Class 12

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 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 machines 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 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 the 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 Solution Book Pdf Chapter 5 Index Numbers Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Index Numbers Miscellaneous Exercise 5

(I) Choose the correct alternative.

Question 1.
Price Index Number by Simple Aggregate method is given by

Answer:
(c) \(\frac{\sum p_{1}}{\sum p_{0}} \times 100\)

Question 2.
Quantity Index Number by Simple Aggregate Method is given by

Answer:
(c) \(\frac{\sum q_{1}}{\sum q_{0}} \times 100\)

Question 3.
Value Index Number by Simple Aggregate Method is given by

Answer:
(b) \(\sum \frac{p_{0} q_{1}}{p_{0} q_{0}} \times 100\)

Question 4.
Price Index Number by Weighted Aggregate Method is given by

Answer:
(c) \(\frac{\sum p_{1} w}{\sum p_{0} w} \times 100\)

Question 5.
Quantity Index Number By Weighted Aggregate Method is given by

Answer:
(c) \(\frac{\sum q_{1} w}{\sum q_{0} w} \times 100\)

Question 6.
Value Index Number by Weighted aggregate Method is given by

Answer:
(d) \(\frac{\sum p_{1} q_{1} w}{\sum p_{0} q_{0} w} \times 100\)

Question 7.
Laspeyre’s Price Index Number is given by

Answer:
(c) \(\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times 100\)

Question 8.
Paassche’s Price Index Number is given by

Answer:
(d) \(\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}} \times 100\)

Question 9.
Dorbish-Bowley’s Price Index Number is given by


Answer:
(c) \(\frac{\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}}+\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}}}{2} \times 100\)

Question 10.
Fisher’s Price Number is given by

Answer:
(a) \(\sqrt{\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times \frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}}} \times 100\)

Question 11.
Marshall-Edge worth’s Price Index Number is given by

Answer:
(a) \(\frac{\sum p_{1}\left(q_{0}+q_{1}\right)}{\sum p_{0}\left(q_{0}+q_{1}\right)} \times 100\)

Question 12.
Walsh’s Price Index Number is given by


Answer:
(a) \(\frac{\sum p_{1} \sqrt{q_{0} q_{1}}}{\sum p_{0} \sqrt{q_{0} q_{1}}} \times 100\)

Question 13.
The Cost of Living Index Number using Aggregate Expenditure Method is given by

Answer:
(a) \(\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times 100\)

Question 14.
The Cost of Living Index Number using Weighted Relative Method is given by

Answer:
(a) \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\)

(II) Fill in the blanks.

Question 1.
Price Index Number by Simple Aggregate Method is given by ____________
Answer:
\(\frac{\sum p_{1}}{\sum p_{0}} \times 100\)

Question 2.
Quantity Index number by Simple Aggregate Method is given by ____________
Answer:
\(\frac{\sum q_{1}}{\sum q_{0}} \times 100\)

Question 3.
Value Index Number by Simple Aggregate Method is given by ____________
Answer:
\(\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{0}} \times 100\)

Question 4.
Price Index Number by Weighted Aggregate Method is given by ____________
Answer:
\(\frac{\sum p_{1} w}{\sum p_{0} w} \times 100\)

Question 5.
Quantity Index Number by Weighted Aggregate Method is given by ____________
Answer:
\(\frac{\sum q_{1} w}{\sum q_{0} w} \times 100\)

Question 6.
Value Index Number by Weighted Aggregate Method is given by ____________
Answer:
\(\frac{\sum p_{1} q_{1} w}{\sum p_{0} q_{0} w} \times 100\)

Question 7.
Laspeyre’s Price Index Number is given by ____________
Answer:
\(\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times 100\)

Question 8.
Paasche’s Price Index Number is given by ____________
Answer:
\(\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}} \times 100\)

