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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

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 3 Linear Regression Ex 3.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 3 Linear Regression Ex 3.1

Question 1.
The HRD manager of the company wants to find a measure which he can use to fix the monthly income of persons applying for the job in the production department. As an experimental project. He collected data of 7 persons from that department referring to years of service and their monthly incomes.

(i) Find the regression equation of income on years of service.
(ii) What initial start would you recommend for a person applying for the job after having served in a similar capacity in another company for 13 years?
Solution:


(i) Regression equation of Y on X is (Y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(Y – 8) = 0.75(x – \(\bar{x}\))
Y = 0.75x + 2
(ii) When x = 13
Y = 0.75(13) + 2 = 11.75
Recommended income for the person is ₹ 11750.

Question 2.
Calculate the regression equations of X on Y and Y on X from the following date:

Solution:


Regression equation of X on Y is (X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 14) = 1(Y – 8)
X – 14 = Y – 8
X = Y + 6
Regression equation Y on X is (Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 8) = 0.87(X – 14)
Y – 8 = 0.87X – 12.18
Y = 0.87X – 4.18

Question 3.
For a certain bivariate data on 5 pairs of observations given
Σx = 20, Σy = 20, Σx² = 90, Σy² = 90, Σxy = 76
Calculate (i) cov(x, y), (ii) b_yx and b_xy, (iii) r
Solution:


Sine byx and bxy are negative, r = -0.4

Question 4.
From the following data estimate y when x = 125

Solution:
Let u = x – 122, v = y – 14

Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 13.5) = -0.21(x – 121.5)
Y – 13.5 = -0.21x + 25.52
Y = -0.21x + 39.02
When x = 125
Y = -0.21(125) + 39.02
= -26.25 + 39.02
= 12.77

Question 5.
The following table gives the aptitude test scores and productivity indices of 10 works selected at workers selected randomly.

Obtain the two regression equation and estimate
(i) The productivity index of a worker whose test score is 95.
(ii) The test score when productivity index is 75.
Solution:


Regression equation of Y on X,
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{y}\))
(Y – 65) = 1.16 (x – 65)
Y – 65 = 1.16x – 75.4
Y = 1.16x – 10.4
(i) When x = 95
Y = 1.16(95) – 10.4
= 110.2 – 10.4
= 99.8
Regression equation of X on Y,
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 65) = 0.59(y – 65)
(X – 65) = 0.59y – 38.35
X = 0.59y + 26.65
(ii) When y = 75
x = 0.59(75) + 26.65
= 44.25 + 26.65
= 70.9

Question 6.
Compute the appropriate regression equation for the following data.

Solution:
Since x is the independent variable, and y is the dependent variable,
we need to find regression equation of y on x


Regression equation of y on x is (y – \(\bar{y}\)) = b_yx (x – \(\bar{x}\))
(y – 10) = -13.4(x – 6)
y – 10 = -1.34x + 8.04
y = -1.34x + 18.04

Question 7.
The following are the marks obtained by the students in Economic (X) and Mathematics (Y)

Find the regression equation of Y and X.
Solution:
Let u = x – 61, v = y – 80

Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 80.4) = 0.3(x – 61)
Y – 80.4 = 0.3x – 18.3
Y = 0.3x + 62.1

Question 8.
For the following bivariate data obtain the equation of two regressions lines:

Solution:


Regression equation of Y on X
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 9) = 2(x – 3)
Y – 9 = 2x – 6
Y = 2x + 3
Regression equation of X on Y
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 3) = 0.5(y – 9)
(X – 3) = 0.5y – 4.5
X = 0.5y – 1.5

Question 9.
Find the following data obtain the equation of two regression lines:

Solution:


Regression of Y on X,
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 8) = 0.65(x – 6)
Y – 8 = -0.65x + 3.9
Y = -0.65x + 11.9
Regression of X on Y
(X – \(\bar{x}\)) = b_xy (Y – \(\bar{y}\))
(X – 6) = -1.3(y – 8)
(X – 6) = -1.3y + 10.4
X = -1.3y + 16.4

Question 10.
For the following data, find the regression line of Y on X

Hence find the most likely value of y when x = 4
Solution:

(Y – 3) = 2(x – 2)
Y – 3 = 2x – 4
Y = 2x – 1
When x = 4
Y = 2(4) – 1
= 8 – 1
= 7

Question 11.
Find the following data, find the regression equation of Y on X, and estimate Y when X = 10.

