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NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

Edited By Ramraj Saini | Updated on Sep 19, 2023 09:36 AM IST | #CBSE Class 12th

NCERT Linear Programming Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming provided here. These NCERT solutions are prepared by expert team at Careers360 considering the latest syllabus of CBSE 2023-24. This ch 12 maths class 12 helps solving and finding optimal solutions to problems related to maximising profit or minimising cost. NCERT linear programming class 12 solutions will help in formulating these real life problems into a mathematical model. You should practise linear programming problems (lpp class 12) for getting command of concepts and in depth understanding of this linear programming class 12 chapter.

In this linear programming class 12 solutions, you are going to deal with problems on linear programming like maximisation and minimization of equations, mathematical and graphical methods to solve problems of linear programming. You can also refer to linear programming class 12 ncert solutions for better understanding of concepts. Below we have given complete Class 12 maths chapter 12 NCERT solutions. Check all NCERT solutions from class 6 to 12 at a single place to prepare better for exams.

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NCERT Class 12 Maths Chapter 12 Question Answer - Important Formulae

>> Feasible Region: The feasible region, or solution region, of a linear programming problem is the common area determined by all the constraints, including the non-negativity constraints (x ≥ 0, y ≥ 0).

>> Infeasible Solution: Any point within or on the boundary of the feasible region represents a feasible solution to the constraints. Points outside the feasible region are considered infeasible solutions.

>> Optimal Solution: An optimal solution is any point within the feasible region that provides the optimal value (maximum or minimum) of the objective function.

Fundamental Theorems in Linear Programming:

>> Optimality at Corner Points: For a linear programming problem with a feasible region represented as a convex polygon, if the objective function Z = ax + by has an optimal value, this optimal value must occur at one of the corner points (vertices) of the feasible region.

>> Existence of Maxima and Minima: If the feasible region R is bounded, then the objective function Z has both a maximum and a minimum value on R, and each of these values occurs at a corner point (vertex) of R. If R is unbounded, a maximum or minimum may not exist. However, if it does exist, it must occur at a corner point of R.

>> Corner Point Method: The corner point method is used to solve a linear programming problem and consists of the following steps:

Find the feasible region of the linear programming problem and determine its corner points (vertices).

Evaluate the objective function Z = ax + by at each corner point. Let M and m represent the largest and smallest values obtained at these points.

If the feasible region is bounded, M and m respectively represent the maximum and minimum values of the objective function.

If the feasible region is unbounded, then:

  • M is the maximum value of the objective function if the open half-plane determined by ax + by > M has no points in common with the feasible region.

  • m is the minimum value of the objective function if the open half-plane determined by ax + by < M has no points in common with the feasible region.

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NCERT Linear Programming Class 12 Questions And Answers (Intext Questions and Exercise)

NCERT linear programming class 12 solutions - Exercise: 12.1

Question:1 Solve the following Linear Programming Problems graphically: Maximise Z = 3x + 4y 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 = 3x + 4y

The value of these points at these corner points are :

Corner points
Z = 3x + 4y

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=-3x+4y Subject to . x+2y\leq 8,3x+2y\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+2y\leq 8,3x+2y\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=-3x+4y

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 = 5x + 3y Subject to 3x + 5y \leq 15 , 5x+2y\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, 3x + 5y \leq 15 , 5x+2y\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 = 5x + 3y

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 = 3x + 5y Such that x+3y\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+3y\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 = 3x + 5y

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 3x + 5y< 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 = 3x + 5y

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 = 3x + 2y Subject to x+2y\leq 10,3x+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+2y\leq 10,3x+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 = 3x + 2y

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 + 2y Subject to 2x+y\geq 3,x+2y\geq 6,x,y\geq 0.

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

Answer:

The region determined by constraints 2x+y\geq 3,x+2y\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 + 2y
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 + 2y , 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 + 2y .

