Bhagya

Students can download Maths Chapter 1 Set Language Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Additional Questions

Choose the correct answer

Question 1.
A = {set of odd natural numbers}, B = {set of even natural numbers}, then A and B are……….
(a) equal set
(b) equivalent sets
(c) overlapping sets
(d) disjoint sets
Solution:
(d) disjoint sets

Question 2.
Number of subsets in set A = {1, 2, 3} is
(a) 3
(b) 6
(c) 8
(d) 9
Solution:
(c) 8

Question 3.
The set does not have a proper subset is
(a) Finite set
(b) Infinite set
(c) Null set
(d) Singleton set
Solution:
(c) Null set

Question 4.
Sets having the same number of elements are called
(a) overlapping sets
(b) disjoints sets
(c) equivalent sets
(d) equal sets
Solution:
(c) equivalent sets

Question 5.
The set (A – B) ∪ (B – A) is
(a) AΔB
(b) A∪B
(c) A∩B
(d) A’∪B’
Solution:
(a) AΔB

Question 6.
The set of (A∪B) – (A∩B) is
(a) (A∪B)’
(b) AΔB
(c) (A∩B)’
(d) A’∪B’
Solution:
(b) AΔB

Question 7.
The set {x : x ∈ A, x ∈ B, x ∉ A∩B} is
(a) A∩B
(b) A∪B
(c) A – B
(d) AΔB
Solution:
(d) AΔB

Question 8.
The number of elements of the set {x : x ∈ Z , x² = 1} is
(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(c) 2

Question 9.
If A is a proper subset of B, then A∩B =…………..
(a) A
(b) B
(c) 0
(d)A∪B
Solution:
(a) A

Question 10.
The shade region with adjoint diagram represents ……….

(a) A – B
(b) B – A
(c) A’
(d) B’
Solution:
(c) A’

Question 11.
From the given venn diagram (A∪B)’ is ………..

(a) {5, 6}
(b) {1, 2, 3, 4, 7}
(c) {1, 2, 3, 4, 5, 6, 7}
(d) {8, 9}
Solution:
(d) {8, 9}

Question 12.
If n(A∪B∪C) = 73, n(A) = 2x, n(B) = 3x, n(C) = 5x, n(A∩B) = 10, n(B∩C) = 15, n(A∩C) = 5 and n(A∩B∩C) = 3, then the value of x is ………
(a) 9
(b) 10
(c) 5
(d) 18
Solution:
(b) 10

Question 13.
For any three sets, n(A∪B∪C) = 60, n(A) = 25, n(B) = 20, n(C) = 15, n(A∩B) = 10, n(B∩C) = 7, n(A∩C) = 3, then n(A∩B∩C) is……….
(a) 10
(b) 15
(c) 20
(d) 25
Solution:
(c) 20

Question 14.
If n(U) = 70, n(A) = 25, n(B) = 30, n(A∩B) = 5, then n(A∪B)’ is……….
(a) 5
(b) 10
(c) 15
(d) 20
Solution:
(d) 20

Question 15.
Which of the following is not correct?
(a) A – (B∪C) = (A – B) ∩ (A – C)
(b) A – (B∩C) = (A – B) ∪ (A – C)
(c) (A∪B)’ = A’∩B’
(d) A’∪B’ = (A – B)’
Solution:
(d) A’∪B’ = (A – B)’

Answer the following questions.

Question 1.
Write the following in “Roster” form?
(a) A = set of the months having 31 days.
(b) B = {x : x is a natural number of 2 digits divisible by 13}
(c) C = {set of vowels in the word “father”}
(d) D = {x : 5 < x ≤ 10; x ∈ N}
(e) E = {x : x is a square natural number less than 16}
Solution:
(a) A = {Jan, March, May, July, Aug, Oct, Dec}
(b) B = {13, 26, 39, 52, 65, 78, 91}
(c) C = {a, e}
(d) D = {6, 7, 8, 9, 10}
(e) E = {1, 4, 9}

Question 2.
Given that A = {1, 3, 5, 7} B = {1, 2, 4, 6, 8}. Find
(i) AΔB and
(ii) BΔA
Solution:
(i) A = {1, 3, 5, 7}; B = {1, 2, 4, 6, 8}
A – B = {1, 3, 5, 7} – {1, 2, 4, 6, 8}
= {3, 5, 7}
B – A = {1, 2, 4, 6, 8} – {1, 3, 5, 7}
= {2, 4, 6, 8}
AΔB = (A – B) ∪ (B – A)
= {3, 5, 7} ∪ {2, 4, 6, 8}
= {2, 3, 4, 5, 6, 7, 8}
(ii) BΔA = (B – A) ∪ (A – B)
= {2, 4, 6, 8} ∪ {3, 5, 7}
= {2, 3, 4, 5, 6, 7, 8}

