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

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

(I) Choose the correct alternative.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

Question 10.
Fisher’s Price Number is given by

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

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

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

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


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

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

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

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

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

(II) Fill in the blanks.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(IV) Solve the following problems.

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

Solution:

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

Solution:

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

Solution:

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

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

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

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

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

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

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

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Balbharati Maharashtra State Board 12th Commerce Maths 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 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 1.
Calculate D₆ and P₈₅ for the following data:
79, 82, 36, 38, 51, 72, 68, 70, 64, 63
Solution:
The given data can be arranged in ascending order as follows:
36, 38, 51, 63, 64, 68, 70, 72, 79, 82
Here, n = 10
D₆ = value of 6\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 6\(\left(\frac{10+1}{10}\right)^{\text {th }}\) observation
= value of (6 × 1.1)th observation
= value of (6.6)th observation
= value of 6th observation + 0.6(value of 7th observation – value of 6th observation)
= 68 + 0.6(70 – 68)
= 68 + 0.6(2)
= 68 + 1.2
∴ D₆ = 69.2
P₈₅ = value of \(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{100}\right)^{\text {th }}\) observation
= value of (85 × 0. 11)th observation
= value of (9.35)th observation
= value of 9th observation + 0.35(value of 10th observation – value of 9th observation)
= 19 + 0.35(82 – 79)
= 79 + 0.35(3)
= 79 + 1.05
∴ P₈₅ = 80.05

Question 2.
The daily wages (in ₹) of 15 labourers are as follows:
230, 400, 350, 200, 250, 380, 210, 225, 375, 180, 375, 450, 300, 350, 250
Calculate D₈ and P₉₀.
Solution:
The given data can be arranged in ascending order as follows:
180, 200, 210, 225, 230, 250, 250, 300, 350, 350, 375, 375, 380, 400, 450
Here, n = 15
D₈ = value of 8\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 8\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (8 × 1.6)th observation
= value of (12.8)th observation
= value of 12th observation – 0.8(value of 13th observation – value of 12th observation)
= 375 + 0.8(380 – 375)
= 375 + 0.8(5)
= 375 + 4
∴ D₈ = 379
P₉₀ = value of 90\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 90\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (90 × 0.16)th observation
= value of (14.4)th observation
= value of 14th observation + 0.4 (value of 15th observation – value of 14th observation)
= 400 + 0.4(450 – 400)
= 400 + 0.4(50)
= 400 + 20
∴ P₉₀ = 420

Question 3.
Calculate 2nd decile and 65th percentile for the following:

Solution:
We construct the less than cumulative frequency table as given below:

Here, n = 200
D₂ = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{200+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 20.1)th observation
= value of (40.2)th observation
Cumulative frequency which is just greater than (or equal to) 40.2 is 58.
∴ D₂ = 120
P₆₅ = value of 65\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 65\(\left(\frac{200+1}{100}\right)^{\text {th }}\) observation
= value of (65 × 2.01)th observation
= value of (130.65)th observation
The cumulative frequency which is just greater than (or equal to) 130.65 is 150.
∴ P₆₅ = 280

Question 4.
From the following data calculate the rent of the 15th, 65th, and 92nd house.

Solution:
Arranging the given data in ascending order.

Here, n = 100
P₁₅ = value of 15
= value of 15\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 15\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (15 × 1.01 )th observation
= value of (15.15)th observation
Cumulative frequency which is just greater than (or equal to) 15.15 is 25.
∴ P₁₅ = 11000
P₆₅ = value of 65\(\left(\frac{n+1}{100}\right)^{\text {th }}\)observation
= value of 65\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (65 × 1.01)th observation
= value of (65.65)th observation
Cumulative frequency which is just greater than (or equal to) 65.65 is 70.
∴ P₆₅ = 14000
P₉₂ = value of 92\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 92\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (92 × 1.01)th observation
= value of (92.92)th observation
Cumulative frequency which is just greater than (or equal to) 92.92 is 98.
∴ P₉₂ = 17000

Question 5.
The following frequency distribution shows the weight of students in a class.

