Class 12

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

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

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

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

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


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

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

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

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

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


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

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

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

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

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

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

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

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

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

(ii)

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

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

(I) Choose the correct alternative.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(II) Fill in the blanks:

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

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

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

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

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

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

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

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

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

Question 10.
b_xy . b_yx = ______
Answer:
r²

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

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

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

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

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

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

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

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

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

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

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

(IV) Solve the following problems.

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

Since b_yx and b_xy are positive, ∴ r = 0.61

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

Question 7.
For a certain bivariate data

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

Solution:


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

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


Sine byx and bxy are negative, r = -0.4

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

Solution:
Let u = x – 122, v = y – 14

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

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

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


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

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

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


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

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

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

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

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

Solution:


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

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

Solution:


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

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

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

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

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

Solution:

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

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

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

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

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

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

(I) Choose the correct alternative.

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

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

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

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

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

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

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

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

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

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

(II) Fill in the blanks.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(IV) Solve the following problems.

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

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

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

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

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

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

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

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

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

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

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

∴ 875 Bedsheets damaged.

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

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

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

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

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

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

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

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

∴ r = 5% p.a.

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

= 4,00,000 × 0.2278
= ₹ 91,120

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

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

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

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

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

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

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

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

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


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

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

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

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


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

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

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

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

∴ C = 0.12 × 20,000
∴ C = ₹ 2,400

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

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

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

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 2 Insurance and Annuity Ex 2.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Insurance and Annuity Ex 2.1

Question 1.
Find the premium on a property worth ₹ 25,00,000 at 3% if
(i) the property is fully insured
(ii) the property is insured for 80% of its value.
Solution:
Case-1
Property value = ₹ 25,00,000
Rate of Premium = 3%
Policy Value = ₹ 25,00,000
∴ Amount of Premium = 3% × 25,00,000 = ₹ 75,000
Case-2
Property Value = ₹ 25,00,000
Policy value = 80% × 25,00,000 = ₹ 20,00,000
Rate of Premium = 3%
∴ Amount of Premium = 3% × 20,00,000 = ₹ 60,000

Question 2.
A shop is valued at ₹ 3,60,000 for 75% of its value. If the rate of premium is 0.9%, find the premium paid by the owner of the shop. Also, find the agents commission if the agent gets commission at 15% of the premium.
Solution:
Property Value = ₹ 3,60,000
Policy Value = 75% × 3,60,000 = ₹ 2,70,000
Rate of Premium = 0.9%
∴ Amount of Premium = 0.9% × 2,70,000 = ₹ 2,430
Rate of Commission = 15%
∴ Amount of Commission = 15% × 2,430 = ₹ 364.5

Question 3.
A person insures his office valued at ₹ 5,00,000 for 80% of its value. Find the rate of premium if he pays ₹ 13,000 as premium. Also, find agent’s commission at 11%.
Solution:
Property Value = ₹ 5,00,000
Policy Value = 80% × 5,00,000 = ₹ 4,00,000
Amount of Premium = ₹ 13000
Let the rate of Premium be x%
Amount of premium = Rate × Policy Value
∴ 13000 = x% × 4,00,000
∴ \(\frac{13,000}{4,00,000}=\frac{x}{100}\)
∴ \(\frac{13,000 \times 100}{4,00,000}\) = x
∴ x = 3.25%
Rate of commission = 11%
∴ Amount of Commission = 11% × 13,000 = ₹ 1,430

Question 4.
A building is insured for 75% of its value. The annual premium at 0.70 percent amounts to ₹ 2625. If the building is damaged to the extent of 60% due to fire, how much can be claimed under the policy?
Solution:
Let the Property Value of building be ₹ x
Policy Value = 75% × x = 0.75x
Rate of Premium = 0.70%
Amount of Policy = Rate × Policy Value
2625 = 0.70% × 0.75x
\(\frac{2625}{0.75}\) = 0.70% × x
3520 = \(\frac{0.70}{100}\) × x
\(\frac{3500 \times 100}{0.70}\) = x
x = ₹ 5,00,000
∴ Damage = 60% × Property Value
= \(\frac{60}{100}\) × 5,00,000
= ₹ 3,00,000
∴ Policy Value = 0.75 × 3,00,000 = ₹ 2,25,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{2,25,000}{5,00,000}\) × 3,00,000
= ₹ 1,35,000

Question 5.
A stock worth ₹ 7,00,000 was insured for ₹ 4,50,000. Fire burnt stock worth ₹ 3,00,000 completely and damaged there remaining stock to the extent of 75% of its value. What amount can be claimed undertaken policy?
Solution:
Property Value = ₹ 7,00,000
Policy Value = ₹ 4,50,000
Complete Loss = 3,00,000
Partial loss = 75% × [7,00,000 – 3,00,000]
= \(\frac{75}{100}\) × 4,00,000
= ₹ 3,00,000
∴ Total loss = ₹ 3,00,000 + ₹ 3,00,000 = ₹ 6,00,000
∴ Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{4,50,000}{7,00,000}\) × 6,00,000
= ₹ 3,85,714.29