Question 3.
From the venn-diagram, list the following:

(i) A
(ii) B
(iii) A∩B
(iv)A∪B
(v) A – B
(vi) B – A
(vii) (A – B) ∩ (B – A)
Solution:
(i) A = {1, 2, 5, 6, 7}
(ii) B = {3, 4, 5, 6}
(iii) A∩B = {5, 6}
(iv) A∪B = {1, 2, 3, 4, 5, 6, 7}
(v) A – B = {1, 2, 7}
(vi) B – A = {3, 4}
(vii) (A – B) ∩ (B – A) = { }

Question 4.
In a class there are 40 students. 26 have opted for Mathematics and 24 have opted for Science. How many student have opted for Mathematics and Science.
Solution:
Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M∪S) = 40, n (M) = 26, n(S) = 24
n(M∪S) = n (M) + n (S)- n(M∩S)
40 = 26 + 24 – n(M∩S)
n (M∩S) = 26 + 24 – 40 = 50 – 40 = 10
∴ Number of students opted for Mathematics and Science = 10.
Another Method:
Let “x” be the number of students opted for Mathematics and Science.
Let M and S represent students opting Mathematics and Science.
n(M∪S) = 40, n(M) = 26, n(S) = 24

By venn-diagram, number of students in a class = 26 – x + x + 24 – x
40 = 50 – x
x = 50 – 40 = 10
x = 10
∴ Number of students opted for Mathematics and Science = 10.

Question 5.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {4, 5, 7, 9}, B = {1, 3, 5, 7, 8}, Verify De Morgan’s Laws for complementation.
De Morgan’s Laws (i) (A∪B)’ = A’∩B’ (ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {4, 5, 7, 9} ∪ {1, 3, 5, 7, 8}
= {1, 3, 4, 5, 7, 8, 9}
(A∪B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 4, 5, 7, 8, 9}
= {2, 6}……….(1)
A’= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {4, 5, 7, 9}
= {1, 2, 3, 6, 8}
B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 3, 5, 7, 8}
= {2, 4, 6, 9}
A’∩B’ = {1, 2, 3, 6, 8} ∩ {2, 4, 6, 9}
= {2, 6}………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’.

(ii) A∩B = {4, 5, 7, 9} ∩ {1, 3, 5, 7, 8}
= {5, 7}
(A∩B)’= {1, 2, 3, 4, 5, 6, 7, 8, 9} – {5, 7}
= {1, 2, 3, 4, 6, 8, 9}………(1)
A’ = {1, 2, 3, 6, 8}
B’ = {2, 4, 6, 9}
A’∪B’ = {1, 2, 3, 6, 8} ∪ {2, 4, 6, 9}
= {1, 2, 3, 4, 6, 8, 9}………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’.

Students can download Maths Chapter 1 Set Language Ex 1.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.7

I. Multiple Choice Questions.

Question 1.
Which of the following is correct?
(a) {7} ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(c) 7 ∉ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(d) {7} ⊄ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Solution:
(b) 1 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Question 2.
The set P = {x | x ∈ Z, -1 < x < 1} is a ……….
(a) Singleton set
(b) Power set
(c) Null set
(d) Subset
Solution:
(a) Singleton set
Hint: P = {0}

Question 3.
If U = {x | x ∈ N, x < 10} and A = {x | x ∈ N, 2 ≤ x < 6} then (A’)’ is………..
{a) {1, 6, 7, 8, 9}
(b) {1, 2, 3, 4}
(c) {2, 3, 4, 5}
(d) { }
Solution:
(c) {2, 3, 4, 5}
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A= {2, 3, 4, 5}; A’ = {1, 6, 7, 8, 9}
(A’)’ = U – A
(A’)’ = {2, 3, 4, 5}

Question 4.
If B⊆A then n(A∩B) is………..
(a) n(A – B)
(b) n(B)
(c) n(B – A)
(d) n(A)
Solution:
(b) n(B)

Question 5.
If A = {x, y, z} then the number of non- empty subsets of A is…………..
(a) 8
(b) 5
(c) 6
(d) 7
Solution:
(d) 7
Hint:
n(A) = 3; P(A) = 2^m = 2³ = 8
Non-empty subsets of A = 8 – 1 = 7

Question 6.
Which of the following is correct?
(a) Ø ⊆ {a, b}
(b) Ø ∈ {a, b}
(c) {a} ∈ {a, b}
(d) a ⊆ {a, b}
Solution:
(a) Ø ⊆ {a, b}
Hint:
‘Q’ is a subset of every set.

