Class 12

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:

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 4 Time Series Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 4 Time Series Ex 4.1

Question 1.
The following data gives the production of bleaching powder (in ‘000 tonnes) for the years 1962 to 1972.

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

Question 2.
Use the method of least squares to fit a trend line to the data in problem 1 above. Also, obtain the trend value for the year 1975.
Solution:


n = 11, let the trend line the
y = a + bu ……..(I)
Σy = na + bΣu ……..(i)
Σuy = aΣu + bΣu² ………(ii)
Substituting the values of Σy, Σu, Σuy, & Σu², we get
46 = 11a + 0
∴ a = 4.18 And
114 = 0 + b(110)
∴ b = 1.04
By (I) the equation of the trends line is
y = 4.18 + 1.04u
Where u = t – 1967 ……..(iii)
For the year 1975 we have u = 8
Substituting in (iii) we get
Y= 4.18 + 1.04(8) = 12.5
Trend value for the year 1975 is 12.5 (in ‘000 tonnes).

Question 3.
Obtain the trend line for the above data using 5 yearly moving averages.
Solution:

Question 4.
The following table shows the index of industrial production for the period from 1976 to 1985, using the year 1976 as the base year.

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

Question 5.
Fit a trend line to the data in problem 4 above by the method of least squares. Also, obtain the trend value for the index of industrial production for the year 1987.
Solution:


u = \(\frac{t-1980.5}{\frac{1}{2}}\), n = 10, Σu = 0, Σy = 42, Σu² = 330, Σuy = 148
Let the trend line be y = a + bu ……(i)
where u = \(\frac{t-1980.5}{\frac{1}{2}}\)
i.e. u = 2t – 3961
Σy = na + bΣu ……(ii)
Σuy = aΣu + bΣu² ……….(iii)
Substituting the values of Σy, n, Σu, Σuy & Σu² We get
42 = 10a + 0
∴ a = 4.2 and
148 = 0 + 5.330
∴ b = 0.4485
∴ by (i) the equation of the trends line is
Y = 4.2 + 0.4485u ………(iv)
where u = 2t – 3961
For the year 1987,
u = 13 by (iv) we have
Y = 4.2 + 0.4485(13) = 10.0305
∴ The trend value for the year 1987 is 10.0305

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

Question 7.
The following table gives the production of steel (in millions of tonnes) for the years 1976 to 1986.

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

Question 8.
Fit a trend line to the data in Problem 7 by the method of least squares. Also, obtain the trend value for the year 1990.
Solution:

u = \(\frac{t-1981}{1}\), n = 10, Σu = 0, ΣY = 62, Σu² = 110, Σuy = 87
Let the equation of the trend line be
Y = a + bu
where u = t – 1981 ……(i)
ΣY = na + bΣu ………(ii)
Σuy = aΣu + bΣu² ………(iii)
Substituting the values of Σy, n, Σu, Σuy, Σu² in (ii) & (iii)
62 = 11a + 0
∴ a = 5.6364 And
87 = 0 + 5(110)
∴ b = 0.7909
∴ by (i) equation of the trend line is y = 5.6364 + 0.7909u
Where u = t – 1981
For the year 1990,
u = 9
∴ y = 5.6364 + 0.7909(9)
∴ y = 12.7545 (in million tonnes)

Question 9.
Obtain the trend values for the above data using 3-yearly moving averages.
Solution:

Question 10.
The following table shows the production of gasoline in the U.S.A. for the years 1962 to 1976.

(i) Obtain trend values for the above data using 5-yearly moving averages.
(ii) Plot the original time series and trend values obtained above on the same graph.
Solution:
(i)

(ii)

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Miscellaneous Exercise 3

(I) Choose the correct alternative.

Question 1.
Regression analysis is the theory of
(a) Estimation
(b) Prediction
(c) Both a and b
(d) Calculation
Answer:
(c) Both a and b

Question 2.
We can estimate the value of one variable with the help of other known variable only if they are
(a) Correlated
(b) Positively correlated
(c) Negatively correlated
(d) Uncorrelated
Answer:
(a) Correlated

Question 3.
There are ________ types of regression equation
(a) 4
(b) 2
(c) 3
(d) 1
Answer:
(b) 2

Question 4.
In the regression equation of Y on X
(a) X is independent and Y is dependent
(b) Y is independent and X is dependent
(c) Both X and Y are independent
(d) Both X and Y are dependent.
Answer:
(a) X is independent and Y is dependent

Question 5.
In the regression equation of X on Y
(a) X is independent and Y is dependent
(b) Y is independent and X is dependent
(c) Both X and Y are independent
(d) Both X and Y are dependent
Answer:
(b) Y is independent and X is dependent

Question 6.
b_xy is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(b) Regression coefficient of X on Y