Question 9.
Dorbish-Bowley’s Price Index Number is given by ____________
Answer:
\(\frac{1}{2}\left[\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}}+\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}}\right] \times 100\)

Question 10.
Fisher’s Price Index Number is given by ____________
Answer:
\(\sqrt{\left[\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times \frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}}\right]} \times 100\)

Question 11.
Marshall-Edgeworth’s Price Index Number is given my ____________
Answer:
\(\frac{\sum p_{1}\left(q_{0}+q_{1}\right)}{\sum p_{0}\left(q_{0}+q_{1}\right)} \times 100\)

Question 12.
Walsh’s Price Index Number is given by ____________
Answer:
\(\frac{\sum p_{1} \sqrt{q_{0} q_{1}}}{\sum p_{0} \sqrt{q_{0} q_{1}}} \times 100\)

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

Question 1
\(\frac{\sum p_{1}}{\sum p_{0}} \times 100\) is the Price Index Number by Simple Aggregate Method.
Answer:
True

Question 2
\(\frac{\sum q_{0}}{\sum q_{1}} \times 100\) is the Quantity Index Number by Simple Aggregate Method.
Answer:
False

Question 3.
\(\sum \frac{p_{0} q_{0}}{p_{1} q_{1}} \times 100\) is value Index Number by Simple Aggregate Method.
Answer:
False

Question 4.
\(\sum \frac{p_{1} q_{0}}{p_{1} q_{1}} \times 100\) Paasche’s Price Index Number.
Answer:
False

Question 5.
\(\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}} \times 100\) is Laspeyre’s Price Index Number.
Answer:
False

Question 6.
\(\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times \frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}} \times 100\) is Dorbish-Bowley’s Index Number.
Answer:
False

Question 7.
\(\frac{1}{2}\left[\sqrt{\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}}}+\frac{\sqrt{p_{1} q_{1}}}{\sqrt{p_{0} q_{1}}}\right] \times 100\) is Fisher’s Price Index Number.
Answer:
False

Question 8.
\(\frac{\sum p_{0}\left(q_{0}+q_{1}\right)}{\sum p_{1}\left(q_{0}+q_{1}\right)} \times 100\) is Marshall-Edgeworth’s Index Number.
Answer:
False

Question 9.
\(\frac{\sum p_{0} \sqrt{q_{0} q_{1}}}{\sum p_{1} \sqrt{q_{0} q_{1}}} \times 100\) is Walsh’s Price Index Number.
Answer:
False

Question 10.
\(\sqrt{\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}}} \times \sqrt{\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}}} \times 100\) is Fisher’s Price Index Number.
Answer:
True

(IV) Solve the following problems.

Question 1.
Find the price Index Number using simple Aggregate Method Consider 1980 as base year.

Solution:

Question 2.
Find the Quantity Index Number using Simple Aggregate Method.

Solution:

Question 3.
Find the Value Index Number using Simple Aggregate Method.

Solution:

= \(\frac{10200}{8400}\) × 100
= 121.43

Question 4.
Find x if the Price Index Number using Simple Aggregate Method is 200.

Solution:

Question 5.
Calculate Laspeyre’s and Paasche’s Price Index Number for the following data.

Solution:

Question 6.
Calculate Dorbish-Bowley’s Price Index Number for the following data.

Solution:

Question 7.
Calculate Marshall-Edge worth’s Price Index Number for the following data.

Solution:

Question 8.
Calculate Walsh’s Price Index Number for the following data.

Solution:

Question 9.
Calculate Laspeyre’s Price Index Number for the following data.

Solution:

Question 10.
Find x if Laseyre’s Price Index Number is same as Paasche’s Price Index Number for the following data.

Solution:

Question 11.
Find x if Walsh’s Price Index Number is 150 for the following data.

Solution:

Question 12.
Find x if Paasche’s Price Index Number is 140 for the following data.

