NCERT Solutions for Miscellaneous Exercise Chapter 12 Class 12 - Linear Programming

# NCERT Solutions for Miscellaneous Exercise Chapter 12 Class 12 - Linear Programming

Edited By Ramraj Saini | Updated on Dec 04, 2023 01:54 PM IST | #CBSE Class 12th

## NCERT Solutions For Class 12 Chapter 12 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 12 class 12 Linear Programming are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 12 miscellaneous exercise gives more knowledge of solving a few types of linear programming problems using graphical methods. Miscellaneous exercise chapter 12 Class 12 are solved in a detailed manner with necessary steps and graphs. Going through NCERT solutions for Class 12 Maths chapter 12 miscellaneous exercise gives a better understanding of the linear programming problems. The questions discussed in the NCERT book Class 12 Maths chapter 12 miscellaneous solutions are a bit higher level as compared to the other two exercises. Class 12 Maths chapter 12 miscellaneous solutions are important for Class 12 CBSE Board Exams.

Miscellaneous exercise class 12 chapter 12 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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Linear Programming Class 12 Chapter 12-Miscellaneous Exercise

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How many packets of each food should be used to maximise the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?

Let diet contain x packets of food P and y packets of food Q. Thus, $x\geq 0,y\geq 0$ .

The mathematical formulation of the given problem is as follows:

Total cost is Z . $Z=6x+3y$

Subject to constraint,

$4x+y\geq 80$

$x+5y\geq 115$

$x\geq 0,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(15,20),B(40,15),C(2,72)$

The value of Z at corner points is as shown :

 corner points $Z=6x+3y$ $A(15,20)$ 150 MINIMUM $B(40,15)$ 285 maximum $C(2,72)$ 228

Hence, Z has a maximum value of 285 at the point $B(40,15)$ .

to maximise the amount of vitamin A in the diet, 40 packets of food P and 15 packets of food Q should be used. The maximum amount of vitamin A is 285 units.

Let farmer mix x bags of brand P and y bags of brand Q. Thus, $x\geq 0,y\geq 0$ .

The given information can be represented in the table as :

 Vitamin A Vitamin B Cost Food P 3 5 60 Food Q 4 2 80 requirement 8 11

The given problem can be formulated as follows:

Therefore, we have

$3x+1.5y\geq 18$

$2.5x+11.25y\geq 45$

$2x+3y\geq 24$

$Z=250x+200y$

Subject to constraint,

$3x+1.5y\geq 18$

$2.5x+11.25y\geq 45$

$2x+3y\geq 24$

$x\geq 0,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of the feasible region are $A(18,0),B(9,2),C(3,6),D(0,12)$

The value of Z at corner points is as shown :

 corner points $Z=250x+200y$ $A(18,0)$ 4500 $B(9,2)$ 2650 $C(3,6)$ 1950 minimum $D(0,12)$ 2400

Feasible region is unbounded, therefore 1950 may or may not be a minimum value of Z. For this, we draw $250x+200y< 1950$ and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with $250x+200y< 1950$ .

Hence, Z has a minimum value 1950 at point $C(3,6)$ .

 Food Vitamin A Vitamin B Vitamin C X 1 2 3 Y 2 2 1

One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?

Let mixture contain x kg of food X and y kg of food Y.

Mathematical formulation of given problem is as follows:

Minimize : $z=16x+20y$

Subject to constraint ,

$x+2y\geq 10$

$x+y\geq 6$

$3x+y\geq 8$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(10,0),B(2,4),C(1,5),D(0,8)$

The value of Z at corner points is as shown :

 corner points $z=16x+20y$ $A(10,0)$ 160 $B(2,4)$ 112 minimum $C(1,5)$ 116 $D(0,8)$ 160

The feasible region is unbounded , therefore 112 may or may not be minimum value of Z .

For this we draw $16x+20y< 112$ and check whether resulting half plane has point in common with feasible region or not.

We can see feasible region has no common point with $16x+20y< 112$ .

Hence , Z has minimum value 112 at point $B(2,4)$

 Types of toys Machines I II III A 12 18 6 B 6 0 9

Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.

Let x and y toys of type A and type B.

Mathematical formulation of given problem is as follows:

Minimize : $z=7.5x+5y$

Subject to constraint ,

$2x+y\leq 60$

$x\leq 20$

$2x+3y \leq 120$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(20,0),B(20,20),C(15,30),D(0,40)$

The value of Z at corner points is as shown :

 corner points $z=7.5x+5y$ $A(20,0)$ 150 $B(20,20)$ 250 $C (15,30)$ 262.5 maximum $D(0,40)$ 200

Therefore 262.5 may or may not be maximum value of Z .