Question 7.
b_yx is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(a) Regression coefficient of Y on X

Question 8.
‘r’ is ________
(a) Regression coefficient of Y on X
(b) Regression coefficient of X on Y
(c) Correlation coefficient between X and Y
(d) Covariance between X and Y
Answer:
(d) Correlation coefficient between X and Y

Question 9.
b_xy . b_yx = _________
(a) v
(b) y_x
(c) r²
(d) (y_y)²
Answer:
(c) r²

Question 10.
If b_yx > 1 then b_xy is ______
(a) > 1
(b) < 1
(c) > 0
(d) < 0
Answer:
(b) < 1

Question 11.
|b_xy + b_yx| > ______
(a) |r|
(b) 2|r|
(c) r
(d) 2r
Answer:
(b) 2|r|

Question 12.
b_xy and b_yx are ________
(a) Independent of change of origin and scale
(b) Independent of change of origin but not of the scale
(c) Independent of change of scale but not of origin
(d) Affected by change of origin and scale
Answer:
(b) Independent of change of origin but not of the scale

Question 13.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_yx = ________
(a) \(\frac{d}{c} b_{v u}\)
(b) \(\frac{c}{d} b_{v u}\)
(c) \(\frac{a}{b} b_{v u}\)
(d) \(\frac{b}{a} b_{v u}\)
Answer:
(a) \(\frac{d}{c} b_{v u}\)

Question 14.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_xy = ________
(a) \(\frac{d}{c} b_{u v}\)
(b) \(\frac{c}{d} b_{u v}\)
(c) \(\frac{a}{b} b_{u v}\)
(d) \(\frac{b}{a} b_{u v}\)
Answer:
(b) \(\frac{c}{d} b_{u v}\)

Question 15.
Corr(x, x) = ________
(a) 0
(b) 1
(c) -1
(d) can’t be found
Answer:
(b) 1

Question 16.
Corr (x, y) = ________
(a) corr(x, x)
(b) corr(y, y)
(c) corr(y, x)
(d) cov(y, x)
Answer:
(c) corr(y, x)

Question 17.
Corr\(\left(\frac{x-a}{c}, \frac{y-b}{d}\right)\) = -corr(x, y) if,
(a) c and d are opposite in sign
(b) c and d are same in sign
(c) a and b are opposite in sign
(d) a and b are same in sign
Answer:
(a) c and d are opposite in sign

Question 18.
Regression equation of X and Y is
(a) y – \(\bar{y}\) = b_yx (x – \(\bar{x}\))
(b) x – \(\bar{x}\) = b_xy (y – \(\bar{y}\))
(c) y – \(\bar{y}\) = b_xy (x – \(\bar{x}\))
(d) x – \(\bar{x}\) = b_yx (y – \(\bar{y}\))
Answer:
(b) x – \(\bar{x}\) = b_xy (y – \(\bar{y}\))

Question 19.
Regression equation of Y and X is
(a) y – \(\bar{y}\) = b_yx (x – \(\bar{x}\))
(b) x – \(\bar{x}\) = b_xy (y – \(\bar{y}\))
(c) y – \(\bar{y}\) = b_xy (x – \(\bar{x}\))
(d) x – \(\bar{x}\) = b_yx (y – \(\bar{y}\))
Solution:
(a) y – \(\bar{y}\) = b_yx (x – \(\bar{x}\))

Question 20.
b_yx = ________
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)
(c) \(\frac{1 \sigma_{y}}{r \sigma_{x}}\)
(d) \(\frac{1 \sigma_{y}}{r \sigma_{y}}\)
Answer:
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)

Question 21.
b_xy = ________
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)
(b) \(r \frac{\sigma_{y}}{\sigma_{x}}\)
(c) \(\frac{1 \sigma_{y}}{r \sigma_{x}}\)
(d) \(\frac{1 \sigma_{y}}{r \sigma_{y}}\)
Answer:
(a) \(r \frac{\sigma_{x}}{\sigma_{y}}\)

Question 22.
Cov (x, y) = ________
(a) Σ(x – \(\bar{x}\))(y – \(\bar{y}\))
(b) \(\frac{\sum(x-\bar{x})(y-\bar{y})}{n}\)
(c) \(\frac{\sum x y}{n}-\bar{x} \bar{y}\)
(d) b and c both
Answer:
(d) b and c both

Question 23.
If b_xy < 0 and b_yx < 0 then ‘r’ is ________
(a) > 0
(b) < 0
(c) > 1
(d) not found
Answer:
(b) < 0

Question 24.
If equation of regression lines are 3x + 2y – 26 = 0 and 6x + y – 31 = 0 then means of x and y are ________
(a) (7, 4)
(b) (4, 7)
(c) (2, 9)
(d) (-4, 7)
Answer:
(b) (4, 7)

