Class 11

Measures of Dispersion Class 11 Commerce Maths 2 Chapter 2 Exercise 2.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.1 Questions and Answers.

Std 11 Maths 2 Exercise 2.1 Solutions Commerce Maths

Question 1.
Find range of the following data:
575, 609, 335, 280, 729, 544, 852, 427, 967, 250
Solution:
Here, largest value (L) = 967, smallest value (S) = 250
∴ Range = L – S
= 967 – 250
= 717

Question 2.
The following data gives number of typing mistakes done by Radha during a week. Find the range of the data.

Solution:
Here, largest value (L) = 21, smallest value (S) = 10
∴ Range = L – S
= 21 – 10
= 11

Question 3.
Find range for the following data:

Solution:
Here, upper limit of the highest class (L) = 72, lower limit of the lowest class (S) = 62
∴ Range = L – S
= 72 – 62
= 10

Question 4.
Find the Q. D. for the following data.
3, 16, 8, 15, 19, 11, 5, 17, 9, 5, 3.
Solution:
The given data can be arranged in ascending order as follows:
3, 3, 5, 5, 8, 9, 11, 15, 16, 17, 19
Here, n = 11
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{11+1}{4}\right)^{\text {th }}\) observation
= value of 3rd observation
∴ Q₁ = 5
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{11+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 3)th observation
= value of 9th observation
= 16
∴ Q.D.= \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)
= \(\frac{16-5}{2}\)
= \(\frac{11}{2}\)
= 5.5

Question 5.
Given below are the prices of shares of a company for the last 10 days. Find Q.D.:
172, 164, 188, 214, 190, 237, 200, 195, 208, 230.
Solution:
The given data can be arranged in ascending order as follows:
164, 172, 188, 190, 195, 200, 208, 214, 230, 237
Here, n = 10
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= value of (2.75)th observation
= value of 2nd observation + 0.75(value of 3rd observation – value of 2nd observation)
= 172 + 0.75(188 – 172)
= 172 + 0.75(16)
= 172 + 12
= 184
∴ Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 2.75)th observation
= value of (8.25)th observation
= value of 8th observation + 0.25(value of 9th observation – value of 8th observation)
= 214 + 0.25(230 – 214)
= 214 + 0.25(16)
= 214 + 4
= 218
∴ Q.D. = \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)
= \(\frac{218-184}{2}\)
= \(\frac{34}{2}\)
= 17

Question 6.
Calculate Q.D. for the following data.

Solution:
Since the given data is arranged in ascending order, we construct less than cumulative frequency table as follows:

Here, n = 30
Q₁ = value of \(\left(\frac{\mathrm{n}+1}{4}\right)^{\mathrm{th}}\) observation
= value of \(\left(\frac{30+1}{4}\right)^{\text {th }}\) observation
= value of (7.75)th observation
Cumulative frequency which is just greater than (or equal to) 7.75 is 11.
∴ Q₁ = 25
Q₃ = value of \(\left[3\left(\frac{\mathrm{n}+1}{4}\right)\right]^{\mathrm{th}}\) observation
= value of \(\left[3\left(\frac{30+1}{4}\right)\right]^{\text {th }}\) observation
= value of (3 × 7.75)th observation
= value of (23.25)th observation
Cumulative frequency which is just greater than (or equal to) 23.25 is 27.
∴ Q₃ = 29
∴ Q.D. = \(\frac{Q_{3}-Q_{1}}{2}\)
= \(\frac{29-25}{2}\)
∴ Q.D. = 2

Question 7.
Following data gives the age distribution of 240 employees of a firm. Calculate Q.D. of the distribution.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 240
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{N}{4}=\frac{240}{4}\) = 60
Cumulative frequency which is just greater than (or equal to) 60 is 70.
∴ Q₁ lies in the class 25 – 30.
∴ L = 25, c.f. = 30, f = 40, h = 5

Cumulative frequency which is just greater than (or equal to) 180 is 180.
∴ Q₃ lies in the class 35-40.
∴ L = 35, c.f. = 130, f = 50, h = 5

Question 8.
Following data gives the weight of boxes. Calculate Q.D. for the data.

Solution:

Here, N = 60
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{60}{4}\) = 15
Cumulative frequency which is just greater than (or equal to) 15 is 26.
∴ Q₁ lies in the class 14 – 16.
∴ L = 14, c.f. = 10, f = 16, h = 2

Cumulative frequency which is just greater than (or equal to) 45 is 58.
∴ Q₃ lies in the class 18 – 20.
∴ L = 18, c.f. = 40, f = 18, h = 2

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 2.1 Solutions
• 11th Commerce Maths Exercise 2.2 Solutions
• 11th Commerce Maths Exercise 2.3 Solutions
• 11th Commerce Maths Miscellaneous Exercise 2 Solutions

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

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

Std 11 Maths 2 Miscellaneous Exercise 7 Solutions Commerce Maths

Question 1.
From a group of 2 men (M₁, M₂) and three women (W₁, W₂, W₃), two persons are selected. Describe the sample space of the experiment. If E is the event in which one man and one woman are selected, then which are the cases favourable to E?
Solution:
Let S be the sample space of the given event.
∴ S = {(M₁, M₂), (M₁, W₁), (M₁, W₂), (M₁, W₃), (M₂, W₁), (M₂, W₂), (M₂, W₃), (W₁, W₂) (W₁, W₃), (W₂, W₃)}
Let E be the event that one man and one woman are selected.
∴ E = {(M₁, W₁), (M₁, W₂), (M₁, W₃), (M₂, W₁), (M₂, W₂), (M₂, W₃)}
Here, the order is not important in which 2 persons are selected e.g. (M₁, M₂) is the same as (M₂, M₁)

Question 2.
Three groups of children contain respectively 3 girls and 1 boy, 2 girls and 2 boys and 1 girl and 3 boys. One child is selected at random from each group. What is the chance that the three selected consist of 1 girl and 2 boys?
Solution:

Let G₁, G₂, G₃ denote events for selecting a girl,
and B₁, B₂, B₃ denote events for selecting a boy from 1st, 2nd and 3rd groups respectively.
Then P(G₁) = \(\frac{3}{4}\), P(G₂) = \(\frac{2}{4}\), P(G₃) = \(\frac{1}{4}\)
P(B₁) = \(\frac{1}{4}\), P(B₂) = \(\frac{2}{4}\), P(B₃) = \(\frac{3}{4}\)
Where G₁, G₂, G₃, B₁, B₂ and B₃ are mutually exclusive events.
Let E be the event that 1 girl and 2 boys are selected
∴ E = (G₁ ∩ B₂ ∩ B₃) ∪ (B₁ ∩ G₂ ∩ B₃) ∪ (B₁ ∩ B₂ ∩ G₃)
∴ P(E) = P(G₁ ∩ B₂ ∩ B₃) + P(B₁ ∩ G₂ ∩ B₃) + P(B₁ ∩ B₂ ∩ G₃)

Question 3.
A room has 3 sockets for lamps. From a collection of 10 light bulbs, 6 are defective. A person selects 3 at random and puts them in every socket. What is the probability that the room, will be lit?
Solution:
Total number of bulbs = 10
Number of defective bulbs = 6
∴ Number of non-defective bulbs = 4
3 bulbs can be selected out of 10 light bulbs in ¹⁰C₃ ways.
∴ n(S) = ¹⁰C₃
Let A be the event that room is lit.
∴ A’ is the event that the room is not lit.
For A’ the bulbs should be selected from the 6 defective bulbs.
This can be done in ⁶C₃ ways.
∴ n(A’) = ⁶C₃
∴ P(A’) = \(\frac{\mathrm{n}\left(\mathrm{A}^{\prime}\right)}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{6} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}\)
∴ P(Room is lit) = 1 – P(Room is not lit)

Question 4.
There are 2 red and 3 black balls in a bag. 3 balls are taken out at random from the bag. Find the probability of getting 2 red and 1 black ball or 1 red and 2 black balls.
Solution:
There are 2 + 3 = 5 balls in the bag and 3 balls can be drawn out of these in
⁵C₃ = \(\frac{5 \times 4 \times 3}{1 \times 2 \times 3}\) = 10 ways.
∴ n(S) = 10
Let A be the event that 2 balls are red and 1 ball is black
2 red balls can be drawn out of 2 red balls in ²C₂ = 1 way
and 1 black ball can be drawn out of 3 black balls in ³C₁ = 3 ways.
∴ n(A) = ²C₂ × ³C₁ = 1 × 3 = 3
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{3}{10}\)
Let B be the event that 1 ball is red and 2 balls are black
1 red ball out of 2 red balls can be drawn in ²C₁ = 2 ways
and 2 black balls out of 3 black balls can be drawn in ³C₂ = \(\frac{3 \times 2}{1 \times 2}\) = 3 ways.
∴ n(B) = ²C₁ × ³C₂ = 2 × 3 = 6
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{10}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B) = P(A) + P(B)
= \(\frac{3}{10}+\frac{6}{10}\)
= \(\frac{9}{10}\)

Question 5.
A box contains 25 tickets numbered 1 to 25. Two tickets are drawn at random. What is the probability that the product of the numbers is even?
Solution:
Two tickets can be drawn out of 25 tickets in ²⁵C₂ = \(\frac{25 \times 24}{1 \times 2}\) = 300 ways.
∴ n(S) = 300
Let A be the event that product of two numbers is even.
This is possible if both numbers are even, or one number is even and other is odd.
As there are 13 odd numbers and 12 even numbers from 1 to 25.
∴ n(A) = ¹²C₂ + ¹²C₁ × ¹³C₁
= \(\frac{12 \times 11}{1 \times 2}\) + 12 × 13
= 66 + 156
= 222
∴ Required probability = P(A)
= \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}\)
= \(\frac{222}{300}\)
= \(\frac{37}{50}\)

Question 6.
A, B and C are mutually exclusive and exhaustive events associated with the random experiment. Find P(A), given that
P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B)
Solution:
P(B) = \(\frac{3}{2}\) P(A) and P(C) = \(\frac{1}{2}\) P(B)
Since A, B, C are mutually exclusive and exhaustive events,

Question 7.
An urn contains four tickets marked with numbers 112, 121, 122, 222, and one ticket is drawn at random. Let A_i (i = 1, 2, 3) be the event that ith digit of the number of the ticket drawn is 1. Discuss the independence of the events A₁, A₂, and A₃.
Solution:
One ticket can be drawn out of 4 tickets in ⁴C₁ = 4 ways.
∴ n(S) = 4
According to the given information,
Let A₁ be the event that 1st digit of the number of tickets is 1
A₂ be the event that the 2nd digit of the number of tickets is 1
A₃ be the event that the 3rd digit of the number of tickets is 1
∴ A₁ = {112, 121, 122}, A₂ = {112}, A₃ = {121}

∴ A₁, A₂, A₃ are not pairwise independent
For mutual independence of events A₁, A₂, A₃
We require to have
P(A₁ ∩ A₂ ∩ A₃) = P(A₁) P(A₂) P(A₃)
and P(A₁) P(A₂) = P(A₁ ∩ A₂),
P(A₂) P(A₃) = P(A₂ ∩ A₃),
P(A₁) P(A₃) = P(A₁ ∩ A₃)
∴ From (iii),
A₁, A₂, A₃ are not mutually independent.