Solution:

Regression equation of Y on X is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 5) = (0.63)(x – 3.5)
Y – 5 = 0.63x – 2.2
Y = 0.63x + 2.8
When x = 10
Y = 0.63(10) + 2.8
= 6.3 + 2.8
= 9.1

Question 12.
The following sample gave the number of hours of study (X) per day for an examination and marks (Y) obtained by 12 students.

Obtain the line of regression of marks on hours of study.
Solution:
Let u = x – 5, v = y – 70

∴ Equation of marks on hours of study is
(Y – \(\bar{y}\)) = b_yx (X – \(\bar{x}\))
(Y – 70.83) = 6.6(x – 4.92)
Y – 70.83 = 6.6x – 32.47
∴ Y = 6.6x + 38.36

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 2 Insurance and Annuity Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Insurance and Annuity Miscellaneous Exercise 2

(I) Choose the correct alternative.

Question 1.
“A contract that pledges payment of an agreed-upon amount to the person (or his/her nominee) on the happening of an event covered against” is technically known as
(a) Death coverage
(b) Saving for future
(c) Life insurance
(d) Provident fund
Answer:
(c) Life insurance

Question 2.
Insurance companies collect a fixed amount from their customers at a fixed interval of time. This amount is called
(a) EMI
(b) Installment
(c) Contribution
(d) Premium
Answer:
(d) Premium

Question 3.
Following are different types of insurance.
I. Life insurance
II. Health insurance
III. Liability insurance
(a) Only I
(b) Only II
(c) Only III
(d) All the three
Answer:
(d) All the three

Question 4.
By taking insurance, an individual
(a) Reduces the risk of an accident
(b) Reduces the cost of an accident
(c) Transfers the risk to someone else
(d) Converts the possibility of large loss to the certainty of a small one
Answer:
Converts the possibility of large loss to the certainty of a small one

Question 5.
You get payments of ₹ 8,000 at the beginning of each year for five years ta 6%, what is the value of this annuity?
(a) ₹ 34,720
(b) ₹ 39,320
(c) ₹ 35,720
(d) ₹ 40,000
Answer:
(c) ₹ 35,720

Question 6.
In an ordinary annuity, payments or receipts occur at
(a) Beginning of each period
(b) End of each period
(c) Mid of each period
(d) Quarterly basis
Answer:
(b) End of each period

Question 7.
The amount of money today which is equal to a series of payments in the future is called
(a) Normal value of the annuity
(b) Sinking value of the annuity
(c) Present value of the annuity
(d) Future value of the annuity
Answer:
(c) Present value of the annuity

Question 8.
Rental payment for an apartment is an example of
(a) Annuity due
(b) Perpetuity
(c) Ordinary annuity
(d) Installment
Answer:
(b) Perpetuity

Question 9.
_________ is a series of constant cash flows over a limited period of time.
(a) Perpetuity
(b) Annuity
(c) Present value
(d) Future value
Answer:
(b) Annuity

Question 10.
A retirement annuity is particularly attractive to someone who has
(a) A severe illness
(b) Risk of low longevity
(c) Large family
(d) Chance of high longevity
Answer:
(d) Chance of high longevity

(II) Fill in the blanks.

Question 1.
An installment of money paid for insurance is called _________
Answer:
premium

Question 2.
General insurance covers all risks except _________
Answer:
life

Question 3.
The value of insured property is called _________
Answer:
property value

Question 4.
The proportion of property value to insured is called _________
Answer:
policy value

Question 5.
The person who receive annuity is called _________
Answer:
Annuitant

Question 6.
The payment of each single annuity is called _________
Answer:
installment

Question 7.
The intervening time between payment of two successive installments is called as _________
Answer:
payment period

Question 8.
An annuity where payments continue forever is called _________
Answer:
perpetuity

Question 9.
If payments of an annuity fall due at the beginning of every period, the series is called _________
Answer:
annuity due

Question 10.
If payments of an annuity fall due at the end of every period, the series is called annuity _________
Answer:
immediate

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

Question 1.
General insurance covers life, fire, and theft.
Answer:
False

Question 2.
The amount of claim cannot exceed the amount of loss.
Answer:
True

Question 3.
Accident insurance has a period of five years.
Answer:
False

Question 4.
Premium is the amount paid to the insurance company every month.
Answer:
True

Question 5.
Payment of every annuity is called an installment.
Answer:
False

Question 6.
Annuity certainly begins on a fixed date and ends when an event happens.
Answer:
True

Question 7.
Annuity contingent begins and ends on certain fixed dates.
Answer:
False

Question 8.
The present value of an annuity is the sum of the present value of all installments.
Answer:
True

Question 9.
The future value of an annuity is the accumulated value of all installments.
Answer:
False

Question 10.
The sinking fund is set aside at the beginning of a business.
Answer:
True

(IV) Solve the following problems.