(II) Fill in the blanks:

Question 1.
If b_xy < 0 and b_yx < 0 then ‘r’ is ________
Answer:
negative

Question 2.
Regression equation of Y on X is ________
Answer:
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))

Question 3.
Regression equation of X on Y is ________
Answer:
(x – \(\bar{x}\)) = b_xy (y – \(\bar{y}\))

Question 4.
There are ______ types of regression equations.
Answer:
2

Question 5.
Corr (x₁ – x) = ______
Answer:
-1

Question 6.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_xy = ______
Answer:
\(\frac{c}{d} b_{u v}\)

Question 7.
If u = \(\frac{x-a}{c}\) and v = \(\frac{y-b}{d}\) then b_yx = ______
Answer:
\(\frac{d}{c} b_{v u}\)

Question 8.
|b_xy + b_yx| ≥ ______
Answer:
2|r|

Question 9.
If b_yx > 1 then b_xy is ______
Answer:
< 1

Question 10.
b_xy . b_yx = ______
Answer:
r²

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

Question 1.
Corr (x, x) = 1.
Answer:
True

Question 2.
Regression equation of X on Y is y – \(\bar{y}\) = b_xy (x – \(\bar{x}\)).
Answer:
False

Question 3.
Regression equation of Y on X is y – \(\bar{y}\) = b_yx (x – \(\bar{x}\)).
Answer:
True

Question 4.
Corr (x, y) = Corr (y, x).
Answer:
True

Question 5.
b_xy and b_yx are independent of change of origin and scale.
Answer:
False

Question 6.
‘r’ is the regression coefficient of Y on X.
Answer:
False

Question 7.
b_yx is the correlation coefficient between X and Y.
Answer:
False

Question 8.
If u = x – a and v = y – b then b_xy = b_uv.
Answer:
True

Question 9.
If u = x – a and v = y – b then r_xy = r_uv.
Answer:
True

Question 10.
In the regression equation of Y on X, b_yx represents the slope of the line.
Answer:
True

(IV) Solve the following problems.

Question 1.
The data obtained on X, the length of time in weeks that a promotional project has been in progress at a small business, and Y the percentage increase in weekly sales over the period just prior to the beginning of the campaign.

Find the equation of regression line to predict percentage increase in sales if the company has been in progress for 1.5 weeks.
Solution:
Let u = x – 3, v = y – 15


∴ Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 14.67) = 2.6(x – 2.5)
y – 14.67 = 2.6x – 6.5
y = 2.6x + 8.17
When x = 1.5
y = (2.6)(1.5) + 8.17
= 3.9 + 8.17
= 12.07

Question 2.
The regression equation of y on x is given by 3x + 2y – 26 = 0. Find b_yx.
Solution:
Given, regression equation of Y on X is
3x + 2y – 26 = 0
∴ 2y = -3x + 26
∴ y = \(\frac{-3}{2}\)x + 13
∴ b_yx = \(\frac{-3}{2}\)

Question 3.
If for a bivariate data \(\bar{x}\) = 10, \(\bar{y}\) = 12, v(x) = 9, σ_y = 4 and r = 0.6. Estimate y when x = 5.
Solution:
Given, V(x) = 9
∴ σ_x = 3
b_yx = \(\frac{r \cdot \sigma_{y}}{\sigma_{x}}\)
= 0.6 × \(\frac{4}{3}\)
= 0.8
∴ Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 12) = 0.8(5 – 10)
y – 12 = 0.8(-5)
y – 12 = -4
y = 8

Question 4.
The equation of the line of regression of y on x is v = \(\frac{2}{9} x\) and x on y is x = \(\frac{y}{2}+\frac{7}{6}\). Find (i) r (ii) \(\sigma_{y}^{2} \text { if } \sigma_{x}^{2}=4\).
Solution:

Question 5.
Identify the regression equations of x on y and y on x from the following equations.
2x + 3y = 6 and 5x + 7y – 12 = 0
Solution:

∴ Our assumption is correct
∴ Regression equation of Y on X is 2x + 3y = 6
∴ Regression equation of X on Y is 5x + 7y – 12 = 0

Question 6.
(i) If for a bivariate data b_yx = -1.2 and b_xy = -0.3 then find r.
(ii) From the two regression equations y = 4x – 5 and 3x = 2y + 5, find \(\bar{x}\) and \(\bar{y}\).
Solution:
r² = b_yx . b_xy
r² = (-1.2) × (-0.3)
r² = 0.36
r = ±0.6
Since, b_yx . b_xy are negative, r = -0.6
Also,(\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
y = 4x – 5, 3x = 2y + 5
8x – 2y = 10
3x – 2y = 5
on subtracting,
5x = 5
x = 1
Substituting x = 1 in y = 4x – 5
y = 4(1) – 5
y = -1
∴ \(\bar{x}\) = 1, \(\bar{y}\) = -1