Question 8.
The odds against a certain event are 5 : 2 and the odds in favour of another independent event are 6 : 5. Find the chance that at least one of the events will happen.
Solution:
Let A and B be two independent events.
Odds against A are 5 : 2
∴ the probability of occurrence of event A is given by
P(A) = \(\frac{2}{5+2}=\frac{2}{7}\)
Odds in favour of B are 6 : 5
∴ the probability of occurrence of event B is given by
P(B) = \(\frac{6}{6+5}=\frac{6}{11}\)
∴ P(at least one event will happen) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= P(A) + P(B) – P(A) P(B) ……[∵ A and B are independent events]

Question 9.
The odds against a husband who is 55 years old living till he is 75 is 8 : 5 and it is 4 : 3 against his wife who is now 48, living till she is 68. Find the probability that
(i) the couple will be alive 20 years hence
(ii) at least one of them will be alive 20 years hence.
Solution:
Let A be the event that husband would be alive after 20 years.
Odds against A are 8 : 5
∴ the probability of occurrence of event A is given by
P(A) = \(\frac{5}{8+5}=\frac{5}{13}\)
∴ P(A’) = 1 – P(A)
= 1 – \(\frac{5}{13}\)
= \(\frac{8}{13}\)
Let B be the event that wife would be alive after 20 years.
Odds against B are 4 : 3
∴ the probability of occurrence of event B is given by
P(B) = \(\frac{3}{4+3}=\frac{3}{7}\)
∴ P(B’) = 1 – P(B)
= 1 – \(\frac{3}{7}\)
= \(\frac{4}{7}\)
Since A and B are independent events
∴ A’ and B’ are also independent events
(i) Let X be the event that both will be alive after 20 years.
∴ P(X) = (A ∩ B)
∴ P(X) = P(A) . P(B)
= \(\frac{5}{13} \times \frac{3}{7}\)
= \(\frac{15}{91}\)

(ii) Let Y be the event that at least one will be alive after 20 years.
∴ P(Y) = P(at least one would be alive)
= 1 – P(both would not be alive)
= 1 – P(A’ ∩ B’)
= 1 – P(A’). P(B’)
= 1 – \(\frac{8}{13} \times \frac{4}{7}\)
= 1 – \(\frac{32}{91}\)
= \(\frac{59}{91}\)

Question 10.
Two throws are made, the first with 3 dice and the second with 2 dice. The faces of each die are marked with the number 1 to 6. What is the probability that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8?
Solution:
When 3 dice are thrown, then the sample space S₁ has 6 × 6 × 6 = 216 sample points.
∴ n(S₁) = 216
Let A be the event that the sum of the numbers is not less than 15.
∴ A = {(3, 6, 6), (4, 5, 6), (4, 6, 5), (4, 6, 6), (5, 4, 6), (5, 5, 5), (5, 5, 6), (5, 6, 4), (5, 6, 5), (5, 6, 6), (6, 3, 6), (6, 4, 5), (6, 4, 6), (6, 5, 4), (6, 5, 5), (6, 5, 6), (6, 6, 3), (6, 6, 4), (6, 6, 5), (6, 6, 6)}
∴ n(A) = 20
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}\left(\mathrm{S}_{1}\right)}=\frac{20}{216}=\frac{5}{54}\)
When 2 dice are thrown, the sample space S₂ has 6 × 6 = 36 sample points.
∴ n(S₂) = 36
Let B be the event that sum of numbers is not less than 8.
∴ B = {(2, 6), (3, 5), (3,6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
∴ n(B) = 15
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}\left(\mathrm{S}_{2}\right)}=\frac{15}{36}=\frac{5}{12}\)
A ∩ B = Event that the total in the first throw is not less than 15 and at the same time the total in the second throw is not less than 8.
∴ A and B are independent events
∴ P(A ∩ B) = P(A) . P(B)
= \(\frac{5}{54} \times \frac{5}{12}\)
= \(\frac{25}{648}\)

Question 11.
Two-thirds of the students in a class are boys and the rest are girls. It is known that the probability of a girl getting first class is 0.25 and that of a boy getting is 0.28. Find the probability that a student chosen at random will get first class.
Solution:
Let A be the event that student chosen is a boy
B be the event that student chosen is a girl
C be the event that student gets first class
∴ P(A) = \(\frac{2}{3}\), P(B) = \(\frac{1}{3}\)
Probability of student getting first class, given that student is boy
Probability of student getting first class given that student is a girl, is
P(C/A) = 0.28 = \(\frac{28}{100}\)
and P(C/B) = 0.25 = \(\frac{25}{100}\)
∴ Required probability = P((A ∩ C) ∪ (B ∩ C))
Since A ∩ C and B ∩ C are mutually exclusive events
∴ Required probability = P(A ∩ C) + P(B ∩ C)
= P(A) . P(C/A) + P(B) . P(C/B)

Question 12.
A number of two digits is formed using the digits 1, 2, 3,……, 9. What is the probability that the number so chosen is even and less than 60?
Solution:
The number of two digits can be formed from the given 9 digits in 9 × 9 = 81 different ways.
∴ n(S) = 81
Let A be the event that the number is even and less than 60.
Since the number is even, the unit place of two digits can be filled in ⁴P₁ = 4 different ways by any one of the digits 2, 4, 6, 8.
Also the number is less than 60, so tenth place can be filled in ⁵P₁ = 5 different ways by any one of the digits 1, 2, 3, 4, 5.
∴ n(A) = 4 × 5 = 20
∴ Required probability = P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{20}{81}\)

Question 13.
A bag contains 8 red balls and 5 white balls. Two successive draws of 3 balls each are made without replacement. Find the probability that the first drawing will give 3 white balls and the second drawing will give 3 red balls.
Solution:
Total number of balls = 8 + 5 = 13.
3 balls can be drawn out of 13 balls in ¹³C₃ ways.
∴ n(S) = ¹³C₃
Let A be the event that all 3 balls drawn are white.
3 white balls can be drawn out of 5 white balls in ⁵C₃ ways.
∴ n(A) = ⁵C₃
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{{ }^{5} C_{3}}{{ }^{13} C_{3}}=\frac{5 \times 4 \times 3}{13 \times 12 \times 11}=\frac{5}{143}\)
After drawing 3 white balls which are not replaced in the bag, there are 10 balls left in the bag out of which 8 are red balls.
Let B be the event that the second draw of 3 balls are red.
∴ Probability of drawing 3 red balls, given that 3 white balls have been already drawn, is given by
P(B/A) = \(\frac{{ }^{8} \mathrm{C}_{3}}{{ }^{10} \mathrm{C}_{3}}=\frac{8 \times 7 \times 6}{10 \times 9 \times 8}=\frac{7}{15}\)
∴ Required probability = P(A ∩ B)
= P(A) . P(B/A)
= \(\frac{5}{143} \times \frac{7}{15}\)
= \(\frac{7}{429}\)

Question 14.
The odds against student X solving a business statistics problem are 8 : 6 and the odds in favour of student Y solving the same problem are 14 : 16
(i) What is the chance that the problem will be solved, if they try independently?
(ii) What is the probability that neither solves the problem?
Solution:
(i) Let A be the event that X solves the problem B be the event that Y solves the problem.
Since the odds against student X solving the problem are 8 : 6
∴ Probability of occurrence of event A is given by
P(A) = \(\frac{6}{8+6}=\frac{6}{14}\)
and P(A’) = 1 – P(A)
= 1 – \(\frac{6}{14}\)
= \(\frac{8}{14}\)
Also, the odds in favour of student Y solving the problem are 14 : 16
∴ Probability of occurrence of event B is given by
P(B) = \(\frac{14}{14+16}=\frac{14}{30}\) and
P(B’) = 1 – P(B)
= 1 – \(\frac{14}{30}\)
= \(\frac{16}{30}\)
Now A and B are independent events.
∴ A’ and B’ are independent events.
∴ A’ ∩ B’ = Event that neither solves the problem
= P(A’ ∩ B’)
= P(A’) . P(B’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)
A ∪ B = the event that the problem is solved
∴ P(problem will be solved) = P(A ∪ B)
= 1 – P(A ∪ B)’
= 1 – P(A’ ∩ B’)
= 1 – \(\frac{32}{105}\)
= 1 – \(\frac{73}{105}\)

(ii) P (neither solves the problem) = P(A’ ∩ B’)
= P(A’) P(B’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)

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

Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.2 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.2 Questions and Answers.

Std 11 Maths 2 Exercise 6.2 Solutions Commerce Maths

Question 1.
Evaluate:
(i) 8!
Solution:
8!
= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
= 40320

(ii) 6!
Solution:
6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

(iii) 8! – 6!
Solution:
8! – 6!
= 8 × 7 × 6! – 6!
= 6! (8 × 7 – 1)
= 6! (56 – 1)
= 6 × 5 × 4 × 3 × 2 × 1 × 55
= 39,600

(iv) (8 – 6)!
Solution:
(8 – 6)!
= 2!
= 2 × 1
= 2

Question 2.
Compute:
(i) \(\frac{12 !}{6 !}\)
Solution:
\(\frac{12 !}{6 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\)
= 12 × 11 × 10 × 9 × 8 × 7
= 665280

(ii) \(\left(\frac{12}{6}\right) !\)
Solution:
\(\left(\frac{12}{6}\right) !\)
= 2!
= 2 × 1
= 2

(iii) (3 × 2)!
Solution:
(3 × 2)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720

(iv) 3! × 2!
Solution:
3! × 2!
= 3 × 2 × 1 × 2 × 1
= 12

Question 3.
Compute:
(i) \(\frac{9 !}{3 ! 6 !}\)
Solution:

(ii) \(\frac{6 !-4 !}{4 !}\)
Solution:

(iii) \(\frac{8 !}{6 !-4 !}\)
Solution:

(iv) \(\frac{8 !}{(6-4) !}\)
Solution:

Question 4.
Write in terms of factorials
(i) 5 × 6 × 7 × 8 × 9 × 10
Solution:
5 × 6 × 7 × 8 × 9 × 10
= 10 × 9 × 8 × 7 × 6 × 5
Multiplying and dividing by 4!, we get

(ii) 3 × 6 × 9 × 12 × 15
Solution:
3 × 6 × 9 × 12 × 15
= 3 × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5)
= (3⁵) (5 × 4 × 3 × 2 × 1)
= 3⁵ (5!)