Question 1.
A house valued at ₹ 8,00,000 is insured at 75% of its value. If the rate of premium is 0.80%. Find the premium paid by the owner of the house. If the agent’s commission is 9% of the premium, find the agent’s commission.
Solution:
Property value = ₹ 8,00,000
Policy value = 75% × 8,00,000 = ₹ 6,00,000
∵ Rate of Premium = 0.80%
∴ Amount of Premium = 0.80% × 6,00,000 = ₹ 4,800
∵ Rate of commission = 9%
∴ Agent commission = 9% × 4800 = ₹ 432

Question 2.
A shopkeeper insures his shop and godown are valued at ₹ 5,00,000 and ₹ 10,00,000 respectively for 80% of their values. If the rate of premium is 8%, find the total annual premium.
Solution:
Property value of shop = ₹ 5,00,000
∴ Policy value = 80% × 5,00,000 = ₹ 4,00,000
∵ Rate of Premium = 8%
∴ Amount of premium = 8% × 4,00,000 = ₹ 32,000
∵ Property value of Godown = ₹ 10,00,000
∴ Policy value = 80% × 10,00,000 = ₹ 8,00,000
∵ Rate of Premium = 8%
∴ Amount of Premium = 8% × 8,00,000 = ₹ 64,000
∴ Total annual Premium = 64,000 + 32,000 = ₹ 96,000

Question 3.
A factory building is insured for \(\left(\frac{5}{6}\right)^{\text {th }}\) of its value at a rate of premium of 2.50%. If the agent is paid a commission of ₹ 2,812.50, which is 7.5% of the premium, find the value of the building.
Solution:
Let the Property value be ₹ x
∴ Policy value = ₹ \(\frac{5 x}{6}\)
∵ Rate of premium = 2.50%
∴ Amount of premium = \(\frac{5 x}{6}\) × 2.50% = ₹ \(\frac{x}{48}\)
∵ Rate of Agent commission = 7.5%
∴ Agent commission = 7.5% × \(\frac{x}{48}\)
∴ 2812.50 = \(\frac{x}{640}\)
∴ 2812.50 × 640 = x
∴ x = ₹ 18,00,000
∴ Value of the building is ₹ 18,00,000.

Question 4.
A merchant takes a fire insurance policy to cover 80% of the value of his stock. Stock worth ₹ 80,000 was completely destroyed in a fire. While the rest of the stock was reduced to 20% of its value. If the proportional compensation under the policy was ₹ 67,200, find the value of the stock.
Solution:
Let the Property value be ₹ x
∴ Policy value 80% × x = ₹ \(\frac{4 x}{5}\)
∵ Complete loss = ₹ 80,000
∴ Partial loss = 20% × (x – 8,00,000) = \(\frac{x-80,000}{5}\)
∴ Total loss = 80,000 + \(\frac{x-80,000}{5}\) = \(\frac{x}{5}\) + 64,000
∵ Claim = ₹ 67,200

∴ x = ₹ 1,00,000
∴ The value of the stock is ₹ 1,00,000.

Question 5.
A 35-year old person takes a policy for ₹ 1,00,000 for a period of 20 years. The rate of premium is ₹ 76 and the average rate of bonus is ₹ 7 per thousand p.a. If he dies after paying 10 annual premiums, what amount will his nominee receive?
Solution:
Policy value = ₹ 1,00,000
Period of Policy = 20 years
∵ Rate of premium = ₹ 76 per thousand
∴ Amount of premium = \(\frac{76}{1,000}\) × 1,00,000 = ₹ 7,600
∴ Total Premium = 7,600 × 10 = ₹ 76,000
∴ Rate of Bonus = ₹ 7 per thousand p.a
∴ Total Bonus = \(\frac{7}{1,000}\) × 1,00,000 = ₹ 7,000
∴ Amount received by Nominee = Policy value + Bonus earned
= 1,00,000 + 7,000
= ₹ 1,07,000

Question 6.
15,000 articles costing ₹ 200 per dozen were insured against fire for ₹ 1,00,000. If 20% of the articles were burnt completely and 2,400 other articles were damaged to the extent of 80% of their value, find the amount that can be claimed under the policy.
Solution:
Total Articles = 15,000
∴ Property value = \(\frac{15,000}{12}\) × 200 = 2,50,000
∵ Policy value = ₹ 1,00,000
∴ Complete loss = 20% × 2,50,000 = ₹ 50,000
∴ Partial loss = 80% × \(\frac{2,400}{12}\) × 200 = ₹ 3,20,000
∴ Total loss = 32,000 + 50,000 = ₹ 82,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{1,00,000}{2,50,000}\) × 82,000
= ₹ 32,800

