CBSE Class 12 Mathematics Linear Programming MCQs Set 05

Question. The sum of two positive integers is at most 5. The difference between two times of second number and first number is at most 4. If the first number is x and second number y, then for maximizing the product of these two numbers, the mathematical formulation is
(a) \( x + y \ge 5, 2y - x \ge 4, x \ge 0, y \ge 0 \)
(b) \( x + y \ge 5, -2x + y \ge 4, x \ge 0, y \ge 0 \)
(c) \( x + y \le 5, 2y - x \le 4, x \ge 0, y \ge 0 \)
(d) None of the options
Answer: (c) \( x + y \le 5, 2y - x \le 4, x \ge 0, y \ge 0 \)

Question. Mohan wants to invest the total amount of Rs. 15,000 in saving certificates and national saving bonds. According to rules, he has to invest at least Rs. 2000 in saving certificates and Rs. 2500 in national saving bonds. The interest rate is 8% on saving certificate and 10% on national saving bonds per annum. He invests Rs. x in saving certificates and Rs. y in national saving bonds. Then the objective function for this problem is
(a) \( 0.08x + 0.10y \)
(b) \( \frac{x}{2000} + \frac{y}{2500} \)
(c) \( 2000x + 2500y \)
(d) \( \frac{x}{8} + \frac{y}{10} \)
Answer: (a) \( 0.08x + 0.10y \)

Question. Two tailors A and B earn Rs. 15 and Rs. 20 per day respectively. A can make 6 shirts and 4 pants in a day while B can make 10 shirts and 3 pants. To spend minimum on 60 shirts and 40 pants, A and B work x and y days respectively. Then linear constraints except \( x \ge 0, y \ge 0 \) are
(a) \( 15x + 20y \ge 0, 60x + 40y \ge 0 \)
(b) \( 15x + 20y \ge 0, 6x + 10y = 10 \)
(c) \( 6x + 10y \ge 60, 4x + 3y \ge 40 \)
(d) \( 6x + 10y \le 60, 4x + 3y \le 40 \)
Answer: (d) \( 6x + 10y \le 60, 4x + 3y \le 40 \)

Question. In the examination of P.E.T. the total marks of mathematics are 300. If the answer is right, marks provided is 3 and if the answer is wrong, marks provided is –1. A student knows the correct answer of 67 questions and remaining questions are doubtful for him. He takes the time \( 1\frac{1}{2} \) minute to give the correct answer and 3 minute that for doubtful. Total time is 3 hour. In the question paper after every two simple questions, one question is doubtful. He solves the questions one by one, then the number of questions solved by him, is
(a) 67
(b) 90
(c) 79
(d) 80
Answer: (b) 90

Question. A shopkeeper wants to purchase two articles A and B of cost price Rs. 4 and Rs. 3 respectively. He thought that he may earn 30 paise by selling article A and 10 paise by selling article B. He has not to purchase total articles of more than Rs. 24. If he purchases the number of articles of A and B, x and y respectively, then linear constraints are
(a) \( 4x + 3y \le 24, x \ge 0, y \ge 0 \)
(b) \( 30x + 10y \le 24, x \ge 0, y \ge 0 \)
(c) \( 4x + 3y \ge 24, x \ge 0, y \ge 0 \)
(d) \( 30x + 40y \ge 24, x \ge 0, y \ge 0 \)
Answer: (a) \( 4x + 3y \le 24, x \ge 0, y \ge 0 \)

Question. A company manufacturers two types of products A and B. The storage capacity of its godown is 100 units. Total investment amount is Rs. 30,000. The cost prices of A and B are Rs. 400 and Rs. 900 respectively. All the products are sold and per unit profit is Rs. 100 and Rs. 120 through A and B respectively. If x units of A and y units of B be produced, then two linear constraints and iso-profit line are respectively
(a) \( x + y = 100, 4x + 9y = 300, 100x + 120y = c \)
(b) \( x + y \le 100, 4x + 9y \le 300, 2x + y = c \)
(c) \( x + y \le 100, 4x + 9y \le 300, 100x + 120y = c \)
(d) \( x + y \ge 100, 4x + 9y \ge 300, 5x + 6y = c \)
Answer: (c) \( x + y \le 100, 4x + 9y \le 300, 100x + 120y = c \)