(iii) 6 × 7 × 8 × 9
Solution:
6 × 7 × 8 × 9
= 9 × 8 × 7 × 6
Multiplying and dividing by 5!, we get

(iv) 5 × 10 × 15 × 20 × 25
Solution:
5 × 10 × 15 × 20 × 25
= (5 × 1) × (5 × 2) × (5 × 3) × (5 × 4) × (5 × 5)
= (5⁵) (5 × 4 × 3 × 2 × 1)
= (5⁵) (5!)

Question 5.
Evaluate: \(\frac{n !}{r !(n-r) !}\) for
(i) n = 8, r = 6
Solution:
n = 8, r = 6

(ii) n = 12, r = 12
Solution:
n = 12, r = 12

Question 6.
Find n, if
(i) \(\frac{n}{8 !}=\frac{3}{6 !}+\frac{1}{4 !}\)
Solution:

(ii) \(\frac{n}{6 !}=\frac{4}{8 !}+\frac{3}{6 !}\)
Solution:

(iii) \(\frac{1}{n !}=\frac{1}{4 !}-\frac{4}{5 !}\)
Solution:

Question 7.
Find n, if
(i) (n + 1)! = 42 × (n – 1)!
Solution:
(n + 1)! = 42(n – 1)!
∴ (n + 1) n (n – 1)! = 42(n – 1)!
∴ n² + n = 42
∴ n(n + 1) = 6 × 7
Comparing on both sides, we get
∴ n = 6

(ii) (n + 3)! = 110 × (n + 1)!
Solution:
(n + 3)! = 110 × (n + 1)!
∴ (n + 3) (n + 2) (n + 1)! = 110 (n + 1)!
∴ (n + 3) (n + 2) = (11) (10)
Comparing on both sides, we get
n + 3 = 11
∴ n = 8

Question 8.
Find n, if:
(i) \(\frac{n !}{3 !(n-3) !}: \frac{n !}{5 !(n-5) !}=5: 3\)
Solution:

∴ 12 = (n – 3)(n – 4)
∴ (n – 3)(n – 4) = 4 × 3
Comparing on both sides, we get
n – 3 = 4
∴ n = 7

(ii) \(\frac{n !}{3 !(n-5) !}: \frac{n !}{5 !(n-7) !}=10: 3\)
Solution:

∴ (n – 5) (n – 6) = 3 × 2
Comparing on both sides, we get
n – 5 = 3
∴ n = 8

Question 9.
Find n, if:
(i) \(\frac{(17-n) !}{(14-n) !}\) = 5!
Solution:
\(\frac{(17-n) !}{(14-n) !}\) = 5!
∴ \(\frac{(17-n)(16-n)(15-n)(14-n) !}{(14-n) !}\) = 5 × 4 × 3 × 2 × 1
∴ (17 – n) (16 – n) (15 – n) = 6 × 5 × 4
Comparing on both sides, we get
17 – n = 6
∴ n = 11

(ii) \(\frac{(15-n) !}{(13-n) !}\) = 12
Solution:
\(\frac{(15-n) !}{(13-n) !}\) = 12
∴ \(\frac{(15-\mathrm{n})(14-\mathrm{n})(13-\mathrm{n}) !}{(13-\mathrm{n}) !}\) = 12
∴ (15 – n) (14 – n) = 4 × 3
Comparing on both sides, we get
15 – n = 4
∴ n = 11

Question 10.
Find n if \(\frac{(2 n) !}{7 !(2 n-7) !}: \frac{n !}{4 !(n-4) !}\) = 24 : 1
Solution:

∴ (2n – 1) (2n – 3) (2n – 5) = 9 × 7 × 5
Comparing on both sides, we get
2n – 1 = 9
∴ n = 5

Question 11.
Show that \(\frac{n !}{r !(n-r) !}+\frac{n !}{(r-1) !(n-r+1) !}=\frac{(n+1) !}{r !(n-r+1)}\)
Solution:

Question 12.
Show that \(\frac{9 !}{3 ! 6 !}+\frac{9 !}{4 ! 5 !}=\frac{10 !}{4 ! 6 !}\)
Solution:

Question 13.
Find the value of:
(i) \(\frac{8 !+5(4 !)}{4 !-12}\)
Solution:

(ii) \(\frac{5(26 !)+(27 !)}{4(27 !)-8(26 !)}\)
Solution:

Question 14.
Show that
\(\frac{(2 n) !}{n !}\) = 2ⁿ (2n – 1) (2n – 3)…5.3.1
Solution:

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

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

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

Std 11 Maths 2 Miscellaneous Exercise 4 Solutions Commerce Maths

Question 1.
Following data gives the coded price (X) and demand (Y) of a commodity.

Classify the data by taking classes 0 – 4, 5 – 9, etc. for X and 5 – 8, 9 – 12, etc. for Y.
Also find
(i) marginal frequency distribution of X and Y.
(ii) conditional frequency distribution of Y when X is less than 10.
Solution:
Given, X = coded price
Y = demand
Bivariate frequency table can be prepared by taking class intervals 0 – 4, 5 – 9,… etc for X and 5 – 8, 9 – 12,… etc for Y.
Bivariate frequency distribution is as follows.

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X < 10:

Question 2.
Following data gives the age in years and marks obtained by 30 students in an intelligence test.


Prepare a bivariate frequency distribution by taking class intervals 16 – 18, 18 – 20,…,etc. for age and 10 – 20, 20 – 30,…, etc. for marks.
Find
(i) marginal frequency distributions.
(ii) conditional frequency distribution of marks obtained when age of students is between 20 – 22.
Solution:
Let X = Age in years
Y = Marks
Bivariate frequency table can be prepared by taking class intervals 16 – 18, 18 – 20,…, etc for X and 10 – 20, 20 – 30,…, etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X is between 20 – 22:

Question 3.
Following data gives Sales (in Lakh ?) and Advertisement Expenditure (in Thousand ₹) of 20 firms.
(115, 61) (120, 60) (128, 61) (121, 63) (137, 62) (139, 62) (143, 63) (117, 65) (126, 64) (141, 65) (140, 65) (153, 64) (129, 67) (130, 66) (150, 67) (148, 66) (130, 69) (138, 68) (155, 69) (172, 68)
(i) Construct a bivariate frequency distribution table for the above data by taking classes 115 – 125, 125 – 135, ….etc. for sales and 60 – 62, 62 – 64, …etc. for advertisement expenditure.
(ii) Find marginal frequency distributions
(iii) Conditional frequency distribution of Sales when the advertisement expenditure is between 64 – 66 (Thousand ₹)
(iv) Conditional frequency distribution of advertisement expenditure when the sales are between 125 – 135 (lakh ₹)
Solution:
(i) Let X = Sales (in lakh ₹)
Y = Advertisement Expenditure (in Thousand ₹)
Bivariate frequency table can be prepared by taking class intervals 115 – 125, 125 – 135, …. etc for X and 60 – 62, 62 – 64, ….etc for Y.
Bivariate frequency distribution is as follows:

(ii) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of X when Y is between 64 – 66:

(iii) Conditional frequency distribution of Y when X is between 125 – 135:

Question 4.
Prepare a bivariate frequency distribution for the following data, taking class intervals for X as 35 – 45, 45 – 55, …. etc and for Y as 115 – 130, 130 – 145, … etc where, X denotes the age in years and Y denotes blood pressure for a group of 24 persons.
(55, 151) (36, 140) (72, 160) (38, 124) (65, 148) (46, 130) (58, 152) (50, 149) (38, 115) (42, 145) (41, 163) (47, 161) (69, 159) (60, 161) (58, 131) (57, 136) (43, 141) (52, 164) (59, 161) (44, 128) (35, 118) (62, 142) (67, 157) (70, 162)
Also find
(i) Marginal frequency distribution of X.
(ii) Conditional frequency distribution of Y when X < 45.
Solution:
Given X = Age in years
Y = Blood pressure
Bivariate frequency table can be prepared by taking class intervals 35 – 45, 45 – 55, …, etc for X and 115 – 130, 130 – 145, ….., etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

(ii) Conditional frequency distribution of Y when X < 45:

Question 5.
Thirty pairs of values of two variables X and Y are given below. Form a bivariate frequency table. Also find marginal frequency distributions of X and Y.

Solution:
Bivariate frequency table can be prepared by taking class intervals 80 – 90, 90 – 100, etc for X and 500 – 600, 600 – 700, …., etc for Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y

Question 6.
The following table shows how the samples of Mathematics and Economics scores of 25 students are distributed:

Find the value of ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{35 \times 25}{50}\) = 17.5
E₁₂ = \(\frac{35 \times 25}{50}\) = 17.5
E₂₁ = \(\frac{15 \times 25}{50}\) = 7.5
E₂₂ = \(\frac{15 \times 25}{50}\) = 7.5
Table of expected frequencies.

Question 7.
Compute ϰ² statistic from the following data:

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 60}{100}\) = 30
E₁₂ = \(\frac{50 \times 40}{100}\) = 20
E₂₁ = \(\frac{50 \times 60}{100}\) = 30
E₂₂ = \(\frac{50 \times 40}{100}\) = 20
Table of expected frequencies.

Question 8.
The attitude of 250 employees towards a proposed policy of the company is as observed in the following table. Calculate ϰ² statistic.

Solution:
Table of observed frequencies

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{150 \times 95}{250}\) = 57
E₁₂ = \(\frac{150 \times 95}{250}\) = 57
E₁₃ = \(\frac{150 \times 60}{250}\) = 36
E₂₁ = \(\frac{100 \times 95}{250}\) = 38
E₂₂ = \(\frac{100 \times 95}{250}\) = 38
E₂₃ = \(\frac{100 \times 60}{250}\) = 24
Table of observed frequencies.

Question 9.
In a certain sample of 1000 families, 450 families are consumers of tea. Out of 600 Hindu families, 286 families consume tea. Calculate ϰ² statistic.
Solution:
The given data can be arranged in the following table.

Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{600 \times 450}{1000}\) = 270
E₁₂ = \(\frac{600 \times 550}{1000}\) = 330
E₂₁ = \(\frac{400 \times 450}{1000}\) = 180
E₂₂ = \(\frac{400 \times 550}{1000}\) = 220
Table of expected frequencies.

Question 10.
A sample of boys and girls were asked to choose their favourite sport, with the following results. Find the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{200 \times 126}{300}\) = 84
E₁₂ = \(\frac{200 \times 90}{300}\) = 60
E₁₃ = \(\frac{200 \times 69}{300}\) = 46
E₁₄ = \(\frac{200 \times 15}{300}\) = 10
E₂₁ = \(\frac{100 \times 126}{300}\) = 42
E₂₂ = \(\frac{100 \times 90}{300}\) = 30
E₂₃ = \(\frac{100 \times 69}{300}\) = 23
E₂₄ = \(\frac{100 \times 15}{300}\) = 5
Table of expected frequencies.

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

Probability Class 11 Commerce Maths 2 Chapter 7 Exercise 7.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 7 Probability Ex 7.3 Questions and Answers.

Std 11 Maths 2 Exercise 7.3 Solutions Commerce Maths

Question 1.
Two dice are thrown together. What is the probability that sum of the numbers on two dice is 5 or the number on the second die is greater than or equal to the number on the first die?
Solution:
When two dice are thrown, the sample space is
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)}
∴ n(S) = 36
Let A be the event that sum of numbers on two dice is 5.
∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)}
∴ n(A) = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{36}\)
Let B be the event that number on second die is greater than or equal to number on first die.
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 5), (5, 6), (6, 6)}
∴ n(B) = 21
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{21}{36}\)
Now, A ∩ B = {(1, 4), (2, 3)}
∴ n(A ∩ B) = 2
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{2}{36}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{4}{36}+\frac{21}{36}-\frac{2}{36}\)
= \(\frac{23}{36}\)

Question 2.
A card is drawn from a pack of 52 cards. What is the probability that,
(i) card is either red or black?
(ii) card is either red or face card?
Solution:
One card can be drawn from the pack of 52 cards in ⁵²C₁ = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that a red card is drawn Red card can be drawn in ²⁶C₁ = 26 ways
∴ n(A) = 26
∴ P(A) = \(\frac{26}{52}\)
Let B be the event that a black card is drawn
∴ Black card can be drawn in ²⁶C₁ = 26 ways.
∴ n(B) = 26
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= \(\frac{26}{52}+\frac{26}{52}\)
= \(\frac{52}{52}\)
= 1

(ii) Let A be the event that a red card is drawn
∴ red card can be drawn in ²⁶C₁ = 26 ways
∴ n(A) = 26
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{26}{52}\)
Let B be the event that a face card is drawn There are 12 face cards in the pack of 52 cards
∴ 1 face card can be drawn in ¹²C₁ = 12 ways
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{52}\)
There are 6 red face cards.
∴ n(A ∩ B) = 6
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{6}{52}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{26}{52}+\frac{12}{52}-\frac{6}{52}\)
= \(\frac{32}{52}\)
= \(\frac{8}{13}\)

Question 3.
Two cards are drawn from a pack of 52 cards. What is the probability that,
(i) both the cards are of the same colour?
(ii) both the cards are either black or queens?
Solution:
Two cards can be drawn from 52 cards in ⁵²C₂ ways.
∴ n(S) = ⁵²C₂
Also, the pack of 52 cards consists of 26 red and 26 black cards.
(i) Let A be the event that both cards are red.
∴ 2 red cards can be drawn in ²⁶C₂ ways.
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Let B be the event that both cards are black.
∴ 2 black cards can be drawn in ²⁶C₂ ways
∴ n(B) = ²⁶C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
∴ P(A ∪ B) = P(A) + P(B)
= \(\frac{25}{102}+\frac{25}{102}\)
= \(\frac{25}{51}\)

(ii) Let A be the event that both cards are black.
∴ 2 black cards can be drawn in ²⁶C₂ ways.
∴ n(A) = ²⁶C₂
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{26} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{26 \times 25}{52 \times 51}=\frac{25}{102}\)
Let B be the event that both cards are queens.
There are 4 queens in a pack of 52 cards
∴ 2 queen cards can be drawn in ⁴C₂ ways.
∴ n(B) = ⁴C₂
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{2}}{{ }^{52} \mathrm{C}_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)
There are two black queen cards.
∴ n(A ∩ B) = ²C₂ = 1

Question 4.
A bag contains 50 tickets, numbered from 1 to 50. One ticket is drawn at random. What is the probability that
(i) number on the ticket is a perfect square or divisible by 4?
(ii) number on the ticket is a prime number or greater than 30?
Solution:
Out of the 50 tickets, a ticket can be drawn in ⁵⁰C₁ = 50 ways.
∴ n(S) = 50
(i) Let A be the event that the number on the ticket is a perfect square.
∴ A = {1, 4, 9, 16, 25, 36, 49}
∴ n(A) = 7
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{7}{50}\)
Let B be the event that the number on the ticket is divisible by 4.
∴ B = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}
∴ n(B) = 12
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{12}{50}\)
Now, A ∩ B = {4, 16, 36}
∴ n(A ∩ B) = 3
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{3}{50}\)
Required probability = P (A u B)
P (A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{7}{50}+\frac{12}{50}-\frac{3}{50}\)
= \(\frac{8}{25}\)

(ii) Let A be the event that the number on the ticket is a prime number.
∴ A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}
∴ n(A) = 15
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{15}{50}\)
Let B be the event that the number is greater than 30.
∴ B = {31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}
∴ n(B) = 20
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{20}{50}\)
Now, A ∩ B = {31, 37, 41, 43, 47}
∴ n(A ∩ B) = 5
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{5}{50}\)
∴ Required probability = P(A ∪ B)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{15}{50}+\frac{20}{50}-\frac{5}{50}\)
= \(\frac{15+20-5}{50}\)
= \(\frac{3}{5}\)

Question 5.
A hundred students appeared for two examinations. 60 passed the first, 50 passed the second, and 30 passed in both. Find the probability that students selected at random
(i) passed at least one examination.
(ii) passed in exactly one examination.
(iii) failed in both examinations.
Solution:
Out of hundred students 1 student can be selected in ¹⁰⁰C₁ = 100 ways.
∴ n(S) = 100
Let A be the event that the student passed in the first examination.
Let B be the event that student passed in second examination.
∴ n(A) = 60, n(B) = 50 and n(A ∩ B) = 30

(i) P(student passed in at least one examination) = P(A ∪ B)
= P(A) + P(B) – P (A ∩ B)
= \(\frac{6}{10}+\frac{5}{10}-\frac{3}{10}\)
= \(\frac{4}{5}\)

(ii) P(student passed in exactly one examination) = P(A) + P(B) – 2.P(A ∩ B)
= \(\frac{6}{10}+\frac{5}{10}-2\left(\frac{3}{10}\right)\)
= \(\frac{1}{2}\)

(iii) P(student failed in both examinations) = P(A’ ∩ B’)
= P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – \(\frac{4}{5}\)
= \(\frac{1}{5}\)

Question 6.
If P(A) = \(\frac{1}{4}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = \(\frac{1}{2}\). Find the values of the following probabilities.
(i) P(A ∩ B)
(ii) P(A ∩ B’)
(iii) P(A’ ∩ B)
(iv) P(A’ ∪ B’)
(v) P(A’ ∩ B’)
Solution:
Here, P(A) = \(\frac{1}{4}\), P(B) = \(\frac{2}{5}\) and P(A ∪ B) = \(\frac{1}{2}\)
(i) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
∴ P(A ∩ B) = P(A) + P(B) – P(A ∪ B)
= \(\frac{1}{4}+\frac{2}{5}-\frac{1}{2}\)
= \(\frac{3}{20}\)

(ii) P(A’ ∩ B’) = P(A) – P(A ∩ B)
= \(\frac{1}{4}-\frac{3}{20}\)
= \(\frac{1}{10}\)

(iii) P(A’ ∩ B) = P(B) – P(A ∩ B)
= \(\frac{2}{5}-\frac{3}{20}\)
= \(\frac{1}{4}\)

(iv) P(A’ ∪ B’) = P(A ∩ B)’ …..[De Morgan’s law]
= 1 – P(A ∩ B)
= 1 – \(\frac{3}{20}\)
= \(\frac{17}{20}\)

(v) P(A’ ∩ B’) = P(A ∪ B)’ …..[De Morgan’s law]
= 1 – P(A ∪ B)
= 1 – \(\frac{1}{2}\)
= \(\frac{1}{2}\)

Question 7.
A computer software company is bidding for computer programs A and B. The probability that the company will get software A is \(\frac{3}{5}\), the probability that the company will get software B is \(\frac{1}{3}\) and the probability that company will get both A and B is \(\frac{1}{8}\). What is the probability that the company will get at least one software?
Solution:
Let A be the event that the company will get software A.
∴ P(A) = \(\frac{3}{5}\)
Let B be the event that company will get software B.
∴ P(B) = \(\frac{1}{3}\)
Also, P(A ∩ B) = \(\frac{1}{8}\)
∴ P(the company will get at least one software) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{3}{5}+\frac{1}{3}-\frac{1}{8}\)
= \(\frac{72+40-15}{120}\)
= \(\frac{97}{120}\)

Question 8.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability of it being a heart or a queen.
Solution:
One card can be drawn from the pack of 52 cards in ⁵²C₁ = 52 ways
∴ n(S) = 52
Also, the pack of 52 cards consists of 13 heart cards and 4 queen cards
Let A be the event that a card drawn is the heart.
A heart card can be drawn from 13 heart cards in ¹³C₁ ways
∴ n(A) = ¹³C₁
∴ P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{13} \mathrm{C}_{1}}{52}=\frac{13}{52}\)
Let B be the event that a card drawn is queen.
A queen card can be drawn from 4 queen cards in ⁴C₁ ways
∴ n(B) = ⁴C₁
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{4} \mathrm{C}_{1}}{52}=\frac{4}{52}\)
There is one queen card out of 4 which is also a heart card
∴ n(A ∩ B) = ¹C₁
∴ P(A ∩ B) = \(\frac{\mathrm{n}(\mathrm{A} \cap \mathrm{B})}{\mathrm{n}(\mathrm{S})}=\frac{{ }^{1} \mathrm{C}_{1}}{52}=\frac{1}{52}\)
∴ P(card is a heart or a queen) = P(A ∪ B)
= P(A) + P(B) – P(A ∩ B)
= \(\frac{13}{52}+\frac{4}{52}-\frac{1}{52}\)
= \(\frac{13+4-1}{52}\)
= \(\frac{16}{52}\)
∴ P(A ∪ B) = \(\frac{4}{13}\)