Question 6.
A cargo of rice was insured at 0.625 % to cover 80% of its value. The premium paid was ₹ 5,250. If the price of rice is ₹ 21 per kg. find the quantity of rice (in kg) in the cargo.
Solution:
Let Property Value be ₹ x
Policy Value = 80% × x = ₹ 0.8x
Rate of Policy = 0.625%
Amount of Premium = Rate × Policy value
∴ 5250 = 0.625% × 0.8x
∴ 5250 = 0.005x
∴ x = \(\frac{5250}{0.005}\)
∴ x = ₹ 10,50,000
Rate of Rice = ₹ 21/kg
∴ Quantity of Rice (in kg) = \(\frac{\text { Total value }}{\text { Rate of Rice }}\)
= \(\frac{10,50,000}{21}\)
= 50,000 kgs

Question 7.
60,000 articles costing ₹ 200 per dozen were insured against fire for ₹ 2,40,000. If 20% of the articles were burnt and 7,200 of the remaining articles were damaged to the extent of 80% of their value, find the amount that can be claimed under the policy.
Solution:
No of articles = 60,000
Cost of articles = ₹ 200/dozen
∴ Property of Value = \(\frac{60,000}{12}\) × 200 = ₹ 1o,oo,ooo
∴ Policy Value = ₹ 2,40,000
Complete Loss = 20% × 10,00,000 = ₹ 2,00,000
Partial loss = \(\frac{7200}{12}\) × 200 × 80% = ₹ 96,000
∴ Total loss = 2,00,000 + 96,000 = ₹ 2,96,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{2,40,000}{10,00,000}\) × 2,96,000
= ₹ 71,040

Question 8.
The rate of premium is 2% and other expenses are 0.075%. A cargo worth ₹ 3,50,100 is to be insured so that all its value and the cost of insurance will be recovered in the event of total loss.
Solution:
Let the Policy Value of Cargo be ₹ 100 which includes insurance and other expenses
∴ Property Value = 100 – [2 + 0.075] = ₹ 97.925
If Policy Value is ₹ 100, then Property Value is ₹ 97.925
If Property Value is ₹ 3,50,100
Then policy Value = \(\frac{100 \times 3,50,100}{97.925}\) = ₹ 3,57,518.51

Question 9.
A property worth ₹ 4,00,000 is insured with three companies. A, B, and C. The amounts insured with these companies are ₹ 1,60,000, ₹ 1,00,000 and ₹ 1,40,000 respectively. Find the amount recoverable from each company in the event of a loss to the extent of ₹ 9,000.
Solution:
Property Value = ₹ 4,00,000
Loss = ₹ 9,000
Total Value of Policies = 1,60,000 + 1,00,000 + 1,40,000 = ₹ 4,00,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
Claim of company A = \(\frac{1,60,000}{40,000}\) × 9,000 = ₹ 3,600
Claim of company B = \(\frac{1,00,000}{4,00,000}\) × 9,000 = ₹ 2,250
Claim of company C = \(\frac{1,40,000}{4,00,000}\) × 9,000 = ₹ 3,150

Question 10.
A car valued at ₹ 8,00,000 is insured for ₹ 5,00,000. The rate of premium is 5% less 20%. How much will the owner bear including the premium if value of the ear is reduced to 60% of its original value.
Solution:
Property Value = ₹ 8,00,000
Policy Value = ₹ 5,00,000
Rate of Premium = 5% less 20%
= 5% – 20% × 5%
= (5 – 1)%
= 4%
Amount of Premium = 4% × 5,00,000 = ₹ 20,000
Loss = [100 – 60]% × Property Value
= 40% × 8,00,000
= ₹ 3,20,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{5,00,000}{8,00,000}\) × 3,20,000
= ₹ 2,00,000
Loss bear by owner = Loss – claim + Premium
= 3,20,000 – 2,00,000 + 20,000
= ₹ 1,40,000

Question 11.
A shop and a godown worth ₹ 1,00,000 and ₹ 2,00,000 respectively were insured through an agent who was paid 12% of the total premium. If the shop was insured for 80% and the godown for 60% of their respective values, find the agent’s commission, given that the rate of premium was 0.80% less 20%.
Solution:
Rate of Premium = 0.80% Less 20%
= 0.80% – 20% × 0.80%
= (0.80 – 0.16)%
= 0.64%
For Shop
Property Value = ₹ 1,00,000
Policy Value = 80% × 1,00,000 = ₹ 80,000
Premium = 0.64% × 80,000 = ₹ 512
For Godown
Property Value = ₹ 2,00,000
Policy Value = 60% × 2,00,000 = ₹ 1,20,000
Premium = 0.64% × 1,20,000 = ₹ 768
∴ Total Premium = 512 + 768 = ₹ 1,280
Rate of Commission = 12%
∴ Agent Commission = 12% × 1,280 = ₹ 153.6