(a) Find the percentage of students whose weight is more than 50 kg.
(b) If the weight column provided is of mid values then find the percentage of students whose weight is more than 50 kg.
Solution:
(a) Let the percentage of students weighing less than 50 kg be x.
∴ Px = 50

From the table, out of 20 students, 84 students have their weight less than 50 kg.
∴ Number of students weighing more than 50 kg = 120 – 84 = 36
∴ Percentage of students having there weight more than 50 kg = \(\frac{36}{120}\) × 100 = 30%

(b) The difference between any two consecutive mid values of weight is 5 kg.
The class intervals must of width 5, with 40, 45,….. as their mid values.
∴ The class intervals will be 37.5 – 42.5, 42.5 – 47.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 120
Let P_x = 50
The value 50 lies in the class 47.5 – 52.5
∴ L = 47.5, h = 5, f = 29, c.f. = 55

∴ x = 58 (approximately)
∴ 58% of students are having weight below 50 kg.
∴ Percentage of students having weight above 50 kg is 100 – 58 = 42
∴ 42% of students are having weight above 50 kg.

Question 6.
Calculate D₄ and P₄₈ from the following data:

Solution:
The difference between any two consecutive mid values is 5, the width of class interval = 5
∴ Class interval with mid-value 2.5 is 0 – 5
Class interval with mid value 7.5 is 5 – 10, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 100
D₄ class = class containing \(\left(\frac{4 \mathrm{N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{4 \mathrm{N}}{10}=\frac{4 \times 100}{10}\) = 40
Cumulative frequency which is just greater than (or equal to) 40 is 50.
∴ D₄ lies in the class 10 – 15.
∴ L = 10,h = 5, f = 25, c.f. = 25
∴ D₄ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{4 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 10 + \(\frac{5}{25}\) (40 – 25)
= 10 + \(\frac{1}{5}\) (15)
= 10 + 3
∴ D₄ = 13
P₄₈ class = class containing \(\left(\frac{48 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{48 \mathrm{~N}}{100}=\frac{48 \times 100}{100}\) = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P₄₈ lies in the class 10 – 15.
∴ L = 10, h = 5, f = 25, c.f. = 25
∴ P₄₈ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{48 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10 + \(\frac{5}{25}\) (48 – 25)
= 10 + \(\frac{1}{5}\) (23)
= 10 + 4.6
∴ P₄₈ = 14.6

Question 7.
Calculate D₉ and P₂₀ of the following distribution.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 240
D₉ class = class containing \(\left(\frac{9 \mathrm{~N}}{10}\right)^{\mathrm{th}}\) observation
∴ \(\frac{9 \mathrm{~N}}{10}=\frac{9 \times 240}{10}\) = 216
Cumulative frequency which is just greater than (or equal to) 216 is 225.
∴ D₉ lies in the class 80 – 100.
∴ L = 80, h = 20, f = 90, c.f. = 135
∴ D₉ = \(L+\frac{h}{f}\left(\frac{9 N}{10}-c . f .\right)\)
= 80 + \(\frac{20}{90}\)(216 – 135)
= 80 + \(\frac{2}{9}\)(81)
= 80 + 18
∴ D₉ = 98
P₂₀ class = class containing \(\left(\frac{20 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{20 \mathrm{~N}}{100}=\frac{20 \times 240}{100}\) = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P₂₀ lies in the class 40 – 60.
∴ L = 40, h = 20, f = 35, c.f. = 15

∴ P₂₀ = 58.86

Question 8.
Weekly wages for a group of 100 persons are given below:

D₃ for this group is ₹ 1100. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 500 – 1000 and class 2000 – 2500 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 …..(i)
Given, D₃ = 1100
∴ D₃ lies in the class 1000 – 1500.
∴ L = 1000, h = 500, f = 25, c.f. = 7 + a
∴ \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 100}{10}=30\)
∴ D₃ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{3 \mathrm{~N}}{10}-\mathrm{c} . \mathrm{f} .\right)\)
∴ 1100 = 1000 + \(\frac{500}{25}\) [30 – (7 + a)]
∴ 1100 – 1000 = 20(30 – 7 – a)
∴ 100 = 20(23 – a)
∴ 100 = 460 – 20a
∴ 20a = 460 – 100
∴ 20a = 360
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18 = 20
∴ 18 and 20 are the missing frequencies of the class 500 – 1000 and class 2000 – 2500 respectively.

