Category: Class 11

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

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Functions Ex 2.1

Question 1.
Check if the following relations are functions.


Solution:
(a) Yes
Reason: Every element of set A has been assigned a unique element in set B.

(b) No
Reason: An element of set A has been assigned more than one element from set B.

(c) No
Reason: Not every element of set A has been assigned an image from set B.

Question 2.
Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}
(ii) {(1, 2), (2, -1), (3, 1), (4, 3)}
(iii) {(1, 3), (4, 1), (2, 2)}
(iv) {(1, 1), (2, 1), (3, 1), (4, 1)}
Solution:
(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.
Reason: (2, -1) and (2, 2) show that element 2 ∈ A has been assigned two images -1 and 2 from set B.

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.
Reason: Every element of set A has a unique image in set B.

(iii) {(1, 3), (4, 1), (2, 2)} does not represent a function.
Reason: 3 ∈ A does not have an image in set B.

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)} represents a function
Reason: Every element of set A has been assigned a unique image in set B.

Question 3.
If f(m) = m² – 3m + 1, find
(i) f(0)
(ii) f(-3)
(iii) f(\(\frac{1}{2}\))
(iv) f(x + 1)
(v) f(-x)
Solution:
f(m) = m² – 3m + 1
(i) f(0) = 0² – 3(0) + 1 = 1

(ii) f(-3) = (-3)² – 3(-3) + 1
= 9 + 9 + 1
= 19

(iii) \(f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{2}\right)+1\)
= \(\frac{1}{4}-\frac{3}{2}+1\)
= \(\frac{1-6+4}{4}\)
= \(-\frac{1}{4}\)

(iv) f(x + 1) = (x + 1)² – 3(x + 1) + 1
= x² + 2x + 1 – 3x – 3 + 1
= x² – x – 1

(v) f(-x) = (-x)² – 3(-x) + 1 = x² + 3x + 1

Question 4.
Find x, if g(x) = 0 where
(i) g(x) = \(\frac{5 x-6}{7}\)
(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
(iii) g(x) = 6x² + x – 2
Solution:
(i) g(x) = \(\frac{5 x-6}{7}\)
g(x) = 0
∴ \(\frac{5 x-6}{7}\) = 0
∴ 5x – 6 = 0
∴ x = \(\frac{6}{5}\)

(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)
g(x) = 0
∴ \(\frac{18-2 x^{2}}{7}\) = 0
∴ 18 – 2x² = 0
∴ x² = 9
∴ x = ±3

(iii) g(x) = 6x² + x – 2
g(x) = 0
∴ 6x² + x – 2 = 0
∴ 6x² + 4x – 3x – 2 = 0
∴ 2x(3x + 2) – 1(3x + 2) = 0
∴ (2x – 1)(3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = \(\frac{1}{2}\) or x = \(\frac{-2}{3}\)

Question 5.
Find x, if f(x) = g(x) where f(x) = x⁴ + 2x², g(x) = 11x².
Solution:
f(x) = x⁴ + 2x², g(x) = 11x²
f(x) = g(x)
∴ x⁴ + 2x² = 11x²
∴ x⁴ – 9x² = 0
∴ x²(x² – 9) = 0
∴ x² = 0 or x² – 9 = 0
∴ x = 0 or x² = 9
∴ x = 0 or x = ±3

Question 6.
If f(x) = \(\begin{cases}x^{2}+3, & x \leq 2 \\ 5 x+7, & x>2\end{cases}\), then find
(i) f(3)
(ii) f(2)
(iii) f(0)
Solution:
f(x) = x² + 3, x ≤ 2
= 5x + 7, x > 2
(i) f(3) = 5(3) + 7 = 15 + 7 = 22
(ii) f(2) = 2² + 3 = 4 + 3 = 7
(iii) f(0) = 0² + 3 = 3

Question 7.
If f(x) = \(\left\{\begin{array}{cl}
4 x-2, & x \leq-3 \\
5, & -3<x<3 \\
x^{2}, & x \geq 3
\end{array}\right.\), then fmd
(i) f(-4)
(ii) f(-3)
(iii) f(1)
(iv) f(5)
Solution:
f(x) = 4x – 2, x ≤ -3
= 5, -3 < x < 3
= x², x ≥ 3
(i) f(-4) = 4(-4) – 2 = -16 – 2 = -18
(ii) f(-3) = 4(-3) – 2 = -12 – 2 = -14
(iii) f(1) = 5
(iv) f(5) = 5² = 25

Question 8.
If f(x) = 3x + 5, g(x) = 6x – 1, then find
(i) (f + g)(x)
(ii) (f – g)(2)
(iii) (fg)(3)
(iv) \(\left(\frac{\mathbf{f}}{\mathbf{g}}\right)(x)\) and its domain
Solution:
f(x) = 3x + 5, g(x) = 6x – 1
(i) (f + g)(x) = f(x) + g(x)
= 3x + 5 + 6x – 1
= 9x + 4

(ii) (f – g) (2) = f(2) – g(2)
= [3(2) + 5] – [6(2) – 1]
= 6 + 5 – 12 + 1
= 0

(iii) (fg)(3) = f(3) g(3)
= [3(3) + 5] [6(3) – 1]
= (14) (17)
= 238

(iv) \(\left(\frac{\mathrm{f}}{\mathrm{g}}\right) x=\frac{\mathrm{f}(x)}{\mathrm{g}(x)}=\frac{3 x+5}{6 x-1}, x \neq \frac{1}{6}\)
Domain = R – {\(\frac{1}{6}\)}

Question 9.
If f(x) = 2x² + 3, g(x) = 5x – 2, then find
(i) fog
(ii) gof
(iii) fof
(iv) gog
Solution:
f(x) = 2x² + 3, g(x) = 5x – 2
(i) (fog)(x) = f(g(x))
= f(5x – 2)
= 2(5x – 2)² + 3
= 2(25x² – 20x + 4) + 3
= 50x² – 40x + 8 + 3
= 50x² – 40x + 11

(ii) (gof)(x) = g(f(x))
= g(2x² + 3)
= 5(2x² + 3) – 2
= 10x² + 15 – 2
= 10x² + 13

(iii) (fof)(x) = f(f(x))
= f(2x² + 3)
= 2(2x² + 3)² + 3
= 2(4x⁴ + 12x² + 9) + 3
= 8x⁴ + 24x² + 18 + 3
= 8x⁴ + 24x² + 21

(iv) (gog)(x) = g(g(x))
= g(5x – 2)
= 5(5x – 2) – 2
= 25x – 10 – 2
= 25x – 12

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

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Sets and Relations Miscellaneous Exercise 1

Question 1.
Write the following sets in set builder form:
(i) {10, 20, 30, 40, 50}
(ii) {a, e, i, o, u}
(iii) {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Solution:
(i) Let A = {10, 20, 30, 40, 50}
∴ A = {x / x = 10n, n ∈ N and n ≤ 5}

(ii) Let B = {a, e, i, o, u}
∴ B = {x / x is a vowel of English alphabets}

(iii) Let C = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
∴ C = {x / x represents days of a week}