Question 12.
The rate of premium on a policy of ₹ 1,00,000 is ₹ 56 per thousand per annum. A rebate of ₹ 0.75 per thousand is permitted if the premium is paid annually. Find the net amount of premium payable if the policy holder pays the premium annually.
Solution:
Policy Value = ₹ 1,00,000
Rate of Premium = ₹ 56 per thousand p.a
Rate of Rebate = ₹ 0.75 per thousand p.a
Premium is paid annually
∴ Net rate of = 56 – 0.75 = ₹ 55.25 per thousand p.a.
∴ Net Amount ot Premium = \(\frac{1,00,000}{1000}\) × 55.25 = ₹ 5,525

Question 13.
A warehouse valued at ₹ 40,000 contains goods worth ₹ 2,40,000. The warehouse is insured against fire for ₹ 16,000 and the goods to the extent of 90% of their value. Goods worth ₹ 80,000 are completely destroyed, while the remaining goods are destroyed to 80% of their value due to a fire. The damage to the warehouse is to the extent of ₹ 8,000. Find the total amount that can be claimed.
Solution:
For Warehouse
Property Value = ₹ 40,000
Policy Value = ₹ 16,000
Loss = ₹ 8,000
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{16,000}{40,000}\) × 8,000
= ₹ 3,200
For Goods
Property Value = ₹ 2,40,000
Policy Value = 90% × 2,40,000 = ₹ 2,16,000
Complete Loss = 80,000
Partial Loss = 80% × (2,16,000 – 80,000)
= 80% × 1,36,000
= ₹ 1,08,800
Claim = \(\frac{\text { Policy value }}{\text { Property value }}\) × Loss
= \(\frac{2,16,000}{24,000}\) × 1,08,800
= ₹ 97,920
∴ Total Claim = 3,200 + 97,920 = ₹ 1,01,120

Question 14.
A person takes a life policy for ₹ 2,00,000 for a period of 20 years. He pays premium for 10 years during which bonus was declared at an average rate of ₹ 20 per year per thousand. Find the paid up value of the policy if he discontinuous paying premium after 10 years.
Solution:
Policy Value = ₹ 2,00,000
Rate of Bonus = ₹ 20 Per thousand p.a.
Total Bonus = \(\frac{2,00,000 \times 20}{1,000}\) = ₹ 4,000
∴ Bonus for 10 years = 4,000 × 10 = ₹ 40,000
Period of Policy = 20 years
∴ Amount of Premium = \(\frac{2,00,000}{20}\) = ₹ 10,000 p.a.
∴ Total Premium for 10 years = 10,000 × 10 = ₹ 1,00,000
∴ Paid up Value of Policy = Total premium + Total Bonus
= 1,00,000 + 40,000
= ₹ 1,40,000

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Commission, Brokerage and Discount Miscellaneous Exercise 1

(I) Choose the correct alternative.

Question 1.
An agent who gives a guarantee to his principal that the party will pay the sale price of goods is called
(a) Auctioneer
(b) Del Credere Agent
(c) Factor
(d) Broker
Answer:
(b) Del Credere Agent

Question 2.
An agent who is given the possession of goods to be sold is known as
(a) Factor
(b) Broker
(c) Auctioneer
(d) Del Credere Agent
Answer:
(a) Factor

Question 3.
The date on which the period of the bill expires is called
(a) Legal Due Date
(b) Grace Date
(c) Nominal Due Date
(d) Date of Drawing
Answer:
(c) Nominal Due Date

Question 4.
The payment date after adding 3 days of grace period is known as
(a) The legal due date
(b) The nominal due date
(c) Days of grace
(d) Date of drawing
Answer:
(a) The legal due date

Question 5.
The sum due is also called as
(a) Face value
(b) Present value
(c) Cash value
(d) True discount
Answer:
(a) Face value

Question 6.
P is the abbreviation of
(a) Face value
(b) Present worth
(c) Cash value
(d) True discount
Answer:
(b) Present worth

Question 7.
Banker’s gain is the simple interest on
(a) Banker’s discount
(b) Face Value
(c) Cash value
(d) True discount
Answer:
(d) True discount

Question 8.
The marked price is also called as
(a) Cost price
(b) Selling price
(c) List price
(d) Invoice price
Answer:
(c) List price

Question 9.
When only one discount is given then
(a) List price = Invoice price
(b) Invoice price = Net selling price
(c) Invoice price = Cost price
(d) Cost price = Net selling price
Answer:
(b) Invoice price = Net selling price

Question 10.
The difference between the face value and present worth is called
(a) Banker’s discount
(b) True discount
(c) Banker’s gain
(d) Cash value
Answer:
(b) True discount

(II) Fill in the blanks.