Question. We have to purchase two articles A and B of cost Rs. 45 and Rs. 25 respectively. I can purchase total article maximum of Rs. 1000. After selling the articles A and B, the profit per unit is Rs. 5 and 3 respectively. If I purchase x and y numbers of articles A and B respectively, then the mathematical formulation of problem is
(a) \( x \ge 0, y \ge 0, 45x + 25y \ge 1000, 5x + 3y = c \)
(b) \( x \ge 0, y \ge 0, 45x + 25y \le 1000, 5x + 3y = c \)
(c) \( x \ge 0, y \ge 0, 45x + 25y \le 1000, 3x + 5y = c \)
(d) None of the options
Answer: (b) \( x \ge 0, y \ge 0, 45x + 25y \le 1000, 5x + 3y = c \)

Question. The L.P. problem Max \( z = x_1 + x_2 \), such that \( -2x_1 + x_2 \le 1, x_1 \le 2, x_1 + x_2 \le 3 \) and \( x_1, x_2 \ge 0 \) has
(a) One solution
(b) Three solution
(c) An infinite number of solutions
(d) None of the options
Answer: (c) An infinite number of solutions

Question. On maximizing \( z = 4x + 9y \) subject to \( x + 5y \le 200, 2x + 3y \le 134 \) and \( x, y \ge 0, z = \)
(a) 380
(b) 382
(c) 384
(d) None of the options
Answer: (b) 382

Question. The point at which the maximum value of \( (3x + 2y) \) subject to the constraints \( x + y \le 2, x \ge 0, y \ge 0 \) is obtained, is
(a) \( (0, 0) \)
(b) \( (1.5, 1.5) \)
(c) \( (2, 0) \)
(d) \( (0, 2) \)
Answer: (c) \( (2, 0) \)

Question. The solution of a problem to maximize the objective function \( z = x + 2y \) under the constraints \( x - y \le 2, x + y \le 4 \) and \( x, y \ge 0, \) is
(a) \( x = 0, y = 4, z = 8 \)
(b) \( x = 1, y = 2, z = 5 \)
(c) \( x = 1, y = 4, z = 9 \)
(d) \( x = 0, y = 3, z = 6 \)
Answer: (a) \( x = 0, y = 4, z = 8 \)

Question. The maximum value of \( P = 6x + 8y \) subject to constraints \( 2x + y \le 30, x + 2y \le 24 \) and \( x \ge 0, y \ge 0 \) is
(a) 90
(b) 120
(c) 96
(d) 240
Answer: (b) 120

Question. The maximum value of \( P = x + 3y \) such that \( 2x + y \le 20, x + 2y \le 20, x \ge 0, y \ge 0 \), is
(a) 10
(b) 60
(c) 30
(d) None of the options
Answer: (c) 30

Question. The point at which the maximum value of \( x + y \), subject to the constraints \( x + 2y \le 70, 2x + y \le 95, x, y \ge 0 \) is obtained, is
(a) \( (30, 25) \)
(b) \( (20, 35) \)
(c) \( (35, 20) \)
(d) \( (40, 15) \)
Answer: (d) \( (40, 15) \)

Question. If \( 3x_1 + 5x_2 \le 15 ; 5x_1 + 2x_2 \le 10 ; x_1, x_2 \ge 0 \) then the maximum value of \( 5x_1 + 3x_2 \) by graphical method is
(a) \( 12\frac{7}{19} \)
(b) \( 12\frac{1}{7} \)
(c) \( 12\frac{3}{5} \)
(d) 12
Answer: (a) \( 12\frac{7}{19} \)

Question. The maximum value of objective function \( c = 2x + 3y \) in the given feasible region (bounded by vertices \( (0, 0), (7, 0), (6, 2), (0, 5) \)), is
(a) 29
(b) 18
(c) 14
(d) 15
Answer: (b) 18