Question 9.
In a group of students, there are 3 boys and 4 girls. Four students are to be selected at random from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.
Solution:
The group consists of 3 boys and 4 girls i.e., 7 students.
4 students can be selected from this group in ⁷C₄
= \(\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}\)
= 35 ways.
∴ n(S) = 35
Let A be the event that 3 boys and 1 girl are selected.
3 boys can be selected in ³C₃ ways while a girl can be selected in ⁴C₁ ways.
∴ n(A) = ³C₃ × ⁴C₁ = 4
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{4}{35}\)
Let B be the event that 3 girls and 1 boy are selected.
3 girls can be selected in ⁴C3 ways and a boy can be selected in ³C₁ ways.
∴ n(B) = ⁴C₃ × ³C₁ = 12
∴ P(B) = \(\frac{n(B)}{n(S)}=\frac{12}{35}\)
Since A and B are mutually exclusive and exhaustive events
∴ P(A ∩ B) = 0
∴ Required probability = P(A ∪ B)
= P(A) + P(B)
= \(\frac{4}{35}+\frac{12}{35}\)
= \(\frac{16}{35}\)

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

Partition Values Class 11 Commerce Maths 2 Chapter 1 Exercise 1.1 Answers Maharashtra Board

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

Std 11 Maths 2 Exercise 1.1 Solutions Commerce Maths

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Permutations and Combinations Class 11 Commerce Maths 2 Chapter 6 Exercise 6.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.1 Questions and Answers.

Std 11 Maths 2 Exercise 6.1 Solutions Commerce Maths

Question 1.
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
Solution:
There are 30 boys and 20 girls in a class.
The teacher wants to select a class monitor from these boys and girls.
A boy can be selected in 30 ways and a girl can be selected in 20 ways.
∴ By using the fundamental principle of addition,
in a number of ways either a boy or a girl is selected as a class monitor = 30 + 20 = 50.

Question 2.
In question 1, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
Solution:
(i) Since there are 30 boys in the class
∴ A boy monitor can be selected in 30 ways.
(ii) Since there are 20 girls in the class
∴ A girl monitor can be selected in 20 ways.

Question 3.
A Signal is generated from 2 flags by putting one flag above the other. If 4 flags of different colours are available, how many different signals can be generated?
Solution:
A signal is generated from 2 flags and there are 4 flags of different colours available.
∴ 1st flag can be any one of the available 4 flags.
∴ It can be selected in 4 ways.
Now, 2nd flag is to be selected for which 3 flags are available for a different signal.
∴ 2nd flag can be anyone from these 3 flags.
∴ It can be selected in 3 ways.
∴ By using the fundamental principle of multiplication,
Total number of ways in which a signal can be generated = 4 × 3 = 12
∴ 12 different signals can be generated.

Question 4.
How many two-letter words can be formed using letters from the word SPACE when repetition of letters
(i) is allowed
(ii) is not allowed
Solution:
A two-letter word is to be formed out of the letters of the word SPACE.
(i) When repetition of the letters is allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 5 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 5 = 25

(ii) When repetition of the letters is not allowed
1st letter can be selected in 5 ways
2nd letter can be selected in 4 ways
∴ By using the fundamental principle of multiplication,
total number of 2-letter words = 5 × 4 = 20

Question 5.
How many three-digit numbers can be formed from the digits 0, 1, 3, 5, 6 if repetitions of digits
(i) are allowed
(ii) are not allowed
Solution:
The three-digit number is to be formed from the digits 0, 1, 3, 5, 6
(i) When repetition of digits is allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways.
0 can appear in 10’s and unit’s place and digits can be repeated.
∴ 10’s place digit can be selected in 5 ways and the unit’s place digit can be selected in 5 ways.
∴ By using the fundamental principle of multiplication,
the total number of three-digit numbers = 4 × 5 × 5 = 100

(ii) When repetition of digits is not allowed:
100’s place digit should be a non-zero number.
Hence, it can be anyone from digits 1, 3, 5, 6
∴ 100’s place digit can be selected in 4 ways
0 can appear in 10’s and unit’s place and digits can’t be repeated.
∴ 10’s place digit can be selected in 4 ways and the unit’s place digit can be selected in 3 ways
∴ By using the fundamental principle of multiplication,
total number of three-digit numbers = 4 × 4 × 3 = 48

Question 6.
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Solution:
A 3-digit number is to be formed from the digits 2, 3, 4, 5, 6 where digits can be repeated.
∴ The unit’s place digit can be selected in 5 ways.
10’s place digit can be selected in 5 ways.
100’s place digit can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of 3-digit numbers = 5 × 5 × 5 = 125

Question 7.
A letter lock has 3 rings and each ring has 5 letters. Determine the maximum number of trials that may be required to open the lock.
Solution:
A letter lock has 3 rings, each ring containing 5 different letters.
∴ A letter from each ring can be selected in 5 ways.
∴ By using fundamental principle of multiplication,
the total number of trials that can be made = 5 × 5 × 5 = 125
Out of these 124 wrong attempts are made and in the 125th attempt,
the lock gets opened, for a maximum number of trials.
∴ A maximum number of trials required to open the lock is 125.

Question 8.
In a test that has 5 true/false questions, no student has got all correct answers and no sequence of answers is repeated. What is the maximum number of students for this to be possible?
Solution:
For a set of 5 true/false questions, each question can be answered in 2 ways.
∴ By using fundamental principle of multiplication,
the total number of possible sequences of answers = 2 × 2 × 2 × 2 × 2 = 32
Since no student has written all the correct answers.
∴ Total number of sequences of answers given by the students in the class = 32 – 1 = 31
Also, no student has given the same sequence of answers.
∴ Maximum number of students in the class = Number of sequences of answers given by the students = 31

Question 9.
How many numbers between 100 and 1000 have 4 in the unit’s place?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 where the unit place digit is 4.
Since Unit’s place digit is 4.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 10 ways.
For 3-digit number 100’s place digit should be a non-zero number.
∴ 100’s place digit can be selected in 9 ways.
∴ By using fundamental principle of multiplication,
total number of numbers between 100 and 1000 which have 4 in the units place = 1 × 10 × 9 = 90

Question 10.
How many numbers between 100 and 1000 have the digit 7 exactly once?
Solution:
Numbers between 100 and 1000 are 3-digit numbers.
A 3-digit number is to be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, where exactly one of the digits is 7.
When 7 is in the unit’s place
The unit’s place digit is 7.
∴ it can be selected in 1 way only.
10’s place digit can be selected in 9 ways.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in the unit’s place = 1 × 9 × 8 = 72
When 7 is in 10’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit is 7
∴ it can be selected in 1 way only.
100’s place digit can be selected in 8 ways.
∴ total number of numbers which have 7 in 10’s place = 9 × 1 × 8 = 72
When 7 is in 100’s place
The unit’s place digit can be selected in 9 ways.
10’s place digit can be selected in 9 ways.
100’s place digit is 7
∴ it can be selected in 1 way.
∴ total numbers which have 7 in 100’s place = 9 × 9 × 1 = 81
∴ total number of numbers between 100 and 1000 having digit 7 exactly once = 72 + 72 + 81 = 225.

Question 11.
How many four-digit numbers will not exceed 7432 if they are formed using the digits 2, 3, 4, 7 without repetition?
Solution:
Among many set’s of digits, the greatest number is possible when digits are arranged in descending order.
∴ 7432 is the greatest number, formed from the digits 2, 3, 4, 7.
∴ Since a 4-digit number is to be formed from the digits 2, 3, 4, 7, where repetition of the digit is not allowed.
∴ 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Total number of numbers not exceeding 7432 that can be formed from the digits 2, 3, 4, 7
= Total number of four-digit numbers formed from the digits 2, 3, 4, 7
= 4 × 3 × 2 × 1
= 24

Question 12.
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Solution:
Case I: Three-digit numbers with 4 occurring in hundred’s place:
100’s place digit can be selected in 1 way.
Ten’s place can be filled by any one of the numbers 2, 3, 5, 6.
∴ 10’s place digit can be selected in 4 ways.
The unit’s place digit can be selected in 3 ways.
∴ total number of numbers which have 4 in 100’s place = 1 × 4 × 3 = 12

Case II: Three-digit numbers more than 500
100’s place digit can be selected in 2 ways.
10’s place digit can be selected in 4 ways.
Unit’s place digit can be selected in 3 ways.
∴ total number of three digit numbers more than 500 = 2 × 4 × 3 = 24

Case III: Number of four digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total four digit numbers formed = 5 × 4 × 3 × 2 = 120

Case IV: Number of five digit numbers formed from 2, 3, 4, 5, 6
Since, repetition of digits is not allowed
∴ total five digit numbers formed = 5 × 4 × 3 × 2 × 1 = 120
∴ total number of numbers that exceed 400 = 12 + 24 + 120 + 120 = 276

Question 13.
How many numbers formed with the digits 0, 1, 2, 5, 7, 8 will fall between 13 and 1000 if digits can be repeated?
Solution:
Case I: 2-digit numbers more than 13, less than 20, formed from the digits 0, 1, 2, 5, 7, 8
Number of such numbers = 3

Case II: 2-digit numbers more than 20 formed from 0, 1, 2, 5, 7, 8
Ten’s place digit is selected from 2, 5, 7, 8.
∴ Ten’s place digit can be selected in 4 ways.
Unit’s place digit is anyone from 0, 1, 2, 5, 7, 8
∴ The unit’s place digit can be selected in 6 ways.
Using the multiplication principle,
the number of such numbers (repetition allowed) = 4 × 6 = 24

Case III: 3-digit numbers formed from 0, 1, 2, 5, 7, 8
100’s place digit is anyone from 1, 2, 5, 7, 8.
∴ 100’s place digit can be selected in 5 ways.
As digits can be repeated, the 10’s place and unit’s place digits are selected from 0, 1, 2, 5, 7, 8
∴ 10’s place and unit’s place digits can be selected in 6 ways each.
Using multiplication principle,
the number of such numbers (repetition allowed) = 5 × 6 × 6 = 180
All cases are mutually exclusive and exhaustive.
∴ Required number = 3 + 24 + 180 = 207