Question 7.
The equation of the two lines of regression are 3x + 2y – 26 = 0 and 6x + y – 31 = 0. Find
(i) Means of X and Y
(ii) Correlation coefficient between X on Y
(iii) Estimate of Y for X = 2
(iv) var (X) if var (Y) = 36
Solution:
(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression lines
3x + 2y = 26
6x + y = 31
3x + 2y = 26 …….(i)
12x + 2y = 62 ……..(ii)
on subtracting,
-9x = -36
x = 4
Substituting x = 4 in equation (i)
3(4) + 2y = 26
2y = 14
y = 7
∴ \(\bar{x}\) = 4, \(\bar{y}\) = 7

Question 8.
Find the line of regression of X on Y for the following data:
n = 8, Σ(x_i – x)² = 36, Σ(y_i – y)² = 44, Σ(x_i – x)(y_i – y) = 24
Solution:

Question 9.
Find the equation of line regression of Y on X for the following data:
n = 8, Σ(x_i – \(\bar{x}\))(y_i – \(\bar{y}\)) = 120, \(\bar{x}\) = 20, \(\bar{y}\) = 36, σ_x = 2, σ_y = 3.
Solution:

Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 36) = 3.75(x – 20)
(y – 36) = 3.75x – 75
y = 3.75x – 39

Question 10.
The following result was obtained from records of age (X) and systolic blood pressure (Y) of a group of 10 men.

and Σ(x_i – \(\bar{x}\))(y_i – \(\bar{x}\)) = 1120. Find the Prediction of blood pressure of a man of age 40 years.
Solution:

Regression equation of Y on X is
(y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 140) = 0.7(40 – 50)
y – 140 = 0.7(-10)
y – 140 = -7
∴ y = 133

Question 11.
The equations of two regression lines are 10x – 4y = 80 and 10y – 9x = -40 Find:
(i) \(\bar{x}\) and \(\bar{y}\)
(ii) b_yx and b_xy
(iii) If var(Y) = 36, obtain var(X)
(iv) r
Solution:
(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression
10x – 4y = 80 ……(i)
-9x + 10y = -40 ……..(ii)
50x – 20y = 400
-18x + 20y = -80
32x = 320
x = 10
x = 10 in equation (i)
10(10) – 4y = 80
4y = 20
y = 5
∴ \(\bar{x}\) = 10, \(\bar{y}\) = 5


(iv) r² = b_yx . b_xy = 0.36
r = ±0.6
Since b_yx and b_xy are positive
∴ r = 0.6

Question 12.
If b_yx = -0.6 and b_xy = -0.216 then find correlation coefficient between X and Y comment on it.
Solution:
r² = b_yx . b_xy
r² = -0.6 × -0.216
r² = 0.1296
r = ±√0.1296
r = ± 0.36
Since b_yx and b_xy are negative
r = -0.36

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.3

Question 1.
From the two regression equations find r, \(\bar{x}\) and \(\bar{y}\).
4y = 9x + 15 and 25x = 4y + 17
Solution:
Given 4y = 9x + 15 and 25x = 4y + 17

Since b_yx and b_xy are positive.
∴ r = \(\frac{3}{5}\) = 0.6
(\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
9x – 4y = -15 …….(i)
25x – 4y = 17 ……….(ii)
-16x = -32
x = 2
∴ \(\bar{x}\) = 2
Substituting x = 2 in equation (i)
9(2) – 4y = -15
18 + 15 = 4y
33 = 4y
y = 33/4 = 8.25
∴ \(\bar{y}\) = 8.25

Question 2.
In a partially destroyed laboratory record of an analysis of regression data, the following data are legible:
Variance of X = 9
Regression equations:
8x – 10y + 66 = 0 And 40x – 18y = 214.
Find on the basis of the above information
(i) The mean values of X and Y.
(ii) Correlation coefficient between X and Y.
(iii) Standard deviation of Y.
Solution:
Given, \(\sigma_{x}{ }^{2}=9, \sigma_{x}=3\)
(i) (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
40x – 50y = -330 …….(i)
40x – 50y = +214 ………(ii)
-32y = -544
y = 17
∴ \(\bar{y}\) = 17
8x – 10(17) + 66 = 0
8x = 104
x = 13
∴ \(\bar{x}\) = 13