Question 1.
A person who draws the bill is called ____________
Answer:
Drawee

Question 2.
An ____________ is an agent who sells the goods by auction.
Answer:
Auctioneer

Question 3.
Trade discount is allowed on the ____________ price.
Answer:
Catalogue/List

Question 4.
The banker’s discount is also called ____________.
Answer:
Commercial Discount

Question 5.
The banker’s discount is always ____________ than the true discount.
Answer:
higher

Question 6.
The diffrence between the banker’s discount and the true discount is called ____________.
Answer:
bankers gain

Question 7.
The date by which the buyer is legally allowed to pay the amount is known as ____________.
Answer:
legal due date

Question 8.
A ____________ is an agent who brings together the buyer and the seller.
Answer:
broker

Question 9.
If buyer is allowed both trade and cash discounts, ____________ discount is fist calculated on ____________ price.
Answer:
Trade, Catalogue/List

Question 10.
____________ = List price (catalogue Price) – Trade Discount.
Answer:
Invoice Price

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

Question 1.
A broker is an agent who gives a guarantee to the seller that the buyer will pay the sale price of goods.
Answer:
False

Question 2.
A cash discount is allowed on the list price.
Answer:
False

Question 3.
Trade discount is allowed on catalogue price.
Answer:
True

Question 4.
The buyer is legally allowed 6 days grace period.
Answer:
False

Question 5.
The date on which the period of the bill expires is called the nominal due date.
Answer:
True

Question 6.
The difference between the banker’s discount and true discount is called sum due.
Answer:
False

Question 7.
The banker’s discount is always lower than the true discount.
Answer:
False

Question 8.
The banker’s discount is also called a commercial discount.
Answer:
True

Question 9.
In general cash, the discount is more than trade discount.
Answer:
False

Question 10.
A person can get both, trade discount and a cash discount.
Answer:
True

(IV) Solve the following problems.

Question 1.
A salesman gets a commission of 6.5% on the total sales made by him and a bonus of 1% on sales over ₹ 50,000. Find his total income on a turnover of ₹ 75,000.
Solution:
Rate of commission = 6.5% on the total sales
∴ Commission on a turnover of ₹ 75,000
= \(\frac{6.5}{100}\) × 75,000
= ₹ 4,875
Rate of bonus = 1% on sales over ₹ 50,000
∴ Amount of bonus = \(\frac{1}{100}\) × (75,000 – 50,000) = ₹ 250
∴ Total income of the sales man = ₹ 4,875 + ₹ 250 = ₹ 5,125

Question 2.
A shop is sold at 30% profit, the amount of brokerage at the rate of \(\frac{3}{4}\)% amounts to ₹ 73,125. Find the cost of the shop.
Solution:
Rate of brokerage = \(\frac{3}{4}\)%
Amount of brokerage = ₹ 73,125
Let the selling price of the shop be ₹ 100 then the brokerage = ₹ \(\frac{3}{4}\)
Thus, if the amount of brokerage is ₹ \(\frac{3}{4}\) then the selling price of the shop is ₹ 100
If the amount of brokerage is ₹ 73,125, then the selling price of the shop is = 73125 × \(\frac{4}{3}\) × 100 = ₹ 97,50,000
The shop is sold at 30% profit
∴ If the cost of the shop is ₹ 100, then it is sold at ₹ 130
Thus, if the shop is sold at ₹ 130, then its cost price is ₹ 100
If the shop is sold at ₹ 97,50,000 then its cost price is = \(\frac{97,50,000 \times 100}{130}\) = ₹ 75,00,000
Then, the cost of the shop is ₹ 75,00,000

Question 3.
A merchant gives 5% commission and 1.5% delcredere to his agents. If the agent sells goods worth ₹ 30,600 how much does he get? How much does the merchant receive?
Solution:
Rate of commission = 5%
Total sales = ₹ 30,600
Amount of commission = \(\frac{5}{100}\) × 30,600
Rate of delcredere = 1.5%
= \(\frac{1.5}{100}\) × 30,600
= ₹ 459
Thus, the agents gets 1,530 + 459 = ₹ 1,989
And the merchant receives = 30,600 – 1,989 = ₹ 28,611

Question 4.
After deducting commission at 7\(\frac{1}{2}\)% on first ₹ 50,000 and 5% on the balance of sales made by him, an agent remits ₹ 93,750 to his principal. Find the value of goods sold by him.
Solution:
Rate of commission = 7\(\frac{1}{2}\)% on first ₹ 50,000
= \(\frac{7.5}{100}\) × 50,000
= ₹ 3,750
Let the total sales be ₹ x
Rate of commission on the balance sales = 5%
Commission on the balance sales = \(\frac{5}{100}\) × (x – 50000) = \(\frac{x}{20}\) – 2,500
Total commission = 3750 + \(\frac{x}{20}\) – 2,500 = \(\frac{x}{20}\) + 1,250
Now, the amount to be remitted to the principal = Value of goods sold – Commission of the agent
= x – (\(\frac{x}{20}\) + 1250)
= \(\frac{19x}{20}\) – 1250
The agents remits ₹ 93,750 to his principal
∴ \(\frac{19x}{20}\) – 1,250 = 93,750
∴ \(\frac{19x}{20}\) = 95,000
∴ x = ₹ 1,00,000
Thus, the value of the goods sold by the agent is ₹ 1,00,000

Question 5.
The present worth of ₹ 11,660 due 9 months hence is ₹ 11,000. Find the rate of interest.
Solution:
Given, PW = ₹ 11,000, SD = ₹ 11,660
n = \(\frac{9}{12}\) year = \(\frac{3}{4}\) year
We have,

∴ The rate of interest is 8% p.a.