Question 14.
A school has three gates and four staircases from the first floor to the second floor. How many ways does a student have to go from outside the school to his classroom on the second floor?
Solution:
A student can go inside the school from outside in 3 ways and from the first floor to the second floor in 4 ways.
∴ Number of ways to choose gates = 3
Number of ways to choose staircase = 4
∴ By using fundamental principle of multiplication,
number of ways in which a student has to go from outside the school to his classroom = 4 × 3 = 12

Question 15.
How many five-digit numbers formed using the digit 0, 1, 2, 3, 4, 5 are divisible by 3 if digits are not repeated?
Solution:
For a number to be divisible by 3.
The sum of digits must be divisible by 3.
Given 6 digits are 0, 1,2, 3, 4, 5.
Sum of 1, 2, 3, 4, 5 = 15, which is divisible by 3.
∴ There are two cases of 5 digit numbers formed from 0, 1, 2, 3, 4, 5 and divisible by 3.
Either 3 is selected in 5 digits (and 0 not selected) or 3 is not selected in 5 digits (and 0 is selected)
Case I:
3 is not selected (and 0 is selected) i.e., the digits are 0, 1, 2, 4, 5.
10000’s place digit can be selected in 4 ways (as 0 cannot appear).
As digits are not repeated, 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Using multiplication theorem,
Number of 5-digit number formed from 0, 1, 2, 4, 5 (with no repetition of digits) = 4 × 4 × 3 × 2 × 1 = 96

Case II:
3 is selected (and 0 is not selected) i.e., 1, 2, 3, 4, 5
10000’s place digit can be selected in 5 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
Using multiplication theorem,
Number of 5-digit numbers formed from 1, 2, 3, 4, 5 = 5 × 4 × 3 × 2 × 1 = 120
Both the cases are mutually exclusive and exhaustive.
∴ Required number = 96 + 120 = 216

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 Exercise 6.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Permutations and Combinations Ex 6.6 Questions and Answers.

Std 11 Maths 2 Exercise 6.6 Solutions Commerce Maths

Question 1.
Find the value of
(i) ¹⁵C₄
Solution:

(ii) ⁸⁰C₂
Solution:

(iii) ¹⁵C₄ + ¹⁵C₅
Solution:

(iv) ²⁰C₁₆ – ¹⁹C₁₆
Solution:

Question 2.
Find n if
(i) ⁶P₂ = n ⁶C₂
Solution:

(ii) ²ⁿC₃ : ⁿC₂ = 52 : 3
Solution:

(iii) ⁿC_n-3 = 84
Solution:
ⁿC_n-3 = 84
∴ \(\frac{n !}{(n-3) ![n-(n-3)] !}\) = 84
∴ \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}{(\mathrm{n}-3) ! \times 3 !}\) = 84
∴ n(n – 1) (n – 2) = 84 × 6
∴ n(n – 1) (n – 2) = 9 × 8 × 7
Comparing on both sides, we get
∴ n = 9

Question 3.
Find r if ¹⁴C_2r : ¹⁰C_2r-4 = 143 : 10
Solution:

∴ \(\frac{14 \times 13 \times 12 \times 11}{2 \mathrm{r}(2 \mathrm{r}-1) \times(2 \mathrm{r}-2)(2 \mathrm{r}-3)}=\frac{143}{10}\)
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 14 × 12 × 10
∴ 2r(2r – 1)(2r – 2)(2r – 3) = 8 × 7 × 6 × 5
Comparing on both sides, we get
∴ r = 4

Question 4.
Find n and r if.
(i) ⁿP_r = 720 and ⁿC_n-r = 120
Solution:

(ii) ⁿC_r-1 : ⁿC_r : ⁿC_r+1 = 20 : 35 : 42
Solution:

Question 5.
If ⁿP_r = 1814400 and ⁿC_r = 45, find r.
Solution:

∴ r! = 40320
∴ r! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
∴ r! = 8!
∴ r = 8

Question 6.
If ⁿC_r-1 = 6435, ⁿC_r = 5005, ⁿC_r+1 = 3003, find ^rC₅.
Solution:

Question 7.
Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 green balls and 7 blue balls so that 3 balls of every colour are drawn.
Solution:
9 balls are to be selected from 6 red, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour.
∴ 3 red balls can be selected from 6 red balls in ⁶C₃ ways.
3 green balls can be selected from 5 green balls in ⁵C₃ ways.
3 blue balls can be selected from 7 blue balls in ⁷C₃ ways.
∴ Number of ways selection can be done if the selection consists of 3 balls of each colour

Question 8.
Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Solution:
There are 6 boys and 4 girls.
A team of 3 boys and 2 girls is to be selected.
∴ 3 boys can be selected from 6 boys in ⁶C₃ ways.
2 girls can be selected from 4 girls in ⁴C₂ ways.
∴ Number of ways the team can be selected = ⁶C₃ × ⁴C₂
= \(\frac{6 !}{3 ! 3 !} \times \frac{4 !}{2 ! 2 !}\)
= \(\frac{6 \times 5 \times 4 \times 3 !}{3 \times 2 \times 1 \times 3 !} \times \frac{4 \times 3 \times 2 !}{2 \times 2 !}\)
= 20 × 6
= 120
∴ The team of 3 boys and 2 girls can be selected in 120 ways.

Question 9.
After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Solution:
Let there be n participants present in the meeting.
A handshake occurs between 2 persons.
∴ Number of handshakes = ⁿC₂
Given 66 handshakes were exchanged.
∴ 66 = ⁿC₂
∴ 66 = \(\frac{n !}{2 !(n-2) !}\)
∴ 66 × 2 = \(\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2) !}{(\mathrm{n}-2) !}\)
∴ 132 = n(n – 1)
∴ n(n – 1) = 12 × 11
Comparing on both sides, we get n = 12
∴ 12 participants were present at the meeting.

Question 10.
If 20 points are marked on a circle, how many chords can be drawn?
Solution:
To draw a chord we need to join two points on the circle.
There are 20 points on a circle.
∴ Total number of chords possible from these points = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 11.
Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when
(i) n = 10
(ii) n = 15
(iii) n = 12
Solution:
In n-sided polygon, there are ‘n’ points and ‘n’ sides. .
∴ Through ‘n’ points we can draw ⁿC₂ lines including sides.
∴ Number of diagonals in n sided polygon = ⁿC₂ – n (∴ n = number of sides)

Question 12.
There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Solution:
There are 20 lines such that no two of them are parallel and no three of them are concurrent.
Since no two lines are parallel
∴ they intersect at a point
∴ Number of points of intersection if no two lines are parallel and no three lines are concurrent = ²⁰C₂
= \(\frac{20 !}{2 ! 18 !}\)
= \(\frac{20 \times 19 \times 18 !}{2 \times 1 \times 18 !}\)
= 190

Question 13.
Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 10 points on a plane.
(i) No three of them are collinear:
Since a line is obtained by joining 2 points,
number of lines passing through these points if no three points are collinear = ¹⁰C₂
= \(\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\)
= 5 × 9
= 45

(ii) When 4 of them arc collinear:
∴ Number of lines passing through these points if 4 points are collinear
= ¹⁰C₂ – ⁴C₂ + 1
= 45 – \(\frac{4 !}{2 ! 2 !}\) + 1
= 45 – \(\frac{4 \times 3 \times 2 !}{2 \times 2 !}\) + 1
= 45 – 6 + 1
= 40

Question 14.
Find the number of triangles formed by joining 12 points if
(i) no three points are collinear
(ii) four points are collinear
Solution:
There are 12 points on the plane
(i) When no three of them are collinear:
Since a triangle can be drawn by joining any three non-collinear points.
∴ Number of triangles that can be obtained from these points = ¹²C₃
= \(\frac{12 !}{3 ! 9 !}\)
= \(\frac{12 \times 11 \times 10 \times 9 !}{3 \times 2 \times 1 \times 9 !}\)
= 220

(ii) When 4 of these points are collinear:
∴ Number of triangles that can be obtained from these points = ¹²C₃ – ⁴C₃
= 220 – \(\frac{4 !}{3 ! \times 1 !}\)
= 220 – \(\frac{4 \times 3 !}{3 !}\)
= 220 – 4
= 216

Question 15.
A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Solution:
Out of 8 consonants, 4 can be selected in ⁸C₄
= \(\frac{8 !}{4 ! 4 !}\)
= \(\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 \times 3 \times 2 \times 1 \times 4 !}\)
= 70 ways
From 3 vowels, 2 can be selected in ³C₂
= \(\frac{3 !}{2 ! 1 !}\)
= \(\frac{3 \times 2 !}{2 !}\)
= 3 ways
Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in ⁶P₆ i.e., 6! ways.
∴ Total number of words that can be formed = 70 × 3 × 6!
= 70 × 3 × 720
= 151200
∴ 151200 words of 4 consonants and 2 vowels can be formed.

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

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

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

Std 11 Maths 2 Miscellaneous Exercise 1 Solutions Commerce Maths

Question 1.
The data gives the number of accidents per day on a railway track. Compute Q₂, P₁₇, and D₇.
4, 2, 3, 5, 6, 3, 4, 1, 2, 3, 2, 3, 4, 3, 2
Solution:
The given data can be arranged in ascending order as follows:
1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6
Here, n = 15
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 4)th observation
= value of 8th observation
∴ Q₂ = 3
P₁₇ = value of 17\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 17\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (17 × 0.16)th observation
= value of (2.72)th observation
= value of 2nd observation + 0.72 (value of 3rd observation – value of 2nd observation)
= 2 + 0.72 (2 – 2)
∴ P₁₇ = 2
D₇ = value of 7\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 7\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (7 × 1.6)th observation
= value of (11.2)th observation
= value of 11th observation + 0.2(value of 12th observation – value of 11th observation)
= 4 + 0. 2(4 – 4)
∴ D₇ = 4

Question 2.
The distribution of daily sales of shoes (size-wise) for 100 days from a certain shop is as follows:

Compute Q₁, D₂, and P₉₅.
Solution:
By arranging the given data in ascending order, we construct the less than cumulative frequency table as given below:

Here, n = 100
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation
= value of (25.25)th observation
Cumulative frequency which is just greater than (or equal) to 25.25 is 27.
∴ Q₁ = 3
D₂ = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{100+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 10.1)th observation
= value of (20.2)th observation
Cumulative frequency which is just greater than (or equal) to 20.2 is 27.
∴ D₂ = 3
P₉₅ = value of 95\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 95\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (95 × 1.01)th observation
= value of (95.95)th observation
The cumulative frequency which is just greater than (or equal) to 95.95 is 100.
∴ P₉₅ = 8

Question 3.
Ten students appeared for a test in Mathematics and Statistics and they obtained the marks as follows:

If the median will be the criteria, in which subject, the level of knowledge of the students is higher?
Solution:
Marks in Mathematics can be arranged in ascending order as follows:
23, 23, 25, 25, 32, 35, 36, 37, 38, 42
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
Median = value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 32 + 0.5 (35 – 32)
= 32 + 0.5(3)
= 32 + 1.5
= 33.5
Marks in Statistics can be arranged in ascending order as follows:
22, 23, 26, 28, 29, 32, 34, 36, 45, 50
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 29 + 0.5(32 – 29)
= 29 + 0.5(3)
= 29 + 1.5
= 30.5
∴ Median marks for Mathematics = 33.5 and
Median marks for Statistics = 30.5
∴ The level of knowledge in Mathematics is higher than that of Statistics.