Solution:

Question 13.
Given that Laspeyre’s and Paasche’s Index Number are 25 and 16 respectively. Find Dorbish-Bowley’s and Fisher’s Price Index Number.
Solution:

= \(\sqrt{25 \times 16}\)
= 20

Question 14.
If Laspeyre’s and Dorbish Price Index Number are 150.2 and 152.8 respectively, find Paasche’s rice Index Number.
Solution:

Question 15.
If Σp₀q₀ = 120, Σp₀q₁ = 160, Σp₁q₁ = 140, and Σp₁q₀ = 200 find Laspeyre’s, Paasche’s, Dorbish-Bowley’s, and Marshall-Edgeworth’s Price Index Numbers.
Solution:

Question 16.
Given that Σp₀q₀ = 130, Σp₁q₁ = 140, Σp₀q₁ = 160, and Σp₁q₀ = 200, find Laspeyare’s, Passche’s, Dorbish-Bowely’s and Mashall-Edegeworth’s Price Inbox Numbers.
Solution:

Question 17.
Given that Σp₁q₁ = 300, Σp₀q₁ = 140, Σp₀q₀ = 120, and Marshall-Edegeworth’s Price Inbox Number is 120, find Laspeyre’s Price Index Number.
Solution:
p₀₁(P) = \(\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}} \times 100\)
= \(\frac{300}{320}\) × 100
= 93.75

Question 18.
Calculate the cost of living number for the following data.

Solution:

Question 19.
Find the cost living index number by the weighted aggregate method.

Solution:

Question 20.
Find the cost of living index number by Family Budget Method for the following data. Also, find the expenditure of a person in the year 2008 if his expenditure in the year 2005 was ₹ 10,000.

Solution:

Question 21.
Find x if cost of living index number is 193 for the following data.

Solution:

Question 22.
The cost of living number for year 2000 and 2003 are 150 and 210 respectively. A person earns ₹ 13,500 per month in the year 2000. What should be his monthly earning in the year 2003 in order to maintain the same standard of living?
Solution:
CLI (2000) = 150
CLI (2003) = 210
Income (2000) = 13500
Income (2003) = ?
Real Income = \(\frac{\text { Income }}{\mathrm{CLI}} \times 100\)
For 2000, Real Income = \(\frac{13500}{150} \times 100\) = ₹ 9000
For 2003, Real Income = \(\frac{\text { Income }}{\mathrm{CLI}} \times 100\)
∴ 9000 = \(\frac{\text { Income }}{210} \times 100\)
∴ Income = \(\frac{9000 \times 210}{100}\) = 18900
∴ Income in 2003 = ₹ 18900

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 5 Index Numbers Ex 5.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Index Numbers Ex 5.3

Calculate the cost of living index in problems 1 to 3.

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Base year weights (W) and current year price relatives (I) are given in problems 4 to 8. Calculate the cost of living index in each case.

Question 4.

Solution:

ΣW = 20, ΣIW = 1540
CLI = \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\)
= \(\frac{1540}{20}\)
= 77

Question 5.

Solution:

ΣW = 17, ΣIW = 3500
CLI = \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\)
= \(\frac{3500}{17}\)
= 205.88

Question 6.

Solution:

ΣW = 150, ΣIW = 25400
CLI = \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\)
= \(\frac{25400}{150}\)
= 169.33

Question 7.

Solution:

ΣW = x + 18, ΣIW = 100x + 3600
CLI = 150
∴ \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\) = 150
∴ \(\frac{100 x+3600}{x+18}\) = 150
∴ 100x + 3600 = 150x + 2700
∴ 50x = 900
∴ x = 18

Question 8.
Find y if the cost of living index is 200

Solution:

ΣW = y + 18, ΣIW = 300y + 2200
CLI = 200
∴ \(\frac{\sum \mathrm{IW}}{\sum \mathrm{W}}\) = 200
∴ \(\frac{300 y+2200}{y+14}\) = 200
∴ 300y + 2200 = 200y + 2800
∴ 100y = 600
∴ y = 6