Question 6.
An article is marked at ₹ 800, a trader allows a discount of 2.5% and gains 20% on the cost. Find the cost price of the article?
Solution:
Marked price of the article = ₹ 800
Rate of discount = 2.5%
Amount of discount = \(\frac{2.5}{100}\) × 800 = ₹ 20
∴ Selling price of the article = 800 – 20 = ₹ 780
Now, given, gain = 20%
Let cost price of the article be ₹ 100, then
The selling price of the article is ₹ 120
Thus if cost price of the articles is ₹ x
Then the selling price is ₹ 780
∴ x = \(\frac{780 \times 100}{120}\)
∴ x = 650
∴ Cost price of the article is ₹ 650

Question 7.
A salesman is paid a fixed monthly salary plus commission on the sales. If on sale of ₹ 96,000 and ₹ 1,08,000 in two successive months he receives in all ₹ 17,600 and ₹ 18,800 respectively. Find his monthly salary and rate of commission paid to him.
Solution:
Let the monthly salary of the salesman be ₹ x
And the rate of commission be y%
Income = monthly salary + commission on the sales
17600 = x + \(\frac{y}{100}\) × 96,000
∴ 17600 = x + 960y ………(1)
and 18800 = x + \(\frac{y}{100}\) × 108000
∴ 18,800 = x + 1080y ………(2)
Subtracting equation (1) from equation (2), we get
1,200 = 120y
∴ y = 10
Substituting y = 10 in (1), we get
17,600 = x + 960(10)
∴ x = 17,600 – 9,600 = 8,000
∴ Salary of the salesman = ₹ 8,000
Rate of commission = 10%

Question 8.
A merchant buys some mixers at a 15% discount on catalogue price. The catalogue price is ₹ 5,500 per price of the mixer. The freight charges amount to 2\(\frac{1}{2}\)% on the catalogue price. The merchant sells each mixer at a 5% discount on the catalogue price. His net profit is ₹ 41,250, Find the number of mixers.
Solution:
Catalogue price of a mixer = ₹ 5,500
Trade discount = 15% on catalogue price
= \(\frac{15}{100}\) × 5,500
= ₹ 825
Freight charges = 2\(\frac{1}{2}\)% of the catalogue price
= \(\frac{5}{2} \times \frac{1}{100} \times 5,500\)
= ₹ 137.5
∴ Cost price of a mixer for the merchant = 5,500 – 825 + 137.5 = 4,812,5
Catalogue price = ₹ 5,500
Rate of discount = 5%
∴ Selling price of one mixer = 5500 – \(\frac{5}{100}\) × 5,500 = ₹ 5,225
∴ Profit on one mixer = 5,225 – 4,812.5 = ₹ 412.5
Now, total profit = ₹ 41,250
∴ Number of mixers = \(\frac{41,250}{412.5}\) = 100
Thus the number of mixers is 100.

Question 9.
A bill is drawn for ₹ 7,000 on 3rd May for 3 months and is discounted on 25th May at 5.5% Find the present worth.
Solution:
Face value of the bill = ₹ 7,000
Date of drawing = 3rd May
Period = 3 months
Normal due date = 3rd August
Legal due date = 6th August
Rate of interest = 5.5%
Date of discounting = 25th May
Unexpired period (number of days from date of discounting to legal due date)

∴ Bankers discount = 7,000 × \(\frac{73}{365} \times \frac{5.5}{100}\) = ₹ 77
Also PW = SD – BD
= 7,000 – 77
= ₹ 6,923
∴ Present worth is ₹ 6,923

Question 10.
A bill was drawn on 14th April 2005 for ₹ 3,500 and was discounted on 6th July 2005 at 5% per annum. The banker paid ₹ 3,465 for the bill. Find the period of the bill.
Solution:
Face value of the bill = ₹ 3,500
Date of drawing = 14/04/2005
Date of discount = 06/07/2005
Rate of interest = 5%
Cash value = ₹ 3,465
Bankers discount = Face value – Cash value
= 3,500 – 3,465
= ₹ 35
Let the unexpired days be n days
∴ BD = \(\frac{\mathrm{FV} \times n \times r}{365 \times 100}\)
∴ 35 = \(\frac{3,500 \times n \times 5}{365 \times 100}\)
∴ n = 73 days
Thus, legal due date is 73 days from the date of discounting

