PYQs Learn Notes Mocks Board Questions Maharashtra HSC Class 12 Mathematics Linear Programming 15 Maharashtra HSC Class 12 Mathematics previous year questions (PYQs) from 14 papers, 2015 to 2026, with answers and worked solutions.
About this chapter Difficulty: 6 easy · 8 moderate · 1 hard. Most-asked subtopics: Graphical Solution of a Linear Programming Problem (13), Formulation of a Linear Programming Problem (1), Linear Inequations in Two Variables (1). Updated 19 September 2026. Try each question before you open its answer. Sign in, and the ones you get wrong come back in Fix your mistakes until you get them right.
Maharashtra HSC Class 12 · 2023 · Q. 8 Easy Add #1 · Linear Inequations in Two Variables
Solve the following inequations graphically and write the corner points of the feasible region: 2 x + 3 y ≤ 6 , x + y ≥ 2 , x ≥ 0 , y ≥ 0 2x + 3y \leq 6,x + y \geq 2,x \geq 0,y \geq 0 2 x + 3 y ≤ 6 , x + y ≥ 2 , x ≥ 0 , y ≥ 0 Maharashtra HSC Class 12 · 2015 · Q. 3. (B) iii. Hard Add #2 · Formulation of a Linear Programming Problem
A diet of a sick person must contain at least 48 units of vitamin A and 64 units of vitamin B. Two foods F 1 F_{1} F 1 and F 2 F_{2} F 2 are available. Food F 1 F_{1} F 1 costs 6 per unit and food F 2 F_{2} F 2 costs 10 per unit. One unit of food F 1 F_{1} F 1 contains 6 units of vitamin A and 7 units of vitamin B. One unit of food F 2 F_{2} F 2 contains 8 units of vitamin A and 12 units of vitamin B. Find the minimum cost for the diet that consists of mixture of these two foods and also meeting the minimal nutritional requirements. #3 · Graphical Solution of a Linear Programming Problem
Solve the linear programming problem (L.P.P.) by graphical method.
Minimize : z = 8 x + 10 y z = 8x + 10y z = 8 x + 10 y
Subject to, 2 x + y ≥ 7 2x + y \geq 7 2 x + y ≥ 7
2 x + 3 y ≥ 15 2x + 3y \geq 15 2 x + 3 y ≥ 15
x ≥ 0 x \geq 0 x ≥ 0
y ≥ 2 y \geq 2 y ≥ 2 #4 · Graphical Solution of a Linear Programming Problem
A firm manufactures two products A A A and B B B on which profit earned per unit are ₹3 and ₹4 respectively. Each product is processed on two machines M 1 M_{1} M 1 and M 2 M_{2} M 2 . The product A requires one minute of processing time on M 1 M_{1} M 1 and two minutes on M 2 M_{2} M 2 , while product B B B requires one minute on M 1 M_{1} M 1 and one minute on M 2 M_{2} M 2 .
Machine M 1 M_{1} M 1 is available for use not more than 450 minutes, while M 2 M_{2} M 2 is available for 600 minutes during any working day.
Find the number of units of products A A A and B B B to be manufactured to get maximum profit. #5 · Graphical Solution of a Linear Programming Problem
Solve the following L.P.P. using graphical method :
Maximize, z = 9 x + 13 y z = 9x + 13y z = 9 x + 13 y
Subject to, 2 x + 3 y ≤ 18 2x + 3y \leq 18 2 x + 3 y ≤ 18 ,
2 x + y ≤ 10 2x + y \leq 10 2 x + y ≤ 10
x ≥ 0 , y ≥ 0 x \geq 0, y \geq 0 x ≥ 0 , y ≥ 0 Done a few? Here’s your next move
#6 · Graphical Solution of a Linear Programming Problem
Solve the L.P.P. graphically :
Minimize : z = 5 x + 2 y z = 5x + 2y z = 5 x + 2 y ,
Subject to, 5 x + y ≥ 10 5x + y \geq 10 5 x + y ≥ 10 , x + y ≥ 6 x + y \geq 6 x + y ≥ 6 ,
x ≥ 0 x \geq 0 x ≥ 0 , y ≥ 0 y \geq 0 y ≥ 0 Maharashtra HSC Class 12 · 2025 · Q. 30 Moderate Add #7 · Graphical Solution of a Linear Programming Problem