Question:7 Solve the following Linear Programming Problems graphically: Minimise and Maximise z=5x+10y Subject to x+2y\leq 120,x+y\geq 60,x-2y\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+2y\leq 120,x+y\geq 60,x-2y\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=5x+10y

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+2y Subject to x+2y\geq 100,2x-y\leq 0,2x+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+2y\geq 100,2x-y\leq 0,2x+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+2y

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+2y Subject to the constraints: x\geq 3,x+y\geq 5,x+2y\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+2y\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+2y

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+2y> 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.


NCERT linear programming class 12 solutions - Exercise: 12.2

Question:1 Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs.80/kg. Food P contains 3 units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.

Answer:

Let mixture contain x kg of food P and y kg of food 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 mixture must contain 8 units of Vitamin A and 11 units of Vitamin B.

Therefore, we have

3x+4y\geq 8

5x+2y\geq 11

Total cost is Z. Z=60x+80y

Subject to constraint,

3x+4y\geq 8

5x+2y\geq 11

x\geq 0,y\geq 0

The feasible region determined by constraints is as follows:

1627041171227

It can be seen that a feasible region is unbounded.

The corner points of the feasible region are A(\frac{8}{3},0),B(2,\frac{1}{2}),C(0,\frac{11}{2})

The value of Z at corner points is as shown :

corner points
Z=60x+80y

A(\frac{8}{3},0)
160
MINIMUM
B(2,\frac{1}{2})
160
minimum
C(0,\frac{11}{2})
440

Feasible region is unbounded, therefore 160 may or may not be the minimum value of Z.

For this, we draw 60x+80y< 160\, \, or \, \, \, 3x+4y< 8 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. \, \, 3x+4y< 8

Hence, Z has a minimum value 160 at line segment joining points A(\frac{8}{3},0) and B(2,\frac{1}{2}) .


Question:2 One kind of cake requires 200g of flour and 25g of fat, and another kind of cake requires 100g of flour and 50g of fat. Find the maximum number of cakes which can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.

Answer:

Let there be x cakes of first kind and y cakes of the second kind.Thus, x\geq 0,y\geq 0 .

The given information can be represented in the table as :


Flour(g)
fat(g)
Cake of kind x
200
25
Cake of kind y
100
50
Availability
5000
1000

Therefore,

200x+100y\leq 5000

\Rightarrow \, \, \, \, 2x+y\leq 50

. \, \, 25x+50y\leq 10000

\Rightarrow \, \, x+2y\leq 400

The total number of cakes, Z. Z=X+Y

Subject to constraint,

\Rightarrow \, \, \, \, 2x+y\leq 50

\Rightarrow \, \, x+2y\leq 400

x\geq 0,y\geq 0

The feasible region determined by constraints is as follows:

1627041262324

The corner points of the feasible region are A(25,0),B(20,10),C(0,20),D(0,0)

The value of Z at corner points is as shown :

corner points
Z=X+Y

A(25,0)
25

B(20,10)
30
maximum
C(0,20)
D(0,0)
20
0

minimum

The maximum cake can be made 30 (20 of the first kind and 10 of the second kind).


Question:3 A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(i) What number of rackets and bats must be made if the factory is to work at full capacity?

Answer:

Let number of rackets be x and number of bats be y.

the machine time availability is not more than 42 hours.

i.e. 1.5x+3y\leq 42

craftsman’s time availability is 24 hours

i.e. 3x+y\leq 24

The factory has to work at full capacity.

Hence, 1.5x+3y= 42...............1

3x+y= 24...............2

Solving equation 1 and 2, we have

x=4\, \, and\, \, \, y=12

Thus, 4 rackets and 12 bats are to be made .

Question:3 A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.

(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity.

Answer:

Let the number of rackets is x and the number of bats is y.

the machine time availability is not more than 42 hours.

craftsman’s time availability is 24 hours

The given information can be repreented in table as shown :


racket
bat
availability
machine time
1.5
3
42
craftman's time
3
1
24

1.5x+3y\leq 42

3x+y\leq 24

x,y\geq 0

The profit on the bat is 10 and on the racket is 20.

