Question 3 - Case Based Questions (MCQ) - Chapter 12 Class 12 Linear Programming

Last updated at April 16, 2024 by Teachoo

A manufacturer manufactures two types of tea-cups, A and B. Three machines are needed for manufacturing the tea cups. The time in minutes required for manufacturing each cup on the machines is given below:
Each machine is available for a maximum of six hours per day. If the profit on each cup of type A is Rs. 1.50 and that on each cup of type B is Rs. 1.00. Then answer the following questions:

Question A manufacturer manufactures two types of tea-cups, A and B. Three machines are needed for manufacturing the tea cups. The time in minutes required for manufacturing each cup on the machines is given below: Each machine is available for a maximum of six hours per day. If the profit on each cup of type A is Rs. 1.50 and that on each cup of type B is Rs. 1.00. Then answer the following questions:
Let Number of cups of Type A = x
Number of cups of Type B = y
Machine I
Time required on
Type A → 12 Min
Type B → 6 Min
Max Available Time
= 360 min
∴ 12x + 6y ≤ 360
2x + y ≤ 60
Machine II
Time required on
Type A → 18 Min
Type B → 0 Min
Max Available Time
= 360 min
∴ 18x + 0.y ≤ 360
x ≤ 20
Machine III
Time required on
Type A → 6 Min
Type B → 9 Min
Max Available Time
= 360 min
∴ 6x + 9y ≤ 360
2x + 3y ≤ 120
As we need to maximize the Profit
Given that the profit on each cup of type A is Rs. 1.50 and that on each cup of type B is Rs. 1.00
∴ Z = 1.50 x + 1.00 y
Combining all constraints :
Max Z = 1.5x + y
Subject to constraints,
2x + y ≤ 60,
2x + 3y ≤ 120,
x ≤ 20,
& x ≥ 0 , y ≥ 0
Question 1 Let x be the number of A type tea cups and y be the number of B type tea cups. Then the objective function associative with the given problem is: (a) Max. Z = 1.50 x + y (b) Max. Z = x + 1.50 y (c) Min. Z = 1.50 x – y (d) Min. Z = x – 1.50 y.
Here, we need to
Maximize Z = 1.5x + y
So, the correct answer is (a)
Question 2 Let x be the number of A type tea cups and y be the number of B type tea cups. Then the constraints associative with the given problem are: (A) ■8(2𝑥+𝑦≥60@𝑥≤20@2𝑥+3𝑦≤120) (B) ■8(2𝑥+𝑦≤60@𝑥≤20@2𝑥+3𝑦≤120) (C) ■8(2𝑥+𝑦≤60@𝑥≥20@2𝑥+3𝑦≤120) (D) ■8(2𝑥+𝑦≤60@𝑥=20@2𝑥+3𝑦≥120)
Constraints in our LPP are
2x + y ≤ 60,
2x + 3y ≤ 120,
x ≤ 20,
So, the correct answer is (b)
Question 3 The non-negative conditions are given as: (a) x ⤶7≥ 0, y ⤶7≥ 0 (b) x ⤶7≥ 0, y ≤ 0 (c) x ≤ 0, y ⤶7≥ 0 (d) x ≤ 0, y ≤ 0
Non-negative conditions are where x, y cannot be negative
Therefore, they are x ⤶7≥ 0, y ⤶7≥ 0
So, the correct answer is (a)
Question 4 Feasible region has how many corner points? (a) 3 (b) 4 (c) 5 (d) 6
Our LPP is
Max Z = 1.5x + y
Subject to constraints,
2x + y ≤ 60,
2x + 3y ≤ 120,
x ≤ 20,
& x ≥ 0 , y ≥ 0
Since feasible region as 5 corner points – A, B, C, D, O
So, the correct answer is (c)
Question 5 The Maximum Profit is : (a) Rs. 40 (b) Rs. 50 (c) Rs. 30 (d) Rs 52.5
Finding values at corner points
Hence,
Maximum Profit = Rs. 52.50
So, the correct answer is (d)

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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