Bhagya

Linear Inequations Class 11 Commerce Maths 2 Chapter 8 Exercise 8.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Linear Inequations Ex 8.3 Questions and Answers.

Std 11 Maths 2 Exercise 8.3 Solutions Commerce Maths

Find the graphical solution for the following system of linear inequations.

Question 1.
x – y ≤ 0, 2x – y ≥ -2
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 2.
2x + 3y ≥ 12, -x + y ≤ 3, x ≤ 4, y ≥ 3
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 3.
3x + 2y ≤ 1800, 2x + 7y ≤ 1400
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 4.
0 ≤ x ≤ 350, 0 ≤ y ≤ 150
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 5.
\(\frac{x}{60}+\frac{y}{90}\) ≤ 1, \(\frac{x}{120}+\frac{y}{75}\) ≤ 1, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 6.
3x + 2y ≤ 24, 3x + y ≥ 15, x ≥ 4
Solution:
To find a graphical solution, construct the table as follows:


The shaded portion represents the graphical solution.

Question 7.
2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0
Solution:
To find a graphical solution, construct the table as follows:


Shaded portion represents the graphical solution.

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

Bivariate Frequency Distribution and Chi Square Statistic Class 11 Commerce Maths 2 Chapter 4 Exercise 4.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Ex 4.1 Questions and Answers.

Std 11 Maths 2 Exercise 4.1 Solutions Commerce Maths

Question 1.
The following table gives income (X) and expenditure (Y) of 25 families:

Find
(i) Marginal frequency distributions of income and expenditure.
(ii) Conditional frequency distribution of X when Y is between 300 – 400.
(iii) Conditional frequency distribution of Y when X is between 200 – 300.
(iv) How many families have their income ₹ 300 and more and expenses ₹ 400 and less?
Solution:
The bivariate frequency distribution table for Income (X) and Expenditure (Y) is as follows:

(i) Marginal frequency distribution of income (X):

Marginal frequency distribution of expenditure ( Y):

(ii) Conditional frequency distribution of X when Y is between 300 – 400:

(iii) Conditional frequency distribution of Y when X is between 200 – 300:

(iv) The cells 300 – 400 and 400 – 500 are having income ₹ 300 and more and the cells 200 – 300 and 300 – 400 are having expenditure ₹ 400 and less.
Now, the following table indicates the number of families satisfying the above condition.

∴ There are 17 families with income ₹ 300 and more and expenditure ₹ 400 and less.

Question 2.
Two dice are thrown simultaneously 25 times. The following pairs of observations are obtained.
(2, 3) (2, 5) (5, 5) (4, 5) (6, 4) (3, 2) (5, 2) (4, 1) (2, 5) (6, 1) (3, 1) (3, 3) (4, 3) (4, 5) (2, 5) (3, 4) (2, 5) (3, 4) (2, 5) (4, 3) (5, 2) (4, 5) (4, 3) (2, 3) (4, 1)
Prepare a bivariate frequency distribution table for the above data. Also, obtain the marginal distributions.
Solution:
Let X = Observation on 1st die
Y = Observation on 2nd die
Now, the minimum value of X is 1 and the maximum value is 6.
Also, the minimum value of Y is 1 and the maximum value is 6.
A bivariate frequency distribution can be prepared by taking X as row and Y as a column.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Question 3.
Following data gives the age of husbands (X) and age of wives (Y) in years. Construct a bivariate frequency distribution table and find the marginal distributions.

Find conditional frequency distribution of age of husbands when the age of wife is 23 years.
Solution:
Given, X = Age of Husbands (in years)
Y = Age of Wives (in years)
Now, the minimum value of X is 25 and the maximum value is 29.
Also, the minimum value of Y is 19 and the maximum value is 23.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of X when Y is 23:

Question 4.
Construct a bivariate frequency distribution table of the marks obtained by students in Statistics (X) and English (Y).

Construct a bivariate frequency distribution table for the above data by taking class intervals 20 – 30, 30 – 40, …. etc. for both X and Y. Also find the marginal distributions and conditional frequency distribution of Y when X lies between 30 – 40.
Solution:
Given, X = Marks in Statistics
Y = Marks in English
A bivariate frequency table can be prepared by taking class intervals 20 – 30, 30 – 40,…, etc for both X and Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when X lies between 30 – 40:

Question 5.
Following data gives height in cms (X) and weight in kgs (Y) of 20 boys. Prepare a bivariate frequency table taking class intervals 150 – 154, 155 – 159,…etc. for X and 35 – 39, 40 – 44,…, etc for Y. Also, find
(i) Marginal frequency distributions.
(ii) Conditional frequency distribution of Y when 155 ≤ X ≤ 159.
(152,40) (160,54) (163,52) (150,35) (154,36) (160,49) (166,54) (157,38)
(159,43) (153,48) (152,41) (158,51) (155,44) (156,47) (156,43) (166,53)
(160,50) (151,39) (153,50) (158,46)
Solution:
Given X = Height in cms.
Y = Weight in kgs.
Bivariate frequency table can be prepared by taking class intervals 150 – 154, 155 – 159, …, etc for X and 35 – 39, 40 – 44,…etc for Y.
The bivariate frequency distribution table is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y:

Conditional frequency distribution of Y when 155 ≤ X ≤ 159:

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 4.1 Solutions
• 11th Commerce Maths Exercise 4.2 Solutions
• 11th Commerce Maths Miscellaneous Exercise 4 Solutions

Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.5 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.5 Questions and Answers.