Z=20x+10y

The mathematical formulation is :

maximise Z=20x+10y

subject to constraints,

1.5x+3y\leq 42

3x+y\leq 24

x,y\geq 0

The feasible region determined by constraints is as follows:

1627370575766

The corner points are A(8,0),B(4,12),C(0,14),D(0,0)

The value of Z at corner points is as shown :

CORNER POINTS
Z=20x+10y

A(8,0)
160

B(4,12)
200
maximum
C(0,14)
140

D(0,0)
0

Thus, the maximum profit of the factory when it works at full capacity is 200.

Question:4 A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?

Answer:

Let packages of nuts be x and packages of bolts be y .Thus, x\geq 0,y\geq 0 .

The given information can be represented in table as :


bolts
nuts
availability
machine A
1
3
12
machine B
3
1
12




Profit on a package of nuts is Rs. 17.5 and on package of bolt is 7.

Therefore, constraint are

x+3y\leq 12

3x+y\leq 12

x\geq 0,y\geq 0

Z= 17.5x+7y

The feasible region determined by constraints is as follows:

1627370632779

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

The value of Z at corner points is as shown :

Corner points
Z= 17.5x+7y

A(4,0)
70

B(3,3)
73.5
maximum
C(0,4)
28

D(0,0)
0

The maximum value of z is 73.5 at B(3,3) .

Thus, 3 packages of nuts and 3 packages of bolts should be manufactured everyday to get maximum profit.

Question:5 A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.

Answer:

Let factory manufactures screws of type A and factory manufactures screws of type B. Thus, x\geq 0,y\geq 0 .

The given information can be represented in the table as :


screw A
screw B
availability
Automatic machine
4
6
4\times 60=240
hand operated machine
6
3
4\times 60=240




Profit on a package of screw A is Rs.7 and on the package of screw B is 10.

Therefore, the constraint is

4x+6y\leq 240

6x+3y\leq 240

x\geq 0,y\geq 0

Z= 7x+10y

The feasible region determined by constraints is as follows:

1627370732345

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

The value of Z at corner points is as shown :

Corner points
Z= 7x+10y

A(40,0)
280

B(30,20)
410
maximum
C(0,40)
400

D(0,0)
0

The maximum value of z is 410 at B(30,20) .

Thus, 30 packages of screw A and 20 packages of screw B should be manufactured every day to get maximum profit.

Question:6 A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?

Answer:

Let the cottage industry manufactures x pedestal lamps and y wooden shades. Thus, x\geq 0,y\geq 0 .

The given information can be represented in the table as :


lamps
shades
availability
machine (h)
2
1
12
sprayer (h)
3
2
20




Profit on a lamp is Rs. 5 and on the shade is 3.

Therefore, constraint is

2x+y\leq 12

3x+2y\leq 20

x\geq 0,y\geq 0

Z= 5x+3y

The feasible region determined by constraints is as follows:

1627370794134

The corner points of the feasible region are A(6,0),B(4,4),C(0,10),D(0,0)

The value of Z at corner points is as shown :

Corner points
Z= 5x+3y

A(6,0)
30

B(4,4)
32
maximum
C(0,10)
30

D(0,0)
0

The maximum value of z is 32 at B(4,4) .

Thus, 4 shades and 4 pedestals lamps should be manufactured every day to get the maximum profit.

Question:7 A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?

Answer:

Let x be Souvenirs of type A and y be Souvenirs of type B .Thus, x\geq 0,y\geq 0 .

The given information can be represented in table as :


Type A
Type B
availability
cutting
5
8
(3\times 60)+20=200
asembling
10
8
4\times 60=240




Profit on type A Souvenirs is Rs. 5 and on type B Souvenirs is 6.

Therefore, constraint are

5x+8y\leq 200

10x+8y\leq 240

x\geq 0,y\geq 0

Z=5x+6y

The feasible region determined by constraints is as follows:

1627370879345

The corner points of feasible region are A(24,0),B(8,20),C(0,25),D(0,0)

The value of Z at corner points is as shown :

Corner points
Z=5x+6y

A(24,0)
120

B(8,20)
160
maximum
C(0,25)
150

D(0,0)
0

The maximum value of z is 160 at B(8,20) .

Thus,8 Souvenirs of type A and 20 Souvenirs of type B should be manufactured everyday to get maximum profit.

Question:8 A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Answer:

Let merchant plans has personal computers x desktop model and y portable model

.Thus, x\geq 0,y\geq 0 .

The cost of desktop model is cost Rs 25000 and portable model is Rs 40000.

Merchant can invest Rs 70 lakhs maximum.

25000x+40000y\leq 7000000

5x+8y\leq 1400

the total monthly demand of computers will not exceed 250 units.

x+y\leq 250

profit on the desktop model is Rs 4500 and on portable model is Rs 5000.

Total profit = Z , Z=4500x+5000y

The mathematical formulation of given problem is :
5x+8y\leq 1400

x+y\leq 250

x\geq 0,y\geq 0

Z=4500x+5000y

The feasible region determined by constraints is as follows:

1627377288762

The corner points of feasible region are A(250,0),B(200,50),C(0,175),D(0,0)

The value of Z at corner points is as shown :

Corner points
Z=4500x+5000y

A(250,0)
1125000

B(200,50)
1150000
maximum
C(0,175)
875000

D(0,0)
0

The maximum value of z is 1150000 at B(200,50) .

Thus, merchant should stock 200 desktop models and 50 portable models to get maximum profit.

Question:9 A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.

Answer:

Let diet contain x unit of food F1 and y unit of foof F2 .Thus, x\geq 0,y\geq 0 .

The given information can be represented in table as :


Vitamin
minerals
cost per unit
foof F1
3
4
4
food F2
6
3
6

80
100

Cost of food F1 is Rs 4 per unit and Cost of food F2 is Rs 6 per unit

Therefore, constraint are

3x+4y\geq 4

6x+3y\geq 6

x\geq 0,y\geq 0

Z= 4x+6y

The feasible region determined by constraints is as follows: 1627377385273

We can see feasible region is unbounded.

The corner points of feasible region are A(\frac{80}{3},0),B(24,\frac{4}{3}),C(0,\frac{100}{3})

The value of Z at corner points is as shown :

Corner points
Z= 4x+6y

A(\frac{80}{3},0)
106.67

B(24,\frac{4}{3}),
104
minimum
C(0,\frac{100}{3})
200
maximum


Feasible region is unbounded , therefore 104 may or may not be minimum value of Z .

For this we draw 4x+6y< 104 or 2x+3y< 52 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 2x+3y< 52 .

Hence , Z has minimum value 104.

Question:10 There are two types of fertilisers F1 and F2 . F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

Answer:

Let farmer buy x kg of fertilizer F1 and y kg of F2 .Thus, x\geq 0,y\geq 0 .

The given information can be represented in table as :


Nitrogen
phosphoric acid
Cost
F1
10
6
6
F2
5
10
5
requirement
14
14

F1 contain 10% nitrogen and F2 contain 5% nitrogen .Farmer requires atleast 14 kg of nitrogen

10\%x+5\%y\geq 14

\frac{x}{10}+\frac{y}{20}\geq 14

2x+y\geq 280

F1 contain 6% phophoric acid and F2 contain 10% phosphoric acid .Farmer requires atleast 14 kg of nitrogen

6\%x+10\%y\geq 14

\frac{6x}{100}+\frac{y}{20}\geq 14

3x+56y\geq 700

Total cost is Z . Z=6x+5y

Subject to constraint ,

2x+y\geq 280

3x+56y\geq 700

x\geq 0,y\geq 0

Z=6x+5y

The feasible region determined by constraints is as follows:

1627377453836

It can be seen that feasible region is unbounded.

The corner points of feasible region are A(\frac{700}{3},0),B(100,80),C(0,280)

The value of Z at corner points is as shown :

corner points
Z=6x+5y

A(\frac{700}{3},0)
1400

,B(100,80)
1000
minimum
C(0,280)
1400

Feasible region is unbounded , therefore 1000 may or may not be minimum value of Z .