Question 3.
For 50 students of a class, the regression equation of marks in statistics (X) on the marks in Accountancy (Y) is 3y – 5x + 180 = 0. The mean marks in accountancy is 44 and the variance of marks in statistics \(\left(\frac{9}{16}\right)^{t h}\) of the variance of marks in accountancy. Find the mean in statistics and the correlation coefficient between marks in two subjects.
Solution:
Given, n = 50, \(\bar{y}\) = 44
\(\sigma_{x}^{2}=\frac{9}{16} \sigma_{y}^{2}\)
∴ \(\frac{\sigma_{x}}{\sigma_{x}}=\frac{3}{4}\)
Since (\(\bar{x}\), \(\bar{y}\)) is the point intersection of the regression line.
∴ (\(\bar{x}\), \(\bar{y}\)) satisfies the regression equation.
3\(\bar{y}\) – 5\(\bar{x}\) + 180 = 0
3(44) – 5\(\bar{x}\) + 180 = 0
∴ 5\(\bar{x}\) = 132 + 180
\(\bar{x}\) = \(\frac{312}{5}\) = 62.4
∴ Mean marks in statistics is 62.4
Regression equation of X on Y is 3y – 5x + 180 = 0
∴ 5x = 3y + 180

Question 4.
For bivariate data, the regression coefficient of Y on X is 0.4 and the regression coefficient of X on Y is 0.9. Find the value of the variance of Y if the variance of X is 9.
Solution:

Question 5.
The equation of two regression lines are 2x + 3y – 6 = 0 and 3x + 2y – 12 = 0
Find (i) Correlation coefficient (ii) \(\frac{\sigma_{x}}{\sigma_{y}}\)
Solution:

Question 6.
For a bivariate data \(\bar{x}\) = 53, \(\bar{y}\) = 28, b_yx =-1.5 and b_xy = -0.2. Estimate Y when X = 50.
Solution:
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 28) = -1.5(50 – 53)
Y – 28 = -1.5(-3)
Y – 28 = 4.5
Y = 32.5

Question 7.
The equation of two regression lines are x – 4y = 5 and 16y – x = 64. Find means of X and Y. Also, find the correlation coefficient between X and Y.
Solution:
Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines.
x – 4y = 5 …..(i)
-x + 16y = 64 …….(ii)
12y = 69
y = 5.75
Substituting y = 5.75 in equation (i)
x – 4(5.75) = 5
x – 23 = 5
x = 28
∴ \(\bar{x}\) = 28, \(\bar{y}\) = 5.75
x – 4y = 5
x = 4y + 5
∴ b_xy = 4
16y – x = 64
16y = x + 64
y = \(\frac{1}{16}\)x + 4
b_yx = \(\frac{1}{16}\)
b_yx . b_xy = \(\frac{1}{16}\) × 4 = \(\frac{1}{4}\) ∈ [0, 1]
∴ Our assumption is correct
∴ r² = b_yx . b_xy
r² = \(\frac{1}{4}\)
r = ±\(\frac{1}{2}\)
Since b_yx and b_xy are positive,
∴ r = \(\frac{1}{2}\) = 0.5

Question 8.
In partially destroyed record, the following data are available variance of X = 25. Regression equation of Y on X is 5y – x = 22 and Regression equation of X on Y is 64x – 45y = 22 Find
(i) Mean values of X and Y.
(ii) Standard deviation of Y.
(iii) Coefficient of correlation between X and Y.
Solution:
Given \(\sigma_{x}^{2}\) = 25, ∴ σ_x = 5
(i) Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of regression lines
-x + 5y = 22 …….(i)
64x – 45y = 22 ………..(ii)
equation (i) becomes
-9x + 45y = 198
64y – 45y = 22
55x = 220
x = 4
Substituting x = 4 in equation (i)
-4 + 5y = 22
5y = 26
∴ y = 5.2
∴ \(\bar{x}\) = 4, \(\bar{y}\) = 5.2
Regression equation of X on Y is
64x – 45y – 22
64x = 45y + 22
x = \(\frac{45}{64} y+\frac{22}{64}\)
bxy = \(\frac{45}{64}\)
(ii) Regression equation of Y on X is
5y – x = 22
5y = x + 22

Question 9.
If the two regression lines for a bivariate data are 2x = y + 15 (x on y) and 4y – 3x + 25 (y on x) find
(i) \(\bar{x}\)
(ii) \(\bar{y}\)
(iii) b_yx
(iv) b_xy
(v) r [Given √0.375 = 0.61]
Solution:
Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression line
2x = y + 15
4y = 3x + 25
2x – y = 15 …….(i)
3x – 4y = -25 ……..(ii)
Multiplying equation (i) by 4
8x – 4y = 60
3x – 4y = -25
on Subtracting,
5x = 85
∴ x = 17
Substituting x in equation (i)
2(17) – y = 15
34 – 15 = y
∴ y = 15

