NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1

Q: What is the number of exercises in the NCERT chapter linear programming?

There are three exercises including miscellaneous.

Q: How many questions are there in exercise 12.1 Class 12 Maths?

Ten questions are explained in the NCERT Class 12 chapter exercise 1

Q: What number of solved examples are given before the NCERT Class 12 chapter linear programming exercise 12.1?

There are 5 solved examples before exercise 12.1

Q: Why to solve NCERT exercises?

Solving NCERT exercise give more conceptual understanding and students will be able to clear their doubts and can understand the are where they have to improve.

Q: How to use NCERT Solutions for Class 12 Maths chapter 12 exercise 12.1?

First understand the concepts and practice solved example. Then move on to the exercise and try to solve it yourself. If you have any doubts look in to the Class 12 Maths chapter 12 exercise 12.1 solutions.

Q: What is the importance of the Class 12 NCERT Maths chapter 12 linear programming for CBSE Class 12 board exams?

For CBSE Class 12 Maths exam one question of 5 marks is expected from the chapter linear programming.

Q: Give the pattern of linear programming questions of Class 12 NCERT exercise 12.1 .

The linear programming questions will have an objective function. Either maximise or minimize the it according to the given constrains.

Q: What method is adopted in the class 12 NCERT chapter 12 problems?

Graphical method is used to solve the problems in Class 12 chapter 12

NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1 - Linear Programming

Edited By Komal Miglani | Updated on May 07, 2025 04:09 PM IST | #CBSE Class 12th

Linear programming involves finding the optimal value of variables that solves a certain problem. This has a wide variety of real life applications such as charting travel paths, buisness and economics, physics based problems and many more. Class 12 maths chapter 12 exercise 12.1 solutions covers graphical methods to solve linear programming problems. The chapter deals with mathematically analysing constraints and conditions to get the best possible solution to day-to-day problems.

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Class 12 Maths Chapter 12 Exercise 12.1 Solutions: Download PDF
NCERT Solutions Class 12 Maths Chapter 12: Exercise 12.1
Topics covered in Chapter 12 Linear Programming: Exercise 12.1
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NCERT solutions for exercise 12.1 Class 12 Maths gives practice questions to understand linear programming problems. These solutions of NCERT are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2025-26. The answers are designed as per the students demand covering comprehensive, step by step solutions of every problem.

Class 12 Maths Chapter 12 Exercise 12.1 Solutions: Download PDF

Students can find all exercise enumerated in NCERT Book together using the link provided below. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts.

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NCERT Solutions Class 12 Maths Chapter 12: Exercise 12.1

Question 1: Solve the following Linear Programming Problems graphically: Maximise $Z = 3 x + 4 y$ Subject to the constraints $x + y \leq 4, x \geq 0, y \geq 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $x + y \leq 4, x \geq 0, y \geq 0.$ is as follows,

1627031435613

The region A0B represents the feasible region

The corner points of the feasible region are $B (4, 0), C (0, 0), D (0, 4)$

Maximize $Z = 3 x + 4 y$

The value of these points at these corner points are :

Corner points	$Z = 3 x + 4 y$
$B (4, 0)$	12
$C (0, 0)$	0
$D (0, 4)$	16	maximum

The maximum value of Z is 16 at $D (0, 4)$

Question 2: Solve the following Linear Programming Problems graphically: Minimise $z = - 3 x + 4 y$ Subject to . $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0.$ Show that the minimum of Z occurs at more than two points

Answer:

The region determined by constraints, $x + 2 y \leq 8, 3 x + 2 y \leq 12, x \geq 0, y \geq 0.$ is as follows,

1627031511366

The corner points of feasible region are $A (2, 3), B (4, 0), C (0, 0), D (0, 4)$

The value of these points at these corner points are :

Corner points	$z = - 3 x + 4 y$
$A (2, 3)$	6
$B (4, 0)$	-12	Minimum
$C (0, 0)$	0
$D (0, 4)$	16

The minimum value of Z is -12 at $B (4, 0)$

Question 3: Solve the following Linear Programming Problems graphically: Maximise $Z = 5 x + 3 y$ Subject to $3 x + 5 y \leq 15$ , $5 x + 2 y \leq 10$ , $x \geq 0, y \geq 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $3 x + 5 y \leq 15$ , $5 x + 2 y \leq 10$ , $x \geq 0, y \geq 0$ is as follows :

1627031555890

The corner points of feasible region are $A (0, 3), B (0, 0), C (2, 0), D (\frac{20}{19}, \frac{45}{19})$

The value of these points at these corner points are :