For this we draw 6x+5y< 1000 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 6x+5y< 1000 .

Hence , Z has minimum value 1000 at point ,B(100,80)

Question:11 The corner points of the feasible region determined by the following system of linear inequalities:

2x+y \leq 10,x+3y \leq 15,x,y\geq 0 are (0,0),(5,0),(3,4) and (0,5) . Let Z=px+qy, where p,q > 0. Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is

(A) p=q

(B)p=2q

(C)p=3q

(D)q=3p

Answer:

The maximum value of Z is unique.

It is given that maximum value of Z occurs at two points (3,4)\, \, and\, \, \, (0,5) .

\therefore Value of Z at (3,4) =value of Z at (0,5)

\Rightarrow \, \, \, p(3)+q(4)=p(0)+q(5)

\Rightarrow \, \, \, 3p+4q=5q

\Rightarrow \, \, \, q=3p

Hence, D is correct option.


NCERT solutions for class 12 maths chapter 12 linear programming-Miscellaneous Exercise

Question:1 Reference of Example 9 (Diet problem): A dietician has to develop a special diet using two foods P and Q. Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A. Each packet of the same quantity of food Q contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 units of iron and at most 300 units of cholesterol.

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?

Answer:

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:

1627377546036

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.

Question:2 A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

Answer:

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:

1627377686090

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) .

Question:3 A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below:

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?

Answer:

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:

1627377734165

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)

Question:4 A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

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.

Answer:

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:

1627377785282

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)

Question:5 An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximise the profit for the airline. What is the maximum profit?

Answer:

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:

1627377831247

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)

Question:6 Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table:

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

Answer:

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:

1627377923876

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:

1627377885300

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)

Question:7 An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distances (in km) between the depots and the petrol pumps is given in the following table:

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?

Answer:

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:

1627377981082

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:

1627378013511

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)

Question:8 A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

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?


Answer:

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:

1627378058480

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)

Question:9 Reference of Que 8 : A fruit grower can use two types of fertilizer in his garden, brand P and brand Q. The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of each brand are given in the table. Tests indicate that the garden needs at least 240 kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine.

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


Answer:

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:

1627378216485

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)

Question:10 A toy company manufactures two types of dolls, A and B. Market research and available resources have indicated that the combined production level should not exceed 1200 dolls per week and the demand for dolls of type B is at most half of that for dolls of type A. Further, the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs 12 and Rs 16 per doll respectively on dolls A and B, how many of each should be produced weekly in order to maximise the profit?

Answer:

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:

1627378272830

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)

If you are looking for ncert exercise solutions of linear programming class 12 then they are listed below.

More about NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming

In this article, you will get NCERT solutions for class 12 and you can also use the webpage pdf downloader tool to download NCERT Solutions for Class 12 maths chapter 12 PDF from here. Generally, one five marks question is asked from Linear programming class 12 in the final examination. This chapter has around 21 lpp class 12 questions, here you will find all solutions very easily. These NCERT Class 12 maths solutions chapter 12 are explained in a detailed manner that will help you in scoring high scores.

Let's take an NCERT problem - A furniture dealer deals in only two items–chairs and tables. Has storage space of at most 60 pieces and He has Rs 50,000 to invest. A chair costs Rs 500 and A table costs Rs 2500. He estimates that from the sale of one chair, he can make a profit of Rs 75 and that from the sale of one table a profit of Rs 250. He wants to know how many chairs and tables he should buy from the available money so as to maximise his total profit, assuming that he can sell all the items which he buys.

The problems that involve maximising cost and minimising profit are called optimization problems. This ch 12 maths class 12 is quite important because it includes such concepts. The above given problem is an example of linear programming.

Also read,

Linear Programming Class 12 - Topics

12.1 Introduction

12.2 Linear Programming Problem and its Mathematical Formulation

12.2.1 Mathematical formulation of the problem

12.2.2 Graphical method of solving linear programming problems

12.3 Different Types of Linear Programming Problems

NCERT solutions for class 12 maths - Chapter Wise

Linear Programming Class 12 NCERT Solutions - Key Features

  1. Detailed explanations: The class 12 maths ch 12 question answer provided in NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming are explained in a detailed and step-by-step manner. This helps students to understand the concepts better and makes it easy for them to solve similar problems.

  2. Easy to understand: The maths chapter 12 class 12 solutions are written in simple language, making it easy for students to understand and learn the concepts. The solutions are designed to cater to the needs of students of all learning levels.

  3. Covers all the topics: The class 12 linear programming solutions cover all the topics in Chapter 12 Linear Programming of Class 12 Maths. This helps students to have a comprehensive understanding of the chapter.

JEE Main high scoring chapters and topics

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JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

Exercise-wise solutions: The maths chapter 12 class 12 solutions are provided exercise-wise, which helps students to focus on specific problems and concepts that they find difficult.

NCERT solutions for class 12 subject wise

class wise NCERT Solutions

Benefits of NCERT Solutions for Class 12 maths chapter 12

  • NCERT Class 12 maths solutions chapter 12 are prepared and explained in detailed form. It makes these Class 12 maths chapter 12 NCERT solutions easy to understand.

  • NCERT Solutions for Class 12 maths chapter 12 PDF is very helpful for the preparation of this chapter.

  • These NCERT Solutions for Class 12 maths chapter 12 will give you a new way to solve questions.

  • Miscellaneous exercise is quite important, so to develop a grip on the concepts. In NCERT Class 12 maths solutions chapter 12 linear programming, you will get solutions for miscellaneous exercise too.

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Frequently Asked Question (FAQs)

1. How can NCERT solutions be helpful in CBSE board exam ?

As most the questions in CBSE board exam are asked directly from NCERT textbook, so one must know the NCERT well but only knowing the answer is not guaranteed to score good marks in the exam. One should know how to answer in board exams in order to get good marks. NCERT solutions are provided by the experts who know how best to write answers in the board exam in order to get good marks. Interested students can study linear programming class 12 solutions pdf both online and offline.

2. How is the Linear Programming chapter helpful in solving real-life problems?

Yes, linear programming is useful in formulating these real-life problems into a mathematical model and then solving them. Some optimisation problems like maximizing the profits and minimizing the cost can be solved by linear programming. concepts of liner programming important in board and engineering exam therefore this chapter is very important. you should solve ncert exercise to get command on concepts.

3. Does CBSE provide the solutions of NCERT for class 12 maths ?

No, CBSE doesn’t provide NCERT solutions for any class or subject. but you can find solutions free from careers360 official website. these solutions are explained in details by our expert team in very simple format so that students can understand these very easily.

4. Where can I find the complete solutions of NCERT for class 12 maths ?

Here, you will get the detailed NCERT solutions for class 12 maths  by clicking on the links that are listed above in this article, for comfortability of students these are listed subjects as well as chapter wise. or you can find these solutions from careers360 official website.

5. What is the weightage of the chapter Linear Programming for CBSE board exam ?

Generally, one question of 5 marks is asked from this chapter in the CBSE 12th board final exam. if you want to score five out of five then it demand practice and in depth understanding of concepts therefore it ncert solutions and ncert exercise are recommended to students.

6. What are the important topics in chapter Linear Programming ?

Linear programming problem and its mathematical formulation, graphical method of solving linear programming problems are the important topics of this chapter. you can refer NCERT solutions to get in depth understanding and getting hold on concepts.

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Questions related to CBSE Class 12th

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Yes, you can give improvement exams for 2-3 subjects in CBSE (Central Board of Secondary Education). CBSE allows students who have appeared for their Class 12 board exams to improve their scores by re-appearing for exams in up to five subjects the following year.


Hope this helps,


Thank you

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

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In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

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  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
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Looking Ahead (2025 Admissions):

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  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

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However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

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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)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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)

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

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

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

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