Hence , Z has maximum value 262.5 at point $C (15,30)$

Let airline sell x tickets of executive class and y tickets of economy class.

Mathematical formulation of given problem is as follows:

Minimize : $z=1000x+600y$

Subject to constraint ,

$x+y\leq 200$

$x\geq 20$

$y-4x\geq 0$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(20,80),B(40,160),C(20,180)$

The value of Z at corner points is as shown :

 corner points $z=1000x+600y$ $A(20,80)$ 68000 $B(40,160)$ 136000 maximum $C (20,180)$ 128000

therefore 136000 is maximum value of Z .

Hence , Z has maximum value 136000 at point $B(40,160)$

 Transportation cost per quintal (in Rs) From/To A B D 6 4 E 3 2 F 2.50 3

How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost?

Let godown A supply x and y quintals of grain to shops D and E respectively. Then , (100-x-y) will be supplied to shop F. Requirements at shop D is 60 since godown A supply x .Therefore remaining (60-x) quintals of grain will be transported from godown B.

Similarly, (50-y) quintals and 40-(100-x-y)=(x+y-60) will be transported from godown B to shop E and F respectively. The problem can be represented diagrammatically as follows:

$x,y\geq 0$ and $100-x-y\geq 0$

$x,y\geq 0$ and $x+y\leq 100$

$60-x\geq 0,50-y\geq 0\, \, \, and\, \, x+y-60\geq 0$

$\Rightarrow \, \, \, \, x\leq 60,y\leq 50,x+y\geq 60$

Total transportation cost z is given by ,

$z=6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)$

$z=2.5x+1.5y+410$

Mathematical formulation of given problem is as follows:

Minimize : $z=2.5x+1.5y+410$

Subject to constraint ,

$x+y\leq 100$

$x\leq 60$

$y\leq 50$

$x+y\geq 60$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(60,0),B(60,40),C(50,50),D(10,50)$

The value of Z at corner points is as shown :

 corner points $z=2.5x+1.5y+410$ $A(60,0)$ 560 $B(60,40)$ 620 $C(50,50)$ 610 $D(10,50)$ 510 minimum

therefore 510 may or may not be minimum value of Z .

Hence , Z has miniimum value 510 at point $D(10,50)$

 Distance in (km.) From/To A B D 7 3 E 6 4 F 3 2

Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost?

Let x and y litres of oil be supplied from A to petrol pump,D and E. Then , (7000-x-y) will be supplied from A to petrol pump F.

Requirements at petrol pump D is 4500 L. since x L A are transported from depot A,remaining 4500-x L will be transported from petrol pump B

Similarly, (3000-y)L and 3500-(7000-x-y)=(x+y-3500) L will be transported from depot B to petrol E and F respectively.

The problem can be represented diagrammatically as follows:

$x,y\geq 0$ and $7000-x-y\geq 0$

$x,y\geq 0$ and $x+y\leq 7000$

$4500-x\geq 0,3000-y\geq 0\, \, \, and\, \, x+y-3500\geq 0$

$\Rightarrow \, \, \, \, x\leq 4500,y\leq 3000,x+y\geq 3500$

Cost of transporting 10 L petrol =Re 1

Cost of transporting 1 L petrol $=\frac{1}{10}$

Total transportation cost z is given by ,

$z=\frac{7}{10}x+\frac{6}{10}y+\frac{3}{10}(7000-x-y)+\frac{3}{10}(4500-x)+\frac{4}{10}(3000-y)+\frac{2}{10}(x+y-3500)$

$z=0.3x+0.1y+3950$

Mathematical formulation of given problem is as follows:

Minimize : $z=0.3x+0.1y+3950$

Subject to constraint ,

$x+y\leq 7000$

$x\leq 4500$

$y\leq 3000$

$x+y\geq 3500$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(3500,0),B(4500,0),C(4500,2500),D(4000,3000),E(500,3000)$

The value of Z at corner points is as shown :

 corner points $z=0.3x+0.1y+3950$ $A(3500,0)$ 5000 $B(4500,0)$ 5300 $C(4500,2500)$ 5550 $E(500,3000)$ 4400 minimum $D(4000,3000)$ 5450

Hence , Z has miniimum value 4400 at point $E(500,3000)$

If the grower wants to minimise the amount of nitrogen added to the garden, how many bags of each brand should be used? What is the minimum amount of nitrogen added in the garden?