Question 9.
The cost of living Index numbers for years 1995 and 1999 are 140 and 200 respectively. A person earns ₹ 11,200 per month in the year 1995. What should be his monthly earning in the year 1999 in order to maintain his standard of living as in the year 1995?
Solution:
CLI (1995) = 140
CLI (1999) = 200
Income (1995) = 11200
Income (1999) = ?
For year 1995
∴ Real Income = \(\frac{\text { Income }}{\text { CLI }} \times 100\)
= \(\frac{11200}{140}\) × 100
= ₹ 8000
For year 1999
∴ Real Income = \(\frac{\text { Income }}{\text { CLI }} \times 100\)
∴ 8000 = \(\frac{\text { Income } \times 100}{200}\)
∴ Income = 16000
∴ Income in 1999 = ₹ 16000

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 5 Index Numbers Ex 5.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Index Numbers Ex 5.2

Calculate Laspeyres, Paasche’s, Dorbish-Bowely’s, and Marshall-Edegworth’s Price Index Numbers in Problems 1 and 2.

Question 1.

Solution:

Question 2.

Solution:

Calculate Walsh’s Price Index Number in Problems 3 and 4.

Question 3.

Solution:

Question 4.

Solution:

Question 5.
If p₀₁(L) = 90, and p₀₁(P) = 40, find p₀₁(D – B) and p₀₁(F)
Solution:

Question 6.
If Σp₀q₀ = 140, Σp₀q₁ = 200, Σp₁q₀ = 350, Σp₁q₁ = 460, find Laspeyre’s Paasche’s Dorbish-Bowley’s and Marshall- Edgeworth’s Price Index Numbers.
Solution:

Question 7.
Given that Laspeyre’s and Dorbish-Bowley’s Price Index Numbers are 160.32 and 164.18 respectively. Find Paasche’s Price Index Number.
Solution:

Question 8.
Given that Σp₀q₀ = 220, Σp₀q₁ = 380, Σp₁q₁ = 350 is Marshall-Edgeworth’s Price Index Number is 150, find Laspeyre’s Price Index Number.
Solution:

Question 9.
Find x in the following table if Laspeyres and Paasche’s Price Index Numbers are equal.

Solution:

Question 10.
If Laspeyre’s Price Index Number is four times Paasche’s Price Index Number, then find the relation between Dorbish-Bowley’s and Fisher’s Price Index Numbers.
Solution:

Question 11.
If Dorbish-Bowley’s and Fisher’s Price Index Numbers are 5 and 4, respectively, then find Laspeyres and Paasche’s Price Index Numbers.
Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 5 Index Numbers Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Index Numbers Ex 5.1

Find the Price Index Number using the Simple Aggregate Method in each of the following examples.

Question 1.
Use 1995 as the base year in the following problem.

Solution:

Question 2.
Use 1995 as the base year in the following problem.

Solution:

Question 3.

Solution:

Question 4.
Use 2000 as the base year in the following problem.

Solution:

Question 5.
Use 1990 as the base year in the following problem.

Solution:

Question 6.
Assume 2000 to be a base year in the following problem.

Solution:

Question 7.
Use 2005 as a year in the following problem.

Solution:

Find the Quantity Index Number using the Simple Aggregate Method in each of the following examples.

Question 8.

Solution:

Question 9.

Solution:

Find the value Index Number using the Simple Aggregate Method in each of the following examples.

Question 10.

Solution:

= \(\frac{3660}{2840}\) × 100
= 128.87

Question 11.

Solution:

Question 12.
Find x if the Price Index Number by Simple Aggregate Method is 125

Solution:

Question 13.
Find y is the Price Index Number by Simple Aggregate Method is 120, taking 1995 as the base year.

Solution:

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 4 Time Series Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Time Series Miscellaneous Exercise 4

(I) Choose the correct alternative.