Corner points	$Z = 5 x + 3 y$
$A (0, 3)$	9
$B (0, 0)$	0
$C (2, 0)$	10
$D (\frac{20}{19}, \frac{45}{19})$	$\frac{235}{19}$	Maximum

The maximum value of Z is $\frac{235}{19}$ at $D (\frac{20}{19}, \frac{45}{19})$

Question 4: Solve the following Linear Programming Problems graphically: Minimise $Z = 3 x + 5 y$ Such that $x + 3 y \geq 3, x + y \geq 2, x, y \geq 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x + 3 y \geq 3, x + y \geq 2, x, y \geq 0.$ is as follows,

1627031646530

The feasible region is unbounded as shown.

The corner points of the feasible region are $A (3, 0), B (\frac{3}{2}, \frac{1}{2}), C (0, 2)$

The value of these points at these corner points are :

Corner points	$Z = 3 x + 5 y$
$A (3, 0)$	9
$B (\frac{3}{2}, \frac{1}{2})$	7	Minimum
$C (0, 2)$	10

The feasible region is unbounded, therefore 7 may or may not be the minimum value of Z .

For this, we draw $3 x + 5 y < 7$ 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. $Z = 3 x + 5 y$

Hence, Z has a minimum value of 7 at $B (\frac{3}{2}, \frac{1}{2})$

Question 5: Solve the following Linear Programming Problems graphically: Maximise $Z = 3 x + 2 y$ Subject to $x + 2 y \leq 10, 3 x + y \leq 15, x, y \geq 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $x + 2 y \leq 10, 3 x + y \leq 15, x, y \geq 0$ is as follows,

1627031733350

The corner points of feasible region are $A (5, 0), B (4, 3), C (0, 5)$

The value of these points at these corner points are :

Corner points	$Z = 3 x + 2 y$
$A (5, 0)$	15
$B (4, 3)$	18	Maximum
$C (0, 5)$	10

The maximum value of Z is 18 at $B (4, 3)$

Question 6: Solve the following Linear Programming Problems graphically: Minimise $Z = x + 2 y$ Subject to $2 x + y \geq 3, x + 2 y \geq 6, x, y \geq 0.$

Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $2 x + y \geq 3, x + 2 y \geq 6, x, y \geq 0.$ is as follows,

1627031776022

The corner points of the feasible region are $A (6, 0), B (0, 3)$

The value of these points at these corner points are :

Corner points	$Z = x + 2 y$
$A (6, 0)$	6
$B (0, 3)$	6

Value of Z is the same at both points. $A (6, 0), B (0, 3)$

If we take any other point like $(2, 2)$ on line $Z = x + 2 y$ , then Z=6.

Thus the minimum value of Z occurs at more than 2 points .

Therefore, the value of Z is minimum at every point on the line $Z = x + 2 y$ .

Question 7: Solve the following Linear Programming Problems graphically: Minimise and Maximise $z = 5 x + 10 y$ Subject to $x + 2 y \leq 120, x + y \geq 60, x - 2 y \geq 0, x, y \geq 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints, $x + 2 y \leq 120, x + y \geq 60, x - 2 y \geq 0, x, y \geq 0$ is as follows,

1627031823610

The corner points of feasible region are $A (40, 20), B (60, 30), C (60, 0), D (120, 0)$

The value of these points at these corner points are :

Corner points	$z = 5 x + 10 y$
$A (40, 20)$	400
$B (60, 30)$	600	Maximum
$C (60, 0)$	300	Minimum
$D (120, 0)$	600	maximum

The minimum value of Z is 300 at $C (60, 0)$ and maximum value is 600 at all points joing line segment $B (60, 30)$ and $D (120, 0)$

Question 8: Solve the following Linear Programming Problems graphically: Minimise and Maximise $z = x + 2 y$ Subject to $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x, y, \geq 0$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x + 2 y \geq 100, 2 x - y \leq 0, 2 x + y \leq 200, x, y, \geq 0$ is as follows,

1627031869634

The corner points of the feasible region are $A (0, 50), B (20, 40), C (50, 100), D (0, 200)$

The value of these points at these corner points are :

Corner points	$z = x + 2 y$
$A (0, 50)$	100	Minimum
$B (20, 40)$	100	Minimum
$C (50, 100)$	250
$D (0, 200)$	400	Maximum

The minimum value of Z is 100 at all points on the line segment joining points $A (0, 50)$ and $B (20, 40)$ .

The maximum value of Z is 400 at $D (0, 200)$ .