Question 9.
The weekly profit (in rupees) of 100 shops are distributed as follows:

Find the limits of the profit of middle 60% of the shops.
Solution:
To find the limits of the profit of the middle 60% of the shops, we have to find P₂₀ and P₈₀.
We construct the less than cumulative frequency table as given below:

Here, N = 100
P₂₀ class = class containing \(\left(\frac{20 \mathrm{N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{20 \mathrm{N}}{100}=\frac{20 \times 100}{100}=20\)
Cumulative frequency which is just greater than (or equal to) 20 is 26.
∴ P₂₀ lies in the class 1000 – 2000.
∴ L = 1000, h = 1000, f = 16, c.f. = 10
∴ P₂₀ = \(L+\frac{h}{f}\left(\frac{20 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 1000 + \(\frac{1000}{16}\) (20 – 10)
= 1000 + \(\frac{125}{2}\) (10)
= 1000 + 625
∴ P₂₀ = 1625
P₈₀ class = class containing \(\left(\frac{80 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{80 \mathrm{~N}}{100}=\frac{80 \times 100}{100}=80\)
Cumulative frequency which is just greater than (or equal to) 80 is 92.
∴ P₈₀ lies in the class 4000 – 5000.
∴ L = 4000, h = 1000, f = 20, c.f. = 72
∴ P₈₀ = \(L+\frac{h}{f}\left(\frac{80 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 4000 + \(\frac{1000}{20}\)(80 – 72)
= 4000 + 50(8)
= 4000 + 400
∴ P₈₀ = 4400
∴ the profit of middle 60% of the shops lie between the limits ₹ 1,625 to ₹ 4,400.

Question 10.
In a particular factory, workers produce various types of output units. The following distribution was obtained:

Find the percentage of workers who have produced less than 82 output units.
Solution:
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be 69.5 – 74.5, 74.5 – 79.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 445
Let P_x = 82
The value 82 lies in the class 79.5 – 84.5
∴ L = 79.5, h = 5, f = 50, c.f. = 85

∴ 24.72% of workers produced less than 82 output units.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Question 1.
Compute all the quartiles for the following series of observations:
16, 14.9, 11.5, 11.8, 11.1, 14.5, 14, 12, 10.9, 10.7, 10.6, 10.5, 13.5, 13, 12.6
Solution:
The given data can be arranged in ascending order as follows:
10.5, 10.6, 10.7, 10.9, 11.1, 11.5, 11.8, 12, 12.6, 13, 13.5, 14, 14.5, 14.9, 16
Here, n = 15
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of 4th observation
∴ Q₁ = 10.9
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 4)th observation
= value of 8th observation
∴ Q₂ = 12
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 4)th observation
= value of 12th observation
∴ Q₃ = 14

Question 2.
The heights (in cm.) of 10 students are given below:
148, 171, 158, 151, 154, 159, 152, 163, 171, 145
Calculate Q₁ and Q₃ for the above data.
Solution:
The given data can be arranged in ascending order as follows:
145, 148, 151, 152, 154, 158, 159, 163, 171, 171
Here, n = 10
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= value of (2.75)th observation
= value of 2nd observation + 0.75 (value of 3rd observation – value of 2nd observation)
= 148 + 0.75 (151 – 148)
= 148 + 0.75(3)
= 148 + 2.25
∴ Q₁ = 150.25
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 2.75)th observation
= value of (8.25)th observation
= value of 8th observation + 0.25 (value of 9th observation – value of 8th observation)
= 163 + 0.25(171 – 163)
= 163 + 0.25(8)
= 163 + 2
∴ Q₃ = 165

Question 3.
The monthly consumption of electricity (in units) of families in a certain locality is given below:
205, 201, 190, 188, 195, 172, 210, 225, 215, 232, 260, 230
Calculate electricity consumption (in units) below which 25% of the families lie.
Solution:
To find the consumption of electricity below which 25% of the families lie, we have to find Q₁.
Monthly consumption of electricity (in units) can be arranged in ascending order as follows:
172, 188, 190, 195, 201, 205, 210, 215, 225, 230, 232, 260.
Here, n = 12
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{12+1}{4}\right)^{\text {th }}\) observation
= value of (3.25)th observation
= value of 3rd observation + 0.25 (value of 4th observation – value of 3rd observation)
= 190 + 0.25(195 – 190)
= 190 + 0.25(5)
= 190 + 1.25
= 191.25
∴ the consumption of electricity below which 25% of the families lie is 191.25.

Question 4.
For the following data of daily expenditure of families (in ₹), compute the expenditure below which 75% of families include their expenditure.

Solution:
To find the expenditure below which 75% of families have their expenditure, we have to find Q₃.
We construct the less than cumulative frequency table as given below:

Here, n = 100
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 25.25)th observation
= value of (75.75)th observation
Cumulative frequency which is just greater than (or equal to) 75.75 is 87.
∴ Q₃ = 650
∴ the expenditure below which 75% of families include their expenditure is ₹ 650.

Question 5.
Calculate all the quartiles for the following frequency distribution:

Solution:
We construct the less than cumulative frequency table as given below:

Here, n = 300
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation
= value of (75.25)th observation
Cumulative frequency which is just greater than (or equal to) 75.25 is 90.
∴ Q₁ = 2
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 75.25)th observation
= value of (150.50)th observation
∴ Cumulative frequency which is just greater than (or equal to) 150.50 is 185.
∴ Q₂ = 3
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 75.25)th observation
= value of (225.75)th observation
Cumulative frequency which is just greater than (or equal to) 225.75 is 249.
∴ Q₃ = 4

Question 6.
The following is the frequency distribution of heights of 200 male adults in a factory:

Find the central height.
Solution:
To find the central height, we have to find Q₂.
We construct the less than cumulative frequency table as given below:

Here, N = 200
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 200}{4}\) = 100
Cumulative frequency which is just greater than (or equal to) 100 is 156.
∴ Q₂ lies in the class 165 – 170.
∴ L = 165, h = 5, f = 64, c.f. = 92
Q₂ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 165 + \(\frac{5}{64}\) (100 – 92)
= 165 + \(\frac{5}{64}\) × 8
= 165 + \(\frac{5}{8}\)
= 165 + 0.625
= 165.625
∴ Central height is 165.625 cm.

Question 7.
The following is the data of pocket expenditure per week of 50 students in a class. It is known that the median of the distribution is ₹ 120. Find the missing frequencies.

Solution:
Let a and b be the missing frequencies of class 50 – 100 and class 150 – 200 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 25 + a + b
Since, N = 50
∴ 25 + a + b = 50
∴ a + b = 25 …..(i)
Given, Median = Q₂ = 120
∴ Q2 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 15, \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 50}{4}\) = 25
∴ Q₂ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
∴ 120 = 100 + \(\frac{50}{15}\) [25 – (7 + a)]
∴ 120 – 100 = \(\frac{10}{3}\) (25 – 7 – a)
∴ 20 = \(\frac{10}{3}\) (18 – a)
∴ \(\frac{60}{10}\) = 18 – a
∴ 6 = 18 – a
∴ a = 18 – 6 = 12
Substituting the value of a in equation (i), we get
12 + b = 25
∴ b = 25 – 12 = 13
∴ 12 and 13 are the missing frequencies of the class 50 – 100 and class 150 – 200 respectively.

Question 8.
The following is the distribution of 160 workers according to the wages in a certain factory:

Determine the values of all quartiles and interpret the results.
Solution:
The given table is a more than cumulative frequency.
We transform the given table into less than cumulative frequency.
We construct the less than cumulative frequency table as given below:

Here, N = 160
∴ Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{160}{4}\) = 40
Cumulative frequency which is just greater than (or equal to) 40 is 69.
∴ Q₁ lies in the class 10000 – 11000
∴ L = 10000, h = 1000, f = 46, c.f. = 23
Q₁ = \(L+\frac{h}{f}\left(\frac{N}{4}-\text { c.f. }\right)\)
= 10000 + \(\frac{1000}{46}\) (40 – 23)
= 10000 + \(\frac{1000}{46}\) (17)
= 10000 + \(\frac{17000}{46}\)
= 10000 + 369.57
= 10369.57
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 160}{4}\) = 80
Cumulative frequency which is just greater than (or equal to) 80 is 103.
∴ Q₂ lies in the class 11000 – 12000.
∴ L = 11000, h = 1000, f = 34, c.f. = 69
∴ Q₂ = \(L+\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
= 11000 + \(\frac{1000}{34}\)(80 – 69)
= 11000 + \(\frac{1000}{34}\)(11)
= 11000 + \(\frac{11000}{34}\)
= 11000 + 323.529
= 11323.529
Q₃ class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 160}{4}\) = 120
Cumulative frequency which is just greater than (or equal to) 120 is 137.
∴ Q₃ lies in the class 12000 – 13000.
∴ L = 12000, h = 1000, f = 34, c.f. = 103
∴ Q₃ = \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 12000 + \(\frac{1000}{34}\) (120 – 103)
= 12000 + \(\frac{1000}{34}\) (17)
= 12000 + \(\frac{1000}{2}\)
= 12000 + 500
= 12500
Interpretation:
Q₁ < Q₂ < Q₃

Question 9.
Following is grouped data for the duration of fixed deposits of 100 senior citizens from a certain bank:

Calculate the limits of fixed deposits of central 50% senior citizens.
Solution:
We construct the less than cumulative frequency table as given below:

To find the limits of fixed deposits of central 50% senior citizens, we have to find Q₁ and Q₃.
Here, N = 100
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{100}{4}\) = 25
Cumulative frequency which is just greater than (or equal to) 25 is 35.
∴ Q1 lies in the class 180 – 360.
∴ L = 180, h = 180, f = 20, c.f. = 15
∴ Q₁ = \(L+\frac{h}{f}\left(\frac{N}{4}-c . f .\right)\)
= 180 + \(\frac{180}{20}\) (25 – 15)
= 180 + 9(10)
= 180 + 90
∴ Q₁ = 270
Q₃ class = class containing \(\left(\frac{3 \mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{N}}{4}=\frac{3 \times 100}{4}\) = 75
Cumulative frequency which is just greater than (or equal to) 75 is 90.
∴ Q₃ lies in the class 540 – 720.
∴ L = 540, h = 180, f = 30, c.f. = 60
∴ Q₃ = \(L+\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 540 + \(\frac{180}{30}\) (75 – 60)
= 540 + 6(15)
= 540 + 90
∴ Q₃ = 630
∴ Limits of duration of fixed deposits of central 50% senior citizens is from 270 to 630.

Question 10.
Find the missing frequency given that the median of the distribution is 1504.

Solution:
Let x be the missing frequency of the class 1550 – 1750.
We construct the less than frequency table as given below:

Here, N = 199 + x
Given, Median (Q₂) = 1504
∴ Q₂ lies in the class 1350 – 1550.
∴ L = 1350, h = 200, f = 100, c.f. = 63,
\(\frac{2 \mathrm{~N}}{4}=\frac{199+x}{2}\)
∴ Q₂ = \(L+\frac{h}{f}\left(\frac{2 N}{4}-c . f .\right)\)
∴ 1504 = 1350 + \(\frac{200}{100}\left(\frac{199+x}{2}-63\right)\)
∴ 1504 – 1350 = 2\(\left(\frac{199+x-126}{2}\right)\)
∴ 154 = 199 + x – 126
∴ 154 = x + 73
∴ x = 81

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Differentiation Miscellaneous Exercise 9 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Differentiation Miscellaneous Exercise 9

I. Differentiate the following functions w.r.t.x.

Question 1.
x⁵
Solution:
Let y = x⁵
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} x^{5}=5 x^{4}\)

Question 2.
x^-2
Solution:
Let y = x^-2
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{-2}\right)=-2 x^{-3}=\frac{-2}{x^{3}}\)

Question 3.
√x
Solution:
Let y = √x
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} \sqrt{x}=\frac{1}{2 \sqrt{x}}\)

Question 4.
x√x
Solution:
Let y = x√x
∴ y = \(x^{\frac{3}{2}}\)
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{3}{2}}=\frac{3}{2} x^{\frac{1}{2}}\)

Question 5.
\(\frac{1}{\sqrt{x}}\)
Solution:
Let y = \(\frac{1}{\sqrt{x}}\)
∴ y = \(x^{\frac{-1}{2}}\)
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{-1}{2} x^{\frac{-3}{2}}=\frac{-1}{2 x^{\frac{3}{2}}}\)

Question 6.
7^x
Solution:
Let y = 7^x
Differentiating w.r.t. x, we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x} 7^{x}=7^{x} \log 7\)

II. Find \(\frac{d y}{d x}\) if

Question 1.
y = x² + \(\frac{1}{x^{2}}\)
Solution:

Question 2.
y = (√x + 1)²
Solution:

Question 3.
y = \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\)
Solution:

Question 4.
y = x³ – 2x² + √x + 1
Solution:

Question 5.
y = x² + 2x – 1
Solution:

Question 6.
y = (1 – x)(2 – x)
Solution:

Question 7.
y = \(\frac{1+x}{2+x}\)
Solution:

Question 8.
y = \(\frac{(\log x+1)}{x}\)
Solution:

Question 9.
y = \(\frac{e^{x}}{\log x}\)
Solution:

Question 10.
y = x log x (x² + 1)
Solution:

III. Solve the following:

Question 1.
The relation between price (P) and demand (D) of a cup of Tea is given by D = \(\frac{12}{P}\). Find
the rate at which the demand changes when the price is ₹ 2/-. Interpret the result.
Solution:
Demand, D = \(\frac{12}{P}\)
Rate of change of demand

When price P = 2,
Rate of change of demand,
\(\left(\frac{\mathrm{dD}}{\mathrm{dP}}\right)_{\mathrm{P}=2}=\frac{-12}{(2)^{2}}=-3\)
∴ When the price is 2, the rate of change of demand is -3.
∴ Here, the rate of change of demand is negative demand would fall when the price becomes ₹ 2.

Question 2.
The demand (D) of biscuits at price P is given by D = \(\frac{64}{P^{3}}\), find the marginal demand
when the price is ₹ 4/-.
Solution:
Given demand D = \(\frac{64}{P^{3}}\)
Now, marginal demand

When P = 4
Marginal demand

Question 3.
The supply S of electric bulbs at price P is given by S = 2p³ + 5. Find the marginal supply when the price is ₹ 5/-. Interpret the result.
Solution:
Given, supply S = 2p³ + 5
Now, marginal supply

∴ When p = 5
Marginal supply = \(\left(\frac{\mathrm{dS}}{\mathrm{dp}}\right)_{\mathrm{p}=5}\)
= 6(5)²
= 150
Here, the rate of change of supply with respect to the price is positive which indicates that the supply increases.

Question 4.
The total cost of producing x items is given by C = x² + 4x + 4. Find the average cost and the marginal cost. What is the marginal cost when x = 7?
Solution:
Total cost C = x² + 4x + 4
Now. Average cost = \(\frac{C}{x}=\frac{x^{2}+4 x+4}{x}\)
= x + 4 + \(\frac{4}{x}\)
and Marginal cost = \(\frac{\mathrm{dC}}{\mathrm{d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\)(x² + 4x + 4)
= \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x²) + 4\(\frac{\mathrm{d}}{\mathrm{d} x}\) (x) + \(\frac{\mathrm{d}}{\mathrm{d} x}\) (4)
= 2x + 4(1) + 0
= 2x + 4
∴ When x = 7,
Marginal cost = \(\left(\frac{\mathrm{d} \mathrm{C}}{\mathrm{d} x}\right)_{x=7}\)
= 2(7) + 4
= 14 + 4
= 18

Question 5.
The demand D for a price P is given as D = \(\frac{27}{P}\), find the rate of change of demand when the price is ₹ 3/-.
Solution:
Demand, D = \(\frac{27}{P}\)
Rate of change of demand = \(\frac{dD}{dP}\)

When price P = 3,
Rate of change of demand,
\(\left(\frac{\mathrm{dD}}{\mathrm{dP}}\right)_{\mathrm{P}=3}=\frac{-27}{(3)^{2}}=-3\)
∴ When price is 3, Rate of change of demand is -3.