Question. The maximum value of the objective function \( P = 5x + 3y \), subject to the constraints \( x \ge 0, y \ge 0 \) and \( 5x + 2y \le 10 \) is
(a) 6
(b) 10
(c) 15
(d) 25
Answer: (c) 15

Question. The maximum value of \( P = 8x + 3y \), subject to the constraints \( x + y \le 3, 4x + y \le 6, x \ge 0, y \ge 0 \) is
(a) 9
(b) 12
(c) 14
(d) 16
Answer: (c) 14

Question. The maximum value of \( P = 6x + 11y \) subject to the constraints \( 2x + y \le 104, x + 2y \le 76 \) and \( x \ge 0, y \ge 0 \) is
(a) 240
(b) 540
(c) 440
(d) None of the options
Answer: (c) 440

Question. For the L.P. problem, Min \( z = -x_1 + 2x_2 \), such that \( -x_1 + 3x_2 \le 10, x_1 + x_2 \le 6, x_1 - x_2 \le 2 \) and \( x_1, x_2 \ge 0 \), then \( x_1 = \)
(a) 2
(b) 8
(c) 10
(d) 12
Answer: (a) 2

Question. For the L.P. problem Min \( z = 2x_1 + 3x_2 \), such that \( -x_1 + 2x_2 \le 4, x_1 + x_2 \le 6, x_1 + 3x_2 \ge 9 \) and \( x_1, x_2 \ge 0 \)
(a) \( x_1 = 1.2 \)
(b) \( x_2 = 2.6 \)
(c) \( z = 10.2 \)
(d) All of the options
Answer: (d) All of the options

Question. For the L.P. problem Min \( z = 2x + y \) subject to \( 5x + 10y \le 50, x + y \ge 1, y \le 4 \) and \( x, y \ge 0, z = \)
(a) 0
(b) 1
(c) 2
(d) 1/2
Answer: (b) 1

Question. For the L.P. problem Min. \( z = 2x - 10y \) subject to \( x - y \ge 0, x - 5y \ge -5 \) and \( x, y \ge 0, z = \)
(a) -10
(b) -20
(c) 0
(d) 10
Answer: (a) -10

Question. The maximum value of objective function \( c = 2x + 2y \) in the given feasible region (bounded by vertices \( (0,0), (67, 0), (10, 38), (0, 40) \)), is
(a) 134
(b) 40
(c) 38
(d) 80
Answer: (a) 134

Question. The Minimum value of \( P = x + 3y \) subject to constraints \( 2x + y \le 20, x + 2y \le 20, x \ge 0, y \ge 0 \) is
(a) 10
(b) 60
(c) 30
(d) None of the options
Answer: (d) None of the options

Question. Min. \( Z = -x_1 + 2x_2 \) subjected to \( x_1 + 3x_2 \le 10, x_1 + x_2 \le 6, x_1 - x_2 \le 2 \) and \( x_1, x_2 \ge 0 \) is:
(a) -4
(b) -2
(c) 2
(d) None of the options
Answer: (b) -2

Question. To maximize the objective function \( z = 2x + 3y \) under the constraints \( x + y \le 30, x - y \ge 0, y \le 12, x \le 20, y \ge 3 \) and \( x, y \ge 0 \), is at
(a) \( x = 12, y = 18 \)
(b) \( x = 18, y = 12 \)
(c) \( x = 12, y = 12 \)
(d) \( x = 20, y = 10 \)
Answer: (b) \( x = 18, y = 12 \)

Question. The point at which the maximum value of \( x + y \) subject to the constraints \( 2x + 5y \le 100, \frac{x}{25} + \frac{y}{49} \le 1, x, y \ge 0 \) is obtained, is
(a) \( (10, 20) \)
(b) \( (20, 10) \)
(c) \( (15, 15) \)
(d) \( \left(\frac{50}{3}, \frac{40}{3}\right) \)
Answer: (d) \( \left(\frac{50}{3}, \frac{40}{3}\right) \)