Since b_yx and b_xy are positive, ∴ r = 0.61

Question 10.
The two regression equation are 5x – 6y + 90 = 0 and 15x – 8y – 130 = 0. Find \(\bar{x}\), \(\bar{y}\), r.
Solution:
Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
5x – 6y + 90 = 0 ……(i)
15x – 8y – 130 = 0
15x – 18y + 270 = 0
15x – 8y – 130 = 0
on subtracting,
-10y + 400 = 0
y = 40
Substituting y = 40 in equation (i)
5x – 6(40) + 90 = 0
5x = 150
x = 30
∴ \(\bar{x}\) = 30, \(\bar{y}\) = 40

Since b_yx and b_xy are positive
∴ r = \(\frac{2}{3}\)

Question 11.
Two lines of regression are 10x + 3y – 62 = 0 and 6x + 5y – 50 = 0 Identify the regression equation equation of x on y. Hence find \(\bar{x}\), \(\bar{y}\), and r.
Solution:

∴ Our assumption is correct.
∴ Regression equation of X on Y is 10x + 3y – 62 = 0
r² = b_yx . b_xy
r² = \(\frac{9}{25}\)
r = ±\(\frac{3}{5}\)
Since, byx and bxy are negative, r = –\(\frac{3}{5}\) = -0.6
Also (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
50x + 15y = 310
18x + 15y = 150
on subtracting
32x = 160
x = 5
Substituting x = 5 in 10x + 3y = 62
10(5) + 3y = 62
3y = 12
∴ y = 4
∴ \(\bar{x}\) = 5, \(\bar{y}\) = 4

Question 12.
For certain X and Y series, which are correlated the two lines of regression are 10y = 3x + 170 and 5x + 70 = 6y. Find the correlation coefficient between them. Find the mean values of X and Y.
Solution:

Since byx and bxy are positive,
r = \(\frac{3}{5}\) = 0.6
Since, (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression lines
3x – 10y = -170 …….(i)
5x – 6y = -70 ………(ii)
9x – 30y = -510
25x – 30y = -350
on subtracting
-16x = -160
x = 10
Substituting x = 10 in equation (i)
3(10) – 10y = -170
30 + 170 = 10y
200 = 10y
y = 20
∴ \(\bar{x}\) = 10, \(\bar{y}\) = 20

Question 13.
Regression equation of two series are 2x – y – 15 = 0 and 4y + 25 = 0 and 3x- 4y + 25 = 0. Find \(\bar{x}\), \(\bar{y}\) and regression coefficients, Also find coefficients of correlation. [Given √0.375 = 0.61]
Solution:
Since (\(\bar{x}\), \(\bar{y}\)) is the point of intersection of the regression line
2x = y + 15
4y = 3x + 25
2x – y = 15 ……(i)
3x – 4y = -15 ……..(ii)
Multiply equation (i) by 4
8x – 4y = 60
3x – 4y = -25
on subtracting,
5x = 85
x = 17
Substituting x in equation (i)
2(17) – y = 15
34 – 15 = y
y = 15
∴ \(\bar{x}\) = 17, \(\bar{y}\) = 19

∴ Our assumption is correct
r² = b_xy . b_yx
r² = \(\frac{3}{8}\) = 0.375
r = ±√o.375 = ±0.61
Since, b_yx and b_xy are positive, ∴ r = 0.61

Question 14.
The two regression lines between height (X) in includes and weight (Y) in kgs of girls are 4y – 15x + 500 = 0 and 20x – 3y – 900 = 0. Find the mean height and weight of the group. Also, estimate the weight of a girl whose height is 70 inches.
Solution:
Since (\(\bar{x}\), \(\bar{y}\)) is the point intersection of the regression lines
15x – 4y = 500 ……(i)
20x – 3y = 900 …….(ii)
60x – 16y – 2000
60x – 9y = 2700
on subtracting,
-7y = -700
y = 100
Substituting y = 100 in equation (i)
15x – 4(100) = 500
15x = 900
x = 60

∴ Our assumption is correct
∴ Regression equation of Y on X is
Y = \(\frac{15}{4}\)x – 125
When x = 70
Y = \(\frac{15}{4}\) × 70 = -125
= 262.5 – 125
= 137.5 kg

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.2

Question 1.
For bivariate data.
\(\bar{x}\) = 53, \(\bar{x}\) = 28, b_yx = -1.2, b_xy = -0.3
Find,
(i) Correlation coefficient between X and Y.
(ii) Estimate Y for X = 50
(iii) Estimate X for Y = 25
Solution:
(i) r² = b_yx . b_xy
r² = (-1.2)(-0.3)
r² = 0.36
r = ±0.6
Since, b_yx and bxy are negative, r = -0.6

(ii) Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
Y – 28 = -1.2(50 – 53)
Y – 28 = -1.2(-3)
Y – 28 = 3.6
Y = 31.6

(iii) Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 53) = -0.3(25 – 28)
X – 53 = -0.3(-3)
X – 53 = 0.9
X = 53.9