Question 6.
If for a commodity; the price demand relation is given as D = \(\left(\frac{P+5}{P-1}\right)\). Find the marginal demand when price is ₹ 2/-
Solution:

Question 7.
The price function P of a commodity is given as P = 20 + D – D² where D is demand. Find the rate at which price (P) is changing when demand D = 3.
Solution:
Given, P = 20 + D – D²
Rate of change of price = \(\frac{dP}{dD}\)
= \(\frac{d}{dD}\)(20 + D – D²)
= 0 + 1 – 2D
= 1 – 2D
Rate of change of price at D = 3 is
\(\left(\frac{\mathrm{dP}}{\mathrm{dD}}\right)_{\mathrm{D}=3}\) = 1 – 2(3) = -5
∴ Price is changing at a rate of -5, when demand is 3.

Question 8.
If the total cost function is given by C = 5x³ + 2x² + 1; find the average cost and the marginal cost when x = 4.
Solution:
Total cost function C = 5x³ + 2x² + 1
Average cost = \(\frac{C}{x}\)
= \(\frac{5 x^{3}+2 x^{2}+1}{x}\)
= 5x² + 2x + \(\frac{1}{x}\)
When x = 4,
Average cost = 5(4)² + 2(4) + \(\frac{1}{4}\)
= 80 + 8 + \(\frac{1}{4}\)
= \(\frac{320+32+1}{4}\)
= \(\frac{353}{4}\)
Marginal cost = \(\frac{\mathrm{dC}}{\mathrm{d} x}\)
= \(\frac{d}{dx}\) (5x³ + 2x² + 1)
= 5\(\frac{d}{dx}\) (x³) + 2 \(\frac{d}{dx}\) (x²) + \(\frac{d}{dx}\) (1)
= 5(3x²) + 2(2x) + 0
= 15x² + 4x
When x = 4, marginal cost = \(\left(\frac{\mathrm{dC}}{\mathrm{d} x}\right)_{x=4}\)
= 15(4)² + 4(4)
= 240 + 16
= 256
∴ The average cost and marginal cost at x = 4 are \(\frac{353}{4}\) and 256 respectively.

Question 9.
The supply S for a commodity at price P is given by S = P² + 9P – 2. Find the marginal supply when the price is 7/-.
Solution:
Given, S = P² + 9P – 2

∴ The marginal supply is 23, at P = 7.

Question 10.
The cost of producing x articles is given by C = x² + 15x + 81. Find the average cost and marginal cost functions. Find the marginal cost when x = 10. Find x for which the marginal cost equals the average cost.
Solution:
Given, cost C = x² + 15x + 81

If marginal cost = average cost, then
2x + 15 = x + 15 + \(\frac{81}{x}\)
∴ x = \(\frac{81}{x}\)
∴ x² = 81
∴ x = 9 …..[∵ x > 0]

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 5 Integration Ex 5.4

Evaluate the following.

Question 1.
\(\int \frac{1}{4 x^{2}-1} d x\)
Solution:

Question 2.
\(\int \frac{1}{x^{2}+4 x-5} d x\)
Solution:

Question 3.
\(\int \frac{1}{4 x^{2}-20 x+17} d x\)
Solution:

Question 4.
\(\int \frac{x}{4 x^{4}-20 x^{2}-3} d x\)
Solution:

Question 5.
\(\int \frac{x^{3}}{16 x^{8}-25} d x\)
Solution:

Question 6.
\(\int \frac{1}{a^{2}-b^{2} x^{2}} d x\)
Solution:

Question 7.
\(\int \frac{1}{7+6 x-x^{2}} d x\)
Solution:

Question 8.
\(\int \frac{1}{\sqrt{3 x^{2}+8}} d x\)
Solution:

Question 9.
\(\int \frac{1}{\sqrt{x^{2}+4 x+29}} d x\)
Solution:

Question 10.
\(\int \frac{1}{\sqrt{3 x^{2}-5}} d x\)
Solution:

Question 11.
\(\int \frac{1}{\sqrt{x^{2}-8 x-20}} d x\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.3 Questions and Answers.