Question. The maximum value of \( Z = 4x + 3y \) subject to the constraints \( 3x + 2y \ge 160, 5x + 2y \ge 200, x + 2y \ge 80, x, y \ge 0 \) is
(a) 320
(b) 300
(c) 230
(d) None of the options
Answer: (d) None of the options

Question. By graphical method, the solution of linear programming problem maximize \( z = 3x_1 + 5x_2 \) subject to \( 3x_1 + 2x_2 \le 18, x_1 \le 4, x_2 \le 6, x_1 \ge 0, x_2 \ge 0 \) is
(a) \( x_1 = 2, x_2 = 0, z = 6 \)
(b) \( x_1 = 2, x_2 = 6, z = 36 \)
(c) \( x_1 = 4, x_2 = 3, z = 27 \)
(d) \( x_1 = 4, x_2 = 6, z = 42 \)
Answer: (b) \( x_1 = 2, x_2 = 6, z = 36 \)

Question. For the L.P. problem Max \( z = 3x_1 + 2x_2 \), such that \( 2x_1 - x_2 \ge 2, x_1 + 2x_2 \le 8 \) and \( x_1, x_2 \ge 0, z = \)
(a) 12
(b) 24
(c) 36
(d) 40
Answer: (b) 24

Question. The maximum value of \( P = 2x + 5y \) subject to the constraints \( x + 4y \le 24, 3x + y \le 21, x + y \le 9 \) and \( x, y \ge 0 \) is
(a) 33
(b) 35
(c) 20
(d) 105
Answer: (a) 33

Question. The maximum value of \( P = 5x + 7y \) subject to the constraints \( x + y \le 4, 3x + 8y \le 24, 10x + 7y \le 35 \) and \( x, y \ge 0 \) is
(a) 14.8
(b) 24.8
(c) 34.8
(d) None of the options
Answer: (b) 24.8

Question. The point which provides the solution to the linear programming problem, Max. \( (2x + 3y) \), subject to constraints: \( x \ge 0, y \ge 0, 2x + 2y \le 9, 2x + y \le 7, x + 2y \le 8 \) is
(a) \( (3, 2.5) \)
(b) \( (2, 3.5) \)
(c) \( (2, 2.5) \)
(d) \( (1, 3.5) \)
Answer: (d) \( (1, 3.5) \)

Question. For maximum value of \( Z = 5x + 2y \), subject to the constraints \( 2x + 3y \ge 6, x - 2y \le 2, 6x + 4y \le 24, -3x + 2y \le 3 \) and \( x, y \ge 0 \) the values of x and y are
(a) 18/7, 2/7
(b) 7/2, 3/4
(c) 3/2, 15/4
(d) None of the options
Answer: (b) 7/2, 3/4

Question. For the following linear programming problem: Minimize \( Z = 4x + 6y \), subject to the constraints \( 2x + 3y \ge 6, x + y \le 8, y \ge 1, x \ge 0 \), the solution is
(a) \( (0, 2) \) and \( (1, 1) \)
(b) \( (0, 2) \) and \( \left(\frac{3}{2}, 1\right) \)
(c) \( (0, 2) \) and \( (1, 6) \)
(d) \( (0, 2) \) and \( (1, 5) \)
Answer: (b) \( (0, 2) \) and \( \left(\frac{3}{2}, 1\right) \)

Question. The minimum value of \( Z = 2x_1 + 3x_2 \) subject to the constraints \( 2x_1 + 7x_2 \ge 22, x_1 + x_2 \ge 6, 5x_1 + x_2 \ge 10 \) and \( x_1, x_2 \ge 0 \) is
(a) 14
(b) 20
(c) 10
(d) 16
Answer: (a) 14

Question. For the L.P. problem Min. \( z = x_1 + x_2 \), such that \( 5x_1 + 10x_2 \le 0, x_1 + x_2 \ge 1, x_2 \le 4 \) and \( x_1, x_2 \ge 0 \)
(a) There is a bounded solution
(b) There is no solution
(c) There are infinite solutions
(d) None of the options
Answer: (c) There are infinite solutions