Question 2.
From the data of 20 pairs of observation on X and Y, following result are obtained \(\bar{x}\) = 199, \(\bar{y}\) = 94, \(\sum\left(x_{i}-\bar{x}\right)^{2}\) = 1200, \(\sum\left(y_{i}-\bar{y}\right)^{2}\) = 300
\(\sum\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)\) = -250
Find
(i) The line of regression of Y on X.
(ii) The line of regression of X on Y.
(iii) Correlation coefficient between X on Y.
Solution:

Question 3.
From the data of 7 pairs of observations on X and Y following results are obtained.
Σ(x_i – 70 ) = -35, Σ(y_i – 60) = -7, Σ(x_i – 70)² = 2989, Σ(y_i – 60)² = 476, Σ(x_i – 70) (y_i – 60) = 1064 [Given √0.7884 = 0.8879]
Obtain
(i) The line of regression of Y on X.
(ii) The line of regression of X on Y.
(iii) The correlation coefficient between X and Y.
Solution:


(i) Line of regression Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 59) = 0.36(x – 65)
(Y – 59) = 0.36x – 23.4
Y = 0.36x + 35.6

(ii) Line of regression X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 65) = 2.19(y – 59)
(X – 65) = 2.19y – 129.21
X = 2.19y – 64.21

(iii) r² = b_yx . b_xy
r² = (0.36) (2.19)
r² = 0.7884
r = ±√0.7884 = ±0.8879
Since b_yx and b_xy are positive.
∴ r = 0.8879

Question 4.
You are given the following information about advertising expenditure and sales.

Correlation coefficient between X and Y is 0.8
(i) Obtain two regression equations.
(ii) What is the likely sales when the advertising budget is ₹ 15 lakh?
(iii) What should be the advertising budget if the company wants to attain sales target of ₹ 120 lakh?
Solution:
Given, \(\bar{x}\) = 10, \(\bar{y}\) = 90, σ_x = 3, σ_y = 12, r = 0.8
byx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.8 \times \frac{12}{3}\) = 3.2
bxy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.8 \times \frac{3}{12}\) = 0.2
(i) Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 90) = 3.2(x – 10)
Y – 90 = 3.2x – 32
Y = 3.2x + 58
Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 10) = 0.2(y – 90)
X – 10 = 0.2y + 18
X = 0.2y – 8

(ii) When x = 15,
Y = 3.2(15) + 58
= 48 + 58
= 106 lakh

(iii) When y = 120
X = 0.2(120) – 8
= 24 – 8
= 16 lakh

Question 5.
Bring out inconsistency if any, in the following:
(i) b_yx + b_xy = 1.30 and r = 0.75
(ii) b_yx = b_xy = 1.50 and r = -0.9
(iii) b_yx = 1.9 and b_xy = -0.25
(iv) b_yx = 2.6 and b_xy = \(\frac{1}{2.6}\)
Solution:
(i) Given, b_yx + b_xy = 1.30 and r = 0.75
\(\frac{b_{y x}+b_{x y}}{2}=\frac{1.30}{2}\) = 0.65
But for regression coefficients b_yx and b_xy
\(\left|\frac{b_{y x}+b_{x y}}{2}\right| \geq r\)
Here, 0.65 < r = 0.75
∴ The data is inconsistent
(ii) The signs of b_yx, b_xy and r must be same (all three positive or all three negative)
∴ The data is inconsistent.

(iii) The signs of b_yx and b_xy should be same (either both positive or both negative)
∴ The data is consistent.

(iv) b_yx . b_xy = 2.6 × \(\frac{1}{2.6}\) = 1
∴ 0 ≤ r² ≤ 1
∴ The data is consistent.

Question 6.
Two sample from bivariate populations have 15 observation each. The sample means of X and Y are 25 and 18 respectively. The corresponding sum of square of deviations from respective means are 136 and 150. The sum of product of deviations from respective means is 123. Obtain the equation of line of regression of X on Y.
Solution:
Given, n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, Σ(x – \(\bar{x}\)) = 136, Σ(y – \(\bar{y}\)) = 150, Σ(x – \(\bar{x}\)) (y – \(\bar{y}\)) = 123
Regression equation of X on Y is (X – \(\bar{x}\)) = bxy (Y – \(\bar{y}\))
(X – 25) = 0.82(y – 18)
(X – 25) = 082y – 14.76
X = 0.82y + 10.24