Question 2.
If U = {x / x ∈ N, 1 ≤ x ≤ 12}, A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}.
Write the sets
(i) A ∪ B
(ii) B ∩ C
(iii) A – B
(iv) B – C
(v) A ∪ B ∪ C
(vi) A ∩ (B ∪ C)
Solution:
U = {x / x ∈ N, 1 ≤ x ≤ 12} = {1, 2, 3, …., 12}
A = {1, 4, 7, 10}, B = {2, 4, 6, 7, 11}, C = {3, 5, 8, 9, 12}
(i) A ∪ B = {1, 2, 4, 6, 7, 10, 11}

(ii) B ∩ C = { }

(iii) A – B = {1, 10}

(iv) B – C = {2, 4, 6, 7, 11}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

(vi) B ∪ C = {2, 3, 4, 5, 6, 7, 8, 9, 11, 12}
∴ A ∩ (B ∪ C) = {4, 7}

Question 3.
In a survey of 425 students in a school, it was found that 115 drink apple juice, 160 drink orange juice, and 80 drink both apple as well as orange juice. How many drinks neither apple juice nor orange juice?
Solution:
Let A = set of students who drink apple juice
B = set of students who drink orange juice
X = set of all students
∴ n(X) = 425, n(A) = 115, n(B) = 160, n(A ∩ B) = 80

No. of students who neither drink apple juice nor orange juice
n(A’ ∩ B’) = n(A ∪ B)’
= n(X) – n(A ∪ B)
= 425 – [n(A) + n(B) – n(A ∩ B)]
= 425 – (115 + 160 – 80)
= 230

Question 4.
In a school, there are 20 teachers who teach Mathematics or Physics. of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teachers teach Physics?
Solution:
Let A = set of teachers who teach Mathematics
B = set of teachers who teach Physics
n(A ∪ B) = 20, n(A) = 12, n(A ∩ B) = 4

Since, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
∴ 20 = 12 + n(B) – 4
∴ n(B) = 12
∴ Number of teachers who teach physics = 12

Question 5.
(i) If A = {1, 2, 3} and B = {2, 4}, state the elements of A × A, A × B, B × A, B × B, (A × B) ∩ (B × A).
(ii) If A = {-1, 1}, find A × A × A.
Solution:
(i) A = {1, 2, 3} and B = {2, 4}
A × A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}
B × A = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}
B × B = {(2, 2), (2, 4), (4, 2), (4, 4)}
(A × B) ∩ (B × A) = {(2, 2)}

(ii) A = {-1, 1}
∴ A × A × A = {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1),(1, 1, 1)}

Question 6.
If A = {1, 2, 3}, B = {4, 5, 6}, which of the following are relations from A to B.
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
(ii) R₂ = {(1, 5), (2, 4), (3, 6)}
(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
(iv) R₄ = {(4, 2), (2, 6), (5, 1), (2, 4)}
Solution:
A = {1, 2, 3}, B = {4, 5, 6}
∴ A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(i) R₁ = {(1, 4), (1, 5), (1, 6)}
Since, R₁ ⊆ A × B
∴ R₁ is a relation from A to B.

(ii) R₂ = {(1, 5), (2, 4), (3, 6)}
Since, R₂ ⊆ A × B
∴ R₂ is a relation from A to B.

(iii) R₃ = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}
Since, R₃ ⊆ A × B
∴ R₃ is a relation from A to B.

(iv) R₄ = {(4,2), (2, 6), (5,1), (2, 4)}
Since, (4, 2) ∈ R₄, but (4, 2) ∉ A × B
∴ R₄ ⊄ A × B
∴ R₄ is not a relation from A to B.

Question 7.
Determine the domain and range of the following relation.
R = {(a, b) / a ∈ N, a < 5, b = 4}
Solution:
R = {(a, b) / a ∈ N, a < 5, b = 4}
∴ Domain (R) = {a / a ∈ N, a < 5} = {1, 2, 3, 4}
Range (R) = {b / b = 4} = {4}

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

Maharashtra State Board 11th Commerce Maths Solutions Chapter 7 Probability Miscellaneous Exercise 7

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’)
= \(\frac{8}{14} \times \frac{16}{30}\)
= \(\frac{32}{105}\)

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

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.5

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Permutations and Combinations Ex 6.4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.7 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.7

Question 1.
Shantanu has a choice to invest in ₹ 10 shares of two firms at ₹ 13 or at ₹ 16. If the first firm pays a 5% dividend and the second firm pays a 6% dividend per annum, find:
(i) Which firm is paying better?
(ii) If Shantanu invests equally in both the firms and the difference between the return from them is ₹ 30. Find how much, in all, does he invest.
Solution:
(i) For firm 1:
Face value of the share (F.V.) = ₹ 10
Market value of the share (M.V.) = ₹ 13
Dividend = 5%
∴ Annual income from the share = \(\frac{5}{100}\) × 10 = ₹ 0.5

For firm 2:
Face value of the share (F.V.) = ₹ 10
Market value of the share (M.V.) = ₹ 16
Dividend = 5%
∴ Annual income from the share = \(\frac{6}{100}\) × 10 = ₹ 0.6

Since, the profit percentage from firm 1 > profit percentage from firm 2, the first firm is paying better.

(ii) Let ‘X’ be the amount Shantanu invests in each of the firms.
Given that difference between the return from them is ₹ 30, we have

In all, Shantanu invests = 2X
= 2 × 31,200
= ₹ 62,400/-

Question 2.
A dividend of 9% was declared on ₹ 100 shares selling at a certain price in the stock market. If the rate of return is 7.5% calculate
(i) The market price of each share, and
(ii) The amount to be invested to obtain an annual dividend of ₹ 630.
Solution:
(i) Given that,
Face value of the share (F.V) = ₹ 100
Dividend = 9%
Rate of return = 7.5%
Annual income from the share = \(\frac{9}{100}\) × 100 = ₹ 9

∴ The market price of the share is ₹ 120.

(ii) Let ‘X’ be the amount to be invested to obtain an annual dividend of ₹ 630.
∴ 7.5% of X is ₹ 630
∴ \(\frac{7.5}{100}\) × X = 630
∴ X = \(\frac{630 \times 100}{7.5}\)
∴ X = 8400
∴ ₹ 8400 need to be invested to obtain an annual dividend of ₹ 630.

Question 3.
Nilesh has the option of investing his money in 8% ₹ 10 shares at a premium of ₹ 3.50 or 7% ₹ 100 shares at a premium of 20%. Which of the two investments will be more profitable for him?
Solution:
For share 1:
Face value of the share (F.V.) = ₹ 10
Premium = ₹ 3.5
∴ Market value of the share (M.V.) = 10 + 3.5 = ₹ 13.5
Dividend = 8 %
∴ Annual income from the share = \(\frac{8}{100}\) × 10 = ₹ 0.8

= \(\frac{800}{135}\)
= 5.93%

For share 2:
Face value of the share (F.V.) = ₹ 100
Premium = 20%
∴ Market value of the share (M.V.) = 100 + (\(\frac{20}{100}\) × 100) = ₹ 120
Dividend = 7%
Annual income from the share = \(\frac{7}{100}\) × 100 = ₹ 7

Since, profit percentage from share 1 > profit percentage from share 2, investing in the first kind of shares will be more profitable for Nilesh.