 Kg per bag Brand P Brand Q Nitrogen 3 3.5 Phosphoric Acid 1 2 Potash 3 1.5 Chlorine 1.5 2

Let fruit grower use x bags of brand P and y bags of brand Q.

Mathematical formulation of given problem is as follows:

Minimize : $z=3x+3.5y$

Subject to constraint ,

$x+2y\geq 240$

$x+0.5y\geq 90$

$1.5x+2y\geq 310$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(140,50),C(40,100),B(20,140)$

The value of Z at corner points is as shown :

 corner points $z=3x+3.5y$ $A(140,50)$ 595 $B(20,140)$ 550 $C(40,100)$ 470 minimum

Therefore 470 is minimum value of Z .

Hence , Z has minimum value 470 at point $C(40,100)$

If the grower wants to maximise the amount of nitrogen added to the garden, how many bags of each brand should be added? What is the maximum amount of nitrogen added?

 Kg per bag Brand A Brand P Nitrogen 3 3.5 Phosphoric Acid 1 2 Potash 3 1.5 Chlorine 1.5 2

Let fruit grower use x bags of brand P and y bags of brand Q.

Mathematical formulation of given problem is as follows:

Maximize : $z=3x+3.5y$

Subject to constraint ,

$x+2y\geq 240$

$x+0.5y\geq 90$

$1.5x+2y\geq 310$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $B(20,140),A(140,50),C(40,100)$

The value of Z at corner points is as shown :

 corner points $z=3x+3.5y$ $A(140,50)$ 595 maximum $B(20,140)$ 550 $C(40,100)$ 470 minimum

therefore 595 is maximum value of Z .

Hence , Z has minimum value 595 at point $A(140,50)$

Let x and y be number of dolls of type A abd B respectively that are produced per week.

Mathematical formulation of given problem is as follows:

Maximize : $z=12x+16y$

Subject to constraint ,

$x+y\leq 1200$

$y\leq \frac{x}{2}\Rightarrow x\geq 2y$

$x-3y\leq 600$

$x,y\geq 0$

The feasible region determined by constraints is as follows:

The corner points of feasible region are $A(600,0),B(1050,150),C(800,400)$

The value of Z at corner points is as shown :

 corner points $z=12x+16y$ $A(600,0)$ 7200 $B(1050,150)$ 15000 $C(800,400)$ 16000 Maximum

Therefore 16000 is maximum value of Z .

Hence , Z has minimum value 16000 at point $C(800,400)$

## More About NCERT Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise:

There are 10 questions in the miscellaneous exercise chapter 12 Class 12. Solving all these questions gives a good knowledge about the NCERT Class 12th chapter linear programming. Students have to solve the NCERT syllabus exercises and solved examples in order to get a good idea of topics discussed in the chapter and to get a good score in the final exam.

Also Read| Linear Programming Class 12th Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 12 Miscellaneous Exercise.

• Important questions to understand the concepts explained in the chapter are given in Class 12 Maths chapter 12 miscellaneous exercise solutions.
• All questions of miscellaneous exercise chapter 12 Class 12 are important and will be helpful in the board exam.

## Key Features Of NCERT Solutions For Class 12 Chapter 12 Miscellaneous Exercise

• Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 chapter 12 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 12 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this class 12 maths ch 12 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
• Inclusive Approach: Solutions for class 12 chapter 12 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
• Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 12 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

## Subject Wise NCERT Exemplar Solutions

1. How many exercises in NCERT Class 12 chapter 12 linear programming?

There are three exercises including miscellaneous exercises.

2. What is the number of questions discussed in the NCERT solutions for Class 12 Maths chapter 12 miscellaneous exercise?

Ten questions are covered in Class 12 Maths chapter 12 miscellaneous solutions

3. What is the expected number of questions from the chapter linear programming for the CBSE Class 12 Maths board exam?

One question of 5 marks can be expected

4. The number of miscellaneous solved questions given in NCERT Class 12 Maths chapter linear programming is ………….

Three

5. What are the types of linear programming problems discussed in the Class 12 chapter linear programming.?

Diet, manufacturing and transportation problems are discussed before the ncert solutions for Class 12 Maths chapter 12 miscellaneous exercise.

6. What are constraints?

Constraints are rules or conditions that are used in optimization problems

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9