Class 12

Index Numbers Class 12 Commerce Maths 2 Chapter 5 Exercise 5.1 Answers Maharashtra Board

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

Std 12 Maths 2 Exercise 5.1 Solutions Commerce Maths

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:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 5.1 Solutions
• 12th Commerce Maths Exercise 5.2 Solutions
• 12th Commerce Maths Exercise 5.3 Solutions
• 12th Commerce Maths Miscellaneous Exercise 5 Solutions

Time Series Class 12 Commerce Maths 2 Chapter 4 Miscellaneous Exercise 4 Answers Maharashtra Board

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

Std 12 Maths 2 Miscellaneous Exercise 4 Solutions Commerce Maths

(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:

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• 12th Commerce Maths Exercise 4.1 Solutions
• 12th Commerce Maths Miscellaneous Exercise 4 Solutions

Time Series Class 12 Commerce Maths 2 Chapter 4 Exercise 4.1 Answers Maharashtra Board

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

Std 12 Maths 2 Exercise 4.1 Solutions Commerce Maths

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)

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• 12th Commerce Maths Exercise 4.1 Solutions
• 12th Commerce Maths Miscellaneous Exercise 4 Solutions

Linear Regression Class 12 Commerce Maths 2 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

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

Std 12 Maths 2 Miscellaneous Exercise 3 Solutions Commerce Maths

(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

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Linear Regression Class 12 Commerce Maths 2 Chapter 3 Exercise 3.3 Answers Maharashtra Board

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

Std 12 Maths 2 Exercise 3.3 Solutions Commerce Maths

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

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Linear Regression Class 12 Commerce Maths 2 Chapter 3 Exercise 3.2 Answers Maharashtra Board

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

Std 12 Maths 2 Exercise 3.2 Solutions Commerce Maths

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

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Linear Regression Class 12 Commerce Maths 2 Chapter 3 Exercise 3.1 Answers Maharashtra Board

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

Std 12 Maths 2 Exercise 3.1 Solutions Commerce Maths

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

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 3.1 Solutions
• 12th Commerce Maths Exercise 3.2 Solutions
• 12th Commerce Maths Exercise 3.3 Solutions
• 12th Commerce Maths Miscellaneous Exercise 3 Solutions

Differential Equation and Applications Class 12 Commerce Maths 1 Chapter 8 Miscellaneous Exercise 8 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Miscellaneous Exercise 8 Questions and Answers.

Std 12 Maths 1 Miscellaneous Exercise 8 Solutions Commerce Maths

(I) Choose the correct option from the given alternatives:

Question 1.
The order and degree of \(\left(\frac{d y}{d x}\right)^{3}-\frac{d^{3} y}{d x^{3}}+y e^{x}=0\) are respectively.
(a) 3, 1
(b) 1, 3
(c) 3, 3
(d) 1, 1
Answer:
(a) 3, 1

Question 2.
The order and degree of \(\left[1+\left(\frac{d y}{d x}\right)^{3}\right]^{\frac{2}{3}}=8 \frac{d^{3} y}{d x^{3}}\) are respectively
(a) 3, 1
(c) 3, 3
(b) 1, 3
(d) 1, 1
Answer:
(c) 3, 3

Question 3.
The differential equation of y = k₁ + \(\frac{k_{2}}{x}\) is
(a) \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
(b) \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}=0\)
(d) \(x \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}=0\)
Answer:
(b) \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

Question 4.
The differential equation of y = k₁ e^x + k₂ e^-x is
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)
(b) \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)
(c) \(\frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=0\)
(d) \(\frac{d^{2} y}{d x^{2}}+y=0\)
Answer:
(a) \(\frac{d^{2} y}{d x^{2}}-y=0\)

Question 5.
The solution of \(\frac{d y}{d x}\) = 1 is
(a) x + y = c
(b) xy = c
(c) x² + y² = c
(d) y – x = c
Answer:
(d) y – x = c

Question 6.
The solution of \(\frac{d y}{d x}+\frac{x^{2}}{y^{2}}=0\) is
(a) x³ + y³ = 7
(b) x² + y² = c
(c) x³ + y³ = c
(d) x + y = c
Answer:
(c) x³ + y³ = c

Question 7.
The solution of x \(\frac{d y}{d x}\) = y log y is
(a) y = ae^x
(b) y = be^2x
(c) y = be^-2x
(d) y = e^ax
Answer:
(d) y = e^ax

Question 8.
Bacterial increases at a rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in
(a) 4 hours
(b) 6 hours
(c) 8 hours
(d) 10 hours
Answer:
(b) 6 hours

Question 9.
The integrating factor of \(\frac{d y}{d x}\) – y = e^x is
(a) x
(b) -x
(c) e^x
(d) e^-x
Answer:
(c) e^x

Question 10.
The integrating factor of \(\frac{d y}{d x}\) – y = e^x is e^-x, then its solution is
(a) ye^-x = x + c
(b) ye^x = x + c
(c) ye^x = 2x + c
(d) ye^-x = 2x + c
Answer:
(a) ye^-x = x + c

(II) Fill in the blanks:

Question 1.
The order of highest derivative occurring in the differential equation is called ________ of the differential equation.
Answer:
order

Question 2.
The power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called ________ of the differential equation.
Answer:
degree

Question 3.
A solution of differential equation that can be obtained from the general solution by giving particular values to the arbitrary constants is called _________ solution.
Answer:
particular

Question 4.
Order and degree of a differential equation are always _________ integers.
Answer:
positive

Question 5.
The integrating factor of the differential equation \(\frac{d y}{d x}\) – y = x is _________
Answer:
e^-x

Question 6.
The differential equation by eliminating arbitrary constants from bx + ay = ab is _________
Answer:
\(\frac{d^{2} y}{d x^{2}}=0\)

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

Question 1.
The integrating factor of the differential equation \(\frac{d y}{d x}\) – y = x is e^-x.
Answer:
True

Question 2.
The order and degree of a differential equation are always positive integers.
Answer:
True

Question 3.
The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.
Answer:
True

Question 4.
The order of highest derivative occurring in the differential equation is called the degree of the differential equation.
Answer:
False

Question 5.
The power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any is called the order of the differential equation.
Answer:
False

Question 6.
The degree of the differential equation \(e^{\frac{d y}{d x}}=\frac{d y}{d x}+c\) is not defined.
Answer:
True

(IV) Solve the following:

Question 1.
Find the order and degree of the following differential equations:
(i) \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given differential equation is \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3 / 2}=\frac{d^{2} y}{d x^{2}}\)
∴ \(\left[\frac{d^{3} y}{d x^{3}}+x\right]^{3}=\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 3
∴ order = 3 and degree = 3

(ii) \(x+\frac{d y}{d x}=1+\left(\frac{d y}{d x}\right)^{2}\)
Solution:
The given differential equation is \(x+\frac{d y}{d x}=1+\left(\frac{d y}{d x}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ order = 1, degree = 2.

Question 2.
Verify that y = log x + c is a solution of the differential equation \(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\).
Solution:
y = log x + c
Differentiating both sides w.r.t. x, we get
\(\frac{d y}{d x}=\frac{1}{x}+0=\frac{1}{x}\)
∴ x\(\frac{d y}{d x}\) = 1
Differentiating again w.r.t. x, we get
\(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \times 1=0\)
∴ \(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)
This shows that y = log x + c is a solution of the D.E.
\(x \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=0\)

Question 3.
Solve the following differential equations:
(i) \(\frac{d y}{d x}\) = 1 + x + y + xy
Solution:
\(\frac{d y}{d x}\) = 1 + x + y + xy
∴ \(\frac{d y}{d x}\) = (1 + x) + y(1 + x) = (1 + x)(1 + y)
∴ \(\frac{1}{1+y}\) dy = (1 + x) dx
Integrating, we get
∫\(\frac{1}{1+y}\) dy = ∫(1 + x) dx
∴ log|1 + y| = x + \(\frac{x^{2}}{2}\) + c
This is the general solution.

(ii) \(e^{d y / d x}=x\)
Solution:

∴ from (1), the general solution is
y = x log x – x + c, i.e. y = x(log x – 1) + c.

(iii) dr = ar dθ – θ dr
Solution:
dr = ar dθ – θ dr
∴ dr + θ dr = ar dθ
∴ (1 + θ) dr = ar dθ
∴ \(\frac{d r}{r}=\frac{a d \theta}{1+\theta}\)
On integrating, we get
\(\int \frac{d r}{r}=a \int \frac{d \theta}{1+\theta}\)
∴ log |r| = a log |1 + θ| + c
This is the general solution.

(iv) Find the differential equation of the family of curves y = e^x (ax + bx²), where a and b are arbitrary constants.
Solution:
y = e^x (ax + bx²)
ax + bx² = ye^-x …….(1)
Differentiating (1) w.r.t. x twice and writing \(\frac{d y}{d x}\) as y₁ and \(\frac{d^{2} y}{d x^{2}}\) as y₂, we get


This is the required differential equation.

Question 4.
Solve \(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\) when x = \(\frac{2}{3}\) and y = \(\frac{1}{3}\).
Solution:

Question 5.
Solve y dx – x dy = -log x dx.
Solution:
y dx – x dy = -log x dx
∴ y dx – x dy + log x dx = 0
∴ x dy = (y + log x) dx
∴ \(\frac{d y}{d x}=\frac{y+\log x}{x}=\frac{y}{x}+\frac{\log x}{x}\)
∴ \(\frac{d y}{d x}-\frac{1}{x} \cdot y=\frac{\log x}{x}\) …….(1)
This is the linear differential equation of the form


This is the general solution.

Question 6.
Solve y log y \(\frac{d x}{d y}\) + x – log y = 0.
Solution:

Question 7.
Solve (x + y) dy = a² dx
Solution:

Question 8.
Solve \(\frac{d y}{d x}+\frac{2}{x} y=x^{2}\)
Solution:
\(\frac{d y}{d x}+\frac{2}{x} y=x^{2}\) ……..(1)
This is a linear differential equation of the form

This is the general solution.

Question 9.
The rate of growth of the population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lakh, when will the city have a population of 400000?
Solution:
Let P be the population at time t years.
Then the rate of growth of the population is \(\frac{d P}{d t}\) which is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant
∴ \(\frac{d P}{P}\)= k dt
On integrating, we get
\(\int \frac{d P}{P}=k \int d t\)
∴ log P = kt + c
The population doubled in 25 years and present population is 1,00,000.
∴ initial population was 50,000
i.e. when t = 0, P = 50000
∴ log 50000 = k × 0 + c
∴ c = log 50000
∴ log P = kt + log 50000
When t = 25, P = 100000
∴ log 100000 = k × 25 + log 50000
∴ 25k = log 100000 – log 50000 = log(\(\frac{100000}{50000}\))
∴ k = \(\frac{1}{25}\) log 2
∴ log P = \(\frac{t}{25}\) log 2 + log 50000
If P = 400000, then
log 400000 = \(\frac{t}{25}\) log 2 + log 50000
∴ log 400000 – log 50000 = \(\frac{t}{25}\) log 2
∴ log(\(\frac{400000}{50000}\)) = \(\log (2)^{t / 25}\)
∴ log 8 = \(\log (2)^{t / 25}\)
∴ 8 = \((2)^{t / 25}\)
∴ \((2)^{t / 25}\) = (2)³
∴ \(\frac{t}{25}\) = 3
∴ t = 75
∴ the population will be 400000 in (75 – 25) = 50 years.

Question 10.
The resale value of a machine decreases over a 10 years period at a rate that depends on the age of the machine. When the machine is x years old, the rate at which its value is changing is ₹ 2200(x – 10) per year. Express the value of the machine as a function of its age and initial value. If the machine was originally worth ₹ 1,20,000 how much will it be worth when it is 10 years old?
Solution:
Let V be the value of the machine after x years.
Then rate of change of the value is \(\frac{d V}{d x}\) which is 2200(x – 10)
∴ \(\frac{d V}{d x}\) = 2200(x – 10)
∴ dV = 2200(x – 10) dx
On integrating, we get
∫dV = 2200∫(x – 10) dx
∴ V = 2200[\(\frac{x^{2}}{2}\) – 10x] + c
Initially, i.e. at x = 0, V = 120000
∴ 120000 = 2200 × 0 + c = c
∴ c = 120000
∴ V = 2200[\(\frac{x^{2}}{2}\) – 10x] + 120000 …….(1)
This gives value of the machine in terms of initial value and age x.
We have to find V when x = 10.
When x = 10, from (1)
V = 2200[\(\frac{100}{2}\) – 100] + 120000
= 2200 [-50] + 120000
= -110000 + 120000
= 10000
Hence, the value of the machine after 10 years will be ₹ 10000.

Question 11.
Solve y² dx + (xy + x²) dy = 0
Solution:

Question 12.
Solve x²y dx – (x³ + y³) dy = 0
Solution:

Question 13.
Solve yx \(\frac{d y}{d x}\) = x² + 2y²
Solution:

Question 14.
Solve (x + 2y³) \(\frac{d y}{d x}\) = y
Solution:
(x + 2y³) \(\frac{d y}{d x}\) = y
∴ x + 2y³ = y \(\frac{d x}{d y}\)

This is the general solution.

Question 15.
Solve y dx – x dy + log x dx = 0
Solution:
y dx – x dy + log x dx = 0
∴ (y + log x) dx = x dy


This is the general solution.

Question 16.
Solve \(\frac{d y}{d x}\) = log x dx
Solution:
\(\frac{d y}{d x}\) = log x dx
∴ dy = log x dx
On integrating, we get
∫dy = ∫log x . 1 dx
∴ y = (log x) ∫1 dx – \(\int\left[\left\{\frac{d}{d x}(\log x)\right\} \cdot \int 1 d x\right] d x\)
∴ y = (log x) . x – \(\int \frac{1}{x} \cdot x d x\)
∴ y = x log x – ∫1 dx
∴ y = x log x – x + c
This is the general solution.

Question 17.
y log y \(\frac{d x}{d y}\) = log y – x
Solution:

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 8.1 Solutions
• 12th Commerce Maths Exercise 8.2 Solutions
• 12th Commerce Maths Exercise 8.3 Solutions
• 12th Commerce Maths Exercise 8.4 Solutions
• 12th Commerce Maths Exercise 8.5 Solutions
• 12th Commerce Maths Exercise 8.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 8 Solutions

Differential Equation and Applications Class 12 Commerce Maths 1 Chapter 8 Exercise 8.6 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.6 Questions and Answers.

Std 12 Maths 1 Exercise 8.6 Solutions Commerce Maths

Question 1.
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, find the number of times the bacteria are increased in 12 hours.
Solution:
Let x be the number of bacteria in the culture at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
∫\(\frac{d x}{x}\) = k∫dt
∴ log x = kt + c
Initially, i.e. when t = 0, let x = x₀
∴ log x₀ = k × 0 + c
∴ c = log x₀
∴ log x = kt + log x₀
∴ log x – log x₀ = kt
∴ log(\(\frac{x}{x_{0}}\)) = kt ……(1)
Since the number doubles in 4 hours, i.e. when t = 4, x = 2x₀

∴ the number of bacteria will be 8 times the original number in 12 hours.

Question 2.
If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? (Given: \(\sqrt{\frac{3}{2}}\) = 1.2247)
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population, is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, k is a constant
∴ \(\frac{d P}{P}\) = k dt
Integrating, we get
∫\(\frac{d P}{P}\) = k∫dt
∴ log P = kt + c
Initially, i.e. when t = 0, P = 40000
∴ log 40000 = 0 + c
∴ c = log 40000
∴ log P = kt + log 40000
∴ log P – log 40000 = kt
∴ log(\(\frac{P}{40000}\)) = kt ………(1)
When t = 40, P = 60000

∴ population after 60 years will be 73482.

Question 3.
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after \(\frac{5}{2}\) hours. [Given: √2 = 1.414]
Solution:
Let x be the number of bacteria at time t.
Then the rate of increase is \(\frac{d x}{d t}\) which is proportional to x.
∴ \(\frac{d x}{d t}\) ∝ x
∴ \(\frac{d x}{d t}\) = kx, where k is a constant
∴ \(\frac{d x}{x}\) = k dt
On integrating, we get
∫\(\frac{d x}{x}\) = k∫dt
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c
∴ c = log 1000
∴ log x = kt + log 1000
∴ log x – log 1000 = kt
∴ log(\(\frac{x}{1000}\)) = kt …….(1)
Now, when t = 1, x = 2 × 1000 = 2000

∴ number of bacteria after \(\frac{5}{2}\) hours = 5656.

Question 4.
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years, the population increased from 30,000 to 40,000.
Solution:
Let P be the population of the city at time t.
Then \(\frac{d P}{d t}\), the rate of increase of population, is proportional to P.
∴ \(\frac{d P}{d t}\) ∝ P
∴ \(\frac{d P}{d t}\) = kP, where k is a constant.
∴ \(\frac{d P}{P}\) = k dt
On integrating, we get
∫\(\frac{1}{P}\)dP = k∫dt
∴ log P = kt + c
Initially, i.e. when t = 0, P = 30000
∴ log 30000 = k x 0 + c
∴ c = log 30000
∴ log P = kt + log 30000
∴ log P – log 30000 = kt

∴ the population of the city at time t = 30000\(\left(\frac{4}{3}\right)^{\frac{t}{40}}\).

