Class 12 Linear Programming Mock Test (CBSE)

Topics in Linear Programming

• Linear programming problem and its formulation
• Objective function and constraints
• Feasible region
• Graphical method of solution
• Corner point method

Linear Programming practice questions with answers

1. Q1Multiple choice·1 mark
   A farmer mixes two brands of cattle feed, P and Q. Brand P costs Rs $50$ per kg and contains $3$ units of nutritional element A and $1$ unit of element B. Brand Q costs Rs $80$ per kg and contains $1$ unit of element A and $5$ units of element B. The minimum daily requirement of nutritional element A is $12$ units. If $x$ kg of brand P and $y$ kg of brand Q are used, the constraint representing the minimum requirement of nutritional element A is:

   1. (a)$3x + y \ge 12$
   2. (b)$3x + y \le 12$
   3. (c)$50x + 80y \ge 12$
   4. (d)$x + 3y \ge 12$
   Show answer and step marks

   1. Correct option: (a) $3x + y \ge 12$. Since brand P contains 3 units of element A per kg and brand Q contains 1 unit per kg, the total units from $x$ kg of P and $y$ kg of Q is $3x + y$. With a minimum requirement of 12 units, the inequality is $3x + y \ge 12$.+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   In a linear programming problem, the optimal (maximum or minimum) value of the objective function $Z = ax + by$ always occurs at:

   1. (a)any point inside the feasible region
   2. (b)the origin
   3. (c)at least one of the corner points of the feasible region
   4. (d)only at the intersection of non-negativity restrictions
   Show answer and step marks

   1. Correct option: (c) at least one of the corner points of the feasible region. According to the fundamental theorem of linear programming, the optimal value of the objective function occurs at the corner points of the feasible region.+1

   Answer: (c)
3. Q3Short answer·2 marks
   A manufacturer produces two types of toys, A and B. Each toy of type A requires 2 hours on machine $M_1$ and 1 hour on machine $M_2$. Each toy of type B requires 1 hour on machine $M_1$ and 2 hours on machine $M_2$. Machine $M_1$ is available for a maximum of 12 hours and machine $M_2$ for a maximum of 10 hours. The profit on toy A is ₹40 and on toy B is ₹50. Formulate this problem as a linear programming problem (write the objective function and constraints only).
   Show answer and step marks

   1. Let $x$ and $y$ be the numbers of toys A and B; objective: Maximize $Z = 40x + 50y$.+0.5
   2. Machine $M_1$ constraint: $2x + y \le 12$.+0.5
   3. Machine $M_2$ constraint: $x + 2y \le 10$.+0.5
   4. Non-negativity constraints: $x \ge 0, y \ge 0$.+0.5

   Answer: Maximize $Z = 40x + 50y$ subject to $2x + y \le 12$, $x + 2y \le 10$, $x \ge 0, y \ge 0$
4. Q4Short answer·2 marks
   A cooperative society of farmers has $50$ hectares of land to grow two crops A and B. The profit from crops A and B is estimated as Rs $10,500$ and Rs $9,000$ per hectare respectively. To control weeds, a liquid herbicide is used at rates of $20$ litres per hectare for crop A and $10$ litres per hectare for crop B. If the total amount of herbicide available is at most $800$ litres, formulate this as a linear programming problem to maximize total profit.
   Show answer and step marks

   1. Let $x$ and $y$ be the hectares allocated to crops A and B; objective: Maximize $Z = 10500x + 9000y$.+0.5
   2. Land constraint: $x + y \le 50$.+0.5
   3. Herbicide constraint: $20x + 10y \le 800$.+0.5
   4. Non-negativity constraints: $x \ge 0, y \ge 0$.+0.5

   Answer: Maximize $Z = 10500x + 9000y$ subject to $x + y \le 50$, $20x + 10y \le 800$, $x \ge 0, y \ge 0$

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