Question 4.
Sudhakar invests ₹ 1344 in buying shares of face value ₹ 24 selling at a 12% premium. The dividend on the shares is 15% per annum. Calculate
(i) The number of shares Sudhakar buys, and
(ii) The dividend he receives annually.
Solution:
Given that,
Face value of the share (F.V.) = ₹ 24
Premium = 12%
∴ Market value of the share (M.V.) = 24 + (\(\frac{12}{100}\) × 24) = ₹ 26.88
(i) Sudhakar invests ₹ 1344 in the shares
∴ Number of shares purchased by Sudhakar = \(\frac{1344}{26.88}\) = 50
∴ Sudhakar buys 50 shares.
(ii) Dividend on the share = 15%
Annual income on one share = \(\frac{15}{100}\) × 24 = ₹ 3.6
∴ The total dividend he receives annually = 50 × 3.6 = ₹ 180
∴ Sudhakar receives ₹ 180 as his annual dividend.

Question 5.
Sameer invests ₹ 5625 in a company paying 7% per annum when the share of ₹ 10 stands for ₹ 12.50. Find Sameer’s income from this investment. If he sells 60% of these shares of ₹ 10 each, find his gain or loss in this transaction.
Solution:
Given:
Face value of the share (F.V.) = ₹ 10
Market value of the share (M.V.) = ₹ 12.5
Amount invested in shares = ₹ 5625
∴ Number of shares purchased by Sameer = \(\frac{5625}{12.5}\) = 450
Dividend = 7%
Annual income from one share = \(\frac{7}{100}\) × 10 = ₹ 0.7
∴ Sameer’s income from this investment = number of shares × annual income from one share
= 450 × 0.7
= ₹ 315
Sameer sells 60 % of these shares = \(\frac{60}{100}\) × 450 = 270 shares
Sameer purchased these shares at ₹ 12.5 per share.
∴ Purchase price for these shares = 270 × 12.5 = ₹ 3375
If he sells these shares at ₹ 10 per share, he would receive 270 × 10 = ₹ 2700
∴ In this transaction, Sameer would incur a loss of 3375 – 2700 = ₹ 675

Question 6.
Geeta buys ₹ 100 shares of a company that pays a 15% dividend. She buys the shares at a price from the market that gives her a 10% return on her investment. At what price did she buy each share?
Solution:
Given that,
Face value of the share (F.V.) = ₹ 100
Dividend = 15%
∴ Annual income from the share = \(\frac{15}{100}\) × 100 = ₹ 15
Rate of return on investment = 10%

∴ Geeta bought each share from the market at ₹ 150.

Question 7.
Tejas invests in 9% ₹ 100 shares at ₹ 145 but Shail invests in 7% ₹ 100 shares at ₹ 116. Whose investment is better?
Solution:
Investment of Tejas:
Given that, the Face value of the share (F.V.) = ₹ 100
The market value of the share (M.V.) = ₹ 145
Dividend = 9%
Annual income from the share = \(\frac{9}{100}\) × 100 = ₹ 9

Investment of Shail:
Face value of the share (F.V.) = ₹ 100
Market value of the share (M.V.) = ₹ 116
Dividend = 7%
Annual income from the share = \(\frac{7}{100}\) × 100 = ₹ 7

Since the rate of return for Tejas’s investment is greater than that for Shail’s, Tejas’s investment is better.

Question 8.
A 6% share yields 8%. Find the market value of a ₹ 100 share.
Solution:
Given that,
Face value of the share = ₹ 100
Dividend = 6%
Yield = 8%
Annual income on the share = \(\frac{6}{100}\) × 100 = ₹ 6

∴ The market value of the share = ₹ 75

Question 9.
Ashwini bought ₹ 40 shares at a premium of 40%. Find the income, if Ashwini invests ₹ 14,000 in these shares and receives a dividend at the rate of 8% on the nominal value of the shares.
Solution:
Given,
Face value of the shares (F.V.) = ₹ 40
Premium = 40%
Market value of the shares (M.V.) = 40 + (40 × \(\frac{40}{100}\))
= 40 + 16
= ₹ 56
Ashwini invests ₹ 14000 in these shares
∴ Number of shares bought by Ashwini = \(\frac{Amount Invested}{Market value of one share}\)
= \(\frac{14000}{56}\)
= 250
Dividend = 8%
∴ Annual income on one share = \(\frac{8}{100}\) × 40 = ₹ 3.2
∴ Income of Ashwini on 250 shares = 250 × 3.2 = ₹ 800
∴ Ashwini earns ₹ 800 on her investment.

Question 10.
Mr. Rutvik invests ₹ 30,000 in buying shares of a company that pays a 12% dividend annually on ₹ 100 shares selling at a premium of ₹ 50. Find
(i) The number of shares bought Mr. Rutvik and
(ii) His annual income from the shares.
Solution:
Given that,
Face value of a share (F.V.) = ₹ 100
Premium = ₹ 50
∴ Market value of a share (M.V.) = 100 + 50 = ₹ 150
Dividend =12%
Mr. Rutvik invests ₹ 30,000 in the shares.
(i) Number of shares bought by Mr. Rutvik = \(\frac{Amount invested}{Market value}\)
= \(\frac{30000}{150}\)
= 200

(ii) Dividend on the share = 12%
∴ Annual income from one share = \(\frac{12}{100}\) × 100 = ₹ 12
∴ His annual income from shares = number of shares × income from one share
= 200 × 12
= ₹ 2400

Question 11.
Rasika bought ₹ 40 shares at a discount of 40%. Find the income, if she invests ₹ 12,000 in these shares and receives a dividend at the rate of 11% on the nominal value of the shares.
Solution:
Given,
Face value of the shares (F.V.) = ₹ 40
Discount = 40%
∴ Market value of the shares (M.V.) = 40 – (40 × \(\frac{40}{100}\))
= 40 – 16
= ₹ 24
Rasika invests ₹ 12,000 in these shares.
∴ Number of shares bought by Rasika = \(\frac{Amount invested}{Market value of one share}\)
= \(\frac{12000}{24}\)
= 500
Dividend = 11%
∴ Annual income on one share = \(\frac{11}{100}\) × 40 = ₹ 4.4
∴ Rasika’s income on 200 such shares = 500 × 4.4 = ₹ 2200
∴ Rasika earns ₹ 2200 from her investment.

Question 12.
Nisha invests ₹ 15,840 in buying shares of nominal value ₹ 24 selling at a premium of 10%. The company pays a 15% dividend annually. Find
(i) The dividend she receives annually, and
(ii) The rate of return from her investment.
Solution:
Given that,
Face value of the share (F.V.) = ₹ 24
Premium = 10%
∴ Market value of the share (M.V.) = 24 + (24 × \(\frac{10}{100}\))
= 24 + 2.4
= ₹ 26.4
Dividend = 15%
∴ Annual income on the share = \(\frac{15}{100}\) × 24 = ₹ 3.6
Nisha invests ₹ 15,840 in these shares.
∴ Number of shares bought by Nisha

(i) Annual dividend received by Nisha = Number of shares × annual income from one share
= 600 × 3.6
= ₹ 2160

(ii) Rate of return from the investment

Question 13.
Ashutosh buys 80, ₹ 100 shares at a discount of 20% and receives a return of 12% on his money. Calculate
(i) The amount invested by Ashutosh.
(ii) The rate of dividend paid by the company.
Solution:
Given
Face value of the shares (F.V.) = ₹ 100
Discount = 20%
∴ Market value of the shares (M.V.) = 100 – (100 × \(\frac{20}{100}\)) = ₹ 80
(i) Amount invested by Ashutosh = number of shares × market value of the shares
= 80 × 80
= ₹ 6400

(ii) Ashutosh receives a return of 12% on his money.
∴ Ashutosh’s income from shares = \(\frac{12}{100}\) × 6400 = ₹ 768
∴ Ashutosh’s annual income from one share = \(\frac{768}{80}\) = ₹ 9.6
Annual income from one share = \(\frac{\text { Dividend }}{100} \times \text { Face value }\)
∴ 9 6 = \(\frac{\text { Dividend }}{100} \times 100\)
∴ Rate of dividend = 9.6%

Question 14.
Vaishnavi bought 1000, ₹ 100 shares from the stock market carrying 8% dividend quoted at ₹ 130. A few days later the market value of the shares went up by 10%. Vaishnavi sold all her shares. What was her total income from this transaction?
Solution:
Given that,
Face value of the shares (F.V.) = ₹ 100
The market value of the shares (M.V.) = ₹ 130
Dividend = 8%
Income from the each share = \(\frac{8}{100}\) × 100 = ₹ 8
Number of shares bought by Vaishnavi = 1000
∴ Vaishnavi’s income from dividend = 1000 × 8 = ₹ 8000
The price of the shares went up by 10%
New market value of the shares = 130 + (130 × \(\frac{10}{100}\)) = ₹ 143
Vaishnavi sold the shares at ₹ 143 which she bought at ₹ 130 each.
∴ Vaishnavi’s profit on one share =143 – 130 = ₹ 13
∴ Vaishnavi’s profit after selling all her shares =1000 × 13 = ₹ 13,000
Vaishnavi’s total income from this transaction = Income from dividend + income from sale of shares
= 8,000 + 13,000
= ₹ 21,000
∴ Vaishnavi’s total income from this transaction was ₹ 21,000.

Question 15.
Mr. Dinesh invests ₹ 20,800 in 6% ₹ 100 shares at ₹ 104, and ₹ 14,300 in 10.5% ₹ 100 shares at ₹ 143. What will be his annual income from the shares?
Solution:
For 1st kind of shares,
Face value of shares (F.V.) = ₹ 100
Dividend = 6%
∴ Annual income from one share = \(\frac{6}{100}\) × 100 = ₹ 6
Market value of the share (M.V.) = ₹ 104
Total amount invested = ₹ 20,800

∴ Total income from 1st kind of shares = 200 × 6 = ₹ 1200
For 2nd kind of shares,
Face value of shares (F.V.) = ₹ 100
Dividend = 10.5%
∴ Annual income from one share = \(\frac{10.5}{100}\) × 100 = ₹ 10.5
Market value of the share (M.V.) = ₹ 143
Total amount invested = ₹ 14300

∴ Total income from 2nd kind of shares = 100 × 10.5 = ₹ 1050
∴ Total annual income of Dinesh from both these shares = 1200 + 1050 = ₹ 2250

Question 16.
A company declares a semi-annual dividend of 5%. Daniel has 400 shares of the company. If Daniel’s annual income from the shares is ₹ 1,000, find the face value of each share.
Solution:
Given that,
Semi-annual dividend = 5%
∴ Annual dividend = 10%
Number of shares with Daniel = 400
Daniel’s annual income from the shares = ₹ 1000
∴ Annual income from one share = \(\frac{1000}{400}\) = ₹ 2.5
But annual income from one share = \(\frac{\text { Annualdividend }}{100} \times \text { Face value }\)
∴ 2.5 = \(\frac{10}{100}\) × Face value of the share
∴ Face value of the share = ₹ 25

Question 17.
Bhargav buys 400, ₹ 20 shares at a premium of ₹ 4 each and receives a dividend of 12%. Find
(i) The amount invested by Bhargav.
(ii) His total income from the shares.
(iii) Percentage return on his money.
Solution:
Given that,
Face value of the shares (F.V.) = ₹ 20
Premium = ₹ 4
∴ Market value of the shares (M.V.) = ₹ 24
Dividend = 12%
∴ Annual income from the share = \(\frac{12}{100}\) × 20 = ₹ 2.4
Bhargav buys 400 shares
(i) The amount invested by Bhargav = number of shares × market value
= 400 × 24
= ₹ 9600

(ii) Bhargav’s income from the shares = number of shares × annual income from one share
= 400 × 2.4
= ₹ 960

(iii) Percentage return on Bhargav’s money

∴ Bhargav gets 10% as the rate of return on his money.

Question 18.
Anil buys 350 ₹ 100 shares of a company at a premium of 20% from the market. The company pays 12% dividend annually. Find
(i) the investment made by Anil,
(ii) his annual income from the shares, and
(iii) the rate of return from the shares.
Solution:
Given that,
Face value of shares (F.V.) = ₹ 100
Premium = 20%
∴ Market value of shares (M.V.) = 100 + (\(\frac{20}{100}\) × 100) = ₹ 120
Dividend = 12%
∴ Annual income from one share = \(\frac{12}{100}\) × 100 = ₹ 12
Anil buys 350 shares.
(i) Amount invested by Anil = number of shares × market value
= 350 × 120
= ₹ 42,000

(ii) Anil’s annual income from the shares = number of shares × annual income from one share
= 350 × 12
= ₹ 4200

(iii) Rate of return from shares

∴ The rate of return from Anil’s shares is 10%.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.6 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.6

Question 1.
M/s Janaseva sweet mart sold sweets of ₹ 3,86,000. What CGST and SGST he will pay if the rate of GST is 5%?
Solution:
Given that M/s Janaseva sweet mart sold sweets of ₹ 3,86,000
∴ Bill amount = ₹ 3,86,000
GST payable at the rate 5%
∴ CGST and SGST applicable is 2.5% each
∴ CGST on the bill = \(\frac{2.5}{100}\) × 3,86,000 = ₹ 9650
and SGST on the bill = \(\frac{2.5}{100}\) × 3,86,000 = ₹ 9650

Question 2.
Janhavi Gas Agency purchased some gas cylinders for ₹ 5,00,000 and sold them to the customers for ₹ 5,90,000. Find the amount of GST payable and the amount of ITC. 5% GST is applicable.
Solution:
Given that, Janhavi Gas Agency purchased some gas cylinders for ₹ 5,00,000 and GST applicable is 5%.
∴ Input tax (ITC) = 5% of 5,00,000
= \(\frac{5}{100}\) × 5,00,000
= ₹ 25,000
Janhavi Gas Agency sold the gas cylinders for ₹ 5,90,000
∴ Output tax for Janhavi Gas Agency = 5% of 5,90,000
= \(\frac{5}{100}\) × 5,90,000
= ₹ 29,500
GST payable = Output tax – Input tax (ITC)
= 29,500 – 25,000
= ₹ 4500
∴ GST payable for Janhavi Gas Agency is ₹ 4,500 and ITC is ₹ 25,000.

Question 3.
A company dealing in mobile phones purchased mobile phones worth ₹ 5,00,000 and sold the same to customers at ₹ 6,00,000. Find the amount of ITC and amount of GST if the rate of GST is 12%.
Solution:
Given that the rate of GST applicable is 12%.
The company purchased mobile phones worth ₹ 5,00,000.
∴ Input tax (ITC) = 12% of 5,00,000
= \(\frac{12}{100}\) × 5,00,000
= ₹ 60,000
The company dealing in mobile phones sold the same to customers at ₹ 6,00,000.
∴ Output tax of the company = 12% of 6,00,000
= \(\frac{12}{100}\) × 6,00,000
= ₹ 72,000
GST payable for the company = Output tax – Input tax (ITC)
= 72,000 – 60,000
= ₹ 12,000
∴ The ITC for the company is ₹ 60,000 and GST payable is ₹ 12,000.

Question 4.
Prepare business to customers (B2C) tax invoice using given information. Write the name of supplier, address, state, Date, Invoice Number, GSTIN etc. as per your choice
Supplier: ___________
Address: ___________
State: ___________
Date: ___________
Invoice No: ___________
GSTIN: ___________
Particular: Rate of Sarees – ₹ 2750
Rate of GST 5% HSN 5407 – 2 pcs
Rate of Kurta – ₹ 750
Rate of GST 12% HSN 5408
Solution:
Supplier: M/s Swaglife Fashions
Address: 143, Shivaji Rasta, Mumbai 400001
Mobile No. 9263692111
Email: abc@gmail.com
State: Maharashtra
Date: 31/08/19
Invoice No: GST/110
GSTIN: 27ABCDE1234HIZS

∴ Rate of 1 saree = ₹ 2750
∴ Rate of 2 sarees = 2 x 2750 = ₹ 5500
∴ GST on sarees = 12% of 5500
= \(\frac{12}{100}\) × 5500
= ₹ 660
∴ CGST = SGST = ₹ 330
∴ Rate of 1 Kurta = ₹ 750
∴ GST on Kurta = 12% of 750
= \(\frac{12}{100}\) × 750
= ₹ 90
∴ CGST = SGST = ₹ 45

Question 5.
Heena Enterprise sold cosmetics worth ₹ 25,000 to Leena traders, a retailer. Leena Traders sold it further to Meena Beauty Products for ₹ 30,000. Meena Beauty Product sold it further to the customers for ₹ 40,000. The rate of GST is 18%. Find
(i) GST Payable by each party
(ii) CGST and SGST
Solution:
The trading chain,

∴ Output tax for Heena Enterprises = 18% of 25,000
= \(\frac{18}{100}\) × 25,000
= ₹ 4,500
∴ GST payable by Heena Enterprises
Now output tax for Leena traders = 18% of 30,000
= \(\frac{18}{100}\) × 30,000
= ₹ 5,400
∴ GST payable by Leena traders = Output tax – Input tax
= 5,400 – 4,500
= ₹ 900
∴ Output tax for Meena beauty products = 18% of 40,000
= \(\frac{18}{100}\) × 40,000
= ₹ 7,200
∴ GST payable by Meena beauty products = Output tax – Input tax
= 7,200 – 5,400
= ₹ 1,800

(ii) Now, CGST = SGST = \(\frac{\text { GST }}{2}\) = 9%
∴ Statement of GST payable at each stage can be tabulated as:

Question 6.
‘Chitra furnishings’ purchased tapestry (curtain cloth) for ₹ 28,00,000 and sold for ₹ 44,80,000. Rate of GST is 5%. Find
(i) Input Tax
(ii) Output Tax
(iii) ITC
(iv) CGST and SGST
Solution:
Given, that ‘Chitra furnishings’ purchased tapestry (curtain cloth) for ₹ 28,00,000 and rate of GST is 5%
(i) Input tax = 5% of 28,00,000
= \(\frac{5}{100}\) × 28,00,000
= ₹ 1,40,000
The tapestry was sold at ₹ 44,80,000

(ii) Output tax = 5% of 44,80,000
= \(\frac{5}{100}\) × 44,80,000
= ₹ 2,24,000

(iii) Now ITC = Input tax = ₹ 1,40,000
GST payable = Output tax – ITC
= 2,24,000 – 1,40,000
= ₹ 84,000

(iv) CGST = SGST = \(\frac{\text { GST Payable }}{2}\)
= \(\frac{84,000}{2}\)
= ₹ 42,000
∴ CGST = SGST = ₹ 42,000

Question 7.
Two friends ‘Aditi’ and ‘Vaishali’ went to a restaurant. They ordered 2 Masala Dosa costing ₹ 90 each 2 coffee costing ₹ 60 each and 1 sandwich costing ₹ 80. GST is charged at 5%. Find the Total amount of the bill including GST.
Solution:
Aditi and Vaishali ordered for 2 Masala Dosas, 2 Coffees and 1 Sandwich
∴ Total price of their order = 2 × 90 + 2 × 60 + 80 = ₹ 380
GST is charged at 5%
∴ GST on the total order = 5% × 380
= \(\frac{5}{100}\) × 380
= ₹ 19
∴ Total bill amount including GST = 380 + 19 = ₹ 399

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.5

Question 1.
Three partners shared the profit in a business in the ratio 5 : 6 : 7. They had partnered for 12 months, 10 months, and 8 months respectively. What was the ratio of their investments?
Solution:
Let the ratio of investments of the three partners be p : q : r.
They partnered for 12 months, 10 months, and 8 months respectively.
∴ The profit shared by the partners will be in proportion to the product of capital invested and their respective time periods.
∴ 12 × p : 10 × q : 8 × r = 5 : 6 : 7

From (i) & (ii), we have
p : q : r = 50 : 72 : 105
∴ The ratio of their investments was 50 : 72 : 105.

Question 2.
Kamala, Vimala and Pramila enter into a partnership. They invest ₹ 40,000, ₹ 80,000 and ₹ 1,20,000 respectively. At the end of the first year, Vimala withdraws ₹ 40,000, while at the end of the second year, Pramila withdraws ₹ 80,000. In what ratio will the profit be shared at the end of 3 years?
Solution:
Given that, Kamala, Vimala, and Pramila invest ₹ 40,000, ₹ 80,000, and ₹ 1,20,000 respectively.
The ratio of profits is to be calculated at the end of 3 years.
Vimala withdraws ₹ 40,000 at the end of the first year.
∴ Vimala invested ₹ 80,000 for one year and 40,000 for 2 years.
Pramila withdraws ₹ 80,000 at the end of the second year.
∴ Pramila invested ₹ 1,20,000 for two years and 40,000 for one year.
Kamala invested ₹ 40,000 for all the 3 years.
∴ The ratio of profits to be shared at the end of 3 years will be
= 40,000 × 3 : 80,000 × 1 + 40,000 × 2 : 1,20,000 × 2 + 40,000 × 1
= 1,20,000 : 1,60,000 : 2,80,000
= 12 : 16 : 28
= 3 : 4 : 7

Alternate Method:
Given that, Kamala, Vimala and Pramila invest ₹ 40,000, ₹ 80,000 & ₹ 1,20,000 respectively.
Given, information can be tabulated as:

∴ The profits to be shared at the end of 3 years will be
= 1,20,000 : 1,60,000 : 2,80,000
= 12 : 16 : 28
= 3 : 4 : 7

Question 3.
Sanjeev started a business investing ₹ 25,000 in 1999. In 2000, he invested an additional amount of ₹ 10,000 and Rajeev joined him with an amount of ₹ 35,000. In 2001, Sanjeev invested another additional amount of ₹ 10,000 and Pawan joined them with an amount of ₹ 35,000. What will be Rajeev’s share in the profit of ₹ 1,50,000 earned at the end of 3rd year from the start of the business in 1999?
Solution:
The given information can be tabulated as:

∴ The ratio of profits to be shared at the end of 3 years will be 1,05,000 : 70,000 : 35,000
i.e. in the proportion 3 : 2 : 1
Given, profit earned ₹ 1,50,000/-
∴ Rajeev’s share in the profit = \(\frac{2}{6}\) × 1,50,000 = ₹ 50,000/-

Question 4.
Teena, Leena, and Meena invest in a partnership in the ratio: 7/2, 4/3, 6/5. After 4 months, Teena increases her share by 50%. If the total profit at the end of one year is ₹ 21,600, then what is Leena’s share in the profit?
Solution:
Investment of Teena, Leena and Meena are in the ratio \(\frac{7}{2}: \frac{4}{3}: \frac{6}{5}\)
After 4 months, Teena’s share increases by 50%.
i.e. \(\frac{7}{2}+\left(\frac{7}{2} \times \frac{50}{100}\right)=\frac{7}{2}+\frac{7}{4}\)
i.e. \(\frac{21}{4}\)
The profit will be shared in the proportion of product of capitals and respective time periods in months.
i.e. \(\frac{7}{2} \times 4+\frac{21}{4} \times 8: \frac{4}{3} \times 12: \frac{6}{5} \times 12\)
i.e. 56 : 16 : \(\frac{72}{5}\)
i.e. 7 : 2 : \(\frac{9}{5}\)
i.e. in the proportion 35 : 10 : 9 …..[Multiplying throughout by 5]
Given that profit at the end of one year = ₹ 21,600/-
∴ Leena’s share in the profit = \(\frac{10}{54}\) × 21,600
= 5 × 800
= 4000
∴ Leena’s share in the profit is ₹ 4000/-.

Question 5.
Dilip and Pradeep invested amounts in the ratio 2 : 1, whereas the ratio between amounts invested by Dilip and Sudip was 3 : 2. If ₹ 1,49,500 was their profit, how much amount did Sudip receive?
Solution:
Let the amounts invested by Dilip, Pradeep and Sudip be ₹ ‘d’, ₹ ‘p’ and ₹ ‘s’ respectively.
Given that, d : p = 2 : 1
∴ d : p = 6 : 3 …..(i)
and d : s = 3 : 2
∴ d : s = 6 : 4 …..(ii)
From (i) and (ii),
d : p : s = 6 : 3 : 4
∴ The ratio of profits to be shared among Dilip, Pradeep and Sudip will be 6 : 3 : 4.
Given, profit earned = ₹ 1,49,500/-
∴ Sudip’s share in the profit = \(\frac{4}{13}\) × 1,49,500
= 4 × 11,500
= ₹ 46,000/-

Question 6.
The ratio of investments of two partners Jatin and Lalit is 11 : 12 and the ratio of their profits is 2 : 3. If Jatin invested the money for 8 months, find for how much time Lalit invested his money.
Solution:
Let ‘x’ be the time in months for which Lalit invested his money
Jatin and Lalit invested their money in the ratio 11 : 12.
Jatin invested his money for 8 months and the ratio of their profits is 2 : 3.
∴ 11 × 8 : 12 × x = 2 : 3
∴ \(\frac{88}{12 x}=\frac{2}{3}\)
∴ x = \(\frac{88 \times 3}{2 \times 12}\)
∴ x = 11
∴ Lalit invested his money for 11 months.

Question 7.
Three friends had dinner at a restaurant. When the bill was received, Alpana paid \(\frac{2}{3}\) as much as Beena paid and Beena paid \(\frac{1}{2}\) as much as Catherin paid. What fraction of the bill did Beena pay?
Solution:
Let ‘T’ be the total bill amount at the restaurant and ‘a’, ‘b’, and ‘c’ be the share of Alpana, Beena, and Catherin respectively.
Given, that Alpana paid \(\frac{2}{3}\) as much as Beena paid
∴ a = \(\frac{2}{3}\) b …..(i)
Also, Beena paid \(\frac{1}{2}\) as much as Catherin paid.
∴ b = \(\frac{1}{2}\) c
∴ c = 2b …….(ii)
∴ Three friends paid the total bill amount.
∴ a + b + c = T …..(iii)
Using (i) and (ii) in (iii), we get

Thus, Beena paid \(\left(\frac{3}{11}\right)^{\text {th }}\) fraction of the total bill amount.

Question 8.
Roy starts a business with ₹ 10,000, Shikha joins him after 2 months with 20% more investment than Roy, after 2 months Tariq joins him with 40% less than Shikha. If the profit earned by them at the end of the year is equal to twice the difference between the investment of Roy and ten times the investment of Tariq. Find the profit of Roy?
Solution:
Given that, Roy starts the business with ₹ 10,000.
Shikha joins him after 2 months with 20% more investment than Roy.
∴ Shikha’s investment = 10,000 + (10,000 × \(\frac{20}{100}\)) = ₹ 12,000
Tariq joins after two more months with an investment 40% less than Shikha.
∴ Tariq’s investment = 12,000 – (12,000 × \(\frac{40}{100}\)) = ₹ 7,200
Now, the profit will be shared in the proportion of product of capitals and respective periods in months.
i.e. 10,000 × 12 : 12,000 × 10 : 7,200 × 8
i.e. in the proportion, 25 : 25 : 12 …..(i) [Dividing throughout by 4,800]
Given that, profit at the end of the year = twice of the difference between investment of Roy and ten times the investment of Tariq.
∴ Profit = 2 [(10 × 7,200) – 10,000]
= 2[72,000 – 10,000]
= 2 × 62,000
= ₹ 1,24,000
∴ Roy’s share of profit = \(\frac{25}{62}\) × 1,24,000 …..[From (i)]
= ₹ 50,000/-

Question 9.
If 4(P’s Capital) = 6(Q’s Capital) = 10 (R’s Capital), then out of the total profit of ₹ 5,580, what is R’s share?
Solution:
Let ‘p’, ‘q’ and ‘r’ be P, Q and R’s Capital for the business respectively.
∴ 4p = 6q = 10r
L.C.M of 4, 6, 10 = 60
∴ We take 4p = 6q = 10r = 60x
∴ p = 15x, q = 10x, r = 6x
∴ p : q : r = 15 : 10 : 6
Given that total profit = ₹ 5580
R’s share in the profit = \(\frac{6}{31}\) × 5580 = ₹ 1080/-

Question 10.
A and B start a business, with A investing the total capital of ₹ 50,000, on the condition that B pays interest at the rate of 10% per annum on his half of the capital. A is a working partner and receives ₹ 1,500 per month from the total profit and any profit remaining is equally shared by both of them. At the end of the year, it was found that the income of A is twice that of B. Find the total profit for the year?
Solution:
Let ‘x’ and ‘y’ be the profits earned by A and B respectively and let ‘z’ be the total profit for the year.
A is the working partner and receives ₹ 1500 per month from the total profit.
i.e. 12 × 1500 = ₹ 18,000 at the end of the year.
The remaining profit is shared between A and B equally.
∴ y = \(\frac{z-18000}{2}\) …..(i)
Thus, profit earned by A at the end of that year is given by
x = 18000 + \(\left(\frac{z-18000}{2}\right)\)
∴ x = \(\frac{z+18000}{2}\) ……(ii)
A invests the entire capital on the condition that B pays A interest at the rate of 10% per annum on his half of the capital.
∴ At the end of the first year,
A will receive \(\frac{10}{100}\) × 25,000 i.e. ₹ 2500/- over and above his share of profit.
∴ A’s income = Profit of A + 2500 = x + 2500
Given that,
income of A = twice the income of B
∴ x + 2500 = 2y …..(iii)
Using (i) and (ii) in (iii), we get
\(\frac{z+18000}{2}\) + 2500 = 2\(\left(\frac{z-18000}{2}\right)\)
z + 18000 + 5000 = 2(z – 18000)
z + 23000 = 2z – 36000
∴ z = 59,000
∴ The total profit for the year = ₹ 59,000/-

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

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 9 Commercial Mathematics Ex 9.2

Question 1.
Mr. Sarad purchased a laptop for ₹ 24,000 and sold it for ₹ 30,000. What was the profit percentage?
Solution:
Cost price (C.P.) = ₹ 24000
Selling price (S. P.) = ₹ 30,000
Profit = S.P. – C.P.
= 30,000 – 24,000
= 6,000

∴ Profit Percentage = 25%

Question 2.
Shraddha purchased a mobile phone and refrigerator for ₹ 18,000 and ₹ 15,000 respectively. She sold the refrigerator at a loss of 20% and the mobile at a profit of 20%. What is her overall profit or loss?
Solution:
C.P. of mobile phone = ₹ 18,000
Profit percentage on mobile phone = 20%
Selling price (S.P.) of mobile phone = 18,000 (1 + \(\frac{20}{100}\))
= 18,000 (1 + \(\frac{1}{5}\))
= 18,000 × \(\frac{6}{5}\)
= ₹ 21,600
C.P. of refrigerator = 15,000
Loss percentage on refrigerator = 20%
∴ Selling price (S.P.) = 15,000(1 – \(\frac{20}{100}\))
= 15,000(1 – \(\frac{1}{5}\))
= 15,000 × \(\frac{4}{5}\)
= ₹ 12,000
∴ Total gelling price for the transaction = 21,600 + 12,000 = ₹ 33,600
Total cost price (purchase price) for the transaction = 18,000 + 15,000 = ₹ 33,000
∴ Overall profit made by Shraddha = Total S.P. – Total C.P.
= 33,600 – 33,000
= ₹ 600
Thus, Shraddha made on overall profit of ₹ 600.

Question 3.
A vendor bought toffees at 6 for ₹ 10. How many for ₹ 10 must he sell to gain 20%?
Solution:
Vendor bought toffees at the rate of 6 for ₹ 10
∴ Cost price of one toffee = \(\frac{10}{6}\)
i.e. C.P. = \(\frac{10}{6}\) …….(i)
Let x be the number of toffees he must sell in ₹ 10 to gain 20%
i.e. S.P. = \(\frac{10}{x}\) …….(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }}\)
Using (i) and (ii) we have

∴ 30(6 – x) = 6x
∴ 180 – 30x = 6x
∴ 36x = 180
∴ x = 5
The vendor must sell 5 toffees for ₹ 10 in order to gain 20%.

Question 4.
The percentage profit earned by selling an article for ₹ 2,880 is equal to the percentage loss incurred by selling the same article for ₹ 1,920. At what price the article should be sold to earn a 25% profit?
Solution:
Let x be C.P. of the article
Let y % be both, the gain and loss made when article is sold at ₹ 2,880 and ₹ 1,920 respectively. Then
x + \(\frac{y}{100}\) x = 2880 ……(i)
x – \(\frac{y}{100}\) x = 1920 …..(ii)
Adding (i) and (ii), we get
2x = 4800
∴ x = 2400
i.e. C.P. of the article = ₹ 2400
Required profit percentage = 25%

∴ The article should be sold at ₹ 3000 to earn 25% profit.

Question 5.
A cloth merchant advertises for selling cloth at a 4% loss. By using a faulty meter scale, he is earning a profit of 20%. What is the actual length of the scale?
Solution:
Let the cost price of the cloth be ₹ ‘x’ per meter
He claims a loss of 4%
∴ Selling price of the cloth
S.P. = C.P.(1 – \(\frac{loss%}{100}\))
= x(1 – \(\frac{4}{100}\))
= 0.96x …..(i)
The actual cost price of the cloth is lower as the cloth is measured by a faulty meter scale.
Given that shopkeeper’s profit = 20%

∴ The actual cost price is 0.8 times the cost price as advertised.
In other words, the meter scale used for the fraud is 0.8 times the meter scale that should have been used.
∴ The length of the faulty meter scale used = 0.8 × 1 = 0.8 meter
∴ The actual length of the scale is 0.8 meters.

Question 6.
Sunil sells his bike worth ₹ 25,000 to Rohit at a profit of 20%. After 6 months Rohit sells the bike back to Sunil at a loss of 20%. Find the total profit percent of Sunil considering both the transactions.
Solution:
Sunil sells his bike to Rohit at 20% profit.
So S.P. of the bike for Sunil
= 25000 + \(\frac{20}{100}\) × 25000
= 25000 + 5000
= 30000
∴ Cost price of bike to Rohit = ₹ 30000
Rohit sells the bike back to Sunil at 20% loss
∴ S.P. of the bike for Rohit = 30000 – \(\frac{20}{100}\) × 30000
= 30000 – 6000
= 24000
∴ In second transaction Sunil pays 24000 to Rohit
In the first transaction, he had received 30000 from Rohit
∴ Sunil made a profit of ₹ (30000 – 24000) = ₹ 6000
Sunil earned this profit on the bike which costed him ₹ 25000
∴ Total profit % that Sunil makes = \(\frac{6000}{25000}\) × 100
= \(\frac{600}{25}\)
= 24
∴ Sunil makes 24% profit considering both the transactions.