Question 1.
Which of the following can’t be a component of a time series?
(a) Seasonality
(b) Cyclical
(c) Trend
(d) Mean
Answer:
(d) Mean

Question 2.
The first step in time series analysis is to
(a) Perform regression calculations
(b) Calculate a moving average
(c) Plot the data on a graph
(d) Identify seasonal variation
Answer:
(c) Plot the data on a graph

Question 3.
Time-series analysis is based on the assumption that
(a) Random error terms are normally distributed.
(b) The variable to be forecast and other independent variable are correlated.
(c) Past patterns in the variable to be forecast will continue unchanged into the future.
(d) The data do not exhibit a trend.
Answer:
(c) Past patterns in the variable to be forecast will continue unchanged into the future.

Question 4.
Moving averages are useful in identifying
(a) Seasonal component
(b) Irregular component
(c) Trend component
(d) Cyclical component
Answer:
(c) Trend component

Question 5.
We can use regression line for past data to forecast future data. We then use the line which
(a) Minimizes the sum of squared deviations of past data from the line.
(b) Minimizes the sum of deviations of past data from the line.
(c) Maximizes the sum of squared deviations of past data from the line.
(d) Maximizes the sum of deviation of past data from the line.
Answer:
(a) Minimizes the sum of squared deviations of past data from the line

Question 6.
Which of the following is a major problem for forecasting, especially when using the method of least squares?
(a) The past cannot be known
(b) The future is not entirely certain
(c) The future exactly follows the patterns of the past
(d) The future may not follow the patterns of the past
Answer:
(d) The future may not follow the patterns of the past

Question 7.
An overall upward or downward pattern in an annual time series would be contained in which component of the time series
(a) Trend
(b) Cyclical
(c) Irregular
(d) Seasonal
Answer:
(a) Trend

Question 8.
The following trend line equation was developed for annual sales from 1984 to 1990 with 1984 as base or zero year. Y1 = 500 + 60X (in 1000 Rs.) The estimated sales for 1984 (in 1000 Rs) is:
(a) ₹ 500
(b) ₹ 560
(c) ₹ 1,040
(d) ₹ 1100
Answer:
(a) ₹ 500

Question 9.
What is a disadvantage of the graphical method of determining a trend line?
(a) Provides quick approximations
(b) Is subject to human error
(c) Provides accurate forecasts
(d) Is too difficult to calculate
Answer:
(b) Is subject to human error

Question 10.
Which component of time series refers to erratic time series movements that follow no recognizable or regular pattern.
(a) Trend
(b) Seasonal
(c) Cyclical
(d) Irregular
Answer:
(a) Trend

(II) Fill in the blanks.

Question 1.
_________ components of time series is indicated by a smooth line.
Answer:
Trend

Question 2.
_________ component of time series is indicated by periodic variation year after year.
Answer:
Seasonal

Question 3.
_________ component of time series is indicated by a long wave spanning two or more years.
Answer:
Cyclical

Question 4.
_________ component of time series is indicated by up and down movements without any pattern.
Answer:
Irregular

Question 5.
Addictive models of time series _________ independence of its components.
Answer:
assume

Question 6.
Multiplicative models of time series _________ independence of its components.
Answer:
does not assume

Question 7.
The simplest method of measuring the trend of time series is _________
Answer:
graphical method

Question 8.
The method of measuring the trend of time series using only averages is _________
Answer:
moving average method

Question 9.
The complicated but ancient method of measuring the trend of time series is _________
Answer:
least-squares method

Question 10.
The graph of time series clearly shows _________ of it is monotone.
Answer:
trend

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

Question 1.
The secular trend component of the time series represents irregular variations.
Answer:
False

Question 2.
Seasonal variation can be observed over several years.
Answer:
True

Question 3.
Cyclical variation can occur several times in a year.
Answer:
False

Question 4.
Irregular variation is not a random component of time series.
Answer:
False