Solve the linear programming problem graphically. Maximize: z = 3 x + 5 y z = 3x + 5y z = 3 x + 5 y Subject to: x + 4 y ≤ 24 x + 4y \leq 24 x + 4 y ≤ 24 , 3 x + y ≤ 21 3x + y \leq 21 3 x + y ≤ 21 , x + y ≤ 9 x + y \leq 9 x + y ≤ 9 , x ≥ 0 , y ≥ 0 x \geq 0,y \geq 0 x ≥ 0 , y ≥ 0 . Also find the maximum value of z z z . Maharashtra HSC Class 12 · 2024 · Q. 30 Easy Add #8 · Graphical Solution of a Linear Programming Problem
Solve the following L.P.P. by graphical method: Maximize z = 10 x + 25 y z = 10x + 25y z = 10 x + 25 y Subject to: 0 ≤ x ≤ 3 0 \leq x \leq 3 0 ≤ x ≤ 3 , 0 ≤ y ≤ 3 0 \leq y \leq 3 0 ≤ y ≤ 3 , x + y ≤ 5 x + y \leq 5 x + y ≤ 5 . Also find the maximum value of z z z . Maharashtra HSC Class 12 · 2023 · Q. 1. iv. Easy Add #9 · Graphical Solution of a Linear Programming Problem
If the corner points of the feasible solution are ( 0 , 10 ) (0,10) ( 0 , 10 ) , ( 2 , 2 ) (2,2) ( 2 , 2 ) and ( 4 , 0 ) (4,0) ( 4 , 0 ) then the point of minimum z = 3 x + 2 y z = 3x + 2y z = 3 x + 2 y is _______. Maharashtra HSC Class 12 · 2022 · Q. 30 Moderate Add #10 · Graphical Solution of a Linear Programming Problem
Solve the L.P.P. by graphical method. Minimize z = 8 x + 10 y z = 8x + 10y z = 8 x + 10 y subject to 2 x + y ≥ 7 2x + y \geq 7 2 x + y ≥ 7 , 2 x + 3 y ≥ 15 2x + 3y \geq 15 2 x + 3 y ≥ 15 , y ≥ 0 y \geq 0 y ≥ 0 , x ≥ 0 x \geq 0 x ≥ 0 . Maharashtra HSC Class 12 · 2020 · Q. 28 Easy Add #11 · Graphical Solution of a Linear Programming Problem
Solve the following linear programming problem: Maximize Z = 150 x + 250 y Z = 150x + 250y Z = 150 x + 250 y Subject to: 4 x + y ≤ 40 4x + y \leq 40 4 x + y ≤ 40 , 3 x + 2 y ≤ 60 3x + 2y \leq 60 3 x + 2 y ≤ 60 , x ≥ 0 x \geq 0 x ≥ 0 , y ≥ 0 y \geq 0 y ≥ 0 . Maharashtra HSC Class 12 · 2019 · Q. 25 Easy Add #12 · Graphical Solution of a Linear Programming Problem
Maximize: Z = 3 x + 5 y Z = 3x + 5y Z = 3 x + 5 y Subject to: x + 4 y ≤ 24 x + 4y \leq 24 x + 4 y ≤ 24 , x + y ≤ 9 x + y \leq 9 x + y ≤ 9 , x ≥ 0 x \geq 0 x ≥ 0 , y ≥ 0 y \geq 0 y ≥ 0 . Maharashtra HSC Class 12 · 2018 · Q. 3. (B) ii. Easy Add #13 · Graphical Solution of a Linear Programming Problem
Minimize Z = 7 x + y Z = 7x + y Z = 7 x + y subject to 5 x + y ≥ 5 5x + y \geq 5 5 x + y ≥ 5 , x + y ≥ 3 x + y \geq 3 x + y ≥ 3 , x ≥ 0 x \geq 0 x ≥ 0 , y ≥ 0 y \geq 0 y ≥ 0 . Maharashtra HSC Class 12 · 2017 · Q. 3. (B) iii. Moderate Add #14 · Graphical Solution of a Linear Programming Problem
Solve the following L.P.P. by graphical method: Maximize Z = 6 x + 4 y Z = 6x + 4y Z = 6 x + 4 y Subject to x ≤ 2 x \leq 2 x ≤ 2 , x + y ≤ 3 x + y \leq 3 x + y ≤ 3 , − 2 x + y ≤ 1 - 2x + y \leq 1 − 2 x + y ≤ 1 , x ≥ 0 x \geq 0 x ≥ 0 , y ≥ 0 y \geq 0 y ≥ 0 . Maharashtra HSC Class 12 · 2016 · Q. 2. (B) i. Moderate Add #15 · Graphical Solution of a Linear Programming Problem
Minimize z = 4 x + 5 y z = 4x + 5y z = 4 x + 5 y subject to 2 x + y ≥ 7 2x + y \geq 7 2 x + y ≥ 7 , 2 x + 3 y ≤ 15 2x + 3y \leq 15 2 x + 3 y ≤ 15 , x ≥ 0 x \geq 0 x ≥ 0 , y ≥ 0 y \geq 0 y ≥ 0 . Solve using graphical method. Studying this with friends? Send them this chapter's questions. It's free for them too.
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