Question 4.
In the frequency distribution of families given below, the number of families corresponding to expenditure group 2000 – 4000 is missing from the table. However, the value of the 25th percentile is 2880. Find the missing frequency.

Solution:
Let x be the missing frequency of expenditure group 2000 – 4000.
We construct the less than cumulative frequency table as given below:

Here, N = 75 + x
Given, P₂₅ = 2880
∴ P₂₅ lies in the class 2000 – 4000.
∴ L = 2000, h = 2000, f = x, c.f. = 14
∴ P₂₅ = L + \(\frac{h}{f}\left(\frac{25 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 2880 = 2000 + \(\frac{2000}{x}\left(\frac{75+x}{4}-14\right)\)
∴ 2880 – 2000 = \(\frac{2000}{x}\left(\frac{75+x-56}{4}\right)\)
∴ 880x = 500(x + 19)
∴ 880x = 500x + 9500
∴ 880x – 500x = 9500
∴ 380x = 9500
∴ x = 25
∴ 25 is the missing frequency of the expenditure group 2000 – 4000.

Question 5.
Calculate Q₁, D₆, and P₁₅ for the following data:

Solution:
Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
∴ the class intervals will be 0 – 50, 50 – 100, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 160.
Q₁ lies in the class 100 – 150.
∴ L = 100, h = 50, f = 80, c.f. = 80
∴ Q₁ = L + \(\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 100 + \(\frac{50}{80}\)(125 – 80)
= 100 + \(\frac{5}{8}\)(45)
= 100 + 28.125
= 128.125
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 500}{10}\) = 300
Cumulative frequency which is just greater than (or equal) to 300 is 410.
∴ D₆ lies in the class 200 – 250.
∴ L = 200, h = 50, f = 150, c.f. = 260
∴ D₆ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{6 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 200 + \(\frac{50}{150}\)(300 – 260)
= 200 + \(\frac{1}{3}\)(40)
= 200 + 13.33
= 213.33
P₁₅ class = class containing \(\left(\frac{15 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{15 \mathrm{~N}}{100}=\frac{15 \times 500}{100}\) = 75
Cumulative frequency which is just greater than (or equal) to 75 is 80.
∴ P15 lies in the class 50 – 100.
∴ L = 50, h = 50, f = 70, c.f. = 10
∴ P₁₅ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{15 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 50 + \(\frac{50}{70}\) (75 – 10)
= 50 + \(\frac{5}{7}\) (65)
= 50 + \(\frac{325}{7}\)
= 50 + 46.4286
= 96.4286
∴ Q₁ = 128.125, D₆ = 213.33, P₁₅ = 96.4286

Question 6.
Daily income for a group of 100 workers are given below:

P₃₀ for this group is ₹ 110. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 50 – 100 and class 200 – 250 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 ……(i)
Given, P30 = 110
∴ P30 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 25, c.f. = 7 + a
\(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
∴ P₃₀ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{30 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 110 = 100 + \(\frac{50}{25}\) [30 – (7 + a)]
∴ 110 – 100 = 2(30 – 7 – a)
∴ 10 = 2(23 – a)
∴ 5 = 23 – a
∴ a = 23 – 5
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18
∴ b = 20
∴ 18 and 20 are the missing frequencies of the class 50 – 100 and class 200 – 250 respectively.

Question 7.
The distribution of a sample of students appearing for a C.A. examination is:

Help C.A. institute to decide cut-off marks for qualifying for an examination when 3% of students pass the examination.
Solution:
To decide cut-off marks for qualifying for an examination when 3% of students pass, we have to find P97.
We construct the less than cumulative frequency table as given below:

Here, N = 1100
P97 class = class containing \(\left(\frac{97 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{97 \mathrm{~N}}{100}=\frac{97 \times 1100}{100}\) = 1067
Cumulative frequency which is just greater than (or equal) to 1067 is 1100.
∴ P₉₇ lies in the class 500 – 600.
∴ L = 500, h = 100, f = 130, c.f. = 970
∴ P₉₇ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{97 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 500 + \(\frac{100}{130}\)(1067 – 970)
= 500 + \(\frac{10}{13}\) (97)
= 500 + 74.62
= 574.62 ~ 575
∴ the cut off marks for qualifying an examination is 575.

Question 8.
Determine graphically the value of median, D₃, and P₃₅ for the data given below:

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 (15, 8), (20, 22), (25, 30), (30, 55), (35, 70), (40, 84), (45, 90).

N = 90
For median, consider \(\frac{\mathrm{N}}{2}=\frac{90}{2}\) = 45
For D3, consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 90}{10}\) = 27
For P35, consider \(\frac{35 \mathrm{~N}}{100}=\frac{35 \times 90}{100}\) = 31.5
∴ We take the values 45, 27 and 31.5 on the Y-axis and draw lines from these points parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendicular on the X-axis.
Foot of the perpendicular represent the values of median, D3 and P35 respectively.
∴ Median ~ 29, D₃ ~ 23.5, P₃₅ ~ 26

Question 9.
The I.Q. test of 500 students of a college is as follows:

Find graphically the number of students whose I.Q. is more than 55 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 (30, 41), (40, 93), (50, 157), (60, 337), (70, 404), (80, 449), (90, 489), (100, 500)

To find the number of students whose I.Q. is more than 55, we consider the value 55 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point this line intersects the less than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of students whose I.Q. is less than 55.
∴ The foot of perpendicular ~ 244
∴ No. of students whose I.Q. is less than 55 ~ 244
∴ No. of Students whose I.Q. is more than 55 = 500 – 244 = 256

Question 10.
Draw an ogive for the following distribution. Determine the median graphically and verify your result by a mathematical formula.

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 (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).

N = 60
∴ \(\frac{\mathrm{N}}{2}=\frac{60}{2}\) = 30
∴ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis.
From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis.
The foot perpendicular gives the value of the median.
∴ Median ~ 164.67
Now, let us calculate the median from the mathematical formula.
∴ \(\frac{\mathrm{N}}{2}\) = 30
The median lies in the class interval 160 – 165.
∴ L = 160, h = 5, f = 15, c.f. = 16
Median = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{2}-\mathrm{c} . \mathrm{f} .\right)\)
= 160 + \(\frac{5}{15}\) (30 – 16)
= 160+ \(\frac{1}{3}\) × 14
= 160 + 4.67
= 164.67

Question 11.
In a group of 25 students, 7 students failed and 6 students got distinction and the marks of the remaining 12 students are 61, 36, 44, 59, 52, 56, 41, 37, 39, 38, 41, 64. Find the median marks of the whole group.
Solution:
n = 25
Median = \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}\) = 13th observation
We have been stated that 7 students failed (assuming passing marks on 35) and 6 students got distinction (assuming distinction as 70+), and the marks of the remaining 12 students (who will be situated between the two groups mentioned above, if arranged in ascending order), we have,
F, F, F, F, F, F, F, 36, 37, 38, 39, 41, 41, 44, 52, 56, 59, 61, 64, D, D, D, D, D, D
∴ median = 13th observation = 41.

Question 12.
The median weight of a group of 79 students is found to be 55 kg. 6 more students are added to this group whose weights are 50, 51, 52, 59.5, 60, 61 kg. What will be the value of the median of the combined group if the lowest and the highest weights were 53 kg and 59 kg respectively?
Solution:
n = 79
Median = 55kg
Lowest observation = 53 kg
Flighest observation = 59 kg
6 new students are added to the group having weights in Kg as follows:
50, 51, 52, 59.5, 60, 61
From the above, we see that of the 6 new students, 3 have weights which are below the lowest weight of the earlier group and 3 have weights which are above the highest weight of the earlier group.
∴ the median remains the same
∴ median = 55 kg.

Question 13.
The median of the following incomplete table is 92. Find the missing frequencies:

Solution:
Let a and b be the missing frequencies of class 50 – 70 and class 110 – 130 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 54 + a + b
Since, N = 80
∴ 54 + a + b = 80
∴ a + b = 26 …..(i)
Given, Median = Q₂ = 92
∴ Q₂ lies in the class 90 – 110.
∴ L = 90, h = 20, f = 20, c.f. = 24 + a
\(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 80}{4}\) = 40
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
∴ 92 = 90 + \(\frac{20}{20}\) [40 – (24 + a)
∴ 92 – 90 = 40 – 24 – a
∴ 2 = 16 – a
∴ a = 14
Substituting the value of a in equation (i), we get
14 + b = 26
∴ b = 26 – 14 = 12
∴ 14 and 12 are the missing frequencies of the class 50 – 70 and class 110 – 130 respectively.