Question 7.
For a certain bivariate data

And r = 0.5 estimate y when x = 10 and estimate x when y = 16
Solution:
Given, \(\bar{x}\) = 25, \(\bar{y}\) = 20, σ_x = 4, σ_y = 3, r = 0.5
b_yx = \(\frac{r \sigma_{y}}{\sigma_{y}}=0.5 \times \frac{3}{4}\) = 0.375
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 20) = 0.375(x – 25)
Y – 20 = 0.375x – 9.375
Y = 0.375x + 10.625
When, x = 10
Y = 0.375(10) + 10.625
= 3.75 + 10.625
= 14.375
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.5 \times \frac{4}{3}\) = 0.67
Regression equation of X on Y is
(X – \(\bar{x}\)) = b_yx (Y – \(\bar{y}\))
(X – 25) = 0.67(y – 20)
(X – 25) = 0.67y – 13.4
X = 0.67y + 11.6
When, Y = 16
x = 0.67(16) + 11.6
= 10.72 + 11.6
= 22.32

Question 8.
Given the following information about the production and demand of a commodity obtain the two regression lines:

Coefficient of correlation between X and Y is 0.6. Also estimate the problem when demand is 100.
Solution:
Given \(\bar{x}\) = 85, \(\bar{y}\) = 90, σ_x = 5, σ_y = 6 and r = 0.6
bxy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.6 \times \frac{5}{6}\) = 0.5
byx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.6 \times \frac{6}{5}\) = 0.72
Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 85) = 0.5(y – 90)
(X – 85 ) = 0.5y – 45
X = 0.5y + 40
When y = 100,
x = 0.5 (100) + 40
= 50 + 40
= 90
Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 90) = 0.72(x – 85)
(Y – 90) = 0.72x – 61.2
Y = 0.72x + 28.8

Question 9.
Given the following data, obtain linear regression estimate of X for Y = 10
Solution:
\(\bar{x}\) = 7.6, \(\bar{y}\) = 14.8, σ_x = 3.2, σ_y = 16 and r = 0.7
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.7 \times \frac{3.2}{16}\) = 0.14
Regression equation of X on Y is
(X – \(\bar{y}\)) = b_xy (Y – \(\bar{y}\))
(X – 7.6) = 0.14(y – 14.8)
X – 7.6 = 0.14y – 2.072
X = 0.14y + 5.528
When y = 10
x = 0.14(10) + 5.528
= 1.4 + 5.528
= 6.928

Question 10.
An inquiry of 50 families to study the relationship between expenditure on accommodation (₹ x) and expenditure on food and entertainment (₹ y) gave the following result:
Σx = 8500, Σy = 9600, σ_x = 60, σ_y = 20, r = 0.6
Estimate the expenditure on food and entertainment when expenditure on accommodation is ₹ 200
Solution:
n = 50 (given)

Regression equation of Y on X is
Y – \(\bar{y}\) = b_yx (X – \(\bar{x}\))
(Y – 192) = 0.2(200 – 170)
Y – 192 = 0.2(30)
Y = 192 + 6
Y = 198

Question 11.
The following data about the sales and advertisement expenditure of a firms is given below (in ₹ crores)

Also correlation coefficient between X and Y is 0.9
(i) Estimate the likely sales for a proposed advertisement expenditure of ₹ 10 crores.
(ii) What should be the advertisement expenditure if the firm proposes a sales target ₹ 60 crores
Let the sales be X and advertisement expenditure be Y
Solution:
Given, \(\bar{x}\) = 40, \(\bar{y}\) = 6, σ_x = 10, σ_y = 1.5, r = 0.9
b_yx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.9 \times \frac{1.5}{10}\) = 0.135
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.9 \times \frac{10}{1.5}\) = 6
(i) Regression equation of X on Y is
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 40) = 6(y – 6)
X – 40 = 6y – 36
X = 6y + 4
When y = 10
x = 6 (10) + 4
= 60 + 4
= 64 crores

(ii) Regression equation Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 6) = 0.135 (x – 40)
Y – 6 = 0.135x – 5.4
Y = 0.135x + 0.6
When x = 60
Y = 0.135 (60) + 0.6
= 8.1 + 0.6
= 8.7 crores

Question 12.
For certain bivariate data the following information are available

Correlation coefficient between x and y is 0.6, estimate x when y = 15 and estimate y when x = 10.
Solution:
Given, \(\bar{x}\) = 13, \(\bar{y}\) = 17, σ_x = 3, σ_y = 2, r = 0.6
b_yx = \(\frac{r \sigma_{y}}{\sigma_{x}}=0.6 \times \frac{2}{3}\) = 0.4
b_xy = \(\frac{r \sigma_{x}}{\sigma_{y}}=0.6 \times \frac{3}{2}\) = 0.9
Regression equation of Y on X
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
Y – 17 = 0.4(x – 13)
Y = 0.4x + 11.8
When x = 10
Y = 0.4(10) + 11.8
= 4 + 11.8
= 15.8
Regression equation of X on Y
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 13) = 0.9(y – 17)
X – 13 = 0.9y – 15.3
X = 0.9y – 2.3
When y = 15
X = 0.9(15) – 2.3
= 13.5 – 2.3
= 11.2