∴ Legal due date = 17/09/2005
∴ Nominal due date = 14/09/2005
∴ The period of the bill is 5 months

Question 11.
The difference between true discount and banker’s discount on 6 months hence at 4% p.a. is ₹ 80. Find the true discount, banker’s discount, and amount of the bill.
Solution:
BG = BD – TD
∴ BG = ₹ 80
Also BG = \(\frac{\mathrm{TD} \times n \times r}{100}\)
∴ 80 = \(\frac{\mathrm{TD} \times 6 \times 4}{12 \times 100}\)
∴ TD = \(\frac{80 \times 100}{2}\)
∴ TD = ₹ 4,000
Now BD = TD + BG
= 4,000 + 80
= ₹ 4,080
Also, BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 4,080 = \(\frac{\mathrm{FV} \times 6 \times 4}{12 \times 100}\)
∴ FV = \(\frac{4,080 \times 100}{2}\)
∴ FV = ₹ 2,04,000
Amount of the bill = ₹ 2,04,000

Question 12.
A manufacturer makes a clear profit of 30% on the cost after allowing a 35% discount. If the cost of production rises by 20%, by what percentage should he reduce the rate of discount so as to make the same rate of profit keeping his list prices unaltered.
Solution:
Rate of discount = 35%
Let the list price be ₹ 100.
Then discount at 35% = ₹ 35
∴ Net selling price = 100 – 35 = ₹ 65 ……..(1)
The manufacturer makes a clear profit of 30% on the cost after allowing a 35% Discount.
Let the cost be ₹ 100.
Then selling price at 30% profit is 100 + 30 = ₹ 130.
Thus, if the net selling price is ₹ 130, then the cost price is ₹ 100.
But, the net selling price is ₹ 130, then the cost price is ₹ 65 ……[from (1)]
∴ The cost price is \(\frac{65}{130} \times 100\) = ₹ 50
Hence, we have,

Now, the cost of production has increased by 20%.
Let the old cost price be ₹ 100.
∴ The new cost price is ₹ 120.
But, the old cost price is ₹ 50.
∴ The new cost price is = \(\frac{50}{100} \times 120\) = ₹ 60.
The old net price is ₹ 65.
Now 20% of ₹ 65 = \(\frac{20}{100} \times 65\) = ₹ 13
∴ New net price = 65 + 13 = ₹ 78
Hence, we have

Now, 100 – 78 = ₹ 22
Thus, the rate of discount should be reduced by 22%, The original rate of discount is 35%.
Hence, the reduction in discount should be (35 – 22)% = 13%
so as to make the same rate of profit, keeping the list price unaltered.

Question 13.
A trader offers a 25% discount on the catalogue price of the radio and yet makes a 20% profit. If he gains ₹ 160 per radio, what must be the catalogue price of the radio?
Solution:
Rate of discount = 25% on the catalogue price of a radio.
Let the catalogue price of the radio be ₹ 100.
Then, the discount on a radio = ₹ 25.
Net selling price = 100 – 25 = ₹ 75.
He makes a profit of 20%.
Let the cost price be ₹ 100.
Then, at 20% profit, net selling price = ₹ 120.
Thus, if net SP is ₹ 120, then cost price is ₹ 100.
But, the net SP is ₹ 75.
∴ The cost price is \(\frac{75}{120}\) × 100 = \(\frac{750}{12}\) = ₹ 62.50
∴ Profit on a radio set = 75 – 62.5 = ₹ 12.50
Thus, if the profit on a radio set is ₹ 12.50 then its catalogue price is ₹ 100.
But the profit on a radio set is ₹ 160.
∴ The catalogue price of radio = \(\frac{160}{12.50}\) × 100
= 12.80 × 100
= ₹ 1,280
∴ Thus, the catalogue price of the radio is ₹ 1280

Question 14.
A bill of ₹ 4,800 was drawn on 9th March 2006 at 6 months and was discounted on 19th April 2006 for 6\(\frac{1}{4}\)% p.a. How much does the banker charge and how much does the holder receive?
Solution:
Face value of the bill = ₹ 4.800
Date of drawing = 09/03/2006
Period of the bill = 6 months
Normal due date = 09/09/2006
Legal due date = 12/09/2006
Rate of discount = 6\(\frac{1}{4}\)% = 6.25%
Now, for the unexpired

Thus the banker charges ₹ 120
Amount received by the holder = 4,800 – 120 = ₹ 4,680

Question 15.
A bill of ₹ 65,700 drawn on July 10 for 6 months was discounted for ₹ 65,160 at 5% p.a. On what day was the bill discounted?
Solution:
BD = FV – Cash value
= 65,700 – 65,160
= ₹ 540
Let the unexpired days be x days
BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 540 = \(\frac{65,700 \times x \times 5}{365 \times 100}\)
∴ x = 60 days
The unexpired days = 60 days
Date-of drawing = 10th July
Period of the bill = 6 months
Nominal due date = 10th January (next year)
Legal due date = 13th January (next year)
Then the date of discount is 60 days before, the legal due date