Question 7.
For what amount should a cargo worth ₹ 25,350 be insured so that in the event of a total loss, its value, as well as the cost of insurance, may be recovered when the rate of premium is 2.5%.
Solution:
Let the policy value be ₹ 100 which includes the cost of insurance and premium
∴ Property value = 100 – 2.50 = ₹ 97.50
If the value of the cargo is ₹ 97.50, then the policy value is ₹ 100.
If the value of the cargo is ₹ 25,350, then
Policy value = \(\frac{100 \times 25,350}{97.50}\) = ₹ 26,000

Question 8.
A cargo of grain is insured at \(\left(\frac{3}{4}\right)\)% to cover 70% of its value. ₹1,008 is the amount of premium paid. If the grain is worth ₹ 12 per kg, how many kg of the grain did the cargo contain?
Solution:
Let the Property value be ₹ x
∴ policy value = 70% × x = ₹ \(\frac{7 x}{10}\)
∵ Rate of premium = \(\frac{3}{4}\)%
∴ Amount of premium = Rate × Policy value

∴ x = ₹ 1,92,000
∵ Rate of Jowar = ₹ 12/kg
∴ Quantity of Jowar = \(\frac{1,92,000}{12}\) = 16,000 kgs

Question 9.
4,000 bedsheets worth ₹ 6,40,000 were insured for \(\left(\frac{3}{7}\right)^{t h}\) of their value. Some of
the bedsheets were damaged in the rainy season and were reduced to 40% of their value. If the amount recovered against damage was ₹ 32,000. Find the number of damaged bedsheets.
Solution:
∵ Property value = ₹ 6,40,000
∴ Policy value = 6,40,000 × \(\frac{3}{7}\) = ₹ \(\frac{19,20,000}{7}\)
∴ Cost of one Bedsheet = \(\frac{6,40,000}{4,000}\) = ₹ 160
Let ‘x’ bedsheets be damaged.
∴ Cost of x bedsheets = ₹ 160x

∴ 875 Bedsheets damaged.

Question 10.
A property valued at ₹ 7,00,000 is insured to the extent of ₹ 5,60,000 at \(\left(\frac{5}{8}\right)\)% less 20%. Calculate the saving made in the premium. Find the amount of loss that the owner must bear, including premium, if the property is damaged to the extent of 40% of its value.
Solution:
∵ Property value = ₹ 7,00,000
∵ Policy value = ₹ 5,60,000
∵ Rate of premium = \(\frac{5}{8}\)%
∴ Amount of premium = \(\frac{5}{8}\)% × 5,60,000 = ₹ 3,500
New rate of premium = \(\frac{5}{8}\)% less 20%
= \(\frac{5}{8}\) – [20% x \(\frac{5}{8}\)]
= \(\frac{5}{8}\) – \(\frac{1}{8}\)
= \(\frac{1}{2}\)%
∴ Amount of premium = \(\frac{1}{2}\)% × 5,60,000 = ₹ 2,800
∴ Saving made in premium = 3,500 – 2,800 = ₹ 700
∴ Loss = 7,00,000 × 40% = 2,80,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{5,60,000}{7,00,000}\) × 2,80,000
= ₹ 2,24,000
∴ Loss bear by owner = loss – claim + premium
= 2,80,000 – 2,24,000 + 2,800
= ₹ 58,800

Question 11.
Stocks in a shop and godown worth ₹ 75,000 and ₹ 1,30,000 respectively were insured through an agent who receive 15% of the premium as commission. If the shop was insured for 80% and godown for 60% of the value, find the amount of agent’s commission when the premium was 0.80% less 20%. If the entire stock in the shop and 20% stock in the godown is destroyed by fire, find the amount that can be claimed under the policy.
Solution:
∵ Rate of premium = 0.80% less 20%
= 0.80 – 20% × 0.80
= 0.80 – 0.16
= 0.64%
For Shop
∵ Property value = ₹ 75,000
∴ Policy value = 80% × 75,000 = ₹ 60,000
∴ Premium = 0.64% × 60,000 = ₹ 384
∵ Loss = ₹ 75,000
∵ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{60,000}{75,000}\) × 75,000
= ₹ 60,000
For Godown
∵ Property value = ₹ 1,30,000
∴ Policy value = 60% × 1,30,000 = ₹ 78,000
∴ Premium = 0.64% × 78,000 = ₹ 499.2
Loss = 20% × 1,30,000 = ₹ 26,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{78,000}{1,30,000}\) × 26,000
= ₹ 15,600
Total claim = 16,600 + 60,000 = ₹ 75,600
∵ Rate of commission = 15%
∴ Agent commission = 15% × [384 + 499.2]
= 15% × 883.2
= ₹ 132.48