Question 5.
The additive model of time series does not require the assumptions of independence of its components.
Answer:
False

Question 6.
The multiplicative model of time series does not require the assumption of independence of its components.
Answer:
True

Question 7.
The graphical method of finding trends is very complicated and involves several calculations.
Answer:
False

Question 8.
Moving the average method of finding trends is very complicated and involves several calculations.
Answer:
False

Question 9.
The least-squares method of finding trends is very simple and does not involve any calculations.
Answer:
False

Question 10.
All three methods of measuring trends will always give the same results.
Answer:
False

(IV) Solve the following problems.

Question 1.
The following table shows the productivity of pig-iron and ferro-alloys (‘000 metric tonnes)

Fit a trend line to the above data by graphical method.
Solution:

Question 2.
Fit a trend line to the data in Problem IV (1) by the method of least squares.
Solution:

u = \(\frac{t-1978}{1}\), Σy = 57, Σu = 0, Σu² = 60, Σuy = 38, n = 9
Let the equation of the trend line be
Y = a + bu where u = t – 1978 ……(i)
Σy = na + bΣu ………(ii)
Σuy = aΣu + bΣu² ………(iii)
Substituting the values of Σu, n, Σuy, Σu² in (ii) & (iii) we get
57 = 9a + 0 ∴ a = 6.3333
38 = 0 + b . 60 ∴ b = 0.6333.
∴ The equation of the trend line is
y = 6.3333 + 0.63333u where u = t – 1978

Question 3.
Obtain the trends values for the data on problem IV (1) using 5 yearly moving averages.
Solution:

Question 4.
The following table shows the amount of sugar production (in lac tonnes) for the years 1971 to 1982.

Fit a trend line to the above data by graphical method.
Solution:

Question 5.
Fit a trend line to data in problem 4 by the method of least squares.
Solution:

u = \(\frac{t-1976.5}{\frac{1}{2}}\), Σy = 38, Σu = 0, Σu² = 572, Σuy = 160, n = 12
Let the equation of the trend line be
y = a + bu ……..(i)
where u = \(\frac{t-1976.5}{\frac{1}{2}}\)
u = 2t – 3953
Σy = na + bΣu ………(ii)
Σuy = aΣu + bΣu² ……..(iii)
by (ii) 38 = 12o + 0 ∴ a = 3.1867
by (iii) 160 = 0 + b . 572 ∴ b = 0.2797
∴ by (i), Equation of the trend line is
Y = 3.1667 + 0.2797u where u = 2t – 3953.

Question 6.
Obtain trend values for data in Problem 4 using 4-yearly centered moving averages.
Solution:

Question 7.
The percentage of girls’ enrollment in total enrollment for years 1960-2005 is shown in the following table.

Fit a trend line to the above data by graphical method.
Solution:

Question 8.
Fit a trend line to the data in Problem 7 by the method of least squares.
Solution:

u = \(\frac{t-1980.5}{5}\), Σy = 51, Σu = 0, Σu² = 330, Σuy = 157, n = 10
Let the equation of the trend line be
Y = a + bu where u = \(\frac{t-1980.5}{5}\) …….(i)
Σy = na + bΣu ……..(ii)
Σuy = aΣu + bΣu² ……..(iii)
Substituting the values of Σy, Σu, n, Σuy, Σu² We get
51 = 10a + 0 ∴ a = 5.1
and 157 = 0 + 6.330 ∴ b = 0.4758
by (i) equation of the trend line is
Y = 5.1 + 0.4758u where u = \(\frac{t-1980.5}{5}\)

Question 9.
Obtain trend values for the data in Problem 7 using 4-yearly moving averages.
Solution:

Question 10.
The following data shows the number of boxes of cereal sold in the years 1977 to 1984.