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

Evaluate the following:

Question 1.
\(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Solution:
Let I = \(\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t\)
Put, Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[\(\frac{d}{d t}\)(4e^2t – 5)]
∴ 3e^2t + 5 = A(4e^2t – 5) + B[4e^2t × 2 – 0]
∴ 3e^2t + 5 = (4A + 8B) e^2t – 5A
Equating the coefficient of e^2t and constant on both sides, we get
4A + 8B = 3
and -5A = 5
∴ A = -1
∴ from (1), 4(-1) + 8B = 3
∴ 8B = 7
∴ B = \(\frac{7}{8}\)

Question 2.
\(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Solution:
Let I = \(\int \frac{20-12 e^{x}}{3 e^{x}-4} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 20 – 12e^x = A(3e^x – 4) + B[\(\frac{d}{d x}\)(3e^x – 4)]
∴ 20 – 12e^x = A(3e^x – 4) + B(3e^x – 0)
∴ 20 – 12e^x = (3A + 3B)e^x – 4A
Equating the coefficient of ex and constant on both sides, we get
3A + 3B = -12 ……(1)
and -4A = 20
∴ A = -5
from (1), 3(-5) + 3B = -12
∴ 3B = 3
∴ B = 1
∴ 20 – 12e^x = -5(3e^x – 4) + (3e^x)

Question 3.
\(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Solution:
Let I = \(\int \frac{3 e^{x}+4}{2 e^{x}-8} d x\)
Put, Numerator = A (Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ 3e^x + 4 = A(2e^x – 8) + B[\(\frac{d}{d x}\)(2e^x – 8)]
∴ 3e^x + 4 = A(2e^x – 8) + B(2e^x – 0)
∴ 3e^x + 4 = (2A + 2B)e^x – 8A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 3 ……..(1)
and -8A = 4
∴ A = \(-\frac{1}{2}\)
∴ from (1), 2(\(-\frac{1}{2}\)) + 2B = 3
∴ 2B = 4
∴ B = 2

Question 4.
\(\int \frac{2 e^{x}+5}{2 e^{x}+1} d x\)
Solution:

Balbharati Maharashtra State Board Std 12 Commerce Statistics Part 1 Digest Pdf Chapter 5 Integration Ex 5.2 Questions and Answers.

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

Evaluate the following.

Question 1.
\(\int x \sqrt{1+x^{2}} d x\)
Solution:

Question 2.
\(\int \frac{x^{3}}{\sqrt{1+x^{4}}} d x\)
Solution:

Question 3.
\(\int\left(e^{x}+e^{-x}\right)^{2}\left(e^{x}-e^{-x}\right) d x\)
Solution:

Question 4.
\(\int \frac{1+x}{x+e^{-x}} d x\)
Solution:

Question 5.
∫(x + 1)(x + 2)⁷(x + 3) dx
Solution:
Let I = ∫(x + 1)(x + 2)⁷(x + 3) dx
= ∫(x + 2)⁷ (x + 1)(x + 3) dx
= ∫(x + 2)⁷ [(x + 2) – 1][(x + 2) + 1] dx
= ∫(x + 2)⁷ [(x + 2)² – 1] dx
= ∫[(x + 2)⁹ – (x + 2 )⁷] dx
= ∫(x + 2 )⁹ dx – ∫(x + 2)⁷ dx
= \(\frac{(x+2)^{10}}{10}\) – \(\frac{(x+2)^{8}}{8}\) + c

Question 6.
\(\int \frac{1}{x \log x} d x\)
Solution:
Put log x = t
∴ \(\frac{1}{x}\) dx = dt
∴ \(\int \frac{d x}{x \cdot \log x}=\int \frac{1}{\log x} \cdot \frac{1}{x} d x\)
= ∫\(\frac{1}{t}\) dt
= log |t| + c
= log|log x| + c.

Question 7.
\(\int \frac{x^{5}}{x^{2}+1} d x\)
Solution:

Question 8.
\(\int \frac{2 x+6}{\sqrt{x^{2}+6 x+3}} d x\)
Solution:

Question 9.
\(\int \frac{1}{\sqrt{x}+x} d x\)
Solution:

Question 10.
\(\int \frac{1}{x\left(x^{6}+1\right)} d x\)
Solution: