Class 11

Linear Inequations Class 11 Commerce Maths 2 Chapter 8 Exercise 8.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.2 Questions and Answers.

Std 11 Maths 2 Exercise 8.2 Solutions Commerce Maths

Question 1.
Solve the following inequations graphically in a two-dimensional plane
(i) x ≤ -4
Solution:
Given, inequation is x ≤ -4
∴ corresponding equation is x = -4
It is a line parallel to Y-axis passing through the point A(-4, 0)
Origin test:
Substituting x = 0 in inequation, we get
0 ≤ -4 which is false.
∴ Points on the origin side of the line do not satisfy the inequation.
So the points on the non-origin side of the line and points on the line satisfy the inequation
∴ all the points on the line and left of it satisfy the given inequation.
The shaded portion represents the solution set.

(ii) y ≥ 3
Solution:
Given, inequation is y ≥ 3
∴ corresponding equation is y = 3
It is a line parallel to X-axis passing through point A(0, 3)
Origin test:
Substituting y = 0 in inequation, we get
0 ≥ 3 which is false.
∴ Points on the origin side of the line do not satisfy the inequation
∴ Points on the non-origin side of the line satisfy the inequation.
∴ all the points on the line and above it satisfy the given inequation.
The shaded portion represents the solution set.

(iii) y ≤ -2x
Solution:
Given, inequation is y ≤ -2x
∴ corresponding equation is y = -2x
It is a line passing through origin O(0, 0).
To draw the line, we need one more point.
To find another point on the line, we can take any value of x,
say, x = 2.
∴ substituting x = 2 in y = -2x, we get
y = -2(2)
∴ y = -4
∴ another point on the line is A(2, -4)
Now, the origin test is not possible as the origin lies on the line y = -2x
So, choose a point which does not lie on the line say, (2, 1)
∴ substituting x = 2, y = 1 in inequation, we get
1 ≤ -2(2)
∴ 1 ≤ -4 which is false.
∴ the points on the side of the line y = -2x, where (2, 1) lies do not satisfy the inequation.
∴ all the points on the line y = -2x and on the opposite side of the line where (2, 1) lies, satisfy the inequation
The shaded portion represents the solution set.

(iv) y – 5x ≥ 0
Solution:
Given, inequation is y – 5x ≥ 0
∴ corresponding equation is y – 5x = 0
It is a line passing through the point O(0, 0)
To draw the line, we need one more point.
To find another point on the line,
we can take any value of x, say, x = 1.
Substituting x = 1 in y – 5x = 0, we get
y – 5(1) = 0
∴ y = 5
∴ Another point on the line is A(1, 5)
Now origin test is not possible as the origin lies on the line y = 5x
∴ choose a point that does not lie on the line, say (3, 2).
∴ substituting x = 3, y = 2 in inequation, we get
2 – 5(3) ≥ 0
∴ 2 – 10 ≥ 0
∴ -8 ≥ 0 which is false.
∴ the points on the side of line y = 5x where (3, 1) lies do not satisfy the inequation.
∴ the points on the line y = 5x and on the opposite of the line where (3, 2) lies, satisfy the inequation.
The shaded portion represents the solution set.

(v) x – y ≥ 0
Solution:
Given, inequation is x – y ≥ 0
∴ Corresponding equation is x – y = 0
It is a line passing through origin O(0, 0)
To draw the line we need one more point.
To find another point on the line, we can take any value of x,
Say, x = 2.
∴ substituting x = 2 in x – y = 0, we get
2 – y = 0
∴ y = 2
∴ another point on the line is A(2, 2)
Now origin test is not possible as the origin lies on the line y = x
∴ choose a point which not lie on the line say (3, 1)
∴ substituting x = 3, y = 1 in inequation, we get
3 – 1 ≥ 0
∴ 2 ≥ 0 which is true.
∴ all the points on line x – y = 0 and the points on the side where (3, 1) lies satisfy the inequation
The shaded portion represents the solution set.

(vi) 2x – y ≤ -2
Solution:
Given, inequation is 2x – y ≤ -2
∴ corresponding equation is 2x – y = -2
∴ \(\frac{2 x}{-2}-\frac{y}{-2}=\frac{-2}{-2}\)
∴ \(\frac{x}{-1}+\frac{y}{2}=1\)
∴ intersection of line with X-axis is A(-1, 0),
intersection of line with Y-axis is B(0, 2)
Origin test:
Substituting x = 0, y = 0 in the given inequation, we get
2(0) – (0) ≤ -2
∴ 0 ≤ -2
which is false.
∴ Points on the origin side of the line do not satisfy the inequation.
∴ Points on the non-origin side of the line satisfy the inequation
∴ all the points on the line and above it satisfy the given inequation.
The shaded portion represents the solution set.

(vii) 4x + 5y ≤ 40
Solution:
Given, inequation is 4x + 5y ≤ 40
∴ Corresponding equation is 4x + 5y = 40
∴ \(\frac{4 x}{40}+\frac{5 y}{40}=\frac{40}{40}\)
∴ \(\frac{x}{10}+\frac{y}{8}=1\)
∴ Intersection of line with X-axis is A(10, 0)
Intersection of line with Y-axis is B(0, 8)
Origin test:
Substituting x = 0, y = 0 in the inequation, we get
4(0) + 5(0) ≤ 40
∴ 0 ≤ 40 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the given inequation.
The shaded portion represents the solution set.

(viii) \(\left(\frac{1}{4}\right) x+\left(\frac{1}{2}\right) y\) ≤ 1
Solution:
Given, inequation is \(\left(\frac{1}{4}\right) x+\left(\frac{1}{2}\right) y\) ≤ 1
∴ corresponding equation is \(\frac{x}{4}+\frac{y}{2}\) = 1
∴ intersection of line with X-axis is A(4, 0),
intersection of line with Y-axis is B(0, 2)
Origin test:
Substituting x = 0, y = 0 in the given inequation, we get
\(\frac{1}{4}(0)+\frac{1}{2}(0)\) ≤ 1
∴ 0 ≤ 1 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the given inequation.
The shaded portion represents the solution set.

Question 2.
Mr. Rajesh has ₹ 1,800 to spend on fruits for the meeting. Grapes cost ₹ 150 per kg. and peaches cost ₹ 200 per kg. Formulate and solve it graphically.
Solution:
Let x and y be the number of kgs. of grapes and peaches bought.
The cost of grapes is ₹ 150/- per kg, cost of peaches is ₹ 200/- per kg.
∴ cost of v kg of grapes is ₹ 150x
and the cost of y kg of peaches is ₹ 200y.
Mr. Rajesh has ₹ 1800 to spend on fruits.
∴ the total cost of grapes and peaches must be less than or equal to ₹ 1800.
∴ required inequation is 150x + 200y ≤ 1800
i.e., 3x + 4y ≤ 36 ……(i)
Since the number of kg of grapes and peaches can not be negative
∴ x ≥ 0, y ≥ 0
Now, corresponding equation is 3x + 4y = 36
∴ \(\frac{3 x}{36}+\frac{4 y}{36}=\frac{36}{36}\)
∴ \(\frac{x}{12}+\frac{y}{9}=1\)
∴ the intersection of the line with the X-axis is A(12, 0)
the intersection of the line with the Y-axis is B(0, 9)
Origin test:
Substituting x = 0, y = 0 in inequation, we get
3(0) + 4(0) ≤ 36
∴ 0 ≤ 36 which is true.
∴ all the points on the origin side of the line and points on the line satisfy the inequation.
Also, x ≥ 0, y ≥ 0
∴ the solution set is the points on the sides of the triangle OAB and in the interior of ∆OAB.
∴ the shaded portion represents the solution set.

Question 3.
The Diet of the sick person must contain at least 4000 units of vitamin. Each unit of food F₁ contains 200 units of vitamin, whereas each unit of food F₂ contains 100 units of vitamins. Write an inequation to fulfill a sick person’s requirements and represent the solution set graphically.
Solution:
Let the diet of the sick person contain, x units of food F₁ and y units of food F₂.
Since each unit of food F₁ contains 200 units of vitamins.
∴ x units of food F₁ contain 200x units of vitamins.
Also, each unit of food F₂ contains 100 units of vitamins.
y units of food F₂ contain 100y units of vitamins.
Now, Diet for a sick person must contain at least 4000 units of vitamins.
∴ he must take food F₁ and F₂ in such away that total vitamins must be greater than or equal to 4000.
∴ required inequation is 200x + 100y ≥ 4000
i.e., 2x + y ≥ 40
Also x and y cannot be negative.
∴ x ≥ 0, y ≥ 0
Corresponding equation is 2x + y = 40
∴ \(\frac{2 x}{40}+\frac{y}{40}=\frac{40}{40}\)
∴ \(\frac{x}{20}+\frac{y}{40}=1\)
∴ intersection of line with X-axis is A(20, 0)
intersection of line with Y-axis is B(0, 40)
Origin test:
Substituting x = 0, y = 0 in inequation, we get
2(0) + (0) ≥ 40
∴ 0 ≥ 40 which is false
∴ all the points on the non origin side of the line and points on the line satisfy the inequation.
Also, x ≥ 0, y ≥ 0
∴ the solution set is as shown in the figure.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 8.1 Solutions
• 11th Commerce Maths Exercise 8.2 Solutions
• 11th Commerce Maths Exercise 8.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 8 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.4 Questions and Answers.

Std 11 Maths 2 Exercise 9.4 Solutions Commerce Maths

Question 1.
Kanchan purchased a Maruti car for ₹ 2,45,000/- and the rate of depreciation is 14\(\frac{2}{7}\)% per annum. Find the value of the car after two years?
Solution:
Given, purchase price of the car = V = ₹ 2,45,000
Rate of depreciation per annum = r
= 14\(\frac{2}{7}\)%
= \(\frac{100}{7}\)%
∴ Value of the car after two years = \(\mathrm{V}\left(1-\frac{\mathrm{r}}{100}\right)^{\mathrm{n}}\)

∴ The value of the car after two years is ₹ 1,80,000.

Question 2.
The value of a machine depreciates from ₹ 32,768 to ₹ 21,952/- in three years. What is the rate of depreciation?
Solution:
Given, initial value of machine = V = ₹ 32,768/-
Depreciated value of the machine = D.V. = ₹ 21,952/-
Numher of years = n = 3

∴ r = 12.5%
∴ Rate of depreciation is 12.5% per annum.

Question 3.
The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. Its present value is ₹ 2,18,700/-. What was the purchase price of the machine?
Solution:
Given, the rate of depreciation per annum = r = 10%
Number of years = n = 3
Present value of the machine = P.V. = ₹ 2,18,700/-
∴ Purchase price of the machine

∴ The purchase price of the machine is ₹ 3,00,000.

Question 4.
Mr. Manish purchased a motorcycle at ₹ 70,000/-. After some years he sold his motorcycle at its exact depreciated value of it that is ₹ 51,030/-. The rate of depreciation was taken as 10%. Find out how many years he sold his motorcycle.
Solution:
Given, purchase price of the motorcycle = V = ₹ 70,000/-
Depreciated value of the motorcycle = D.V. = ₹ 51,030/-
∴ Rate of depreciation = r = 10%

∴ n = 3
∴ Manish sold his motorcycle after 3 years.

Question 5.
Mr. Chetan purchased furniture for his home at ₹ 5,12,000/-. Considering the rate of depreciation as 12.5%, what will be the value of furniture after 3 years.
Solution:
Given, purchase price of furniture = V = ₹ 5,12,000/-
Rate of depreciation = r = 12.5%
Number of years = n = 3 years
∴ Value of furniture after 3 years = \(\mathrm{V}\left(1-\frac{\mathrm{r}}{100}\right)^{\mathrm{n}}\)
= 5,12,000 \(\left(1-\frac{12.5}{100}\right)^{3}\)
= 5,12,000 \(\left(1-\frac{1}{8}\right)^{3}\)
= 5,12,000 \(\left(\frac{7}{8}\right)^{3}\)
= 5,12,000 × \(\frac{343}{512}\)
= 3,43,000
∴ The value of furniture will be ₹ 3,43,000/-

Question 6.
Grace Fashion Boutique purchased a sewing machine at ₹ 25,000/-. After 3 years machine was sold at depreciated value of ₹ 18,225/-. Find the rate of depreciation.
Solution:
Given, purchase price of sewing machine = V = ₹ 25,000/-
Selling price of machine = D.V. = ₹ 18,225/-
Number of years = n = 3 years

∴ 100 – r = 90
∴ r = 10%
∴ Rate of depreciation is 10% per annum.

Question 7.
Mr. Pritesh reduced the value of his assets by 5% each year, which were purchased for ₹ 50,00,000/-. Find the value of assets after 2 years.
Solution:
Given, initial value of assets = V = ₹ 50,00,000/-
Rate of depreciation per annum = r = 5%
Number of years = n = 2 years
∴ Value of assets aftertwo years = \(V\left(1-\frac{r}{100}\right)^{n}\)

= 12,500 × 361
= 45,12,500
∴ The value of assets after two years is ₹ 45,12,500/-.

Question 8.
A manufacturing company is allowed to charge 10% depreciation on its stock. The initial value of the stock was ₹ 60,000/-. After how many years value of the stock will be ₹ 39366?
Solution:
Given, rate of depreciation = r = 10%
Initial value of stock = V = ₹ 60,000
Depreciated value of stock = D.V. = ₹ 39,366/-
By using,

∴ n = 4
∴ The value of the stock will be ₹ 39,366/- after 4 years.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Skewness Class 11 Commerce Maths 2 Chapter 3 Exercise 3.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Skewness Ex 3.1 Questions and Answers.

Std 11 Maths 2 Exercise 3.1 Solutions Commerce Maths

Question 1.
For a distribution, mean = 100, mode = 127 and S.D. = 60. Find the Pearson coefficient of skewness Sk_p.
Solution:
Given, Mean = 100, Mode = 127, S.D. = 60

Question 2.
The mean and variance of a distribution are 60 and 100 respectively. Find the mode and the median of the distribution if Sk_p = -0.3.
Solution:
Given, Mean = 60, Variance = 100, Sk_p = -0.3
∴ S.D. = √Variance = √100 = 10
Sk_p = \(\frac{\text { Mean }-\text { Mode }}{\text { S.D. }}\)
∴ -0.3 = \(\frac{60-\text { Mode }}{10}\)
∴ -3 = 60 – Mode
∴ Mode = 60 + 3 = 63
Mean – Mode = 3 (Mean – Median)
∴ 60 – 63 = 3(60 – Median)
∴ -3 = 180 – 3Median
∴ 3Median = 180 + 3 = 183
∴ Median = \(\frac{183}{3}\)
∴ Median = 61

Question 3.
For a data set, sum of upper and lower quartiles is 100, difference between upper and lower quartiles is 40 and the median is 30. Find the coefficient of skewness.
Solution:
Given, Q₃ + Q₁ = 100 ……(i)
Q₃ – Q₁ = 40 …..(ii)
Median = Q₂ = 30
Adding (i) and (ii), we get
2Q₃ = 140
∴ Q₃ = 70
Substituting the value of Q₃ in (i), we get
70 + Q₁ = 100
∴ Q₁ = 100 – 70 = 30

Question 4.
For a data set with an upper quartile equal to 55 and median equal to 42, if the distribution is symmetric, find the value of the lower quartile.
Solution:
Upper quartile = Q₃ = 55
Median = Q₂ = 42
Since, the distribution is symmetric.
∴ Sk_b = 0
Sk_b = \(\frac{Q_{3}+Q_{1}-2 Q_{2}}{Q_{3}-Q_{1}}\)
∴ 0 = \(\frac{Q_{3}+Q_{1}-2 Q_{2}}{Q_{3}-Q_{1}}\)
∴ 0 = Q₃ + Q₁ – 2Q₂
∴ Q₁ = 2Q₂ – Q₃
∴ Q₁ = 2(42) – 55
∴ Q₁ = 84 – 55
∴ Q₁ = 29

Question 5.
Obtain coefficient of skewness by formula and comment on the nature of the distribution.

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

Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{82}{4}\) = 20.5
Cumulative frequency which is just greater than (or equal) to 20.5 is 30.
∴ Q₁ lies in the class 60 – 64.
∴ L = 60, h = 4, f = 20, c.f. = 10

Q₂ class = class containing \(\left(\frac{\mathrm{N}}{2}\right)^{\mathrm{th}}\) observation
∴ \(\frac{\mathrm{N}}{2}=\frac{82}{2}\) = 41
Cumulative frequency which is just greater than (or equal) to 41 is 70.
∴ Q₂ lies in the class 64 – 68.
∴ L = 64, h = 4, f = 40, c.f. = 30

Q₃ class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 82}{4}\) = 61.5
Cumulative frequency which is just greater than (or equal) to 61.5 is 70.
∴ Q₃ lies in the class 64 – 68.
∴ L = 64, h = 4, f = 40, c.f. = 30


∴ Sk_b = -0.1881
Since, Sk_b < 0, the distribution is negatively skewed.

Question 6.
Find Sk_p for the following set of observations.
17, 17, 21, 14, 15, 20, 19, 16, 13, 17, 18
Solution:
Σx_i = 17 + 17 + 21 + 14 + 15 + 20 + 19 + 16 + 13 + 17 + 18 = 187
Mean = \(\frac{\sum x_{i}}{n}=\frac{187}{11}\) = 17
Mode = Observation that occurs most frequently in the data = 17

Question 7.
Calculate Sk_b for the following set of observations of the yield of wheat in kg from 13 plots:
4.6, 3.5, 4.8, 5.1, 4.7, 5.5, 4.7, 3.6, 3.5, 4.2, 3.5, 3.6, 5.2
Solution:
The given data can be arranged in ascending order as follows:
3.5, 3.5, 3.5, 3.6, 3.6, 4.2, 4.6, 4.7, 4.7, 4.8, 5.1, 5.2, 5.5
Here, n = 13
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{13+1}{4}\right)^{\text {th }}\) observation
= value of (3.50)th observation
= value of 3rd observation + 0.50(value of 4th observation – value of 3rd observation)
= 3.5 + 0.50(3.6 – 3.5)
= 3.5 + 0.50(0.1)
= 3.5 + 0.05
∴ Q₁ = 3.55
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{13+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 3.50)th observation
= value of 7th observation
∴ Q₂ = 4.6
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{13+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 3.50)th observation
= value of (10.50)th observation
= value of 10th observation + 0.50 (value of 11th obseration – value of 10th observation)
= 4.8 + 0.50(5.1 – 4.8)
= 4.8 + 0.50(0.3)
∴ Q₃ = 4.95

∴ Sk_b = -0.5

Question 8.
For a frequency distribution Q₃ – Q₂ = 90 and Q₂ – Q₁ = 120. Find Sk_b.
Solution:
Given, Q₂ – Q₁ = 90, Q₂ – Q₁ = 120

∴ Sk_b = -0.1429

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• 11th Commerce Maths Exercise 3.1 Solutions
• 11th Commerce Maths Miscellaneous Exercise 3 Solutions

Correlation Class 11 Commerce Maths 2 Chapter 5 Miscellaneous Exercise 5 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Correlation Miscellaneous Exercise 5 Questions and Answers.

Std 11 Maths 2 Miscellaneous Exercise 5 Solutions Commerce Maths

Question 1.
Two series of x and y with 50 items each have standard deviations of 4.8 and 3.5 respectively. If the sum of products of deviations of x and y series from respective arithmetic means is 420, then find the correlation coefficient between x and y.
Solution:

Question 2.
Find the number of pairs of observations from the following data,
r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40.
Solution:
Given, r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40

Question 3.
Given that r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900. Find the number of pairs of observations.
Solution:
Given, r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900

Question 4.
Given the following information, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5, where x_i and y_i are the deviations from their respective means, find the number of items.
Solution:
Here, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5
Here, x_i and y_i are the deviations from their respective means.
∴ If X_i, Y_i are elements of x and y series respectively, then
X_i – \(\bar{x}\) = x_i and Y_i – \(\bar{y}\) = y_i
∴ Σx_iy_i = \(\Sigma\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)\left(\mathrm{Y}_{\mathrm{i}}-\bar{y}\right)\) = 60, \(\sum x_{\mathrm{i}}^{2}=\sum\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)^{2}\) = 90

Question 5.
A sample of 5 items is taken from the production of a firm. The length and weight of 5 items are given below. [Given: √0.8823 = 0.9393]

Calculate the correlation coefficient between length and weight and interpret the result.
Solution:
Let length = x_i (in cm), Weight = y_i (in gm)


∴ the value of r indicates a high degree of positive correlation between length and weight of items.

Question 6.
Calculate the correlation coefficient from the following data and interpret it.

Solution:


∴ the value of r indicates a perfect negative correlation between x and y.

Question 7.
Calculate the correlation coefficient from the following data and interpret it.

Solution:



∴ the value of r indicates a perfect positive correlation between x and y.

Question 8.
If the correlation coefficient between X and Y is 0.8, what is the correlation coefficient between
(i) 2X and Y
(ii) \(\frac{X}{2}\) and Y
(iii) X and 3Y
(iv) X – 5 and Y – 3
(v) X + 7 and Y + 9
(vi) \(\frac{X-5}{7}\) and \(\frac{Y-3}{8}\)?
Solution:
The correlation coefficient remains unaffected by the change of origin and scale.
i.e., if u_i = \(\frac{x_{i}-\mathrm{a}}{\mathrm{h}}\) and v_i = \(\frac{y_{i}-\mathrm{b}}{\mathrm{k}}\), then Corr(U, V) = ±Corr(X, Y).
according to the same or opposite signs of h and k.
(i) u_i = \(\frac{2\left(x_{i}-0\right)}{1}\), v_i = \(\frac{y_{i}-0}{1}\)
Here, h = 1 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(ii) u_i = \(\frac{x_{i}-0}{2}\), v_i = \(\frac{y_{\mathrm{i}}-0}{1}\)
Here, h = 2 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(iii) Corr (X, 3Y) = Corr (X, Y) = 0.8

(iv) Corr (X – 5, Y – 3) = Corr(X, Y) = 0.8

(v) Corr (X + 7, Y + 9) = Corr(X, Y) = 0.8

(vi) Corr(\(\frac{X-5}{7}, \frac{Y-3}{8}\)) = Corr(X, Y) = 0.8

Question 9.
In the calculation of the correlation coefficient between the height and weight of a group of students of a college, one investigator took the measurements in inches and pounds while the other investigator took the measurements in cm. and kg. Will they get the same value of the correlation coefficient or different values? Justify your answer.
Solution:
The coefficient of correlation is a ratio of covariance and standard deviations.
Since covariance and standard deviations are independent of units of measurement.
∴ coefficient of correlation is also independent of units of measurement.
∴ values of coefficient of correlation obtained by first and second investigators are the same.

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• 11th Commerce Maths Exercise 5.1 Solutions
• 11th Commerce Maths Miscellaneous Exercise 5 Solutions

Skewness Class 11 Commerce Maths 2 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 3 Skewness Miscellaneous Exercise 3 Questions and Answers.

Std 11 Maths 2 Miscellaneous Exercise 3 Solutions Commerce Maths

Question 1.
For u distribution, mean = 100, mode = 80 and S.D. = 20. Find Pearsonian coefficient of skewness Sk_p.
Solution:
Given, Mean = 100, Mode = 80, S.D. = 20

Question 2.
For a distribution, mean = 60, median = 75 and variance = 900. Find Pearsonian coefficient of skewness Sk_p.
Solution:
Given. Mean = 60, Median = 75, Variance = 900
∴ S.D. = √Variance = √900 = 30

Question 3.
For a distribution, Q₁ = 25, Q₂ = 35 and Q₃ = 50. Find Bowley’s coefficient of skewness Sk_b.
Solution:
Given Q₁ = 25, Q₂ = 35, Q₃ = 50

Question 4.
For a distribution Q₃ – Q₂ = 40, Q₂ – Q₁ = 60. Find Bowlev’s coefficient of skewness Sk_b.
Solution:
Given, Q₃ – Q₂ = 40, Q₂ – Q₁ = 60

Question 5.
For a distribution, Bowley’s coefficient of skewness is 0.6. The sum of upper and lower quartiles is 100 and median is 38. Find the upper and lower quartiles.
Solution:
Given, Sk_b = 0.6, Q₃ + Q₁ = 100,
Median = Q₂ = 38
Skb = \(\frac{Q_{3}+Q_{1}-2 Q_{2}}{Q_{3}-Q_{1}}\)
∴ 0.6 = \(\frac{100-2(38)}{Q_{3}-Q_{1}}\)
∴ 0.6(Q₃ – Q₁) = 100 – 76 = 24
∴ Q₃ – Q₁ = 40 ….(i)
Q₃ + Q₁ = 100 …..(ii) (given)
Adding (i) and (ii), we get
2Q₃ = 140
∴ Q₃ = 70
Substituting the value of Q₃ in (ii), we get
70 + Q₁ = 100
∴ Q₁ = 100 – 70 = 30
∴ upper quartile = 70 and lower quartile = 30

Question 6.
For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian’s coefficient of skewness is 0.3. Find the mode and median of the distribution.
Solution:
Mean = \(\bar{x}\) = 200
Coefficient of variation, C.V. = 8%, Skp = 0.3
C.V. = \(\frac{\sigma}{\bar{x}} \times 100\), where σ = standard deviation
∴ 8 = \(\frac{\sigma}{200} \times 100\)
∴ σ = \(\frac{8 \times 200}{100}\) = 16
Now, Sk_p = \(\frac{\text { Mean – Mode }}{\text { S.D. }}\)
∴ 0.3 = \(\frac{200-\text { Mode }}{16}\)
∴ 0.3 × 16 = 200 – Mode
∴ Mode = 200 – 4.8 = 195.2
Since, Mean – Mode = 3(Mean – Median)
∴ 200 – 195.2 = 3(200 – Median)
∴ 4.8 = 600 – 3Median
∴ 3Median = 600 – 4.8 = 595.2
∴ Median = 198.4

Question 7.
Calculate Karl Pearsonian’s coefficient of skewness Skp from the follow ing data:

Solution:
The given table is the cumulative frequency table of more than type.
From this table, we have to prepare the frequency distribution table and then calculate the value of Sk_p.
Construct the following table:

From the table, N = 120, Σf_ix_i = 5490 and \(\sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}^{2}\) = 284600
Mean = \(\bar{x}=\frac{\sum f_{i} x_{i}}{N}=\frac{5490}{120}\) = 45.75
Maximum frequency 42 is of the class 50 – 60
∴ Mode lies in the class 50 – 60
∴ L = 50, f₁ = 42, f₀ = 25, f₂ = 13, h = 10

Alternate Method:
Let u = \(\frac{x-45}{10}\)

\(\overline{\mathrm{u}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\mathrm{N}}=\frac{9}{120}\) = 0.075
∴ \(\bar{x}\) = 45 + 10(\(\bar{u}\))
= 45 + 10(0.075)
= 45 + 0.75
= 45.75
Var(u) = \(\sigma_{u}^{2}=\frac{\sum f_{i} u_{i}{ }^{2}}{N}-(\bar{u})^{2}\)
= \(\frac{335}{120}\) – (0.075)²
= 2.7917 – 0.0056
= 2.7861
Var(X) = h² × Var(u)
= 100 × 2.7861
= 278.61
S.D. = √278.61 = 16.6916
Maximum frequency 42 is of the class 50 – 60.
∴ Mode lies in the class 50 – 60.
∴ L = 50, f₁ = 42, f₀ = 25, f₂ = 13, h = 10

Question 8.
Calculate Bowley’s coefficient of skewness Skb from the following data.

Solution:
To calculate Bowley’s coefficient of skewness Skb, we construct the following table:

Here, N = 120
Q₁ class = class containing the \(\left(\frac{N}{4}\right)^{t h}\) observation
∴ \(\frac{N}{4}=\frac{120}{4}\) = 30
Cumulative frequency which is just greater than (or equal to) 30 is 35.
∴ Q₁ lies in the class 30-40.
∴ L = 30, h = 10, f = 13, c.f. = 22

Q₂ class = class containing the \(\left(\frac{N}{2}\right)^{t h}\) observation
∴ \(\frac{\mathrm{N}}{2}=\frac{120}{2}\) = 60
Cumulative frequency which is just greater than (or equal to) 60 is 60.
∴ Q₂ lies in the class 40-50.
∴ L = 40, h = 10, f = 25, c.f. = 35

Q₃ class = class containing the \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 90
Cumulative frequency which is just greater than (or equal to) 90 is 102.
∴ Q₃ lies in the class 50 – 60
∴ L = 50, h = 10, f = 42, c.f. = 60

Question 9.
Find Sk_p for the following set of observations:
18, 27, 10, 25, 31, 13, 28
Solution:
The given data can be arranged in ascending order as follows:
10, 13, 18, 25, 27, 28, 31
Here, n = 7
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{7+1}{2}\right)^{\text {th }}\) observation
= value of 4th observation
= 25
For finding standard deviation, we construct the following table:

Question 10.
Find Skb for the following set of observations:
18, 27, 10, 25, 31, 13, 28
Solution:
The given data can be arranged in ascending order as follows:
10, 13, 18, 25, 27, 28, 31
Here, n = 7
∴ Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of 2nd observation
∴ Q₁ = 13
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 2)th observation
= value of 4th observation
∴ Q₂ = 25
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{7+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 2)th observation
= value of 6th observation
∴ Q₃ = 28
Coefficient of skewness,

∴ Sk_b = -0.6

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• 11th Commerce Maths Exercise 3.1 Solutions
• 11th Commerce Maths Miscellaneous Exercise 3 Solutions

Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.4 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.4 Questions and Answers.

Std 11 Maths 2 Exercise 6.4 Solutions Commerce Maths

Question 1.
Find the number of permutations of letters in each of the following words:
(i) DIVYA
(ii) SHANTARAM
(iii) REPRESENT
(iv) COMBINE
Solution:
(i) There are 5 letters in the word DIVYA which can be arranged in 5! Way = 120 ways

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ repeats 3 times.
∴ Number of permutations of the letters of the word SHANTARAM = \(\frac{9 !}{3 !}\)
= 9 × 8 × 7 × 6 × 5 × 4
= 60480

(iii) There are 9 letters in the word REPRESENT in which ‘E’ repeats 3 times and ‘R’ repeats 2 times.
∴ Number of permutations of the letters of the word REPRESENT = \(\frac{9 !}{3 ! 2 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}\)
= 30240

(iv) There are 7 distinct letters in the word COMBINE which can be arranged among themselves in 7! = 5040 ways

Question 2.
You have 2 identical books on English, 3 identical books on Hindi and 4 identical books on Mathematics. Find the number of distinct ways of arranging them on a shelf.
Solution:
There are total 9 books to be arranged on a shelf.
Out of these 9 books, 2 books on English, 3 books on Hindi and 4 books on Mathematics are identical.
∴ Total number of arrangements = \(\frac{9 !}{2 ! 3 ! 4 !}\)
= \(\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 3 \times 2 \times 4 !}\)
= 9 × 4 × 7 × 5
= 1260
∴ In 1260 distinct ways the books can be arranged on a shelf.

Question 3.
A coin is tossed 8 times. In how many ways can we obtain
(i) 4 heads and 4 tails?
(ii) at least 6 heads?
Solution:
A coin is tossed 8 times. All heads are identical and all tails are identical.
(i) We can obtain 4 heads and 4 tails in \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2}\)
= 70 ways
∴ In 70 different ways we can obtain 4 heads and 4 tails.

(ii) When at least 6 heads are to be obtained
∴ Outcome can be (6 heads and 2 tails) or (7 heads and 1 tail) or (8 heads)
∴ Number of ways in which it can be obtained = \(\frac{8 !}{6 ! 2 !}+\frac{8 !}{7 ! 1 !}+\frac{8 !}{8 !}\)
= \(\frac{8 \times 7}{2}\) + 8 + 1
= 28 + 8 + 1
= 37
∴ In 37 different ways we can obtain at least 6 heads.

Question 4.
A bag has 5 red, 4 blue, and 4 green marbles. If all are drawn one by one and their colours are recorded, how many different arrangements can be found?
Solution:
There is a total of 13 marbles in a bag.
Out of these 5 are Red, 4 Blue, and 4 are Green marbles.
All balls of the same colour are taken to be identical.
∴ Required number of arrangements = \(\frac{13 !}{5 ! 4 ! 4 !}\)

Question 5.
Find the number of ways of arranging letters of the word MATHEMATICAL. How many of these arrangements have all vowels together?
Solution:
There are 12 letters in the word MATHEMATICAL in which ‘M’ repeats 2 times, ‘A’ repeats 3 times, and ‘T’ repeats 2 times.
∴ Total number of arrangements = \(\frac{12 !}{2 ! 3 ! 2 !}\)
When all the vowels
i.e., ‘A’, ‘A’, ‘A’, ‘E’, T are to be kept together
Number of arrangements of these vowels = \(\frac{5 !}{3 !}\) ways.
Let us consider these vowels together as one unit.
This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.
∴ Number of arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ Total number of arrangements = \(\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}\)

Question 6.
Find the number of different arrangements of letters in the word MAHARASHTRA. How many of these arrangements have
(i) letters M and T never together?
(ii) all vowels together?
Solution:
There are 11 letters in the word MAHARASHTRA in which ‘A’ is repeated 4 times, ‘H’ repeated 2 times, and ‘R’ repeated 2 times.
∴ Total number of arrangements is \(\frac{11 !}{4 ! 2 ! 2 !}\)
∴ \(\frac{11 !}{4 ! 2 ! 2 !}\) different words can be formed from the letters of the word MAHARASHTRA.
(i) Other than M and T. there are 9 letters in which A repeats 4 times, H repeats twice, R repeats twice
The number of arrangements of the a letter = \(\frac{9 !}{4 ! 2 ! 2 !}\)
These 9 letters create 10 gaps in which M and T are to be arranged
The number of arrangements of M and T = ¹⁰P₂
∴ Total number arrangement having M and T never together = \(\frac{9 ! \times{ }^{10} \mathrm{P}_{2}}{4 ! 2 ! 2 !}\)

(ii) When all vowels are together.
There are 4 vowels in the word MAHARASHTRA i.e., A, A, A, A
Let us consider these 4 vowels as one unit, they themselves can be arranged in \(\frac{4 !}{4 !}\) = 1 way.
This unit is to be arranged with 7 other letters which can be done in 8! ways
∴ Total number of arrangements = \(\frac{8 !}{2 ! 2 !}\)
∴ \(\frac{8 !}{2 ! 2 !}\) different words can be formed if vowels are always together.

Question 7.
How many different words are formed if the letter R is used thrice and letters S and T are used twice each?
Solution:
When ‘R’ is used thrice, ‘S’ is used twice and ‘T’ is used twice,
∴ Total number of letters available = 7, of which ‘S’ and ‘T’ repeat 2 times each, ‘R’ repeats 3 times.
∴ Required number of arrangements = \(\frac{7 !}{2 ! 2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 1 \times 2 \times 1 \times 3 !}\)
= 7 × 6 × 5
= 210
∴ 210 different words can be formed with the letter R is used thrice and letters S and T are used twice each.

Question 8.
Find the number of arrangements of letters in the word MUMBAI so that the letter B is always next to A.
Solution:
There are 6 letters in the word MUMBAI.
These letters are to be arranged in such a way that ‘B’ is always next to ‘A’.
Let us consider AB as one unit. This unit with other 4 letters in which ‘M’ repeats twice, is to be arranged.
∴ Total number of arrangements when B is always next to A = \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60

Question 9.
Find the number of arrangements of letters in the word CONSTITUTION that begin and end with N.
Solution:
There are 12 letters in the word CONSTITUTION, in which ‘O’, ‘N’, T repeat two times each, ‘T’ repeats 3 times.
The arrangement starts and ends with ‘N’, 10 letters other than N can be arranged between two N, in which ‘O’ and ‘I’ repeat twice each and ‘T’ repeats 3 times.
∴ Total number of arrangements with the letter N at the beginning and at the end = \(\frac{10 !}{2 ! 2 ! 3 !}\)

Question 10.
Find the number of different ways of arranging letters in the word ARRANGE. How many of these arrangements have two R’s and two A’s not together?
Solution:
(i) There are 7 letters in the word ARRANGE in which A is repeated 2 times and R is repeated 2 times
∴ The number of arrangements = \(\frac{7 !}{2 ! 2 !}\) = 1260

(ii) A: set of words having 2A together
B: set of words having 2R together
Number of words having both A and both R not together
= 1260 – n(A ∪ B)
= 1260 – [n(a) + n(B) – n(A ∩ B)] ……(i)
n(A) = number of ways in which (AA) R, R, N, G, E are to be arranged
∴ n(A) = \(\frac{6 !}{2 !}\) = 360
n(B) = number of ways in which (RR), A, A, N, G, E are to be arranged
∴ n(B) = \(\frac{6 !}{2 !}\) = 360
n(A ∩ B) = number of ways in which (AA), (RR), N, G, E are to be arranged
∴ n(A ∩ B) = 5! = 120
Substituting n(A), n(B), n(A ∩ B) in (i), we get
Number of words having both A and both R not together
= 1260 – [360 + 360 – 120]
= 1260 – 600
= 660

Question 11.
How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Solution:
5 digit numbers are to be formed from 2, 3, 2, 3, 4, 5.
Case I: Numbers formed from 2, 2, 3, 4, 5 OR 2, 3, 3, 4, 5
Number of such numbers = \(\frac{5 !}{2 !}\) × 2
= 5!
= 120

Case II: Numbers formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers = \(\frac{5 !}{2 ! 2 !}\) × 2 = 60
Required number = 120 + 60 = 180
∴ 180 distinct 5 digit numbers can be formed using the digit 3, 2, 3, 2, 4, 5.

Question 12.
Find the number of distinct numbers formed using the digits 3, 4, 5, 6, 7, 8, 9, so that odd positions are occupied by odd digits.
Solution:
A number is to be formed with digits 3, 4, 5, 6, 7, 8, 9 such that odd digits always occupy the odd places.
There are 4 odd digits i.e. 3, 5, 7, 9.
They can be arranged at 4 odd places among themselves in 4! ways = 24 ways
3 even places of the number are occupied by even digits (i.e. 4, 6, 8).
∴ They can be arranged in 3! ways = 6 ways
∴ Total number of arrangements = 24 × 6 = 144
∴ 144 numbers can be formed so that odd digits always occupy the odd positions.

Question 13.
How many different 6-digit numbers can be formed using digits in the number 659942? How many of them are divisible by 2?
Solution:
A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Total number of arrangements = \(\frac{6 !}{2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\)
= 360
∴ 360 different 6-digit numbers can be formed.
For a number to be divisible by 2,
Last digits can be selected in 3 ways
Remaining 5 digit in which, 9 appears twice are arranged in \(\frac{5 !}{2 !}\) ways
∴ Total number of arrangements = \(\frac{5 !}{2 !}\) × 3 = 180
∴ 180 numbers are divisible by 2.

Question 14.
Find the number of distinct words formed from letters in the word INDIAN. How many of them have the two N’s together?
Solution:
There are 6 letters in the word INDIAN in which I and N repeat twice.
Number of different words that can be formed using the letters of the word INDIAN = \(\frac{6 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 \times 2 !}\)
= 180
∴ 180 different words can be formed with the letters of the word INDIAN.
When two N’s are together.
Let us consider the two N’s as one unit.
They can be arranged with 4 other letters in \(\frac{5 !}{2 !}\)
= \(\frac{5 \times 4 \times 3 \times 2 !}{2 !}\)
= 60 ways.
∴ 2N can be arranged in 1 way
∴ Total number of arrangements = 60 × 1 = 60 ways
∴ 60 words are such that two N’s are together.

Question 15.
Find the number of different ways of arranging letters in the word PLATOON if
(i) the two O’s are never together.
(ii) consonants and vowels occupy alternate positions.
Solution:
(i) When the two O’s are never together:
Let us arrange the other 5 letters first, which can be done in 5! = 120 ways.
The letters P, L, A, T, N create 6 gaps, in which O’s are arranged.
∴ Two O’s in 6 gaps can be arranged in \(\frac{{ }^{6} \mathrm{P}_{2}}{2 !}\) ways
= \(\frac{\frac{6 !}{(6-2) !}}{2 !}\) ways
= \(\frac{6 \times 5 \times 4 !}{4 ! \times 2 \times 1}\) ways
= 3 × 5 ways
= 15 ways
∴ Total number of arrangements if the two O’s are never together = 120 × 15 = 1800

(ii) When consonants and vowels occupy alternate positions:
There are 4 consonants and 3 vowels in the word PLATOON.
∴ At odd places consonants occur and at even places vowels occur.
4 consonants can be arranged among themselves in 4! ways
3 vowels in which O occurs twice and A occurs once.
∴ They can be arranged in \(\frac{3 !}{2 !}\) ways
∴ Required number of arrangements if the consonants and vowels occupy alternate positions = 4! × \(\frac{3 !}{2 !}\)
= 4 × 3 × 2 × \(\frac{3 \times 2 !}{2 !}\)
= 72

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 6.1 Solutions
• 11th Commerce Maths Exercise 6.2 Solutions
• 11th Commerce Maths Exercise 6.3 Solutions
• 11th Commerce Maths Exercise 6.4 Solutions
• 11th Commerce Maths Exercise 6.5 Solutions
• 11th Commerce Maths Exercise 6.6 Solutions
• 11th Commerce Maths Exercise 6.7 Solutions
• 11th Commerce Maths  Miscellaneous Exercise 6 Solutions

Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Miscellaneous Exercise 6 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Miscellaneous Exercise 6 Questions and Answers.

Std 11 Maths 2 Miscellaneous Exercise 6 Solutions Commerce Maths

Question 1.
Find the value of r if ⁵⁶C_r+6 : ⁵⁴P_r-1 = 30800 : 1
Solution:

Question 2.
How many words can be formed by writing letters in the word CROWN in a different order?
Solution:
Five Letters of the word CROWN are to be permuted.
Number of different words = 5! = 120

Question 3.
Find the number of words that can be formed by using all the letters in the word REMAIN. If these words are written in dictionary order, what will be the 40th word?
Solution:
There are 6 letters A, E, I, M, N, R
The number of words starting with A = 5!
The number of words starting with E = 5!
The number of words starting with I = 5!
The number of words starting with M = 5!
The number of words starting with N = 5!
The number of words starting with R = 5!
Total number of words = 6 × 5! = 720
Number of words starting with AE = 4! = 24
Number of words starting with AIE = 3! = 6
Number of words starting with AIM = 3! = 6
The number of words starting with AINE = 2!
Total words = 24 + 6 + 6 + 2 = 38
39th word is AINMER
40th word is AINMRE

Question 4.
Find the number of ways of distributing n balls in n cells. What will be the number of ways if each cell must be occupied?
Solution:
There are n balls and n cells
(i) Every ball can be put in any of the n cells.
Number of distributions = n × n × …… × n = (n)ⁿ
(ii) For filling the first cell, n balls are available.
The first cell is filled in n ways.
The second cell is filled in (n – 1) ways
The third cell is filled in (n – 2) ways and so on.
the nth cell is tilled in one way.
Required number = n(n – 1)(n – 2) …… 1 = n!

Question 5.
Thane is the 20th station from C.S.T. If a passenger can purchase a ticket from any station to any other station, how many different tickets must be available at the booking window?
Solution:
Taking CST as the first station and Thane as 20th,
Let us name CST as A₀ next station as A₁ and so on, Thane is A₂₀
From station A₀, 20 different journeys are possible
From station A₁, 20 different journeys are possible.
From station A₂₀, 20 different journeys are possible.
Total number of different tickets of different journeys = 21 × 20 = 420

Question 6.
English alphabet has 11 symmetric letters that appear the same when looked at in a mirror. These letters are A, H, I, M, O, T, U, V, W, X, and Y. How many symmetric three letters passwords can be formed using these letters?
Solution:
Number of 3 Letter passwords = ¹¹P₃
= 11 × 10 × 9
= 990

Question 7.
How many numbers formed using the digits 3, 2, 0, 4, 3, 2, 3 exceed one million?
Solution:
A number that exceeds one million is to be formed from the digits 3, 2, 0, 4, 3, 2, 3.
Then the numbers should be any number of 7 digits which can be formed from these digits.
Also among the given numbers 2 repeats twice and 3 repeats thrice.
∴ Required number of numbers = Total number of arrangements possible among these digits – number of arrangements of 7 digits which begin with 0.
= \(\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 \times 3 !}{2 \times 3 !}-\frac{6 \times 5 \times 4 \times 3 !}{2 \times 3 !}\)
= 7 × 6 × 5 × 2 – 6 × 5 × 2
= 6 × 5 × 2(7 – 1)
= 60 × 6
= 360
∴ 360 numbers that exceed one million can be formed with the digits 3, 2, 0, 4, 3, 2, 3.

Question 8.
Ten students are to be selected for a project from a class of 30 students. There are 4 students who want to be together either in the project or not in the project. Find the number of possible selections.
Solution:

Required number = ²⁶C₆ + ²⁶C₁₀

Question 9.
A student finds 7 books of his interest but can borrow only three books. He wants to borrow the Chemistry part II book only if Chemistry Part I can also be borrowed. Find the number of ways he can choose three books that he wants to borrow.
Solution:

Required Number = 5 + 10 = 15

Question 10.
30 objects are to be divided into three groups containing 7, 10, 13 objects. Find the number of distinct ways of doing so. Solution:
Required number = ³⁰C₇ × ²³C₁₀ × ¹³C₁₃

Question 11.
A student passes an examination if he secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
Solution:
Every subject a student may pass or fail.
∴ Total number of outcomes = 2⁷ = 128
This number includes one case when the student passes in all subjects.
∴ Required number = 128 – 1 = 127

Question 12.
Nine friends decide to go for a picnic in two groups. One group decides to go by car and the other group decides to go by train. Find the number of different ways of doing so if there must be at least 3 friends in each group.
Solution:

Question 13.
Five balls are to be placed in three boxes, where each box can contain upto five balls. Find the number of ways if no box is to remain empty.
Solution:
Let boxes be named as I, II, III
Let sets A, B, C represent cases in which boxes I, II, III remain empty
Then A ∪ B ∪ C represent the cases in which at least one box remains empty.
Then we use method of indirect counting
Required number = Total number of distributions – n(A ∪ B ∪ C) …..(i)
n(A ∪ B ∪ C) represent the number of undesirable cases
Total number of distributions = 3 × 3 × 3 × 3 × 3 = 3⁵ = 243 …….(ii)
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C) …..(iii)
In box I is empty then every ball has two places (boxes) to go.
Similarly for box II and III.
∴ n(A) + n(B) + n(C) = 3 × 2⁵ ……(iv)
If boxes I and II remain empty then all balls go to box III
Similarly we would have two more cases.
∴ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 3 × 1⁵ ……(v)
∴ n(A ∩ B ∩ C) = 0 …….(vi) [as all boxes cannot be empty]
Substitute from (iv), (v), (vi) to (iii) to get
n(A ∪ B ∪ C) = 3 × 2⁵ – 3 × 1⁵
= 96 – 3
= 93
Substitute n(A ∪ B ∪ C) and from (ii) to (i), we get
Required number = 243 – 93 = 150

Question 14.
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
Solution:
Every lamp is either ON or OFF.
There are 12 lamps
Number of instances = 2¹²
This number includes the case in which all 12 lamps are OFF.
∴ Required Number = 2¹² – 1 = 4095

Question 15.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficients if a coefficient can be repeated in an equation?
Solution:
Let the quadratic equation be ax² + bx + c = 0, a ≠ 0

∴ Required number = 3 × 4 × 4 = 48

Question 16.
How many six-digit telephone numbers can be formed if the first two digits are 45 and no digit can appear more than once?
Solution:
Let the telephone number be 45abcd

∴ Required number = ⁸P₄ = 1680

Question 17.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
Solution:
Every question is ‘SOLVED’ or ‘NOT SOLVED’.
There are 6 question.
Number of outcomes = 2⁶
This number includes the case when the student solves NONE of the question.
Required number = 2⁶ – 1
= 64 – 1
= 63

Question 18.
Find the number of ways of dividing 20 objects in three groups of sizes 8, 7 and 5.
Solution:
Select 8 object out of 20 in ²⁰C₈ ways
Select 7 object from remaining 12 in ¹²C₇ ways and 5 objects from remaining 5 in ⁵C₅ ways
Required number is = ²⁰C₈ × ¹²C₇ × ⁵C₅

Question 19.
There are 8 doctors and 4 lawyers in a panel. Find the number of ways for selecting a team of 6 if at least one doctor must be in the team.
Solution:
There are 8 doctors and 4 lawyers.
We need to select a team of 6 which contains at least one doctor.
Since, there are only 4 lawyers any team of 6 will contain at least two doctors.
Required number = ¹²C₆ = 924

Question 20.
Four parallel lines intersect another set of five parallel lines. Find the number of distinct parallelograms that can be formed.
Solution:
We need 2 lines from each set.
Required number = ⁴C₂ × ⁵C₂
= 6 × 10
= 60

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 6.1 Solutions
• 11th Commerce Maths Exercise 6.2 Solutions
• 11th Commerce Maths Exercise 6.3 Solutions
• 11th Commerce Maths Exercise 6.4 Solutions
• 11th Commerce Maths Exercise 6.5 Solutions
• 11th Commerce Maths Exercise 6.6 Solutions
• 11th Commerce Maths Exercise 6.7 Solutions
• 11th Commerce Maths  Miscellaneous Exercise 6 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.3 Questions and Answers.

Std 11 Maths 2 Exercise 9.3 Solutions Commerce Maths

Question 1.
What would be the simple interest on an amount of ₹ 9,600 at the rate of 6% per annum after 3 years?
Solution:
Given Principal P = ₹ 9600
Rate of interest R = 6% p.a.
Number of years = T = 3
Simple Interest I = \(\frac{\text { PRT }}{100}\)
= \(\frac{9600 \times 3 \times 6}{100}\)
= 96 × 18
= 1728
∴ Simple interest after 3 years would be ₹ 1728

Question 2.
What would be the simple interest at the rate of 9\(\frac{1}{2}\)% per annum on ₹ 6,000 for 2\(\frac{1}{2}\) years?
Solution:
Rate of interest per annum R = 9\(\frac{1}{2}\)% = \(\frac{19}{2}\)%
Principal P = ₹ 6000
Duration T = 2\(\frac{1}{2}\) = \(\frac{5}{2}\) years
∴ Simple Interest, I = \(\frac{\text { PRT }}{100}\)
= 6000 × \(\frac{19}{2} \times \frac{5}{2} \times \frac{1}{100}\)
= 15 × 19 × 5
= 1425
∴ Simple interest would be ₹ 1425.

Question 3.
What would be the simple interest on ₹ 8,400 in 9 months at the rate of 8.25 percent per annum?
Solution:
Principal P = ₹ 8400
Rate of interest R = 8.25%
Duration T = 9 months = \(\frac{3}{4}\) years

∴ Simple interest would be ₹ 519.75.

Question 4.
What would be the compound interest on ₹ 4200 for 18 months at 10% per annum compounded half yearly?
Solution:
Principal P = ₹ 4200
Rate of interest R = 10%
Duration T = 18 months = 1.5 years
compounding is done half yearly

= \(\frac{4200 \times 9261}{2000}\)
= 4862.025
I = A – P
= 4862.025 – 4200
= 662.025
∴ Compound interest would be ₹ 662.025.

Question 5.
Find compound interest on ₹ 10,000 for 2 years at 8% per annum compounded half yearly.
Solution:
Principal P = ₹ 10,000
Rate of interest R = 8% p.a. compounded half yearly
Duration T = 2 years

I = A – P
= 11648.58 – 10000
= 1698.58
∴ Compound interest is ₹ 1698.58.

Question 6.
In how many years ₹ 1,00,000 will become ₹ 1,33,100 at compound interest rate of 10% per annum?
Solution:
Principal P = ₹ 1,00,000
Amount A = ₹ 1,33,100
Rate of interest R = 10% p.a.

∴ ₹ 1,00,000 will become ₹ 1,33,100 after 3 years.

Question 7.
A certain sum of money becomes three times of itself in 20 years at simple interest. In how many does it become double of itself at the same rate of simple interest?
Solution:
Given that, sum of money triples itself in 20 years
∴ P + I = 3P
∴ I = 2P
and T = 20 years
Now simple interest I = \(\frac{\text { PRT }}{100}\)
∴ 2P = \(\frac{\mathrm{P} \times \mathrm{R} \times 20}{100}\)
∴ R = 10
∴ Rate of interest = 10% per annum
The time period is to be calculated for the condition that the sum doubles itself i.e. for the condition
P + I = 2P
i.e. I = P
\(\frac{\mathrm{P} \times \mathrm{R} \times \mathrm{T}}{100}\) = P
∴ \(\frac{10 \times T}{100}\) = 1
∴ T = 10
∴ The sum will become double of itself in 10 years.

Question 8.
A person borrows 10,000 for 2 year at 4% p.a. simple interest he immediately lends it to another person at 6.5% p.a. for 2 years. Find his total gain in the transaction.
Solution:
Person borrows money at 4% per annum and lends it at 6.5% per annum.
∴ His gain is (6.5 – 4) = 2.5% on ₹ 10000 for 2 years
i.e. gain = \(\frac{1000 \times 2.5 \times 2}{100}\)
= 100 × 5
= ₹ 500
∴ The person will gain ₹ 500 in this transaction.

Question 9.
A man deposits X 200 at the end of each year in recurring account at 5% compound interest. How much will it become at the end of 3 years?
Solution:
At end of 1st year, 2nd year and 3rd year ₹ 200 were deposited.
Rate of interest R = 5% p.a.
At end of 3 years, amount
= 200 + \(200\left[1+\frac{5}{100}\right]+200\left[1+\frac{5}{100}\right]^{2}\)
= 200 [1 + 1.05 + (1.05)2]
= 200 [2.05 + 1.1025]
= 200 [3.1525]
= 630.5
At end of 3 years, the account will have a balance of ₹ 630.5.

Question 10.
A man gets a simple interest of ₹ 2,000 on a certain principal at the rate of 5% p.a. in 4 years. What compound interest will the man get on twice the principal in 2 years at the same rate.
Solution:
Let Principal amount = P
Simple Interest I = ₹ 2000
Rate of interest R = 5% p.a.
Time duration T = 4 years
I = \(\frac{\text { PRT }}{100}\)
∴ 2000 = \(\frac{\mathrm{P} \times 5 \times 4}{100}\)
∴ P = 10000
Twice the principal was invested for compound interest with the same rate of interest for 2 years.
Here, P = 2 × 10,000 = ₹ 20,000
∴ Amount received,

I = A – P = 22050 – 20000 = 2050
The man will receive ₹ 2050 as compound interest.

Question 11.
The difference between simple interest and compound interest on a certain sum of money is ₹ 32 at 8% per annum for 2 years. Find the amount.
Solution:
Compound Interest = A – P = \(\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{\mathrm{T}}-\mathrm{P}\)
Simple interest = \(\frac{\text { PRT }}{100}\)
Given R = 8%, T = 2 years and
compound interest – simple interest = ₹ 32

∴ The man will receive a compound interest of ₹ 5000.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Partition Values Class 11 Commerce Maths 2 Chapter 1 Exercise 1.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.3 Questions and Answers.

Std 11 Maths 2 Exercise 1.3 Solutions Commerce Maths

Question 1.
The following table gives the frequency distribution of marks of 100 students in an examination.

Determine D₆, Q₁, and P₈₅ graphically.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (20, 9), (25, 21), (30, 44), (35, 75), (40, 85), (45, 93), (50, 100).

Here, N = 100
For D₆, \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 100}{10}\) = 60
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₈₅, \(\frac{85 \mathrm{~N}}{100}=\frac{85 \times 100}{100}\) = 85
∴ We take the points having Y co-ordinates 60, 25 and 85 on Y-axis.
From these points, we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X co-ordinates of these points give the values of D₆, Q₁ and P₈₅.
∴ D₆ = 32.5, Q₁ = 26, P₈₅ = 40

Question 2.
The following table gives the distribution of daily wages of 500 families in a certain city.

Draw a ‘less than’ ogive for the above data. Determine the median income and obtain the limits of income of central 50% of the families.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (100, 50), (200, 200), (300, 380), (400, 430), (500, 470), (600, 490) and (700, 500).

Here, N = 500
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
For Q₂, \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 500}{4}\) = 375
∴ We take the points having Y co-ordinates 125, 250 and 375 on Y-axis.
From these points we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X-Co-ordinates of these points give the values of Q₁, Q₂ and Q₃.
∴ Q₁ ~ 150, Q₂ ~ 228, Q₃ ~ 297
∴ Median = 228
50% families lie between Q₁ and Q₃
∴ Limits of income of central 50% families are from ₹ 150 to ₹ 297

Question 3.
From the following distribution, determine the median graphically.

Solution:
To draw an ogive curve, we construct the less than and more than cumulative frequency table as given below:

The points to be plotted for less than ogive are (400, 50), (500, 121), (600, 310), (700, 415), (800, 475), (900, 513) and (1000, 520) and that for more than ogive are (300, 520), (400, 470), (500, 399), (600, 210), (700, 105), (800, 45), (900, 7).

From the point of intersection of two ogives, we draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.
∴ Median ~ 574

Question 4.
The following frequency distribution shows the profit (in ₹) of shops in a particular area of the city.

Find graphically
(i) the Unfits of middle 40% shops.
(ii) the number of shops having a profit of fewer than 35,000 rupees.
Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (10, 12), (20, 30), (30, 57), (40, 77), (50, 94), (60, 100).

The Middle 40% value lies in between P₃₀ and P₇₀.
N = 100
For P₃₀ = \(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
For P₇₀ = \(\frac{70 \mathrm{~N}}{100}=\frac{70 \times 100}{100}\) = 70
∴ We take the points having Y co-ordinates 30 and 70 on Y-axis. From these points we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X-Co-ordinates of these points give the values of P₃₀ and P₇₀.
∴ P₃₀ ~ 20, P₇₀ ~ 36
Limits of middle 40% shops lie between ₹ 20,000 to ₹ 36,000
To find the number of shops having a profit of less than ₹ 35,000, we take the value 35 on the X-axis.
From this point, we draw a line parallel to Y-axis, and from the point where it intersects the less than ogive we draw a perpendicular on Y-axis. It intersects the Y-axis at approximately 67.
∴ No. of shops having profit less than ₹ 35,000 is 67.

Question 5.
The following is the frequency distribution of overtime (per week) performed by various workers from a certain company. Determine the values of D₂, Q₂, and P₆₁ graphically.

Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (8, 4), (12, 12), (16, 28), (20, 46), (24, 66) and (28, 80)
Here, N = 80

For D₂, we have to consider \(\frac{2 \mathrm{~N}}{10}=\frac{2 \times 80}{10}\) = 16
For Q₂, we have to consider \(\frac{\mathrm{N}}{2}=\frac{80}{2}\) = 40
and for P₆₁, we have to consider \(\frac{61 \mathrm{~N}}{100}=\frac{61 \times 80}{100}\) = 48.8
∴ We consider the values 16, 40 and 48.8 on the Y-axis.
From these points, we draw the lines which are parallel to the X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars to X-axis.
The values at the foot of perpendiculars represent the values of D₂, Q₂, and P₆₁ respectively.
∴ D₂ ~ 13, Q₂ ~ 19, P₆₁ ~ 20.5

Question 6.
Draw ogive for the following data and hence find the values of D₁, Q₁, and P₄₀.

Solution:
N = 100
To draw the less than ogive we have to plot the points (10, 4), (20, 6), (30, 24), (40, 46), (50, 67), (60, 86), (70, 96), (80, 99), (90, 100).

For D₁, we have to consider \(\frac{\mathrm{N}}{10}=\frac{100}{10}\) = 10
For Q₁, we have to consider \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₄₀, we have to consider \(\frac{40 \mathrm{~N}}{100}=\frac{40 \times 100}{100}\) = 40
∴ We consider the values 10, 25 and 40 on the Y-axis. From these points we draw lines parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis.
The values at the foot of perpendicular represent the values of D₁, Q₁ and P₄₀ respectively.
∴ D₁ ~ 22, Q₁ ~ 30.5, P₄₀ ~ 37

Question 7.
The following table shows the age distribution of heads of the families in a certain country. Determine the third, fifth, and eighth decile of the distribution graphically.

Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (35, 46), (45, 131), (55, 195), (65, 270), (75, 360), (85, 400).

N = 400
For D₃, we have to consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 400}{10}\) = 120
For D₅, we have to consider \(\frac{5 \mathrm{~N}}{10}=\frac{5 \times 400}{10}\) = 200
For D₈, we have to consider \(\frac{8 \mathrm{~N}}{10}=\frac{8 \times 400}{10}\) = 320
∴ We consider the values 120, 200 and 320 on Y-axis. From these points we draw the lines parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis.
The foot of perpendicular represent the values of D₃, D₅ and D₈.
∴ D₃ ~ 44, D₅ ~ 55.5 and D₈ ~ 70

Question 8.
The following table gives the distribution of females in an Indian village. Determine the median age graphically.

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:


Points to be plotted are (10, 175), (20, 275), (30, 343), (40, 391), (50, 416), (60, 466), (70, 489), (80, 497), (90, 499), (100, 500).

N = 500
For median we have to consider \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
∴ We consider the value 250 on Y-axis. From this point, we draw a line parallel to X-axis.
From the point it intersects the less than ogive, we draw a perpendicular to X-axis.
The foot perpendicular represents the value of the median.
∴ Median ~ 17.5

Question 9.
Draw ogive for the following distribution and hence find graphically the limits of the weight of middle 50% fishes.

Solution:
Since the given data is not continuous, we have to convert it into the continuous form by subtracting 5 from the lower limit and adding 5 to the upper limit of every class interval.
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (895, 8), (995, 24), (1095, 44), (1195, 69), (1295, 109), (1395, 115), (1495, 120).

N = 120
For Q₁ and Q₃ we have to consider
\(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30
\(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 90
For finding Q₁ and Q₃ we consider the values 30 and 90 on the Y-axis.
From these points, we draw the lines which are parallel to X-axis.
From the points where these lines intersect the less than ogive, we draw perpendicular on X-axis.
The feet of perpendiculars represent the values Q₁ and Q₂.
∴ Q₁ ~ 1025 and Q₃ ~ 1248
∴ the limits of the weight of the middle 50% of fishes lie between 1025 to 1248.

Question 10.
Find graphically the values of D₃ and P₆₅ for the data given below:

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.
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (69.5, 20), (79.5, 60), (89.5, 110), (99.5, 160), (109.5, 180), (119.5, 190), (129.5, 200).

N = 200
For D₃, \(\frac{3 \mathrm{N}}{10}=\frac{3 \times 200}{10}\) = 60
For P₆₅, \(\frac{65 \mathrm{N}}{100}=\frac{65 \times 200}{100}\) = 130
∴ We take the values 60 and 130 on the Y-axis.
From these points we draw lines parallel to X-axis and from the points where these lines intersect less than ogive, we draw perpendiculars on X-axis.
The foot of perpendiculars represents the median of the values, D₃ and P₆₅.
∴ D₃ = 79.5, P₆₅ = 93.5

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 1.1 Solutions
• 11th Commerce Maths Exercise 1.2 Solutions
• 11th Commerce Maths Exercise  1.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 1 Solutions

Correlation Class 11 Commerce Maths 2 Chapter 5 Exercise 5.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Correlation Ex 5.1 Questions and Answers.

Std 11 Maths 2 Exercise 5.1 Solutions Commerce Maths

Question 1.
Draw a scatter diagram for the data given below and interpret it.

Solution:

Since all the points lie in a band rising from left to right.
Therefore, there is a positive correlation between the values of X and Y respectively.

Question 2.
For the following data of marks of 7 students in Physics (x) and Mathematics (y), draw scatter diagram and state the type of correlation.

Solution:
We take marks in Physics on X-axis and marks in Mathematics on Y-axis and plot the points as below.

We get a band of points rising from left to right. This indicates the positive correlation between marks in Physics and marks in Mathematics.

Question 3.
Draw a scatter diagram for the data given below. Is there any correlation between Aptitude score and Grade points?

Solution:

The points are completely scattered i.e., no trend is observed.
∴ there is no correlation between Aptitude score (X) and Grade point (Y).

Question 4.
Find correlation coefficient between x andy series for the following data:
n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122
Solution:
Here, n = 15, \(\bar{x}\) = 25, \(\bar{y}\) = 18, σ_x = 3.01, σ_y = 3.03, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 122

Question 5.
The correlation coefficient between two variables x and y are 0.48. The covariance is 36 and the variance of x is 16. Find the standard deviation of y.
Solution:
Given, r = 0.48, Cov(X, Y) = 36
Since \(\sigma_{X}^{2}\) = 16
∴ σ_x = 4

∴ the standard deviation of y is 18.75.

Question 6.
In the following data, one of the values of y is missing. Arithmetic means of x and y series are 6 and 8 respectively. (√2 = 1.4142)

(i) Estimate missing observation.
(ii) Calculate correlation coefficient.
Solution:
(i) Let X = x_i, Y = y_i and missing observation be ‘a’.
Given, \(\bar{x}\) = 6, \(\bar{y}\) = 8, n = 5
∴ 8 = \(\frac{35+a}{5}\)
∴ 40 = 35 + a
∴ a = 5
(ii) We construct the following table:

Question 7.
Find correlation coefficient from the following data. [Given: √3 = 1.732]

Solution:

Question 8.
The correlation coefficient between x and y is 0.3 and their covariance is 12. The variance of x is 9, find the standard deviation of y.
Solution:
Given, r = 0.3, Cov(X, Y) = 12,
\(\sigma_{X}^{2}\) = 9
∴ \(\sigma_{\mathrm{X}}\) = 3

∴ the standard deviation of y is 13.33.

Read More

• 11th Commerce Maths Exercise 5.1 Solutions
• 11th Commerce Maths Miscellaneous Exercise 5 Solutions