Question 12.
A person holding a life policy of ₹ 1,20,000 for a term of 25 years wants to discontinue after paying a premium for 8 years at the rate of ₹ 58 per thousand p.a. Find the amount of paid-up value he will receive on the policy. Find the amount he will receive if the surrender value granted is 35% of the premium paid, excluding the first year’s premium.
Solution:
Policy value = ₹ 1,20,000
∵ Rate of premium = ₹ 58 per thousand p.a.
∴ Premium for 8 years = \(\frac{8 \times 58}{1000}\) × 1,20,000 = ₹ 55,680
∴ Amount of 1st premium = \(\frac{55,680}{8}\) = ₹ 6,960
∵ Paid-up value of policy = \(\frac{\text { No of Premium paid }}{\text { Terms of policy }}\) × Policy value
= \(\frac{8}{25}\) × 1,20,000
= ₹ 38,400
∵ Surrender value = 35% × [Total premium – 1st year premium]
= 35% × [55,680 – 6,960]
= 35% × 48,720
= ₹ 17,052

Question 13.
A godown valued at ₹ 80,000 contained stock worth ₹ 4,80,000. Both were insured against fire. Godown for ₹ 50,000 and stock for 80% of its value. A part of stock worth ₹ 60,000 was completely destroyed and the rest was reduced to 60% of its value. The amount of damage to the godown is ₹ 40,000. Find the amount that can be claimed under the policy.
Solution:
For Godown
∵ Property value = ₹ 80,000
∵ Policy value = ₹ 50,000
∵ Loss = ₹ 40,000
∵ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{50,000}{80,000}\) × 40,000
= ₹ 25,000
For stock
∵ Property value = ₹ 4,80,000
∵ Policy value = 80% × 4,80,000 = ₹ 3,84,000
∵ Complete loss = ₹ 60,000
∴ Partial loss = (100 – 60)% × [4,80,000 – 60,000]
= 40% × 4,20,000
= ₹ 1,68,000
∴ Total loss = 1,68,000 + 60000 = ₹ 2,28,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{3,84,000}{4,80,000}\) × 2,28,000
= ₹ 1,82,400
∴ Total claim = 25,000 + 1,82,400 = ₹ 2,07,400

Question 14.
Find the amount of an ordinary annuity if a payment of ₹ 500 is made at the end of every quarter for 5 years at the rate of 12% per annum compounded quarterly. [Given: (1.03)²⁰ = 1.8061]
Solution:
∵ C = ₹ 500
∵ r = 12% p.a. compounded quarterly,
∴ r = \(\frac{12}{4}\) = 3%
∵ n = 5 years
But, payment is made quarterly
∴ n = 5 × 4 = 20

Question 15.
Find the amount a company should set aside at the end of every year if it wants to buy a machine expected to cost ₹ 1,00,000 at the end of 4 years and interest rate is 5% p.a. compounded annually.
Solution:
∵ A = ₹ 1,00,000
∵ r = 5% p.a.
∴ i = \(\frac{r}{100}=\frac{5}{100}\) = 0.05
∵ n = 4 years
∵ A = \(\frac{C}{i}\left[(1+\mathrm{i})^{n}-1\right]\)
∴ 1,00,000 = \(\frac{C}{0.05}\)[(1 + 0.05)⁴ – 1]
∴ 1,00,000 × 0.05 = C [(1.05)⁴ – 1]
∴ 5,000 = C(1.2155 – 1)
∴ 5,000 = C × 0.2155
∴ \(\frac{5,000}{0.2155}\) = C
∴ C = ₹ 23,201.86

Question 16.
Find the least number of years for which an annuity of ₹ 3,000 per annum must run in order that its amount exceeds ₹ 60,000 at 10%compounded annually. [Given: (1.1)¹¹ = 2,8531, (1.1)¹² = 3.1384]
Solution:
∵ A = ₹ 60,000
∵ C = ₹ 3,000
∵ r = 10% p.a.
∴ i = \(\frac{r}{100}=\frac{10}{100}\) = 0.1
∵ A = \(\frac{C}{i}\left[(1+i)^{n}-1\right]\)
∴ 60,000 = \(\frac{3,000}{0.1}\left[(1+0.1)^{n}-1\right]\)
∴ 60,000 = 30,000[(1.1)ⁿ – 1]
∴ \(\frac{60,000}{30,000}\) + 1 = (1.1)ⁿ
∴ 2 + 1 = (1.1)ⁿ
∴ 3 = (1.1)ⁿ
Taking log
∴ log 3 = log (1.1)ⁿ
∴ log 3 = n log(1.1)
∴ \(\frac{\log 3}{\log 1.1}\) = n
∴ n = \(\frac{0.4771}{0.0414}\) = 11.52 ~ 12 years