∴ The date of discounting is 14th November

Question 16.
An agent sold a car and charged a 3% commission on the sale value. If the owner of the car received ₹ 48,500, find the sale value of the car. If the agent charged 2% from the buyer, find his total remuneration.
Solution:
Let the sale value of the car be ₹ x
Rate of commission of the agent = 3%
Since the owner received ₹ 48,500 after agent has charged his commission
x – \(\frac{3 x}{100}\) = 48500
∴ \(\frac{97 x}{100}\) = 48500
∴ x = \(\frac{48,500 \times 100}{97}\)
∴ x = ₹ 50,000
∴ Sale value of the car = ₹ 50,000
Against commission received from the owner = \(\frac{3}{100}\) × 50,000 = ₹ 1500
Against commission received from the buyer = \(\frac{2}{100}\) × 50,000 = ₹ 1000
∴ Agents total remuneration = 1,500 + 1,000 = ₹ 2,500

Question 17.
An agent is paid a commission of 4% on cash sales and 6% on credit sales made by him. If on the sale of ₹ 51,000 the agent claims a total commission of ₹ 2,700, find the sales made by him for cash and on credit.
Solution:
Total sales = ₹ 51,000
Let eash sales be ₹ x
∴ Credit sales = ₹ (51,000 – x)
Agent’s commission on cash sales = 4%
= \(\frac{4}{100}\) × x
= \(\frac{4x}{100}\)
Commission on credit sales = 6%
= \(\frac{6}{100}\)(51,000 – x)
Given total commission = ₹ 2,700

∴ Cash sales = ₹ 18,000
∴ Credit sales = 51,000 – 18,000 = ₹ 33,000

Balbharati Maharashtra State Board 12th Commerce Maths Digest Pdf Chapter 1 Commission, Brokerage and Discount Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Commission, Brokerage and Discount Ex 1.2

Question 1.
What is the present worth of a sum of ₹ 10,920 due six months hence at 8% p.a simple interest?
Solution:
Given, SD = ₹ 10,920
n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
r = 8%
We have,

Thus the present worth is ₹ 10,500

Question 2.
What is the sum due of ₹ 8,000 due 4 months at 12.5% simple interest?
Solution:
Given, PW = ₹ 8,000, n = \(\frac{4}{12}\) year = \(\frac{1}{3}\) year, r = 12.5%
We have,

Thus, the sum due is ₹ 8,333.33

Question 3.
The true discount on the sum due 8 months hence at 12% p.a. is ₹ 560. Find the sum due and present worth of the bill.
Solution:
Given, TD = ₹ 560, n = \(\frac{8}{12}\) year = \(\frac{2}{3}\) year, r = 12%
We have,
TD = \(\frac{\mathrm{PW} \times n \times r}{100}\)
∴ 560 = \(\frac{\mathrm{PW} \times 2 \times 12}{3 \times 100}\)
∴ PW = 560 × \(\frac{25}{2}\) = ₹ 7,000
Now, SD = PW + TD
= 7,000 + 560
= ₹ 7,560

Question 4.
The true discount on a sum is \(\frac{3}{8}\) of the sum due at 12% p.a. Find the period of the bill.
Solution:

8 × n × 12 = 3(100 + n × 12)
96n = 300 + 36n
60n = 300
∴ n = 5
∴ Period of the bill = 5 years.

Question 5.
20 copies of a book can be purchased for a certain sum payable at the end of 6 months and 21 copies for the same sum in ready cash. Find the rate of interest.
Solution:
Given, n = \(\frac{6}{12}\) year = \(\frac{1}{2}\) year
Let the sum payable be ₹ x
Let the rate of interest be r%
According to given condition,
PW of one book = \(\frac{x}{21}\)
SD of one book = \(\frac{x}{20}\)

Thus, the rate of interest is 10%.

Question 6.
Find the true discount, Banker’s discount, and Banker’s gain on a bill of ₹ 4,240 due 6 months hence at 9% p.a.
Solution:

And, Banker’s Gain (BG) = BD – TD
= 190.80 – 182.58
= ₹ 8.22

Question 7.
The true discount on a bill is ₹ 2,200 and bankers discount is ₹ 2,310. If the bill is due 10 months, hence, find the rate of interest.
Solution:
Given, TD = ₹ 2,200, BD = ₹ 2,310
n = \(\frac{10}{12}=\frac{5}{6}\) year

∴ \(\frac{r}{120}=\frac{1}{20}\)
∴ r = 6%
Thus, rate of interest is 6%

Question 8.
A bill of ₹ 6,395 drawn on 19th January 2015 for 8 months was discounted on 28th February 2015 at 8% p.a. interest. What is the banker’s discount? What is the cash value of the bill?
Solution:
Face value = ₹ 6,395
Date of drawing = 19/01/2015
Period of the bill = 8 months
Nominal Due date = 19/09/2015
Legal due date = 22/09/2015
Date of discounting = 28/02/2015
Now, the unexpired period = Legal due date – Date of discounting
= 22/09/2015 – 28/02/2015
= days (as shown below)

Cash Value = FV – BD
= 6,395 – 313.12
= ₹ 6,621.38

Question 9.
A bill of ₹ 8,000 drawn on 5th January 1998 for 8 months was discounted for ₹ 7,680 on a certain date. Find the date on which it was discounted at 10% p.a.
Solution:
Bankers discount (BD) = FV – cash value
= 8,000 – 7,680
= ₹ 320
Let the unexpired period be x days
∴ BD = \(\frac{\mathrm{FV} \times x \times r}{365 \times 100}\)
∴ 320 = \(\frac{8,000 \times x \times 10}{365 \times 100}\)
∴ x = 146 days
∴ The unexpired days = 146 days
Date of drawing = 05/01/1998
Period of bill = 8 months
Nominal due date = 05/09/1998
Legal due date = 08/09/1998
Thus, the date of discounting is 146 days before the legal due date

∴ Date of discounting of the bill is 15th April 1998

Question 10.
A bill drawn on 5th June for 6 months was discounted at the rate of 5% p.a. on 19th October. If the cash value of the bill is ₹ 43,500, find the face value of the bill.
Solution:
Date of drawing = 5th June
Period of bill = 6 months
Nominal due date = 5th December
Legal due date = 8th December
Date of discounting = 19th October
Rate of interest = 5% p.a.
Let the face value of the bill be ₹ x
The unexpired period

Question 11.
A bill was drawn on 14th April for ₹ 7,000 and was discounted on 6th July at 5% p.a. The Banker paid ₹ 6,930 for the bill. Find the period of the bill.
Solution:
Face value = ₹ 7,000, cash value = ₹ 6,930
∴ Banker’s discount = 7,000 – 6,930 = ₹ 70
Date of drawing = 14/04
Date of discounting = 06/07
Rate of interest = 5%
Let the unexpired period = x days
∴ BD = \(\frac{7,000 \times x \times 5}{365 \times 100}\)
∴ 70 = \(\frac{70 \times x}{73}\)
∴ x = 73 days
∴ Legal due date of the bill is 73 days after the date of discounting.

∴ Legal due date = 17/09
∴ Nominal due date = 14/09
∴ Period of the bill = 5 months

Question 12.
If the difference between true discount and banker’s discount on a sum due 4 months hence is ₹ 20. Find true discount, banker’s discount and amount of bill, the rate of simple interest charged is 5% p.a.
Solution:
Banker’s gain (BG) = Banker’s discount (BD) – True Discount (TD)
∴ BG = ₹ 20
Also, BG = \(\frac{\mathrm{TD} \times n \times r}{100}\)
∴ 20 = \(\frac{\mathrm{TD} \times 4 \times 5}{12 \times 100}\)
∴ 20 = \(\frac{\mathrm{TD}}{60}\)
∴ TD = ₹ 1200
Now, BD = BG + TD
= 20 + 1,200
= ₹ 1,220
Also, BD = \(\frac{\mathrm{FV} \times n \times r}{100}\)
∴ 1,220 = \(\frac{\mathrm{FV} \times 4 \times 5}{12 \times 100}\)
∴ FV = 1,200 × 60 = ₹ 73,200
∴ Amounting the bill = ₹ 73,200

Question 13.
A bill of ₹ 51,000 was drawn on 18th February 2010 for 9 months. It was encashed on 28th June 2010 at 5% p.a. Calculate the banker’s gain and true discount.
Solution:
Face Value = ₹ 51,000
Date of drawing = 18/02/2010
Period of the bill = 9 months
Nominal due date = 18/11/2010
Legal due date = 21/11/2010
Date of discounting = 28/06/2010
Unexpired period

∴ TD = ₹ 1,000
∴ BG = BD – TD
= 1,020 – 1,000
= ₹ 20

Question 14.
A certain sum due 3 months hence is \(\frac{21}{20}\) of the present worth, what is the rate of interest.
Solution:

Question 15.
A bill of a certain sum drawn on 28th February 2007 for 8 months was encashed on 26th March 2007 for ₹ 10,992 at 14% p.a. Find the face value of the bill.
Solution:
Date drawing = 28/02/2007
Period of the bill = 8 months
Nominal due date = 28/10/2007
Legal due date = 31/10/2007
Date of discounting = 26/03/2007
Cash value = ₹ 10,992
Rate of interest = 14%
Let face value of the bill = ₹ x
Bankers discount = Face value – Cash value = x – 10,992
Also, Banker s discount = \(\frac{F V \times n \times r}{365 \times 100}\)
Where n is the unexpired days

Thus face value of the bill = ₹ 12,000