Std 11 Maths 2 Exercise 6.5 Solutions Commerce Maths

Question 1.
In how many different ways can 8 friends sit around a table?
Solution:
We know that ‘n’ persons can sit around a table in (n – 1)! ways
∴ 8 friends can sit around a table in 7! ways
= 7 × 6 × 5 × 4 × 3 × 2 × 1
= 5040 ways.
∴ 8 friends can sit around a table in 5040 ways.

Question 2.
A party has 20 participants and a host. Find the number of distinct ways for the host to sit with them around a circular table. How many of these ways have two specified persons on either side of the host?
Solution:
A party has 20 participants.
All of them and the host (i.e., 21 persons) can be seated at a circular table in (21 – 1)! = 20! ways.
When two particular participants be seated on either side of the host.
Host takes chair in 1 way.
These 2 persons can sit on either side of host in 2! ways
Once host occupies his chair, it is not circular permutation any more.
Remaining 18 people occupy their chairs in 18! ways.
∴ Total number of arrangement possible if two particular participants be seated on either side of the host = 2! × 18!

Question 3.
Delegates from 24 countries participate in a round table discussion. Find the number of seating arrangements where two specified delegates are
(i) always together.
(ii) never together.
Solution:
(i) Delegates of 24 countries are to participate in a round table discussion such that two specified delegates are always together.
Let us consider these 2 delegates as one unit.
They can be arranged among themselves in 2! ways.
Also, these two delegates are to be seated with 22 other delegates (i.e. total 23) which can be done in (23 – 1)! = 22! ways.
∴ The total number of arrangements if two specified delegates are always together = 22! × 2!

(ii) When 2 specified delegates are never together then, the other 22 delegates can participate in a round table discussion in (22 – 1)! = 21! ways.
∴ There are 22 places of which any 2 places can be filled by those 2 delegates who are never together.
∴ Two specified delegates can be arranged in ²²P₂ ways.
∴ Total number of arrangements if two specified delegates are never together = ²²P₂ × 21!
= \(\frac{22 !}{(22-2) !}\) × 21!
= \(\frac{22 !}{20 !}\) × 21!
= 22 × 21 × 21!
= 21 × 22 × 21!
= 21 × 22!

Question 4.
Find the number of ways for 15 people to sit around the table so that no two arrangements have the same neighbours.
Solution:
There are 15 people to sit around a table.
∴ They can be arranged in (15 – 1)! = 14! ways.
But, they should not have the same neighbour in any two arrangements.
Around the table, arrangements (i.e. clockwise and anticlockwise) coincide.

∴ Number of arrangements possible for not to have same neighbours = \(\frac{14 !}{2}\)

Question 5.
A committee of 20 members sits around a table. Find the number of arrangements that have the president and the vice president together.
Solution:
A committee of 20 members sits around a table.
But, President and Vice-president sit together.
Let us consider President and Vice-president as one unit.
They can be arranged among themselves in 2! ways.
Now, this unit with the other 18 members of the committee is to be arranged around a table, which can be done in (19 – 1)! = 18! ways.
∴ The total number of arrangements possible if President and Vice-president sit together = 18! × 2!

Question 6.
Five men, two women, and a child sit around a table. Find the number of arrangements where the child is seated
(i) between the two women.
(ii) between two men.
Solution:
(i) 5 men, 2 women, and a child sit around a table
When a child is seated between two women
∴ The two women can be seated on either side of the child in 2! ways.
Let us consider these 3 (two women and a child) as one unit.
Also, these 3 are to be seated with 5 men, (i.e. a total of 6 units) which can be done in (6 – 1)! = 5! ways.
∴ The total number of arrangements if the child is seated between two women = 5! × 2!

(ii) Two men out of 5 men can sit on either side of the child in 5P2 ways.
Let us take two men and a child as one unit.
Now these are to be arranged with the remaining 3 men and 2 women
i.e., a total of 6 events (3 + 2 + 1) is to be arranged around a round table which can be done in (6 – 1)! = 5! ways.
∴ The total number of arrangements, if the child is seated between two men = ⁵P₂ × 5!

Question 7.
Eight men and six women sit around a table. How many sitting arrangements will have no two women together?
Solution:
8 men can be seated around a table in (8 – 1)! = 7! ways.
There are 8 gaps created by 8 men’s seats.
∴ 6 Women can be seated in 8 gaps in ⁸P₆ ways
∴ Total number of arrangements so that no two women are together = 7! × ⁸P₆

Question 8.
Find the number of seating arrangements for 3 men and 3 women to sit around a table so that exactly two women are together.
Solution:

Two women sit together and one woman sits separately.
Women sitting separately can be selected in 3 ways.
The other two women occupy two chairs in one way (as it is a circular arrangement).
They can be seated on those two chairs in 2 ways. Suppose two chairs are chairs 1 and 2 shown in the figure.
Then the third woman has only two options viz chairs 4 or 5.
∴ The third woman can be seated in 2 ways. 3 men are seated in 3! ways
∴ Required number = 3 × 2 × 2 × 3!
= 12 × 6
= 72

Question 9.
Four objects in a set of ten objects are alike. Find the number of ways of arranging them in a circular order.
Solution:
Ten things can be arranged in a circular order of which 4 are alike in \(\frac{9 !}{4 !}\) ways.
∴ Required total number of arrangements = \(\frac{9 !}{4 !}\)

Question 10.
Fifteen persons sit around a table. Find the number of arrangements that have two specified persons not sitting side by side.
Solution:
Since 2 particular persons can’t be sitting side by side.
The other 13 persons can be arranged around the table in (13 – 1)! = 12!
13 people around a table create 13 gaps in which 2 people are to be seated
Number of arrangements of 2 people = ¹³P₂
∴ The total number of arrangements in which two specified persons not sitting side by side = 12! × ¹³P₂
= 12! × 13 × 12
= 13 × 12! × 12
= 12 × 13!

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

Probability Class 11 Commerce Maths 2 Chapter 7 Exercise 7.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.1 Questions and Answers.

Std 11 Maths 2 Exercise 7.1 Solutions Commerce Maths

Question 1.
State the sample space and n(S) for the following random experiments.
(i) A coin is tossed twice. If a second throw results in a tail, a die is thrown.
(ii) A coin is tossed twice. If a second throw results in a head, a die is thrown, otherwise, a coin is tossed.
Solution:
(i) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
A coin is tossed twice and if the second throw results in a tail, a die is thrown.
Out of the above 4 possibilities, on second throw tail occurs in two cases only i.e., HT, TT.
∴ S = {HH, TH, HT1, HT2, HT3, HT4, HT5, HT6, TT1, TT2, TT3, TT4, TT5, TT6}
∴ n(S) = 14

(ii) When a coin is tossed twice, the outcomes are HH, HT, TH, TT.
Let A be the event that the second throw results in the head when a coin is tossed twice followed by a die is thrown.
∴ A = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6}
The remaining outcomes i.e., HT, TT are followed by the tossing of a coin.
Let us consider this as event B.
∴ B = {HTT, HTH, TTT, TTH}
The sample space S of the experiment is A ∪ B.
∴ S = {HH1, HH2, HH3, HH4, HH5, HH6, TH1, TH2, TH3, TH4, TH5, TH6, HTT, HTH, TTT, TTH}
∴ n(S) = 16

Question 2.
In a bag, there are three balls; one black, one red, and one green. Two balls are drawn one after another with replacement. State sample space and n(S).
Solution:
The bag contains 3 balls out of which one is black (B), one is red (R) and the other one is green (G).
Two balls are drawn one after the other, with replacement, from the bag.
∴ the sample space S is given by
S = {BB, BR, BG, RB, RR, RG, GB, GR, GG}
∴ n(S) = 9

Question 3.
A coin and die are tossed. State sample space of following events.
(i) A: Getting a head and an even number.
(ii) B: Getting a prime number.
(iii) C: Getting a tail and perfect square.
Solution:
When a coin and a die are tossed the sample space S is given by
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
(i) A: getting a head and an even number
∴ A = {H2, H4, H6}

(ii) B: getting a prime number
∴ B = {H2, H3, H5, T2, T3, T5}

(iii) C: getting a tail and a perfect square.
∴ C = {T1, T4}

Question 4.
Find the total number of distinct possible outcomes n(S) for each of the following random experiments.
(i) From a box containing 25 lottery tickets and 3 tickets are drawn at random.
(ii) From a group of 4 boys and 3 girls, any two students are selected at random.
(iii) 5 balls are randomly placed into five cells, such that each cell will be occupied.
(iv) 6 students are arranged in a row for photographs.
Solution:
(i) Let S be the event that 3 tickets are drawn at random from 25 tickets
∴ 3 tickets can be selected in ²⁵C₃ ways
∴ n(S) = ²⁵C₃
= \(\frac{25 \times 24 \times 23}{3 \times 2 \times 1}\)
= 2300

(ii) There are 4 boys and 3 girls i.e., 7 students.
2 students can be selected from these 7 students in ⁷C₂ ways.
∴ n(S) = ⁷C₂
= \(\frac{7 \times 6}{2 \times 1}\)
= 21

(iii) 5 balls have to be placed in 5 cells in such a way that each cell is occupied.
∴ The first ball can be placed into one of the 5 cells in 5 ways, the second ball placed into one of the remaining 4 cells in 4 ways.
Similarly, the third, fourth, and fifth balls can be placed in 3 ways, 2 ways, and 1 way, respectively.
∴ a total number of ways of filling 5 cells such that each cell is occupied = 5!
= 5 × 4 × 3 × 2 × 1
= 120
∴ n(S) = 120

(iv) Six students can be arranged in a row for a photograph in ⁶P₆ = 6! ways.
∴ n(S) = 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

Question 5.
Two dice are thrown. Write favourable outcomes for the following events.
(i) P: The sum of the numbers on two dice is divisible by 3 or 4.
(ii) Q: sum of the numbers on two dice is 7.
(iii) R: sum of the numbers on two dice is a prime number.
Also, check whether
(a) events P and Q are mutually exclusive and exhaustive.
(b) events Q and R are mutually exclusive and exhaustive.
Solution:
When two dice are thrown, all possible outcomes are
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(i) P: sum of the numbers on two dice is divisible by 3 or 4.
∴ P = {(1, 2), (1, 3), (1, 5), (2, 1), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (3, 6), (4, 2), (4, 4), (4, 5), (5, 1), (5, 3), (5, 4), (6, 2), (6, 3), (6, 6)}

(ii) Q: sum of the numbers on two dice is 7.
∴ Q = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}

(iii) R: sum of the numbers on two dice is a prime number.
∴ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)}
(a) P and Q are mutually exclusive events as P ∩ Q = φ and
P ∪ Q = {(1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2), (2, 4), (2, 5), (2, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 6)} ≠ S
∴ P and Q are not exhaustive events as P ∪ Q ≠ S.

(b) Q ∩ R = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
∴ Q ∩ R ≠ φ
Also, Q ∪ R = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)} ≠ S
∴ Q and R are neither mutually exclusive nor exhaustive events.

Question 6.
A card is drawn at random from an ordinary pack of 52 playing cards. State the number of elements in the sample space, if
consideration of suits
(i) is not taken into account.
(ii) is taken into account.
Solution:
(i) If consideration of suits is not taken into account, then one card can be drawn from the pack of 52 playing cards in ⁵²C₁ = 52 ways.
∴ n(S) = 52

(ii) If consideration of suits is taken into account, then one card can be drawn from each suit in ¹³C₁ × ⁴C₁
= 13 × 4
= 52 ways.
∴ n(S) = 52

Question 7.
Box-I contains 3 red (R11, R12, R13) and 2 blue (B11, B12) marbles while Box-II contains 2 red (R21, R22) and 4 blue (B21, B22, B23, B24) marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box-I; if it turns up tails, a marble is chosen from Box-II. Describe the sample space.
Solution:
Box I contains 3 red and 2 blue marbles i.e., (R11, R12, R13, B11, B12)
Box II contains 2 red and 4 blue marbles i.e., (R21, R22, B21, B22, B23, B24)
It is given that a fair coin is tossed and if a head comes then marble is chosen from box I otherwise it is chosen from box II
∴ the sample space is
S = {(H, R11), (H, R12), (H, R13), (H, B11), (H, B12), (T, R21), (T, R22), (T, B21), (T, B22), (T, B23), (T, B24)}

Question 8.
Consider an experiment of drawing two cards at random from a bag containing 4 cards marked 5, 6, 7, and 8. Find the sample space, if cards are drawn
(i) with replacement
(ii) without replacement.
Solution:
The bag contains 4 cards marked 5, 6, 7, and 8.
Two cards are to be drawn from this bag.
(i) If the two cards are drawn with replacement, then the sample space is
S = {(5, 5), (5, 6), (5, 7), (5, 8), (6, 5), (6, 6), (6, 7), (6,8), (7, 5), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), (8, 8)}

(ii) If the two cards are drawn without replacement, then the sample space is
S = {(5, 6), (5, 7), (5, 8), (6, 5), (6, 7), (6, 8), (7, 5), (7, 6), (7, 8), (8, 5), (8, 6), (8, 7)}

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 7.1 Solutions
• 11th Commerce Maths Exercise 7.2 Solutions
• 11th Commerce Maths Exercise 7.3 Solutions
• 11th Commerce Maths Exercise7.4 Solutions
• 11th Commerce Maths Miscellaneous Exercise 7 Solutions

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

Read More

• 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.

Read More

• 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

Read More

• 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