Question 5.
The rate of depreciation \(\frac{d V}{d t}\) of a machine is inversely proportional to the square of t + 1, where V is the value of the machine t years after it was purchased. The initial value of the machine was ₹ 8,00,000 and its value decreased ₹ 1,00,000 in the first year. Find the value after 6 years.
Solution:
Let V be the value of the machine at the end of t years.
Then \(\frac{d V}{d t}\), the rate of depreciation, is inversly proportional to (t + 1)².

Initially, i.e. when t = 0, V = 800000
∴ 800000 = \(\frac{k}{1}\) + c = k + c ………(1)
Now, when t = 1, V = 800000 – 100000 = 700000
∴ 700000 = \(\frac{k}{1+1}\) + c = \(\frac{k}{2}\) + c ……(2)
Subtracting (2) from (1), we get
100000 = \(\frac{1k}{2}\)
∴ k = 200000
∴ from (1), 800000 = 200000 + c
∴ c = 600000 200000
∴ V = \(\frac{200000}{t+1}\) + 600000
When t = 6,
V = \(\frac{200000}{7}\) + 600000
= 28571.43 + 600000
= 628571.43 ~ 628571
Hence, the value of the machine after 6 years will be ₹ 6,28,571.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 8.1 Solutions
• 12th Commerce Maths Exercise 8.2 Solutions
• 12th Commerce Maths Exercise 8.3 Solutions
• 12th Commerce Maths Exercise 8.4 Solutions
• 12th Commerce Maths Exercise 8.5 Solutions
• 12th Commerce Maths Exercise 8.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 8 Solutions

Differential Equation and Applications Class 12 Commerce Maths 1 Chapter 8 Exercise 8.5 Answers Maharashtra Board

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 8 Differential Equation and Applications Ex 8.5 Questions and Answers.

Std 12 Maths 1 Exercise 8.5 Solutions Commerce Maths

Solve the following differential equations.

Question 1.
\(\frac{d y}{d x}+y=e^{-x}\)
Solution:
\(\frac{d y}{d x}+y=e^{-x}\) …….(1)
This is the linear differential equation of the form

This is the general solution.

Question 2.
\(\frac{d y}{d x}\) + y = 3
Solution:
\(\frac{d y}{d x}\) + y = 3
This is the linear differential equation of the form

This is the general solution.

Question 3.
x\(\frac{d y}{d x}\) + 2y = x² . log x.
Solution:
x\(\frac{d y}{d x}\) + 2y = x² . log x
∴ \(\frac{d y}{d x}+\left(\frac{2}{x}\right) \cdot y=x \cdot \log x\) …….(1)
This is the linear differential equation of the form


This is the general solution.

Question 4.
(x + y)\(\frac{d y}{d x}\) = 1
Solution:
(x + y) \(\frac{d y}{d x}\) = 1
∴ \(\frac{d x}{d y}\) = x + y
∴ \(\frac{d x}{d y}\) – x = y
∴ \(\frac{d x}{d y}\) + (-1) x = y ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 5.
y dx + (x – y²) dy = 0
Solution:
y dx + (x – y²) dy = 0
∴ y dx = -(x – y²) dy
∴ \(\frac{d x}{d y}=-\frac{\left(x-y^{2}\right)}{y}=-\frac{x}{y}+y\)
∴ \(\frac{d x}{d y}+\left(\frac{1}{y}\right) \cdot x=y\) ……(1)
This is the linear differential equation of the form

This is the general solution.

Question 6.
\(\frac{d y}{d x}\) + 2xy = x
Solution:
\(\frac{d y}{d x}\) + 2xy = x ………(1)
This is the linear differential equation of the form

This is the general solution.

Question 7.
(x + a) \(\frac{d y}{d x}\) = -y + a
Solution:
(x + a) \(\frac{d y}{d x}\) + y = a
∴ \(\frac{d y}{d x}+\left(\frac{1}{x+a}\right) y=\frac{a}{x+a}\) ……..(1)
This is the linear differential equation of the form

This is the general solution.

Question 8.
dy + (2y) dx = 8 dx
Solution:
dy + (2y) dx = 8 dx
∴ \(\frac{d y}{d x}\) + 2y = 8 …….(1)
This is the linear differential equation of the form

This is the general solution.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 8.1 Solutions
• 12th Commerce Maths Exercise 8.2 Solutions
• 12th Commerce Maths Exercise 8.3 Solutions
• 12th Commerce Maths Exercise 8.4 Solutions
• 12th Commerce Maths Exercise 8.5 Solutions
• 12th Commerce Maths Exercise 8.6 Solutions
• 12th Commerce Maths Miscellaneous Exercise 8 Solutions