Question 7.
By selling a book at ₹ 405 bookseller incurs a loss of 25%. Find the cost price of the book.
Solution:
S.P. = ₹ 405
Loss% = 25
S.P. when there is a loss is given by
S.P. = C.P. × \(\frac{\text { Loss } \%}{100}\)

∴ The cost price of the book is ₹ 540.

Question 8.
A cloth costs ₹ 675. If it is sold at a loss of 20%, what is its cost price as a percentage of its selling price?
Solution:
C.P. = ₹ 675
Loss% = 20%
∴ Loss made in selling = \(\frac{20}{100}\) × 675 = ₹ 135
S.P. = C.P. – Loss
= 675 – 135
= ₹ 540
Let C.P. be x % S.P.,
Then 675 = \(\frac{x}{100}\) × 540
∴ x = \(\frac{675 \times 100}{540}\) = 125
∴ Cost price is 125% of the selling price.

Question 9.
Ashwin buys an article for ₹ 500. He marks it for sale at 75% more than the cost price. He offers a 25% discount on the marked price to his customer. Calculate the actual percentage of profit made by Ashwin.
Solution:
C.P. = ₹ 500
Marked price = C.P. + \(\frac{75}{100}\) × C.P.
= \(\frac{75}{100}\) × 500
= 500 + 75 × 5
= 500 + 375
= 875
25% discount was given on marked price
∴ Discount = \(\frac{25}{100}\) × 875 = \(\frac{875}{4}\)
Selling price = marked price – discount

∴ Ashwin makes 31.25% profit.

Question 10.
The combined cost price of a refrigerator and a mixer is ₹ 12,400. If the refrigerator costs 600% more than the mixer, find the cost price of the mixer.
Solution:
Let ₹ x be the cost price of the mixer.
The cost price of the refrigerator = x + \(\frac{600}{100}\) x
= x + 6x
= 7x
Total cost price =12400 …..[Given]
i.e. x + 7x = 12400
i.e. 8x = 12400
∴ x = \(\frac{12400}{8}\) = 1550
∴ The cost price of mixer is ₹ 1550.

Question 11.
Find the single discount equivalent to the discount series of 5%, 7%, and 9%.
Solution:
Let the marked price be ₹ 100
After 1st discount the price = 100(1 – \(\frac{5}{100}\)) = 95
After 2nd discount the price = 95(1 – \(\frac{7}{100}\)) = \(\frac{95 \times 93}{100}\)
After 3rd discount the price = \(\frac{95 \times 93}{100}\left(1-\frac{9}{100}\right)\)
= \(\frac{95 \times 93 \times 91}{100 \times 100}\)
= \(\frac{803985}{10000}\)
= 80.3985 ~ 80.4
Selling price after 3 discounts is ₹ 80.4
Single equivalent discount = Marked price – Selling price
= 100 – 80.4
= ₹ 19.6
∴ Single equivalent discount is 19.6%.

Question 12.
The printed price of a shirt is ₹ 390. Lokesh pays ₹ 175.50 for it after getting two successive discounts. If the first discount is 10%, find the second discount.
Solution:
Marked price = ₹ 390
After the first discount of 10%, the price of the shirt

∴ x = 50
∴ Second discount is 50%

Question 13.
Amar, a manufacturer, gives a discount of 25% on the list price to his distributor Akbar, Akbar sells at a 10% discount on the list price to his customer Anthony. Anthony paid ₹ 540 for the article. What is the profit percentage of Akbar on his cost price?
Solution:
Let ₹ ‘x’ be the list price of the article.
Amar gives a discount of 25% on the list price.
∴ Selling price for Amar = \(x\left(1-\frac{25}{100}\right)\)
= \(x\left(1-\frac{1}{4}\right)\)
= ₹ \(\frac{3 x}{4}\)
Amar sells the article to Akbar
Cost price of article for Akbar = ₹ \(\frac{3 x}{4}\) ……(i)
Akbar sells the article to Anthony at 10% discount on list price
∴ Selling price for Akbar = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(1-\frac{1}{10}\right)\)
= ₹ \(\frac{9 x}{10}\) …..(ii)
Profit percentage = \(\frac{\text { S.P. }-\text { C.P. }}{\text { C.P. }} \times 100\)
Using (i) and (ii), we have the profit percentage for Akbar as,

∴ Akbar gets a profit of 20% on his cost price.

Question 14.
A man sells an article at a profit of 25%. If he had bought it at a 10% loss and sold it for ₹ 7 less, he would have gained 35%. Find the cost price of the article.
Solution:
Let ₹ ‘x’ be the C.P. of the article
∴ Article was sold at 25% profit
∴ S.P. of the article = \(x\left(1+\frac{25}{100}\right)\)
= \(x\left(1+\frac{1}{4}\right)\)
= 1.25x
If the article was bought at 10% loss
i.e., the new C.P. = \(x\left(1-\frac{10}{100}\right)\)
= \(x\left(\frac{9}{10}\right)\)
= 0.9x
and sold at ₹ 7 less
∴ New S.P. = 1.25x – 7
Then, the profit would have been 35%
Using profit percentage = \(\frac{\text { S.P.-C.P. }}{\text { C.P. }} \times 100\)

∴ Cost price of the article is ₹ 200

Question 15.
Mr. Mehta sold his two luxury cars at ₹ 39,10,000 each. On one he gains 15% but on the other, he loses 15%. How much does he gain or lose in the whole transaction?
Solution:
Let x, y be the C.P. of two cars.
S.P. of both the cars = 39,10,000 …..[Given]
∴ One car is sold at 15% loss
∴ S.P. of the first car = x – \(\frac{15}{100}\)x
∴ \(\frac{85}{100}\)x = 39,10,000
∴ x = \(\frac{39,10,000 \times 100}{85}\)
∴ x = 46,000 × 100
∴ x = 46,00,000
Other car is sold at 15% gain
∴ S.P. of second car = y + \(\frac{15}{100}\) y
∴ y + \(\frac{15}{100}\) y = 39,10,000
∴ \(\frac{115}{100}\)y = 39,10,000
∴ y = \(\frac{39,10,000 \times 100}{115}\)
∴ y = 34,000 × 100
∴ y = 34,00,000
x + y = Total C.P. of two cars
= 46,00,000 + 34,00,000
= 80,00,000
Total S.P. = 39,10,000 + 39,10,000 = 78,20,000
∴ S.P. < C.P.
∴ There is a loss of ₹ (80,00,000 – 78,20,000) = ₹ 1,80,000
∴ Loss % = \(\frac{1,80,000}{80,00,000} \times 100\)
= \(\frac{18}{8}\)
= 2.25
∴ Mr. Mehta bears a 2.25% loss in the whole transaction.