. 50 at X = 3, y = 3 Ans 05. Let the manufacturer produce x packages of screw A and y packages Screw B. Z = 7x +10y 4x + 6y = 240 2 3 120 6 3 240 x y x y ? + = + = 2 80 x y ? + = x=0,y =0 profit is maximum = 410 When 30 packages of screw A and 20 package of screw B. Ans 06. Let x be pedestal lamps and y wooden shades Z = 5x + 3y 2x + y = 12 3x + 2y = 20 x = 0, y = 0 profit maximum when 4 pedestal lamps 4 wooden shades Ans 07. Let x souvenirs of type A and y souvenirs of type B Z = 5x + 6y 5x + 8y = 200 10x + 8y = 240 ? 5x + 4y = 120 x = 0, y = 0. Maximum profit is Rs 160 When 8 souvenirs of Type A and 20 souvenirs of type B. Page 3 CBSE TEST PAPER-01 CLASS - XII MATHEMATICS (Linear Programming) Topic: - Linear Programming [ANSWERS] Ans 01. Let food P consist of x Kg and food Q consists of Y Kg. Z = 60x + 80y 3x + 4y = 8 5x + 2y = 11 x = 0 y = 0 Hence, Cost is minimum = Rs 160 When x = 2, y = 1 2 Ans 02. Let x be number of cakes of first kind and y the number of cakes of other kind. Z = x + y 200x + 100y = 5000 ? 2x + y = 50 25x + 50y = 1000 ? x + 2y = 40 x = 0, y = 0 Maximum number of cakes Z = 30 When x = 20, y = 10. Ans 03. Let the number of cricket and the number of cricket bats to be made in a day be x and y respectively. Z = x + y and also P = 20x + 10y 3 3 42 2 2 28 3 24 0, 0 x y x y x y x y + = ? + = + = = = (i) Maximum Z = 16 at x = 4 y = 12 (ii) P = 20×4 +10×12 = 200 Ans 04. Let the manufacture produce x nuts and y bolts Z = 17.50x + 7y x + 3y = 12 3x + y = 12 X 1 y = 0 Maximum profit Z = Rs 73. 50 at X = 3, y = 3 Ans 05. Let the manufacturer produce x packages of screw A and y packages Screw B. Z = 7x +10y 4x + 6y = 240 2 3 120 6 3 240 x y x y ? + = + = 2 80 x y ? + = x=0,y =0 profit is maximum = 410 When 30 packages of screw A and 20 package of screw B. Ans 06. Let x be pedestal lamps and y wooden shades Z = 5x + 3y 2x + y = 12 3x + 2y = 20 x = 0, y = 0 profit maximum when 4 pedestal lamps 4 wooden shades Ans 07. Let x souvenirs of type A and y souvenirs of type B Z = 5x + 6y 5x + 8y = 200 10x + 8y = 240 ? 5x + 4y = 120 x = 0, y = 0. Maximum profit is Rs 160 When 8 souvenirs of Type A and 20 souvenirs of type B. Ans 08. Let the merchant stock x desktop computers and y portable computer. Z = 4500x + 5000y x + y =250 25000x + 40000y = 700000 ? 5x + 8y = 1400 x = 0, y = 0. Profit is maximum = 1150000 When 250 desktop computers and 50 portable computers are stocked. Ans 09. Let the diet contain x unit of food F 1 and y units of food F 2. Z = 4x +6y 3x + 6y = 80 4x + 3y = 100 X = 0, y = 0 Z is minimum when 24 units of food F 1 and 4 3 unit of food F 2 are mixed minimum cost = 104 . Ans 10. Let the farmer use x Kg of F 1 and y Kg of F 2. Z = 6x + 5y 10 5 14 100 100 6 10 14 100 100 0, 0. x y x y x y + = + = = = Minimum cost Z = 100 at x = 100 y = 80Read More

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