Question 9: Solve the following Linear Programming Problems graphically: Maximise $Z = - x + 2 y$ Subject to the constraints: $x \geq 3, x + y \geq 5, x + 2 y \geq 6, y \geq 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x \geq 3, x + y \geq 5, x + 2 y \geq 6, y \geq 0.$ is as follows,

1627032034587

The corner points of the feasible region are $A (6, 0), B (4, 1), C (3, 2)$

The value of these points at these corner points are :

Corner points	$Z = - x + 2 y$
$A (6, 0)$	- 6	minimum
$B (4, 1)$	-2
$C (3, 2)$	1	maximum

The feasible region is unbounded, therefore 1 may or may not be the maximum value of Z.

For this, we draw $- x + 2 y > 1$ and check whether resulting half plane has a point in common with a feasible region or not.

We can see the resulting feasible region has a common point with a feasible region.

Hence , Z =1 is not maximum value , Z has no maximum value.

Question 10: Solve the following Linear Programming Problems graphically: Maximise $Z = x + y,$ Subject to $x - y \leq - 1, - x + y \leq 0, x, y, \geq 0.$ Show that the minimum of Z occurs at more than two points.

Answer:

The region determined by constraints $x - y \leq - 1, - x + y \leq 0, x, y, \geq 0.$ is as follows,

1627032109317

There is no feasible region and thus, Z has no maximum value.

Topics covered in Chapter 12 Linear Programming: Exercise 12.1

Linear programming is generally defined as the technique for maximising or minimising a linear function of several variables, like input or output cost. The following are some of the basic terminology used in linear programming problems.

Objective Function: Let Z = ax + by be a linear function, where a, b are the constants. Linear objective function goes by the computation of the maximum or the minimum of X
Decision Variables: Let Z = ax + by be a linear objective function. Then the variables x and y are called decision variables.
Constraints: Constraints are the limitations or restrictions imposed on the decision variables.
Optimization problem: A problem that asks to maximise or minimise a linear function limited to certain constraints.
Optimal (feasible) solution: Any point in the feasible region that gives the maximum or minimum value of the objective function is called an optimal solution.

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Theorems

Theorem 1: Let R be the feasible region (convex polygon) for a linear programming problem and let $Z = a x +$ by be the objective function. When Z has an optimal value (maximum or minimum), where the variables $x$ and $y$ are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region.
Theorem 2: Let R be the feasible region for a linear programming problem, and let $Z = a x + b y$ be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R.

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Method Of Solving A Linear Problem

Find the feasible region of the problem and find the vertices.
Find the objective function Z = ax + by. Let M and m be the largest and the smallest points of the problem
When the area is bounded. "M" and "m" are maximum and minimum values. If a feasible area is unbounded then

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ax + by > M, no common points with the feasible region.
ax + by < m, no common points with the feasible region.

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Frequently Asked Questions (FAQs)

1. What is the number of exercises in the NCERT chapter linear programming?

There are three exercises including miscellaneous.

2. How many questions are there in exercise 12.1 Class 12 Maths?

Ten questions are explained in the NCERT Class 12 chapter exercise 1

3. What number of solved examples are given before the NCERT Class 12 chapter linear programming exercise 12.1?

There are 5 solved examples before exercise 12.1

4. Why to solve NCERT exercises?

Solving NCERT exercise give more conceptual understanding and students will be able to clear their doubts and can understand the are where they have to improve.

5. How to use NCERT Solutions for Class 12 Maths chapter 12 exercise 12.1?

First understand the concepts and practice solved example. Then move on to the exercise and try to solve it yourself. If you have any doubts look in to the Class 12 Maths chapter 12 exercise 12.1 solutions.

6. What is the importance of the Class 12 NCERT Maths chapter 12 linear programming for CBSE Class 12 board exams?

For CBSE Class 12 Maths exam one question of 5 marks is expected from the chapter linear programming.

7. Give the pattern of linear programming questions of Class 12 NCERT exercise 12.1 .

The linear programming questions will have an objective function. Either maximise or minimize the it according to the given constrains.

8. What method is adopted in the class 12 NCERT chapter 12 problems?

Graphical method is used to solve the problems in Class 12 chapter 12

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

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Manasvini Gupta 30 January,2025

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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.

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Ashvadha 12 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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NCERT Solutions for Class 12 Maths Chapter 12 Exercise 12.1 - Linear Programming

Class 12 Maths Chapter 12 Exercise 12.1 Solutions: Download PDF

NCERT Solutions Class 12 Maths Chapter 12: Exercise 12.1

Topics covered in Chapter 12 Linear Programming: Exercise 12.1

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