Fit a trend line to the above data by graphical method.
Solution:

Question 11.
Fit a trend line to data in Problem 10 by the method of least squares.
Solution:

u = \(\frac{t-1980.5}{\frac{1}{2}}\), Σy = 39, Σu = 0, Σu² = 168, Σuy = 79, n = 8
Let the equation of the trend line by
Y = a + bu
Where u = 2t – 3961 …….(i)
Σy = na + bΣu …….(ii)
Σuy = aΣu + bΣu² ……..(iii)
Substituting the values of Σy, n, Σu, Σuy, Σu², in (ii) & (iii)
39 = 8a + 0 ∴ a = 4.875
79 = 0 + b (168) ∴ b = 0.4702
by (i) the equation of the trend line is
Y = 4.875 + 0.4702u Where u = 2t – 3961.

Question 12.
Obtain trend values for data in Problem 10 using 3-yearly moving averages.
Solution:

Question 13.
The following table shows the number of trade fatalities (in a state) resulting from drunken driving for the years 1975 to 1983.

Fit a trend line to the above data by graphical method.
Solution:

Question 14.
Fit a trend line to data in Problem 13 by the method of least squares.
Solution:

u = \(\frac{t-1979}{1}\), Σy = 47, Σu = 0, Σu² = 60, Σuy = 40, n = 9
Let the equation of the trends line be
Y = a + bu where u = t – 1979 …….(i)
Σy = na + bΣu ………(ii)
Σuy = aΣu + bΣu² ……..(iii)
Substituting values of Σy, n, Σu, Σuy, Σu² in (ii) & (iii)
We get 47 = 9a + 0 ∴ a = 5.2222
and 40 = 0 + b(60) ∴ b = 0.6667
∴ by (i) the equation of the trend line is
Y = 5.2222 + 0.6667u Where u = t – 1979.

Question 15.
Obtain trend values for data in Problem 13 using 4-yearly moving averages.
Solution:

Question 16.
The following table shows the all India infant mortality rates (per ‘000) for the years 1980 to 2000.

Fit a trend line to the above data by graphical method.
Solution:

Question 17.
Fit a trend line to data in Problem 16 by the method of least squares.
Solution:

u = \(\frac{t-1995}{5}\), Σy = 30, Σu = 0, Σu² = 70, Σuy = -70, n = 7
Let the equation of the trend line be
Y = a + bu Where u = \(\frac{t-1995}{5}\) …..(i)
Σy = na + bΣu ……(ii)
Σuy = aΣu + bΣu² …….(iii)
Substituting values of Σy, n, Σu, Σuy & Σu² in (ii) & (iii) we get
30 = 7a + 0 ∴ a = 4.2857
-70 = 0 + 6(70) ∴ b = -1
∴ by (i) the equation of the trend line is
y = 4.2857 – 1(u) Where u = \(\frac{t-1995}{5}\)

Question 18.
Obtain trend values for data in Problem 16 using 3-yearly moving averages.
Solution:

Question 19.
the following table shows the wheat yield (‘000 tonnes) in India for the years 1959 to 1968.

Fit a trend line to the above data by the method of least squares.
Solution:

u = \(\frac{t-1963.5}{\frac{1}{2}}\), Σy = 24, Σu = 0, Σu² = 330, Σuy = 94, n = 10
Let the equation of the trend line be
y = a + bu where u = \(\frac{t-1963.5}{\frac{1}{2}}\) ……(i)
i.e. u = 2t – 3927
Σy = na + bΣu …….(ii)
Σuy = aΣu + Σu² …..(iii)
Substituting values of Σy, n, Σu, Σuy & Σu² in (ii) & (iii) we get
24 = 10a + 0 ∴ a = 2.4
94 = 0 + 6.330 ∴ b = 0.2848
∴ Equation of the trend line is
y = 2.4 + (0.2848)u where u = 2t – 3927

Question 20.
Obtain trend values for data in problem 19 using 3-yearly moving averages.
Solution: