Class 12

Maharashtra State Board OCM 12th Commerce Textbook Solutions Digest

• Chapter 1 Principles of Management
• Chapter 2 Functions of Management
• Chapter 3 Entrepreneurship Development
• Chapter 4 Business Services
• Chapter 5 Emerging Modes of Business
• Chapter 6 Social Responsibilities of Business
• Chapter 7 Consumer Protection
• Chapter 8 Marketing

Maharashtra State Board Class 12 Textbook Solutions

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• Chapter 1 नवनिर्माण
• Chapter 2 निराला भाई
• Chapter 3 सच हम नहीं; सच तुम नहीं
• Chapter 4 आदर्श बदला
• Chapter 5.1 गुरुबानी
• Chapter 5.2 वृंद के दोहे
• Chapter 6 पाप के चार हथियार
• Chapter 7 पेड़ होने का अर्थ
• Chapter 8 सुनो किशोरी
• Chapter 9 चुनिंदा शेर
• Chapter 10 ओजोन विघटन का संकट
• Chapter 11 कोखजाया
• Chapter 12 लोकगीत

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• Chapter 13 कनुप्रिया

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• Chapter 14 पल्लवन
• Chapter 15 फीचर लेखन
• Chapter 16 मैं उद्घोषक
• Chapter 17 ब्लॉग लेखन
• Chapter 18 प्रकाश उत्पन्न करने वाले जीव

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• मुहावरे
• भावार्थ : गुरुबानी, वृंद के दोहे
• भावार्थ : सुनु रे सखिया और कजरी
• पारिभाषिक शब्दावली
• अपठित बोध
• व्याकरण

Maharashtra State Board Class 12 Textbook Solutions

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• Chapter 1 वेगवशता
• Chapter 2 रोज मातीत
• Chapter 3 आयुष्य… आनंदाचा उत्सव
• Chapter 4 रे थांब जरा आषाढघना
• Chapter 5 वीरांना सलामी
• Chapter 6 रंग माझा वेगळा
• Chapter 6.1 आत्मविश्वासासारखी शक्ती नाही

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• Chapter 7 विंचू चावला…
• Chapter 8 रेशीमबंध
• Chapter 9 समुद्र कोंडून पडलाय
• Chapter 10 दंतकथा
• Chapter 11 आरशातली स्त्री
• Chapter 12 रंगरेषा व्यंगरेषा
• Chapter 12.1 जयपूर फूटचे जनक

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• Bhag 3 कथा-साहित्यप्रकार-परिचय
• Bhag 3.1 शोध
• Bhag 3.2 गढी

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• Bhag 4.1 मुलाखत
• Bhag 4.2 माहितीपत्रक
• Bhag 4.3 अहवाल
• Bhag 4.4 वृत्तलेख (फिचर रायटिंग)

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व्याकरण

• वाक्यप्रकार
• वाक्यरूपांतर
• समास
• प्रयोग
• अलंकार
• निबंध लेखन

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Maharashtra State Board Class 12 Textbook Solutions

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12th English Digest 2021-2022 Section 1 (Prose)

• Chapter 1.1 An Astrologer’s Day
• Chapter 1.2 On Saying “Please”
• Chapter 1.3 The Cop and the Anthem
• Chapter 1.4 Big Data-Big Insights
• Chapter 1.5 The New Dress
• Chapter 1.6 Into the Wild
• Chapter 1.7 Why We Travel
• Chapter 1.8 Voyaging Towards Excellence

English Yuvakbharati 12th Full Digest Section 2 (Poetry)

• Chapter 2.1 Song of the Open Road
• Chapter 2.2 Indian Weavers
• Chapter 2.3 The Inchcape Rock
• Chapter 2.4 Have you Earned Your Tomorrow
• Chapter 2.5 Father Returning Home
• Chapter 2.6 Money
• Chapter 2.7 She Walks in Beauty
• Chapter 2.8 Small Towns and Rivers

Yuvakbharati English 12th Digest Guide Section 3 (Writing Skills)

• Chapter 3.1 Summary Writing
• Chapter 3.2 Do Schools Really Kill Creativity? (Mind-Mapping)
• Chapter 3.3 Note–Making
• Chapter 3.4 Statement of Purpose
• Chapter 3.5 Drafting a Virtual Message
• Chapter 3.6 Group Discussion

Yuvakbharati English 12th Textbook Answers Solutions Section 4 (Genre-Drama)

• Chapter 4.1 History of Novel
• Chapter 4.2 To Sir, with Love
• Chapter 4.3 Around the World in Eighty Days
• Chapter 4.4 The Sign of Four

Maharashtra State Board Class 12 Textbook Solutions

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Ex 7.1

Question 1.
A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machine P, Q, R and S. The processing cost of each job for each machine is given in the following table:

Find the optimal assignment to minimize the total processing cost.
Solution:
The cost matrix is given by

Subtracting row minimum from all the elements in that row we get

Subtracting column minimum from all the elements in that column we get the same matrix.
As all the rows and columns have single zeros the allotment can be done as follows.

As per the table, the job allotments are
P → II, Q → IV, R → I, S → III
The total minimum cost = 25 + 21 + 19 + 34 = ₹ 99

Question 2.
Five wagons are available at stations 1, 2, 3, 4, and 5. These are required at 5 stations I, II, III, IV, and V. The mileage between various stations are given in the table below. How should the wagons be transported so as to minimize the mileage covered?

Solution:
The mileage matrix is given by

Subtracting row minimum from all elements in that row we get

Subtracting column minimum from all elements in that column we get

Draw minimum lines covering all the zeros

The number of lines covering all the zeros (3) is less than the order of the matrix (5). Hence an assignment is not possible. The modification is required. The minimum uncovered value 1 is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same we get

Drawing a minimum number of lines covering all the zeros.
No. of lines covering all the zeros (4) is less than the order of the matrix (5).
Hence assignment is not possible.
Again modification is required. The minimum uncovered value 3 is subtracted from the uncovered values and added to the values at the intersection.
The numbers on the lines remain the same we get

No. of lines covering all the zeros (5) are equal to the order of the matrix so the assignment is possible.

According to the table the assignment is
1 → I, 2 → II, 3 → IV, 4 → II, 5 → V
Total minimum mileage = 10 + 6 + 4 + 9 + 10 = 39 units

Question 3.
Five different machines can do any of the five required jobs, with different profits five required jobs, with different profits resulting from each assignment as shown below:

Find the optimal assignment schedule.
Solution:
This profit matrix has to be reduced to cost matrix by subtracting all the values of the matrix from the largest value (62) we get

Subtracting row minimum value from all the elements in that column we get

Subtracting column minimum from all the elements in that column we get

Drawing minimum lines covering all zeros we get

No. of lines (4) is less than the order of the matrix (5). Hence assignment is not possible. The modification is required. The minimum uncovered value (4) is subtracted from the uncovered value and added to the values at the intersection. The values on the lines remain the same, we get

No. of lines (5) are equal to the order of the matrix (5). So assignments are possible
1 → C, 2 → E, 3 → A, 4 → D, 5 → B
For the minimum profit look at the corresponding in the profit matrix given.
Maximum profit = 40 + 36 + 40 + 36 + 62 = 214 units

Question 4.
Four new machines M₁, M₂, M₃, and M₄ are to be installed in a machine shop. There are five vacant places A, B, C, D, and E available. Because of limited space, machine M₂ cannot be placed at C and M₁ cannot be placed at A. The cost matrix is given below.

Find the optimal assignment schedule.
Solution:
This is a restricted assignment so we assign a very high cost ‘∞’ to the prohibited all.
Also as it is an unbalanced problem we add a dummy row M₅ with all values as ‘0’, we get

Subtracting row minimum from all the elements in that row, we get

Subtracting column minimum from all the elements in that column we get the same matrix.

As minimum no. of lines covering all zeros (5) is equal to the order of the matrix, Assignment is possible

The assignments are given by
M₁ → A, M₂ → B, M₃ → E, M₄ → D, M₅ → C
As M₅ is dummy no machine is installed at C
For minimum cost taking the corresponding values in the cost matrix we get
Minimum cost = 4 + 4 + 2 + 2 = 12 units

Question 5.
A company has a team of four salesmen and there is four districts where the company wants to start its business. After taking into account the capabilities of salesmen and the nature of districts, the company estimates that the profit per day in rupees for each salesman in each district is as below:

Find the assignments of a salesman to various districts which will yield maximum profit.
Solution:
The profit matrix has to be reduced to the cost matrix. Subtracting all the values from the maximum value (16) we get

Subtracting row minimum from all values in that row we get

Subtracting column minimum from each column we get the same matrix

As minimum no. of lines covering all zeros (4) is equal to the order of the matrix (4) Assignment is possible

∴ A → 1, B → 3, C → 2, D → 4
For maximum profit, we take the corresponding values in the profit matrix. We get
Maximum profit = 16 + 15 + 15 + 15 = ₹ 61

Question 6.
In the modification of a plant layout of a factory four new machines M₁, M₂, M₃, and M₄ are to be installed in a machine shop. There are five vacant places A, B, C, D, and E available. Because of limited space, machine M₂ can not be placed at C and M₃ can not be placed at A the cost of locating a machine at a place (in hundred rupees) is as follows.

Find the optimal assignment schedule.
Solution:
This is an unbalanced problem so we add a dummy row M5 with all values as ‘0’.
Also, this is on restricted assignment problem. So we assign a very high-cost W to the prohibited cells we have

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get the same matrix

As minimum no. of lines covering all zeros (5) is equal to the order of the matrix (5) assignment is possible.

The assignment is
M₁ → A, M₂ → B, M₃ → E, M₄ → D, M₅ → C
As M5 is dummy, no machine is installed at C.
The minimum cost is found by taking the corresponding values in the cost matrix
Minimum cost = 9 + 9 + 7 + 7 + 0 = 32 (in hundred ₹)

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Ex 7.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Ex 7.2

Question 1.
A machine operator has to perform two operations, turning and threading on 6 different jobs. The time required to perform these operations (in minutes) for each job is known. Determine the order in which the jobs should be processed in order to minimize the total time required to complete all the jobs. Also, find the total processing time and idle times for turning and threading operations.

Solution:
Let turning to be A and threading be B.

∴ Observe Min{A, B} = 1 for job 6 on B.

Then the problem reduces to

∴ Now Min {A, B} = 2 for job 4 on A

Then the problem reduce to

Now Min {A, B} = 3 for job 1 on A and job 5 on B

Then the problem reduces to

Now Min {A, B) = 5 for job 3 on A

Only job 2 is left so the optimal sequence is

Worktable is given by

Total elapsed time = 43 minutes
Idle time for A (turning) = 43 – 42 = 1 min
Idle time for B (threshing) = 2 + 4 = 6 min

Question 2.
A company has three jobs on hand, Each of these must be processed through two departments, in the AB where
Department A: Press shop and
Department B: Finishing
The table below gives the number of days required by each job each department

Find the sequence in which the three jobs should be processed so as to take minimum time to finish all the three jobs. Also find idle time for both the departments.
Solution:
Observe Min {A, B} = 3 for job II on B.

Then the problem is reduced to

Now Min {A, B} = 4 for job III at B

Now only job I in left
∴ the optimal sequence is given by

The work table is

Total elapsed time = 23 days
Idle time for A = 23 – 19 = 4 days
Idle time for B = 8 days

Question 3.
An insurance company receives three types of policy application bundles daily from its head office for data entry and filing. The time (in minutes) required for each type for these two operations is given in the following table:

Find the sequence that minimizes the total time required to complete the entire task. Also, find the total elapsed time and idle times for each operation.
Solution:
Let Data entry be A and filing be B. So

Observe min {A, B} = 90 for policy 1 at A

Then the problem reduces to

Observe min {A, B} = 100 for policy 3 at B

Now only policy 2 is left
∴ The optimal sequence is

Worktable

So Total elapsed time = 490 min
Idle time for A (data entry) = 490 – 390 = 100 min
Idle time for B (filing) = 140 min.

Question 4.
There are five jobs, each of which must go through two machines in the order XY. Processing times (in hours) are given below. Determine the sequence for the jobs that will minimize the total elapsed time. Also, find the total elapse time and idle time for each machine.

Solution:
Observe min {x, y} = 2 for job B on x

The problem reduces to

Now min [x, y] = 4 for job A on x

The problem reduces to

Now min [x, y] = 6 for job D on x

The problem reduces to

Now min [x, y] = 8 for job E on y

Now only job C in left
∴ The optimal sequence is

Worktable

Total elapsed time = 60 hrs
Idle time for X = 60 – 56 = 4 hrs
Idle time for Y = 6 hrs

Question 5.
Find the sequence that minimizes the total elapsed time to complete the following jobs in the order AB. Find the total elapsed time and idle times for both machines.

Solution:
Observe min {A, B} = 5 for job VI for B and job VII for A

The problem reduces to

Now min {A, B] = 7 for job I on A

The problem reduces to

Now min {A, B] = 10 for job IV on A and B so we have two options.

Or

we take the 1st one.
The problem reduces to

Now min {A, B} = 14 for job V on A and job II and III for job B.
∴ We have

Or

We take the optimal sequence as.
VII – I – IV – V – III – II – VI
Worktable

Total elapsed time = 91 units
Idle time for A = 91 – 86 = 5 units
Idle time for B = 13 units

Question 6.
Find the optimal sequence that minimizes the total time required to complete the following jobs in the order ABC. The processing times are given in hrs.


Solution:
(i) Min A = 5, Max B = 5
As min A ≥ max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C we get

Now min {G, H} = 7 for job III & V for G and job I for H
∴ We have two options

Or

We take the first one
The problem reduces to

Min {G, H} = 9 for job IV on H

The problem reduces to

Now min {G, H} = 10 for job II for G and job VII for H

Now job VI is left
∴ The optimal sequence is

The work table is

Total elapsed time = 61 hrs
Idle time for A = 61 – 54 = 7 hrs
Idle time for B = 35 + [61 – 58] = 38 hrs
Idle time for C = 15 hrs

(ii) Min A = 5, Max B = 5
Min A ≥ Max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C we get

Now min {G, H} = 5 for job 1 for H.

The problem reduces to

Now min {G, H} = 8 for job 2 for G and job H also job 5 for G
∴ We have two options

Or

We take the first one
The problem reduces to

Now min {G, H} = 9 for job 3 for H

Now only job 4 is left
∴ The optimal sequence is

Worktable

Total elapsed time = 40 hrs
Idle time for A = 40 – 32 = 8 hrs
Idle time for B = 19 + [40 – 34] = 25 hrs
Idle time for C = 12 hrs

Question 7.
A publisher produces 5 books on Mathematics. The books have to go through composing, printing, and binding was done by 3 machines P, Q, E. The time schedule for the entire task in the proper unit is as follows.

Determine the optimum time required to finish the entire task.
Solution:
Min R = 6, Max Q = 6
As min R ≥ max Q.
The problem can be converted into a two-machine problem.
Let G and H be two fictitious machines such that G = P + Q and H = Q + R we get

Min {G, H} = 9 for books A, D, E for G.
∴ We have more than one option we take

The problem reduces to

Min {G, H} = 8 for book C on H

Now only B is left. So the optimal sequence is

Worktable

Total elapsed time = 51 units
Idle time for P = 51 – 32 = 19 units
Idle time for Q = 14 + [51 – 34] = 31 units
Idle time for R = 9 units

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Miscellaneous Exercise 6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Miscellaneous Exercise 6

(I) Choose the correct alternative.

Question 1.
The value of objective function is maximized under linear constraints.
(a) at the centre of feasible region
(b) at (0, 0)
(c) at any vertex of feasible region.
(d) The vertex which is at maximum distance from (0, 0).
Answer:
(a) at the centre of feasible region

Question 2.
Which of the following is correct?
(a) Every LPP has on optional solution
(b) Every LPP has unique optional solution
(c) If LPP has two optional solutions then it has infinitely many solutions
(d) The set of all feasible solutions LPP may not be a convex set.
Answer:
(c) If LPP has two optional solutions then it has infinitely many solutions

Question 3.
Objective function of LPP is
(a) a constraint
(b) a function to be maximized or minimized
(c) a relation between the decision variables
(d) a feasible region.
Answer:
(b) a function to be maximized or minimized

Question 4.
The maximum value of z = 5x + 3y. subject to the constraints 3x + 5y = 15; 5x + 2y ≤ 10, x, y ≥ 0 is
(a) 235
(b) \(\frac{235}{9}\)
(c) \(\frac{235}{19}\)
(d) \(\frac{235}{3}\)
Answer:
(c) \(\frac{235}{19}\)

Question 5.
The maximum value of z = 10x + 6y. subject to the constraints 3x + y ≤ 12, 2x + 5y ≤ 34, x ≥ 0, y ≥ 0 is.
(a) 56
(b) 65
(c) 55
(d) 66
Answer:
(a) 56

Question 6.
The point at which the maximum value of z = x + y subject to the constraint x + 2y ≤ 70, 2x + y ≤ 15, x ≥ 0, y ≥ 0 is
(a) (36, 25)
(b) (20, 35)
(c) (35, 20)
(d) (40, 15)
Answer:
(d) (40, 15)

Question 7.
Of all the points of the feasible region the optimal value of z is obtained at a point
(a) Inside the feasible region
(b) at the boundary of the feasible region
(c) at vertex of feasible region
(d) on x -axis
Answer:
(c) at vertex of feasible region

Question 8.
Feasible region; the set of points which satisfy
(a) The objective function
(b) All of the given function
(c) Some of the given constraints
(d) Only non-negative constraints
Answer:
(b) All of the given function

Question 9.
Solution of LPP to minimize z = 2x + 3y subjected to x ≥ 0, y ≥ 0, 1 ≤ x + 2y ≤ 10 is
(a) x = 0, y = \(\frac{1}{2}\)
(b) x = \(\frac{1}{2}\), y = 0
(c) x = 1, y = -2
(d) x = y = \(\frac{1}{2}\)
Answer:
(a) x = 0, y = \(\frac{1}{2}\)

Question 10.
The corner points of the feasible region given by the inequation x + y ≤ 4, 2x + y ≤ 7, x ≥ 0, y ≥ 0, are
(a) (0, 0), (4, 0), (3, 1), (0, 4)
(b) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 4)
(c) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (5, 7)
(d) (6, 0), (4, 0), (3, 1), (0, 7)
Answer:
(b) (0, 0), (\(\frac{7}{2}\), 0), (3, 1), (0, 4)

Question 11.
The corner point of the feasible region are (0, 0), (2, 0), (\(\frac{12}{7}\), \(\frac{3}{7}\)) and (0, 1) then the point of maximum z = 6.5x + y = 13
(a) (0, 0)
(b) (2, 0)
(c) (\(\frac{11}{7}\), \(\frac{3}{7}\))
(d) (0, 1)
Answer:
(b) (2, 0)

Question 12.
If the corner points of the feasible region are (0, 0), (3, 0), (2, 1) and (0, \(\frac{7}{3}\)) the maximum value of z = 4x + 5y is
(a) 12
(b) 13
(c) \(\frac{35}{2}\)
(d) 0
Answer:
(b) 13

Question 13.
If the comer points of the feasible region are (0, 10), (2, 2), and (4, 0) then the point of minimum z = 3x + 2y is.
(a) (2, 2)
(b) (0, 10)
(c) (4, 0)
(d) (2, 4)
Answer:
(a) (2, 2)

Question 14.
The half plane represented by 3x + 2y ≤ 0 contains the point.
(a) (1, \(\frac{5}{2}\))
(b) (2, 1)
(c) (0, 0)
(d) (5, 1)
Answer:
(c) (0, 0)

Question 15.
The half plane represented by 4x + 3y ≥ 14 contains the point
(a) (0, 0)
(b) (2, 2)
(c) (3, 4)
(d) (1, 1)
Answer:
(c) (3, 4)

(II) Fill in the blanks.

Question 1.
Graphical solution set of the in equations x ≥ 0, y ≥ 0 is in _________ quadrant.
Answer:
First

Question 2.
The region represented by the in equations x ≥ 0, y ≥ 0 lines in _________ quadrants.
Answer:
First

Question 3.
The optimal value of the objective function is attained at the _________ points of feasible region.
Answer:
End

Question 4.
The region represented by the inequality y ≤ 0 lies in _________ quadrants
Answer:
Third and Fourth

Question 5.
The constraint that a factory has to employ more women (y) than men (x) is given by _________
Answer:
y > x

Question 6.
A garage employs eight men to work in its showroom and repair shop. The constants that there must be not least 3 men in showroom and repair shop. The constrains that there must be at least 3 men in showroom and at least 2 men in repair shop are _________ and _________ respectively.
Answer:
x ≥ 3 and y ≥ 2

Question 7.
A train carries at least twice as many first class passengers (y) as second class passangers (x). The constraint is given by _________
Answer:
x ≥ 2y

Question 8.
A dishwashing machine hold up to 40 pieces of large crockery (x) this constraint is given by _________
Answer:
x ≤ 40

(III) State whether each of the following is True or False.

Question 1.
The region represented by the inequalities x ≥ 0, y ≥ 0 lies in first quadrant.
Answer:
True

Question 2.
The region represented by the inqualities x ≤ 0, y ≤ 0 lies in first quadrant.
Answer:
False

Question 3.
The optimum value of the objective function of LPP occurs at the center of the feasible region.
Answer:
False

Question 4.
Graphical solution set of x ≤ 0, y ≥ 0 in xy system lies in second quadrant.
Answer:
True

Question 5.
Saina wants to invest at most ₹ 24000 in bonds and fixed deposits. Mathematically this constraints is written as x + y ≤ 24000 where x is investment in bond and y is in fixed deposits.
Answer:
True

Question 6.
The point (1, 2) is not a vertex of the feasible region bounded by 2x + 3y ≤ 6, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0.
Answer:
True

Question 7.
The feasible solution of LPP belongs to only quadrant I. The Feasible region of graph x + y ≤ 1 and 2x + 2y ≥ 6 exists.
Answer:
True

(IV) Solve the following problems.

Question 1.
Maximize z = 5x₁ + 6x₂, Subject to 2x₁ + 3x₂ ≤ 18, 2x₁ + x₂ ≤ 12, x ≥ 0, y ≥ 0
Solution:


OAED is the feasible region O(0, 0) A(0, 6) D (6, 0)
and E is the intersection of 2x₁ + 3x₂ = 18 and 2x₁ + x₂ = 12
For E, Solving, 2x₁ + 3x₂ = 18 …….(i)
2x₁ + x₂ = 12 ……(ii)
We get x₁ = 4.5, x₂ = 3
∴ E = (4.5, 3)

∴ Maximum value of z = 40.5 at E(4.5, 3)

Question 2.
Minimize z = 4x + 2y, Subject to 3x + y ≥ 27, x + y ≥ 21, x ≥ 0, y ≥ 0
Solution:

AED is the feasible region A(0, 27), D(21, 0)
and E is the point of intersection of 3x + y = 27 and x + y = 21
For E, Solving 3x + y = 27 ………(i)
x + y = 21 …….(ii)
We get x = 3, y = 18
∴ E(3, 18)


∴ Minimum value of z = 48 at (3, 18)

Question 3.
Maximize z = 6x + 10y, subject to 3x + 5y ≤ 10, 5x + 3y ≤ 15, x ≥ 0, y ≥ 0
Solution:


OAED is the feasible region;
O(0, 0), A (0, 2) D (3, 0) and E is the point of intersection of 3x + 5y = 10 and 5x + 3y = 15
For E, Solving 3x + 5y = 10
5x + 3y = 15
We get, x = \(\frac{45}{16}\), y = \(\frac{5}{16}\)
∴ E(\(\frac{45}{16}\), \(\frac{5}{16}\))

Since the maximum value of z = 20 at two points i.e, at A(0, 2) and E(\(\frac{45}{16}\), \(\frac{5}{16}\)).
z is maximum at all points on segment AE.
Hence it has infinite number of solutions.

Question 4.
Minimize z = 2x + 3y, Subject to x – y ≤ 1, x + y ≥ 3, x ≥ 0, y ≥ 0
Solution:


Shaded portion CE is the feasible region
Where C = (0, 3) and E is the point of intersection of x – y = 1 and x + y = 3
For E, Solving x – y = 1 …….(i)
x + y = 3 ………(ii)
We get, x = 2, y = 1
∴ E(2, 1)

∴ Minimum value z = 7 at E(2, 1)

Question 5.
Maximize z = 4x₁ + 3x₂, Subject to 3x₁ + x₂ ≤ 15, 3x₁ + 4x₂ ≤ 24, x ≥ 0, y ≥ 0
Solution:


OCEB is the feasible region where O(0, 0), C(0, 6), B(5, 0)
E is the point of intersection of 3x₁ + x₂ = 15 and 3x₁ + 4x₂ = 24
For E, Solving 3x₁ + x₂ = 15 ……..(i)
3x₁ + 4x₂ = 24 …….(ii)
We get, x₁ = 4, x₂ = 3
∴ E(4, 3)

∴ Maximum value of z = 25 at (4, 3)

Question 6.
Maximize z = 60x + 50y, Subject to x + 2y ≤ 40, 3x + 2y ≤ 60, x ≥ 0, y ≥ 0
Solution:


OAED is the feasible region O(0, 0), A(0, 20), D(20, 0)
E is x + 2y = 40 and 3x + 2y = 60
For E, Solving x + 2y = 40
3x + 2y = 60
We get, x = 10, y = 15
∴ E = (10, 15)

∴ Maximum value of z = 1350 at E(10, 15).

Question 7.
Minimize z = 4x + 2y, Subject to 3x + y ≥ 27, x + y ≥ 21, x + 2y ≥ 30, x ≥ 0, y ≥ 0
Solution:


AGHF is the feasible region where A(0, 27) F(30, 0)
G is the point of intersection of 3x + y = 27 and x + y = 21
H is the point of intersection of x + y = 21 and x + 2y = 30
For G, Solving 3x + y = 27 …….(i)
x + y = 21 ………(ii)
We get, x = 3, y = 18
∴ G(3, 8)
For H, Solving x + y = 21 …….(i)
x + 2y = 30 ……..(ii)
We get, x = 12, y = 9
∴ G = (12, 9)

∴ Maximum value of z = 48 at G(3, 18)

Question 8.
A carpenter makes chairs and table profit are ₹ 140 per chair and ₹ 210 per table Both products are processed on three machines, Assembling, Finishing and Polishing the time required for each product in hours and availability of each machine is given by the following table.

Formulate and solve the following Linear programming problem using graphical method.
Solution:
Let z be the profit which can be made by selling x chair and y table.
∴ x ≥ 0, y ≥ 0
Total profit = 140x + 210y
According to the table, the constraints can be written as
3x + 3y ≤ 36
5x + 2y ≤ 50
2x + 6y ≤ 60
∴ The given LPP can be formulated as.
Maximize z = 140x + 210y
Subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.


OEGH is the feasible region
O(0, 0), D(10.4, 0), E(0, 10)
For G, Solving 3x + 3y = 36 ……..(i)
2x + 6y = 60 …….(ii)
∴ G = (3, 9)
For H, Solving 5x + 2y = 52 ………(i)
3x + 3y = 36 ……..(ii)
We get, x = \(\frac{28}{3}\), y = \(\frac{8}{3}\)
∴ H(\(\frac{28}{3}\), \(\frac{8}{3}\))

∴ z in maximum at (3, 9) and maximum profit = ₹ 2310.

Question 9.
A company manufactures bicycles and tricycles, each of which must be processed through two machines A and B Maximum availability of machines A and B are respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B. If profits are ₹ 180 for a bicycle and ₹ 220 on a tricycle, determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.
Solution:
Let x number of bicycle and y number of tricycle has to be manufactured and to be sold to get the profit (z)
∴ x ≥ 0, y ≥ 0
Total profit = 180x + 220y.
The given LPP can be tabulated as follows:

∴ The given LPP can be formulated as
Maximize z = 180x + 220y
Subject to 6x + 4y ≤ 120, 3x + 10y ≤ 180, x ≥ 0, y ≥ 0


OCEB is the feasible region where O(0, 0) C(0, 18) B(20, 0)
E is the point of intersection of 6x + 4y = 120 and 3x + 10y = 180
For E, Solving 6x + 4y = 120 ……..(i)
3x + 10y = 180 …….(ii)
We get, x = 10, y = 15
∴ E(10, 15)

∴ Maximum value of z is 5100 at E(10, 15)
Hence 10 bicycles and 15 tricycles should be produced to get maximum profit.

Question 10.
A factory produced two types of chemicals A and B. The following table gives the units of ingredients P and Q (per kg) of chemicals A and B as well as minimum requirements of P and Q and also cost per kg. Chemicals A and B:

Find the number of units of chemicals A and B should be produced sp as to minimize the cost.
Solution:
Let x be the no. of units of chemicals, A produced and y be the no. of units of chemical B produced.
Total cost is 4x + 6y
The LPP is. Minimise z = 4x + 6y
Subject to x + 2y ≥ 80, 3x + y ≥ 75, x ≥ 0, y ≥ 0.


The shaded region BEC in the feasible region B(80, 0) C(0, 75)
and E is the point of intersection of x + 2y = 80 and 3x + y = 75
For E, Solving x + 2y = 80 …….(i)
3x + y = 75 …….(ii)
We get, x = 14, y = 33
∴ E(14, 33)

∴ z is minimum at E(14, 33) and the minimum value of z = 254.
Hence 14 units of chemical A and 33 units of chemical B are produced to get a minimum cost of ₹ 254.

Question 11.
A company produces mixers and food processors. Profit on selling one mixer and one food processor is ₹ 2,000/- and ₹ 3,000/- respectively. Both the products are processed through three Machines A, B, C. The time required in hours by each product and total time available in hours per week on each machine are as follows:

How many mixes and food processors should be produced to maximize profit?
Solution:
Let x be the no. of mixers produced and y be the no.of food processors produced.
The profit is 2000x + 3000y
The LPP is Maximize 2 = 2000x + 3000y
Subject to 3x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60 x, y ≥ 0.
Let 3x + 3y = 36,
i.e. x + y = 12
x = 0, y = 12, (0, 12)
y = 0, x = 12, (12, 0)
5x + 2y = 50
x = 0, y = 25, (0, 25)
y = 0, x = 10, (10, 0)
Let 2x + 6y = 60, x + 3y = 30
x = 0, y = 10 (0, 10)
y = 0, x = 30 (30, 0)

The Shaded region OABCD is the feasible region.
O(0, 0) A (10, 0) D(0, 10)
B is the intersection of x + y = 12 and 5x + 2y = 50
Solving we get x = 8.67, y = 3.33
∴ B(8.67, 3.33)
C is the intersection of x + y = 12 and x + 3y = 30
Solving we get x = 3, y = 9
∴ C(3, 9)

∴ Maximum value of z is 330000 at C(3, 9)
Hence 3 mixers and 9 food processors should be produced to get a maximum profit of ₹ 33,000.

Question 12.
A chemical company produces a chemical containing three basic elements A, B, C so that it has at least 16 liters of A, 24 liters of B, and 18 liters of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 liters of A, 12 liters of B, 2 liters of C. Each unit of compound II has 2 liters of A, 2 liters of B and 6 liters of C. The cost per unit of compound is ₹ 800/- and that of compound II is ₹ 640/- Formulate the problem as L.P.P and solve it to minimize the cost.
Solution:
Let x be the no. of units of compound I used and y be the no of units of compound II used.
The data can be tabulated as

The LPP is minimize z = 800x + 640y
Subject to 4x + 2y ≥ 16, 12x + 2y ≥ 24, 2x + 6y ≥ 18, x, y ≥ 0.
Let 4x + 2 y = 16,
2x + y = 8
x = 0, y = 8, (0, 8)
y = 0, x = 4, (4, 0)
12x + 2y = 24,
6x + y = 12
x = 0, y = 12, (0, 12)
y = 0, x = 2, (2, 0)
2x + 6y = 18
x + 3y = 9
x = 0, y = 3, (0, 3)
y = 0, x = 9, (9, 0)

The Shaded region ABCD is the feasible region.
A(0, 12) D(9, 0)
B is the intersection of 6x + y = 12 and 2x + y = 8
Solving we get x = 1, y = 6
∴ B(1, 6)
C is the intersection of 2x + y = 8 and x + 3y = 9
Solving we get x = 3, y = 2
∴ C(3, 2)

∴ The minimum value of z = 3680 at (3, 2)
Hence 3 unit of compound I and 2 units of compound II should be used to get the minimum cost of ₹ 3680.

Question 13.
A person who makes two types of gift items A and B requires the services of a cutter and a finisher. Gift item A requires 4 hours of the cutter’s time and 2 hours of the finisher’s time. B required 2 hours of the cutter’s time and 4 hours of finisher‘s time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is ₹ 75/- and on gift item B is ₹ 125/-. Assuming that the person can sell all the gift items produced, determine how many gift items of each type should he make every month to obtain the best returns?
Solution:
Let x be the no. of gift A produced and y be the no. of gift B produced.
The data can be tabulated as

The LLP is maximize z = 75x + 125y
Subject to 4x + 2y ≤ 208, 2x + 4y ≤ 152, x, y ≥ 0
Let 4x + 2y = 208, 2x + 4y = 152,
2x + y = 104, x + 2y = 76,
x = 0, y = 104; (0, 104), x = 0, y = 38; (0, 38)
y = 0, x = 52; (52, 0), y = 0, x = 76; (76, 0)

The shaded region OABC is the feasible region.
O(0,0) A (52, 0) C(0, 38)
B is the intersection of 2x + y = 104 and x + 2y = 76
Solving we get x = 44, y = 16
∴ B(44, 16)

∴ Maximum value of z = 5300 at B(44, 16)
Hence he should produce 44 gifts of type A & 16 of type B to get a maximum profit of ₹ 5300.

Question 14.
A firm manufactures two products A and B on which profit is earned per unit ₹ 3/- and ₹ 4/- respectively. Each product is processed on two machines M₁ and M₂. Product A requires one minute of processing time on Mx and two minutes of processing time on M₂. B requires one minute of processing time on M₁ and one minute processing time on M₂ Machine M₁ is available for use for 450 minutes while M₂ is available for 600 minutes during any working day. Final the number of units of products A and B to be manufactured to get the maximum profit.
Solution:
Let x denote the number of units of product A and y denote the number of units of product B.
The total profit in ₹ 3x + 4y,
The given statements can be tabulated as

The constraints are x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0
∴ The LPP can be formulated as
Maximize z = 3x + 4y
Subject to x + y ≤ 450, 2x + y ≤ 600, x ≥ 0, y ≥ 0.


OAED in the feasible region O(0, 0) A(0, 450) D(300, 0)
E is the intersection of x + y = 450 and 2x + y = 600
For E, Solving x + y = 450 ……..(i)
2x + y = 600 ……..(ii)
Solving x = 150, y = 300
∴ E(150, 300)

∴ Maximum value of z = 1800 at (0, 450)

Question 15.
A firm manufacturing two types of electrical items A and B, can make a profit of ₹ 20/- per unit of A and ₹ 30/- per unit of B. Both A and B make use of two essential components a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each unit of B required 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should they manufacture per month to maximize profit? How much is the maximum profit?
Solution:
Let x be the no. of electrical item A and y be the no. of electrical items to be manufactured per month to maximize the profit.
Total profit is ₹ 20x + 30y
The given condition can be tabulated as.

The given LPP can be formulated as
Maximize z = 20x + 30y
Subject of 3x + 2y ≤ 210, 2x + 4y ≤ 300, x ≥ 0, y ≥ 0.


BOCE is the feasible region
B(70, 0) O(0, 0) C(0, 75)
E is the intersection of 3x + 2y = 210 and 2x + 4y = 300
For E, Solving 3x + 2y = 210 ……….(i)
2x + 4y = 300 …….(ii)
We get x = 30, y = 60
∴ E(30, 60)

∴ Maximum value of z = ₹ 2400 at (30, 60)
Hence 30 units of A and 60 units of B should be manufactured per month to get the maximum profit of ₹ 2400

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Ex 6.2

Solve the following LPP by graphical method.

Question 1.
Maximize z = 11x + 8y, Subject to x ≤ 4 ,y ≤ 6 x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution:


The feasible solution is AOBE
Where A(4, 0) O(0, 0) B(0, 6)
E is the point of intersection of x + y = 6 and x = 4.
∴ 4 + y = 6
∴ y = 2
∴ E = (4, 2)

∴ z is maximum at (4, 12) and the maximum value of z = 60

Question 2.
Maximize z = 4x + 6y, Subject to 3x + 2y ≤ 12, x + y ≥ 4 x, y ≥ 0.
Solution:


From figure, ABC is the feasible region
Where A(0, 6) B(4, 0) C(0, 4)

Maximum value of z = 36 at A(0, 6)

Question 3.
Maximize z = 7x + 11y, Subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:


∴ AODE is the feasible region where
A(0, 5.2) O(0, 0) D(6, 0) and E is the intersection of 3x + 5y = 26 and 5x + 3y = 30
For E,
Solving 3x + 5y = 26 ……(i)
5x + 3y = 30 ……(ii)
We get, x = 4.5, y = 2.5
∴ E = (4.5, 2.5)

∴ Maximum value of z = 59 at E(4.5, 2.5)

Question 4.
Maximize z = 10x + 25y, Subject to 0 ≤ x ≤ 3, 0≤ y ≤ 3, x + y ≤ 5.
Solution:
The constraints can be written as, x ≤ 3, x ≥ 0, y ≥ 0, y ≤ 3, x + y ≤ 5


ABCDE is the feasible region where A(3, 0) B(0, 0) and C(0, 3) D is the intersection of y = 3 and x + 5y = 5 and E is the intersection of x = 3 and x + 7 = 5
For D,
Solving y = 3 ………(i)
x + y = 5 ……..(ii)
We get x = 2, y = 3
∴ D = (2, 3)
For E,
Solving x = 3 …….(i)
x + y = 5 ……….(ii)
We get x = 3, y = 2
∴ E = (3, 2)

∴ Maximum value of z = 90 at D(2, 3)

Question 5.
Maximize z = 3x + 5y, Subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Solution:


OAGHD is the feasible region where O(0, 0), A(0, 6), D(7, 0) G is the intersecting point of x + 4y = 24 and x + y = 9
H is the intersecting points of 3x + y = 21 and x + y = 9.
For G, Solving x + 4y = 24 …….(i)
x + y = 9 ………(ii)
We get, x = 4, y = 5
∴ G (4, 5)
For H, Solving x + y = 9 ………(i)
3x + y = 21 ……..(ii)
We get x = 6, y = 3
∴ H(6, 3)

∴ Maximum value of z = 37 at the point G(4, 5)

Question 6.
Minimize z = 7x + y Subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
Solution:


AED is the feasible region where A(0, 5) D(3, 0) and E is the point of intersection of 5x + y = 5 and x + y = 3.
For E, Solving 5x + y = 5 ………(i)
x + y = 3 ……..(ii)
We get, x = \(\frac{1}{2}\), y = \(\frac{5}{2}\)
∴ E(\(\frac{1}{2}\), \(\frac{5}{2}\))

∴ Minimum value of z = 5 at A(0, 5)

Question 7.
Minimize z = 8x + 10y, Subject to 2x + y ≥ 7, 2x + 3y ≥ 15 ,y ≥ 2, x ≥ 0, y ≥ 0
Solution:


AEG is the feasible solution where A(0, 7)
F is the point of intersection of 2x + y = 7 and 2x + 3y = 15
G is the point of intersection of y = 2 and 2x + 3y = 15
For F, Solving 2x + y = 7 ……..(i)
2x + 3y = 15 ……..(ii)
We get x = \(\frac{3}{2}\), y = 4
∴ F = (\(\frac{3}{2}\), 4)
For G, Solving 2x + 3y = 15 ……..(i)
y = 2 …….(ii)
We get x = 4.5, y = 2
∴ G = (4.5, 2)

∴ Minimum value of z = 52 at F(\(\frac{3}{2}\), 4)

Question 8.
Minimize z = 6x + 2y, Subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution:


DGHB is the feasible region where D(0, 3), B(4, 0)
G is the point of intersection of 3x + y = 3 and x + 2y = 3 and H is the point of intersection of x + 2y = 3 and x + 4y = 4
For G, Solving 3x + y = 3 ……(i)
x + 2y = 3 ………(ii)
W e get x = \(\frac{3}{5}\), y = \(\frac{6}{5}\)
∴ G(\(\frac{3}{5}\), \(\frac{6}{5}\))
For H, Solving x + 2y = 3 ……..(i)
x + 4y = 4 ………(ii)
We get, x = 2, y = \(\frac{1}{2}\)
∴ H(2, \(\frac{1}{2}\))

∴ Minimum value of z = 22.5 at H(2, \(\frac{1}{2}\))

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 6 Linear Programming Ex 6.1

Question 1.
A manufacturing firm produces two types of gadgets A and B, which are first processed in the foundry and then sent to a machine shop for finishing. The number of man-hours of labour required in each shop for production of A and B and the number of man-hours available for the firm is as follows.

Profit on the sale of A is ₹ 30 and B ₹ 20 Per unit Formulate the LPP to have maximum profit.
Solution:
Let the manufacturing firm produce x gadgets of type A and y gadgets of type B.
On selling x gadgets of type A the firm gets ₹ 30 and that on type B is ₹ 20.
∴ Total profit is z = ₹ 30x + 20y.
Since x and y are the numbers of gadgets, x ≥ 0, y ≥ 0
From the given table, the availability of man-hours of labour required in each shop and for the firm is given as 60 and 35.
∴ The inequation are 10x + 6y ≤ 60 and 5x + 4y ≤ 35.
Hence the given LPP can be formulated as Maximize z = 30x + 20y
Subject to 10x + 6y ≤ 60, 5x + 4y ≤ 35, x ≥ 0, y ≥ 0.

Question 2.
In a cattle breeding farm, it is prescribed that the food ratio for one animal must contain 14, 22, and 1 unit of nutrients A, B, and C respectively. Two different kinds of fodder are available. Each unit weight of these two contains the following:

The cost of fodder 1 is ₹ 3 per unit and that of fodder 2 is ₹ 2 per unit. Formulate the LPP to minimize the cost.
Solution:
Let x unit of fodder 1 and y unit of fodder 2 be included in the ration of an animal
The cost of 1 unit of fodder 1 is ₹ 3 and the cost of 1 unit of fodder 2 is ₹ 2.
∴ The total cost is ₹ 3x + 2y.
The minimum requirement of the nutrients A, B, and C is given as 14 units, 22 units, and 1 unit.
∴ From the given table, the daily food ration will include (2x + 2y) unit of Nutrient A, (2x + 3y) unit of Nutrient B, and (x + y) of Nutrient C.
The total cost is 2 = ₹ 3x + 2y
Hence the given LPP can be formulated as Minimize z = 3x + 2y
subject to 2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1, x ≥ 0, y ≥ 0.

Question 3.
A Company manufactures two types of chemicals A and B. Each chemical requires two types of raw materials P and Q. The table below shows a number of units of P and Q required to manufacture one unit of A and one unit of B.

The company gets profits of ₹ 350/- and ₹ 400/- by selling one unit of A and one unit of B respectively. Formulate the problem as LPP to maximize the profit.
Solution:
∴ Let the company manufactures x unit of chemical A and y unit of chemical B.
The availability of the raw materials for the production of chemicals A and B are given as 120 and 160 units.
The company gets ₹ 350 as profit on selling one unit of chemical A and ₹ 400 as profit on selling one unit of chemical B.
∴ Total profit is ₹ (350x + 400y).
The inequation can be written as.
3x + 2y ≤ 120
2x + 5y ≤ 160
and x & y cannot be negative
Hence the LPP can be formulated as follows,
Maximize z = 350x + 400y
Subject to 3x + 2y ≤ 120, 2x + 5y ≤ 160, x ≥ 0, y ≥ 0.