Question 14.
A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

estimate the number of defective tubes in the central batch.
Solution:
To find the number of defective tubes in the central batch, we have to find Q₂.
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 4.5, 4.5 – 9.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 251
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 251}{4}\) = 125.5
Cumulative frequency which is just greater than (or equal to) 125.5 is 180.
∴ Q₂ lies in the class 9.5 – 14.5.
∴ L = 9.5, h = 5, f = 84, c.f. = 96
∴ Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 9.5 + \(\frac{5}{84}\) (125.5 – 96)
= 9.5 + \(\frac{5}{84}\) × 29.5
= 9.5 + \(\frac{147.5}{84}\)
= 9.5 + 1.76
= 11.26

Question 15.
In a college, there are 500 students in junior college, 5% score less than 25 marks, 68 scores from 26 to 30 marks, 30% score from 31 to 35 marks, 70 scores from 36 to 40 marks, 20% score from 41 to 45 marks and the rest score 46 and above marks. What are the median marks?
Solution:
Given data can be written in tabulated form as follows:

Since the given data is not continuous, we have to convert it into the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 25.5, 25.5 – 30.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 500}{4}\) = 250
Cumulative frequency which is just greater than (or equal to) 250 is 313.
∴ Q₂ lies in the class 35.5 – 40.5.
∴ L = 35.5, h = 5, f = 70, c.f. = 243
∴ Median = Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 35.5 + \(\frac{5}{70}\) (250 – 243)
= 35.5 + \(\frac{1}{14}\) (7)
= 35.5 + 0.5
= 36

Question 16.
Draw a cumulative frequency curve more than typical for the following data and hence locate Q₁ and Q₃. Also, find the number of workers with daily wages
(i) Between ₹ 170 and ₹ 260
(ii) less than ₹ 260

Solution:
For more than ogive points to be plotted are (100, 200), (150, 188), (200, 160), (250, 124), (300, 74), (350, 49), (400, 31), (450, 15), (500, 5)

Here, N = 200
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{200}{4}\) = 4
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 200}{4}\) = 150
We take the points having Y co-ordinates 50 and 150 on Y-axis.
From these points, we draw lines which are parallel to X-axis.
From the points of intersection of these lines with the curve, we draw perpendicular on X-axis.
X-Co-ordinates of these points gives the values of Q₁ and Q₃.
Since X-axis has daily wages more than and not less than the given amounts.
∴ Q₁ = Q₃ and Q₃ = Q₁
∴ Q₂ ~ 215 , Q₃ ~ 348

(i) To find the number of workers with daily wages between ₹ 170 and ₹ 260,
Take the values 170 and 260 on X-axis. From these points, we draw lines parallel to Y-axis.
From the point where they intersect the more than ogive, we draw perpendiculars on Y-axis.
The points where they intersect the Y-axis gives the values 178 and 114.
∴ Number of workers having daily wages between ₹ 170 and ₹ 260 = 178 – 114 = 64

(ii) To find the number of workers having daily wages less than ₹ 260, we consider the value 260 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point where the line intersects the more than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of workers having daily wages of more than 260.
The foot of perpendicular ~ 114
∴ No. of workers whose daily wages are more than ₹ 260 ~ 114
∴ No. of workers whose daily wages are less than ₹ 260 = 200 – 114 = 86

Question 17.
Draw ogive of both the types for the following frequency distribution and hence find the median.

Solution:

For less than given points to be plotted are (10, 5), (20, 10), (30, 18), (40, 30), (50, 46), (60, 61), (70, 71), (80, 79), (90, 84), (100, 86)
For more than given points to be plotted are (0, 86), (10, 81), (20, 76), (30, 68), (40, 56), (50, 40), (60, 25), (70, 15), (80, 7), (90, 2)

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.

Question 18.
Find Q1, D6 and P78 for the following data:

Solution:
Since the given data is not in the form of a continuous frequency distribution, we have to convert it into that form by subtracting 0.025 from the lower limit and adding 0.025 to the upper limit of each class interval.
∴ the class intervals will be 7.975 – 8.975, 8.975 – 9.975, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 50
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{50}{4}\) = 12.5
Cumulative frequency which is just greater than (or equal) to 12.5 is 15.
∴ Q₁ lies in the class 8.975 – 9.975.
∴ L = 8.975, h = 1, f = 10, c.f. = 5
Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 8.975 + \(\frac{1}{10}\) (12.5 – 5)
= 8.975 + 0.1(7.5)
= 8.975 + 0.75
= 9.725
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 50}{10}\) = 30
Cumulative frequency which is just greater than (or equal) to 30 is 35.
∴ D6 lies in the class 9.975 – 10.975.
∴ L = 9.975, h = 1, f = 20, c.f. = 15
D₆ = L + \(\frac{h}{f}\left(\frac{6 N}{10}-\text { c.f. }\right)\)
= 9.975 + \(\frac{1}{20}\) (30 – 15)
= 9.975 + 0.05(15)
= 9.975 + 0.75
= 10.725
P₇₈ class = class containing \(\left(\frac{78 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
\(\frac{78 \mathrm{~N}}{100}=\frac{78 \times 50}{100}\) = 39
Cumulative frequency which is just greater than (or equal) to 39 is 45.
∴ P₇₈ lies in the class 10.975 – 11.975.
∴ L = 10.975, h = 1, f = 10, c.f. = 35
∴ P₇₈ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{78 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10.975 + \(\frac{1}{10}\) (39 – 35)
= 10.975 + 0.1(4)
= 10.975 + 0.4
= 11.375

Question 19.

For the above data, find all quartiles and number of persons weighing between 57 kg and 72 kg.
Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 111
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{111}{4}\) = 27.75
Cumulative frequency which is just greater than (or equal) to 27.75 is 39.
∴ Q₁ lies in the class 50 – 55.
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 50 + \(\frac{5}{20}\) (27.75 – 19)
= 50 + \(\frac{1}{4}\) × 8.75
= 50 + 2.1875
= 52.1875
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 N}{4}=\frac{2 \times 111}{4}\) = 55.5
Cumulative frequency which is just greater than (or equal) to 55.5 is 69.
∴ Q₂ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 30, c.f. = 39
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
= 55 + \(\frac{5}{30}\) (55.5 – 39)
= 55 + \(\frac{1}{6}\) × 16.5
= 55 + 2.75
= 57.75
Q₃ class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 111}{4}\) = 83.25
Cumulative frequency which is just greater than (or equal) to 83.25 is 89.
∴ Q₃ lies in the class 60 – 65.
∴ L = 60, h = 5, f = 20, c.f. = 69
∴ Q₃ = L + \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 60 + \(\frac{5}{20}\) (83.25 – 69)
= 60 + \(\frac{1}{4}\) × 14.25
= 60 + 3.5625
= 63.5625
In order to find the number of persons between 57 kg and 72 kg,
We need to find x in P_x, where P_x = 57 kg and y in P_y, where P_y = 72 kg
Then (y – x) would be the % of persons weighing between 57 kg and 72 kg
P_x = 57
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{x \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 57
∴ 55 + \(\frac{5}{30}\) (1.11x – 39) = 57
∴ \(\frac{1}{6}\) (1.11x – 39) = 2
∴ 1.11x – 39 = 12
∴ 1.11x = 51
∴ x = 45.95
∴ P_y = 72
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{y \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 72
∴ 70 + \(\frac{5}{8}\) (1.11y – 99) = 72
∴ 0.625(1.11y – 99) = 2
∴ 1.11y – 99 = 3.2
∴ 1.11y = 102.2
∴ y = 92.07
∴ % of people weighing between 57 kg and 72 kg = 92.07 – 45.95 = 46.12 %
∴ No. of people weighing between 57 kg and 72 kg = 111 × 46.12% = 51.1932 ~ 51

Question 20.
For the following data showing weights of 100 employees, find the maximum weight of the lightest 25% of employees.

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

Here, N = 100
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
Cumulative frequency which is just greater than (or equal) to 25 is 29.
∴ Q₁ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 15, c.f. = 14
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 55 + \(\frac{5}{15}\) (25 – 14)
= 55 + \(\frac{1}{3}\) × 11
= 55 + 3.67
= 58.67
∴ Maximum weight of the lightest 25% of employees is 58.67 kg.

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

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.1 Questions and Answers.

Std 11 Maths 2 Exercise 9.1 Solutions Commerce Maths

Question 1.
Find 77% of 580 + 34% of 390.
Solution:

Question 2.
240 candidates appeared for an examination, of which 204 passed. What is the pass percentage?
Solution:
We find the pass percentage using the unitary method

∴ The pass percentage for the examination is 85%.

Question 3.
What percent of 8.4 kg are 168 grams?
Solution:
Let 168 gms be x% of 8.4 kg
i.e., let 168 gms be \(\frac{x}{100}\) of 8400 gms
∴ 168 = \(\frac{x}{100}\) × 8400
∴ x = \(\frac{168}{84}\) = 2
∴ 168 gms is 2% of 8.4 kg.

Question 4.
If the length of a rectangle is decreased by 20%, what should be the increase in the breadth of the rectangle so that the area remains the same?
Solution:
Let x and y represent the length and breadth of the rectangle respectively.
∴ The original area of the rectangle = xy
There is a 20% decrease in length.

Let k % be the required increase in breadth

Given that the new and old areas should be equal.

∴ 100 + k = 125
∴ k = 125 – 100 = 25
∴ Breadth should be increased by 25% so that the area remains same.

Question 5.
The price of rice increased by 20%, as a result, a person can have 5kg rice for ₹ 600. What was the initial price of rice per kg?
Solution:
A person can buy 5 kg of rice for ₹ 600 after the increase in price
∴ New price of rice = \(\frac{600}{5}\) = ₹ 120/kg …..(i)
Let ‘x’ be the initial price per kg of rice.
There is a 20% increase in the price of rice.
Thus the new price of the rice will be given as

∴ The initial price of rice is ₹ 100 per kg

Question 6.
What percent is 3% of 5%?
Solution:
Let 3% be x % of 5%.
Then \(\frac{3}{100}=\frac{x}{100} \times \frac{5}{100}\)
∴ x = \(\frac{3 \times 100}{5}\) = 60
∴ 3% is 60% of 5%.

Question 7.
After availing of two successive discounts of 20% each, Madhavi paid ₹ 64 for a book. If she would have got only one discount of 20%, how much additional amount would she have paid?
Solution:
Let the price of the book be ₹ x.
After the first 20% discount, the price of the book becomes

After another 20% discount, the price of the book becomes

This price = ₹ 64 …..[Given]
∴ \(\frac{16}{25}\)x = 64
∴ x = 4 × 25 = 100
Thus, Amount of the book after one discount = \(\frac{4}{5}\)(100) = 80 …..[from (i)]
∴ The additional amount that Madhavi would have paid = 80 – 64 = ₹ 16

Question 8.
The price of the table is 40% more than the price of a chair. By what percent price of a chair is less than the price of a table?
Solution:
Let ₹ x and ₹ y be the price of a table and chair respectively.
The price of the table is 40% more than the price of a chair
∴ \(\frac{x-y}{y}\) × 100 = 40

We need to find by how much percent is the price of a chair less than that of a table.

∴ The price of a chair is 28.57% less than the price of a table.

Question 9.
A batsman scored 92 runs which includes 4 boundaries 5 sixes. He scored other runs by running between the wickets. What percent of his total score did he make by running between the wickets?
Solution:
Batsman scores 4 fours (boundaries) and 5 sixes in 92 runs.
Number of runs scored by fours and sixes = 4 × 4 + 5 × 6 = 46
∴ 92 – 46 = 46
Let 46 be x% of 92.
Then 46 = \(\frac{x}{100}\) × 92
∴ x = \(\frac{46 \times 100}{92}=\frac{100}{2}\) = 50
∴ 50% of the total runs were scored by running between the wickets.

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