Question 17.
Find the rate of interest compounded annually if an ordinary annuity of ₹ 20,000 per year amounts to ₹ 41,000 in 2 years.
Solution:
∵ C = ₹ 20,000
∵ A = ₹ 41,000
∵ n = 2 years

∴ r = 5% p.a.

Question 18.
A person purchases a television by paying ₹ 20,000 in cash and promising to pay ₹ 1,000 at the end of every month for the next 2 years. If money is worth 12% p.a., converted monthly. Find the cash price of the television. [Given: (1.01)^-24 = 0.7880]
Solution:
Down payment = ₹ 20,000
∵ n = 2 years
But, EMI Payable monthly
∴ n = 2 × 12 = 24
∵ r = 12% p.a. compounded monthly

∴ P = 1,00,00 × 0.2120
∴ P = ₹ 21,200
Cash price = Present value + Down payment
= 21,200 + 20,000
= ₹ 41,200

Question 19.
Find the present value of an annuity immediate of ₹ 20,000 per annum for 3 years at 10% p.a. compounded annually. [Given: (1.1)^-3 = 0.7513]
Solution:
∵ C = ₹ 20,000
∵ n = 3 years
∵ r = 10% p.a.

∴ P = 2,00,000 [1 – 0.7513]
∴ P = 2,00,000 [0.2487]
∴ P = ₹ 49,740

Question 20.
A man borrowed some money and paid it back in 3 equal installments of ₹ 2,160 each. What amount did he borrow if the rate of interest was 20% per annum compounded annually? Also, find the total interest charged. [Given: (1.2)^-3 = 0.5788]
Solution:
∵ C = ₹ 2,160
∵ n = 3
∵ r = 20% p.a.

∴ P = ₹ 6,251.04
∴ Total amount paid = 2,160 × 3 = ₹ 6,480
∴ Interest = 6,480 – 6,251.04 = ₹ 228.96

Question 21.
A company decides to set aside a certain amount at the end of every year to create a sinking fund that should amount to ₹ 9,28,200 in 4 years at 10% p.a. Find the amount to be set aside every year. [Given: (1.1)⁴ = 1.4641]
Solution:
∵ A = ₹ 9,28,200
∵ n = 4 years
∵ r = 10% p.a.

∴ 9,28,200 × 0.1 = C[1.4641 – 1]
∴ 92,820 = C × 0.4641
∴ \(\frac{92,820}{0.4641}\) = C
∴ C = ₹ 2,00,000

Question 22.
Find the future value after 2 years if an amount of ₹ 12,000 is invested at the end of every half-year at 12% p.a. compounded half-yearly. [Given: (1.06)⁴ = 1.2625]
Solution:
∵ n = 2 years
Payable half yearly, n = 2 × 2 = 4
∵ C = ₹ 12,000
∵ r = 12% p.a. Compounded half yearly


∴ A = 1,00,000 [1.2625 – 1]
∴ A = 1,00,000 × 0.2625
∴ A = ₹ 26,250

Question 23.
After how many years would an annuity due of ₹ 3,000 p.a. accumulated ₹ 19,324.80 at 20% p.a. compounded annually? [Given: (1.2)⁴ = 2.0736]
Solution:
∵ C = ₹ 3,000
∵ A = ₹ 9,324.80
∵ r = 20% p.a.

∴ 19,324.80 = 15,000 × 1.2[(1.2)ⁿ – 1]
∴ 19,324.80 = 18,000[(1.2)ⁿ – 1]
∴ \(\frac{19,324.80}{18,000}\) + 1 = (1.2)ⁿ
∴ 1.0736 + 1 = (1.2)ⁿ
∴ 2.0736 = (1.2)ⁿ
∴ (1.2)⁴ = (1.2)ⁿ
∴ n = 4 years

Question 24.
Some machinery is expected to cost 25% more over its present cost of ₹ 6,96,000 after 20 yeas. The scrap value of the machinery will realize ₹ 1,50,000. What amount should be set aside at the end of every year at 5% p.a. compound interest for 20 years to replace the machinery? [Given: (1.05)²⁰ = 2655]
Solution:
Present cost = ₹ 6,96,000
Expected cost = 25% × 6,96,000 + 6,96,000
= 1,74,000 + 6,96,000
= ₹ 8,70,000
∴ Scrap value = ₹ 1,50,000
∴ Sinking fund = 8,70,000 – 1,50,000 = ₹ 7,20,000
∴ A = ₹ 7,20,000, n = 20 years, r = 5% p.a.

∴ 7,20,000 × 0.05 = C[(1.05)²⁰ – 1]
∴ 36,000 = C[2.655 – 1]
∴ 36,000 = C × 1.655
∴ \(\frac{36,000}{1.655}\) = C
∴ C = ₹ 21,752.27

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 2 Insurance and Annuity Ex 2.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Insurance and Annuity Ex 2.2

Question 1.
Find the accumulated (future) value of annuity of ₹ 800 for 3 year at interest rate 8% compounded annually. [Given: (1.08)³ = 1.2597]
Solution:
∵ C = ₹ 800
∵ n = 3 years
∵ r = 8% p.a.

∴ A = 10,000[(1.08)³ – 1]
∴ A = 10,000[1.2597 – 1]
∴ A = 10,000 × 0.2597
∴ A = ₹ 2,597

Question 2.
A person invested ₹ 5,000 every year in finance company that offered him interest compounded at 10% p.a., what is the amount accumulated after 4 years? [Given: (1.1)⁴ = 1.4641]
Solution:
∵ C = ₹ 5,000
∵ r = 10% p.a.

= 50,000[(1.1)⁴ – 1]
= 50,000[1.4641 – 1]
= 50,000 × 0.4641
= ₹ 23,205

Question 3.
Find the amount accumulated after 2 years if a sum of ₹ 24,000 is invested every six months at 12% p.a. compounded half yearly. [Given: (1.06)⁴ = 1.2625]
Solution:
∵ C = ₹ 24,000
∵ n = 2 years
But invested half yearly
∴ n = 2 × 2 = 4
∵ r = 12% p.a. compounded half yearly

= 4,00,000[(1.06)⁴ – 1]
= 4,00,000[1.2625 – 1]
= 4,00,000 × 0.2625
= ₹ 1,05,000

Question 4.
Find the accumulated value after 1 year of an annuity immediate in which ₹ 10,000 are invested every quarter at 16% p.a. compounded quarterly. [Given: (1.04)⁴ = 1.1699]
Solution:
∵ C = ₹ 10,000
∵ n = 1 year
But invested every quarterly
∴ n = 1 × 4 = 4
∴ r = 16% p.a. compounded quarterly

= 2,50,000 [1.1699 – 1]
= 2,50,000 × 0.1699
= ₹ 42,475

Question 5.
Find the present value of an annuity immediate of ₹ 36,000 p.a. for 3 years at 9% p.a. compounded annually. [Given: (1.09)^-3 = 0.7722]
Solution:
∵ C = ₹ 36,000
∵ n = 3 years
∵ r = 9% p.a.

= 4,00,000 × 0.2278
= ₹ 91,120

Question 6.
Find the present value of ordinary annuity of ₹ 63,000 p.a. for 4 years at 14% p.a. compounded annually. [Given: (1.14)^-4 = 0.5921]
Solution:
∵ C = ₹ 63,000
∵ n = 4 years
∵ r = 14% p.a.

= 4,50,000[1 – 0.5921]
= 4,50,000 × 0.4079
= ₹ 1,83,555

Question 7.
A lady plans to save for her daughter’s marriage. She wishes to accumulate a sum of ₹ 4,64,100 at the end of 4 years. What amount should she invest every year if she get an interest of 10%p.a. compounded annually? [Given: (1.1)⁴ = 1.4641]
Solution:
∵ A = ₹ 4,64,100
∵ n = 4 years
∵ r = 10% p.a.

∴ 46,410 = C[1.4641 – 1]
∴ 46,410 = C × 0.4641
∴ \(\frac{46,410}{0.4641}\) = C
∴ C = ₹ 1,00,000

Question 8.
A person wants to create a fund of ₹ 6,96,150 after 4 years at the time of his retirement. He decides to invest a fixed amount at the end of every year in a bank that offers him interest of 10% p.a. compounded annually. What amount should he invest every year? [Given: (1.1)⁴ = 1.4641]
Solution:
∵ A = ₹ 6,96,150
∵ n = 4 years
∵ r = 10% p.a

∴ 69,615 = C[1.4641 – 1]
∴ 69,615 = C × 0.4641
∴ \(\frac{69,615}{0.4641}\) = C
∴ C = ₹ 1,50,000

Question 9.
Find the rate of interest compounded annually if an annuity immediate at ₹ 20,000 per year amounts to ₹ 2,60,000 in 3 years.
Solution:
∵ C = ₹ 20,000
∵ A = ₹ 2,60,000
∵ n = 3 years

∴ 13i = 3i + 3 i² + i³
∴ 13i = i(3 + 3i + i²)
∴ 13 = 3 + i + i²
∴ i² + 3i + 3 – 13 = 0
∴ i² + 3i – 10 = 0
∴ (i + 5) (i – 2) = 0
∴ i + 5 = 0 or i – 2 = 0
∴ i = -5 or i = 2
∵ Rate of interest cannot be negative
∴ i = 2 is accepted
∴ \(\frac{r}{100}\) = 2
∴ r = 200% p.a.

Question 10.
Find the number of years for which an annuity of ₹ 500 is paid at the end of every years, if the accumulated amount works out to be ₹ 1,655 when interest is compounded annually at 10% p.a.
Solution:
∵ C = 7500
∵ A = 71,655
∵ r = 10% p.a.


∴ 0.331 + 1 = (1.1)ⁿ
∴ 1.331 = (1.1)ⁿ
∴ (1.1)³ = (1.1)ⁿ
∴ n = 3 years

Question 11.
Find the accumulated value of annuity due of ₹ 1,000 p.a. for 3 years at 10% p.a. compounded annually. [Given: (1.1)³ = 1.331]
Solution:
∵ C = ₹ 1,000
∵ n = 3 years
∵ r = 10% p.a.

∴ A’ = 10,000 × 1.1[(1.1)³ – 1]
∴ A’ = 11,000 [1.331 – 1]
∴ A’ = 11,000 × 0.331
∴ A’ = ₹ 3,641

Question 12.
A person plans to put ₹ 400 at the beginning of each year for 2 years in a deposit that gives interest at 2% p.a. compounded annually. Find the amount that will be accumulated at the end of 2 years. [Given: (1.02)² = 1.0404]
Solution:
∵ C = ₹ 400
∵ r = 2% p.a.


= 20,000 (1.02) (1.0404 – 1)
= 20,400 [0.0404]
= ₹ 824.16

Question 13.
Find the present value of an annuity due of ₹ 600 to be paid quarterly at 32% p.a. compounded quarterly. [Given (1.08)^-4 = 0.7350]
Solution:
∵ C = ₹ 600
∵ n = 1 year
∴ But invested every quarterly
∴ n = 1 × 4 = 4
∵ r = 32% p.a. compounded quarterly

= 7,500(1.08) [1 – 0.7350]
= 8,100 [0.2650]
= ₹ 2,146.5

Question 14.
An annuity immediate is to be paid for some years at 12% p.a. The present value of the annuity is ₹ 10,000 and the accumulated value is ₹ 20,000. Find the amount of each annuity payment.
Solution:
∵ r = 12% p.a.
∴ i = \(\frac{r}{100}=\frac{12}{100}\) = 0.12
∵ P = ₹ 10,000
∵ A = ₹ 20,000
∵ \(\frac{1}{P}-\frac{1}{A}=\frac{i}{C}\)

∴ C = 0.12 × 20,000
∴ C = ₹ 2,400

Question 15.
For an annuity immediate paid for 3 years with interest compounded at 10% p.a. the present value is ₹ 24,000. What will be the accumulated value after 3 years? [Given (1.1)³ = 1.331]
Solution:
∵ n = 3 years
∵ P = ₹ 24,000
∵ r = 10% p.a.
∴ i = \(\frac{r}{100}=\frac{10}{100}\) = 0.1
∵ A = P(1 + i)ⁿ
∴ A = 24,000 [1 + 0.1]³
∴ A = 24,000 × (1.1)³
∴ A = 24,000 × 1.331
∴ A = ₹ 31,944

Question 16.
A person sets up a sinking fund in order to have ₹ 1,00,000 after 10 years. What amount should be deposited bi-annually in the account that pays him 5% p.a. compounded semi-annually? [Given: (1.025)²⁰ = 1.675]
Solution:
∴ A = ₹ 1,00,000
∴ n = 10 years
But, invested half yearly
∴ n = 10 × 2 = 20
∵ r = 5% p.a. compounded half yearly
∴ r = \(\frac{r}{2}=\frac{5}{2}\) = 2.5%
∴ i = \(\frac{r}{100}=\frac{2.5}{100}\) = 0.025

∴ 2,500 = C[1.675 – 1]
∴ 2,500 = C × 0.675
∴ \(\frac{2,500}{0.675}\) = C
∴ C = ₹ 3,703.70