Question 4.
A printing company prints two types of magazines A and B. The company earns ₹ 10 and ₹ 15 on magazines A and B per copy. These are processed on three machines I, II, III. Magazine A requires 2 hours on the machine I, 5 hours on machine II, and 2 hours on machine III. Magazine B requires 3 hours on machine 1, 2 hours on machine II and 6 hours on machine III. Machines I, II, III are available for 36, 50, 60 hours per week respective. Formulate the linear programming problem to maximize the profit.
Solution:
Let the company print x magazine of type A and y magazines of type B.
Then the total earnings of the company are ₹ 10x + 15y.
The given problem can be tabulated as follows.

From the table, the total time required for Machine I is (2x + 3y) hours, for machine II is (5x + 2y) hours, and for machine III is (2x + 6y) hours.
The machine I, II, and III are available for 36, 50, and 60 hours per work.
∴ The constraints are 2x + 3y ≤ 36, 5x + 2y ≤ 50 and 2x + 6y ≤ 60.
Since x and y cannot be negative, we have x ≥ 0, y ≥ 0.
Hence the given LPP can be formulated as
Maximize z = 10x + 15y
Subject to 2x + 3y ≤ 36, 5x + 2y ≤ 50, 2x + 6y ≤ 60, x ≥ 0, y ≥ 0.

Question 5.
Manufacture produces bulbs and tubes. Each of these must be processed through two machines M₁ and M₂. A package of bulbs requires 1 hour of work on machine M₁ and 3 hours of work on M₂. A package of tubes requires 2 hours on machine M₁ and 4 hours on machine M₂. He earns a profit of ₹ 13.5 per package of bulbs and ₹ 55 per package of tubes. Formulate the LPP to maximize the profit. He operates M₁ for at most 10 hours and M₂ for at most 12 hours a day.
Solution:
Let the manufacturer produce x packages of bulbs and y packages of tubes.
He earns a profit of ₹ 13.5 per packages of bulbs and ₹ 55 per package of tubes.
∴ His total profit = ₹ (13.5x + 55y).
The given problem can be tabulated as follows.

From the above table, the total time required for M₁ is (x + 2y), and that of M₂ is (3x + 4y).
M₁ and M₂ are available for at most 10 hrs per day and 12 hours per day.
∴ The constraint for the objective function is x + 2y ≤ 10, 3x + 4y ≤ 12
Hence the give LPP can be formulated as
Maximize z = 13.5x + 55y
Subject to x + 2y ≤ 10, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0.

Question 6.
A Company manufactures two types of fertilizers F₁ and F₂. Each type of fertilizer requires two raw materials A and B. The number of units of A and B required to manufacture one unit of fertilizer F₁ and F₂ and availability of the raw materials A and B per day are given in the table below

By selling one unit of F₁ and one unit of F₂, the company gets a profit of ₹ 500 and ₹ 750 respectively. Formulate the problem as LPP to maximize the profit.
Solution:
Let the company manufacture x units of Fertilizers F₁ and y units of fertilizer F₂.
The company gets a profit of ₹ 500 and ₹ 750 by selling a unit of F₁ and F₂.
∴ Total profit = ₹ (500x + 750y)
The availability of raw materials A and B per day is given as 40 and 70.
∴ From the given table the constraints can be written as 2x + 3y ≤ 40 and x + 4y ≤ 70.
Since x & y cannot be negative, x ≥ 0, y ≥ 0
Hence the given LPP can be formulated as
Maximize z = 500x + 750y
Subject to 2x + 3y ≤ 40, x + 4y ≤ 70, x ≥ 0, y ≥ 0.

Question 7.
A doctor has prescribed two different kinds of feeds A and B to form a weekly diet for a sick person. The minimum requirement of fats, carbohydrates, and proteins are 18, 28,14 units respectively. One unit of food A has 4 units of fat, 14 units of carbohydrates, and 8 units of protein. One unit of food B has 6 units of fat, 12 units of carbohydrates and 8 units of protein. The price of food A is ₹ 4.5 per unit and that of food B is ₹ 3.5 per unit. Form the LPP so that the sick person’s diet meets the requirements at minimum cost.
Solution:
Let x unit of food A and y unit of food B be consumed by a sick person.
The cost of food A in ₹ 4.5 per unit and food B is ₹ 3.5 per unit.
∴ Total cost = ₹ (4.5x + 3.5y)
The given conditions can be tabulated as follows.

∴ The given LPP can be formulated as
Minimise z = 4.5x + 3.5y
Subject to 4x + 6y ≥ 18, 14x + 12y ≥ 28, 8x + 8y ≥ 14, x ≥ 0, y ≥ 0.

Question 8.
If John drives a car at a speed of 60 kms/hour he has to spend ₹ 5 per km on petrol. If he drives at a faster speed of 90 km/ hour, the cost of petrol increases to ₹ 8 per km. He has ₹ 600 to spend on petrol and wishes to travel the maximum distance within an hour. Formulate the above problem as LPP.
Solution:
Let John drive x km at a speed of 60 km/hr and y km at a speed of 90 km/hr.
∴ Time required to drive a distance of x km is \(\frac{x}{60}\) hours and the time require to drive at a distance of y km is \(\frac{y}{90}\) hours.
∴ Total time required \(\left(\frac{x}{60}+\frac{y}{90}\right)\) hours.
Since he wishes to drive maximum distance within an hour,
\(\frac{x}{60}+\frac{y}{90} \leq 1\)
He has to spend ₹ 5 per km at a speed of 60 km/hr and ₹ 8 per km at a speed of 90 km/hr.
He has ₹ 600 on petrol to spend, 5x + 8y ≤ 600
The total distance he wishes to travel is (x + y) hours.
∴ The given LPP can be formulated as
Maximize z = x + y
Subject to \(\frac{x}{60}+\frac{y}{90}\) ≤ 1, 5x + 8y ≤ 600, x ≥ 0, y ≥ 0.

Question 9.
The company makes concrete bricks made up of cement and sand. The weight of a concrete brick has to be at least 5 kg. Cement costs ₹ 20 per kg. and sand costs ₹ 6 per kg. Strength considerations dictate that a concrete brick should contain a minimum of 4 kg of cement and not more than 2 kg of sand. Formulate the LPP for the cost to be minimum.
Solution:
Let the concrete brick contain x kg of cement and y kg of sand.
The cost of cement is ₹ 20 per kg and sand is ₹ 6 per kg.
∴ The total cost = ₹ (20x + 6y)
Since the weight of the concrete brick has to be at least 5 kg, therefore, x + y ≥ 5
Also, the concrete brick should contain a minimum of 4 kg of cement, i.e. x ≥ 4, and not more than 2 kg of sand, i.e, y ≤ 2.
∴ The LPP can be formulated as
Minimize z = 20x + 6y
Subject to x + y ≥ 5, x ≥ 4, y ≤ 2, x ≥ 0, y ≥ 0.

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Commission, Brokerage and Discount Ex 1.2

Question 1.
What is the present worth of a sum of ₹ 10,920 due six months hence at 8% p.a simple interest?
Solution:
Given, SD = ₹ 10,920
n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
r = 8%
We have,

Thus the present worth is ₹ 10,500

Question 2.
What is the sum due of ₹ 8,000 due 4 months at 12.5% simple interest?
Solution:
Given, PW = ₹ 8,000, n = \(\frac{4}{12}\) year = \(\frac{1}{3}\) year, r = 12.5%
We have,

Thus, the sum due is ₹ 8,333.33

Question 3.
The true discount on the sum due 8 months hence at 12% p.a. is ₹ 560. Find the sum due and present worth of the bill.
Solution:
Given, TD = ₹ 560, n = \(\frac{8}{12}\) year = \(\frac{2}{3}\) year, r = 12%
We have,
TD = \(\frac{\mathrm{PW} \times n \times r}{100}\)
∴ 560 = \(\frac{\mathrm{PW} \times 2 \times 12}{3 \times 100}\)
∴ PW = 560 × \(\frac{25}{2}\) = ₹ 7,000
Now, SD = PW + TD
= 7,000 + 560
= ₹ 7,560

Question 4.
The true discount on a sum is \(\frac{3}{8}\) of the sum due at 12% p.a. Find the period of the bill.
Solution:

8 × n × 12 = 3(100 + n × 12)
96n = 300 + 36n
60n = 300
∴ n = 5
∴ Period of the bill = 5 years.

Question 5.
20 copies of a book can be purchased for a certain sum payable at the end of 6 months and 21 copies for the same sum in ready cash. Find the rate of interest.
Solution:
Given, n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
Let the sum payable be ₹ x
Let the rate of interest be r%
According to given condition,
PW of one book = \(\frac{x}{21}\)
SD of one book = \(\frac{x}{20}\)

Thus, the rate of interest is 10%.

Question 6.
Find the true discount, Banker’s discount, and Banker’s gain on a bill of ₹ 4,240 due 6 months hence at 9% p.a.
Solution:

And, Banker’s Gain (BG) = BD – TD
= 190.80 – 182.58
= ₹ 8.22

Question 7.
The true discount on a bill is ₹ 2,200 and bankers discount is ₹ 2,310. If the bill is due 10 months, hence, find the rate of interest.
Solution:
Given, TD = ₹ 2,200, BD = ₹ 2,310
n = \(\frac{10}{12}=\frac{5}{6}\) year

∴ \(\frac{r}{120}=\frac{1}{20}\)
∴ r = 6%
Thus, rate of interest is 6%

Question 8.
A bill of ₹ 6,395 drawn on 19th January 2015 for 8 months was discounted on 28th February 2015 at 8% p.a. interest. What is the banker’s discount? What is the cash value of the bill?
Solution:
Face value = ₹ 6,395
Date of drawing = 19/01/2015
Period of the bill = 8 months
Nominal Due date = 19/09/2015
Legal due date = 22/09/2015
Date of discounting = 28/02/2015
Now, the unexpired period = Legal due date – Date of discounting
= 22/09/2015 – 28/02/2015
= days (as shown below)

Cash Value = FV – BD
= 6,395 – 313.12
= ₹ 6,621.38

Question 9.
A bill of ₹ 8,000 drawn on 5th January 1998 for 8 months was discounted for ₹ 7,680 on a certain date. Find the date on which it was discounted at 10% p.a.
Solution:
Bankers discount (BD) = FV – cash value
= 8,000 – 7,680
= ₹ 320
Let the unexpired period be x days
∴ BD = \(\frac{\mathrm{FV} \times x \times r}{365 \times 100}\)
∴ 320 = \(\frac{8,000 \times x \times 10}{365 \times 100}\)
∴ x = 146 days
∴ The unexpired days = 146 days
Date of drawing = 05/01/1998
Period of bill = 8 months
Nominal due date = 05/09/1998
Legal due date = 08/09/1998
Thus, the date of discounting is 146 days before the legal due date

∴ Date of discounting of the bill is 15th April 1998

Question 10.
A bill drawn on 5th June for 6 months was discounted at the rate of 5% p.a. on 19th October. If the cash value of the bill is ₹ 43,500, find the face value of the bill.
Solution:
Date of drawing = 5th June
Period of bill = 6 months
Nominal due date = 5th December
Legal due date = 8th December
Date of discounting = 19th October
Rate of interest = 5% p.a.
Let the face value of the bill be ₹ x
The unexpired period

Question 11.
A bill was drawn on 14th April for ₹ 7,000 and was discounted on 6th July at 5% p.a. The Banker paid ₹ 6,930 for the bill. Find the period of the bill.
Solution:
Face value = ₹ 7,000, cash value = ₹ 6,930
∴ Banker’s discount = 7,000 – 6,930 = ₹ 70
Date of drawing = 14/04
Date of discounting = 06/07
Rate of interest = 5%
Let the unexpired period = x days
∴ BD = \(\frac{7,000 \times x \times 5}{365 \times 100}\)
∴ 70 = \(\frac{70 \times x}{73}\)
∴ x = 73 days
∴ Legal due date of the bill is 73 days after the date of discounting.

∴ Legal due date = 17/09
∴ Nominal due date = 14/09
∴ Period of the bill = 5 months

Question 12.
If the difference between true discount and banker’s discount on a sum due 4 months hence is ₹ 20. Find true discount, banker’s discount and amount of bill, the rate of simple interest charged is 5% p.a.
Solution:
Banker’s gain (BG) = Banker’s discount (BD) – True Discount (TD)
∴ BG = ₹ 20
Also, BG = \(\frac{\mathrm{TD} \times n \times r}{100}\)
∴ 20 = \(\frac{\mathrm{TD} \times 4 \times 5}{12 \times 100}\)
∴ 20 = \(\frac{\mathrm{TD}}{60}\)
∴ TD = ₹ 1200
Now, BD = BG + TD
= 20 + 1,200
= ₹ 1,220
Also, BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 1,220 = \(\frac{\mathrm{FV} \times 4 \times 5}{12 \times 100}\)
∴ FV = 1,200 × 60 = ₹ 73,200
∴ Amounting the bill = ₹ 73,200

Question 13.
A bill of ₹ 51,000 was drawn on 18th February 2010 for 9 months. It was encashed on 28th June 2010 at 5% p.a. Calculate the banker’s gain and true discount.
Solution:
Face Value = ₹ 51,000
Date of drawing = 18/02/2010
Period of the bill = 9 months
Nominal due date = 18/11/2010
Legal due date = 21/11/2010
Date of discounting = 28/06/2010
Unexpired period

∴ TD = ₹ 1,000
∴ BG = BD – TD
= 1,020 – 1,000
= ₹ 20

Question 14.
A certain sum due 3 months hence is \(\frac{21}{20}\) of the present worth, what is the rate of interest.
Solution:

Question 15.
A bill of a certain sum drawn on 28th February 2007 for 8 months was encashed on 26th March 2007 for ₹ 10,992 at 14% p.a. Find the face value of the bill.
Solution:
Date drawing = 28/02/2007
Period of the bill = 8 months
Nominal due date = 28/10/2007
Legal due date = 31/10/2007
Date of discounting = 26/03/2007
Cash value = ₹ 10,992
Rate of interest = 14%
Let face value of the bill = ₹ x
Bankers discount = Face value – Cash value = x – 10,992
Also, Banker s discount = \(\frac{F V \times n \times r}{365 \times 100}\)
Where n is the unexpired days

Thus face value of the bill = ₹ 12,000