Question 7.
If A∪B = A∩B then ……………
(a) A ≠ B
(b) A = B
(c) A ⊂ B
(d) B ⊂ A
Solution:
(b) A = B

Question 8.
If B – A is B, then A∩B is………….
(a) A
(b) B
(c) U
(d) Ø
Solution:
(d) Ø

Question 9.
From the adjacent diagram n[P(AΔB)] is………….

(a) 8
(b) 16
(c) 32
(d) 64
Solution:
(c) 32
Hint:
A – B = {60, 85, 75}
B – A = {90, 70}
AΔB = (A – B) ∪ (B – A)
= {60, 70, 75, 85, 90}
n[P(AΔB)] = 2⁵ = 32

Question 10.
If n(A) = 10 and n(B) = 15 then the minimum and maximum number of elements in A∩B is …………
(a) (10, 15)
(b) (15, 10)
(c) (10, 0)
(d) (0, 10)
Solution:
(d) (0, 10)

Question 11.
Let A = {Ø} and B = P(A) then A∩B is………..
(a) {Ø, {Ø}}
(b) {Ø}
(c) Ø
{d) {0}
Solution:
(b) {Ø}
Hint:
A = {Ø}, B = {{ }, {Ø}}
A∩B = {Ø}

Question 12.
In a class of 50 boys, 35 boys play carrom and 20 boys play chess then the number of boys play both games is……..
(a) 5
(b) 30
(c) 15
(d) 10
Solution:
(a) 5
Hint:
n(A∪B) = 50, n(A) = 35, n(B) = 20
n(A∩B) = n(A) + n(B) – n(A∪B)
= 35 + 20 – 50 = 5

Question 13.
If U = {x : x ∈ N and x < 10}, A = {1, 2, 3, 5, 8} and B = (2, 5, 6, 7, 9}, then n [(A∪B)’] is
(a) 1
(b) 2
(c) 4
(d) 8
Solution:
(a) 1
Hint:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A∪B ={1, 2, 3, 5, 8} ∪ {2, 5, 6, 7, 9}
= {1, 2, 3, 5, 6, 7, 8, 9}
(A∪B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 5, 6, 7, 8, 9}
= {4}
n(A∪B)’ = 1

Question 14.
For any three sets P, Q and R, P – (Q∩R) is ………
(a) P – (Q∪R)
(i)(P∩Q) – R
(c) (P – Q) ∪ (P – R)
(d) (P – Q) ∩ (P – R)
Solution:
(c)(P – Q) ∪ (P – R)

Question 15.
Which of the following is true?
(a) A – B = A∩B
(b) A – B = B – A
(c) (A∪B)’ = A’∪B’
(d) (A∩B)’ = A’∪B’
Solution:
(d) (A∩B)’ = A’∪B’

Question 16.
If n(A∪B∪C) = 100, n(A) = 4x, n(B) = 6x, n(C) = 5x, n(A∩B) = 20, n(B∩C) = 15 and n(A∩C) = 25, then n(A∩B∩C) = 10, then the value of x is……….
(a) 10
(b) 15
(c) 25
(d) 30
Solution:
(b) 15
Hint:
n(A∪B∪C) = n( A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)
100 = 4x + 6x + 5x – 20 – 15 – 25 + 10
100 = 15x – 50 ⇒ 150 = 15x ⇒ x = \(\frac{150}{15}\) = 10

Question 17.
For any three sets A, B and C, (A – B) ∩ (B – C) is equal to
(a) A only
(b) B only
(c) C only
(d) \(\phi \)
Solution:
(d) \(\phi \)

Question 18.
If J = Set of three sided shapes, K = Set of shapes with two equal sides and L = Set of shapes with right angle, then J∩K∩L is………
(a) Set of isosceles triangles
(b) Set of equilateral triangles
(c) Set of isosceles right triangles
(d) Set of right angled triangles
Solution:
(c) Set of isosceles right triangles
J = Set of three sided shapes; It is a triangle
K = Set of shapes with two equal sides (Isosceles triangle)
L = Set of shapes with right angle (One angle is right angle)
∴ J∩K∩L = Set of isosceles right triangles

Question 19.
The shaded region in the Venn diagram is

(a) Z – (X∪Y)
(b) (X∪Y) ∩ Z
(c) Z – (X∩Y)
(d) Z ∪ (X∩Y)
Solution:
(c) Z – (X∩Y)

Question 20.
In a city, 40% people like only one fruit, 35% people like only two fruits, 20% people like all the three fruits. How many percentage of people do not like any one of the above three fruits?
(a) 5
(b) 8
(c) 10
(d) 15
Solution:
(a) 5

Students can download Maths Chapter 1 Set Language Ex 1.5 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.5

Question 1.
Using the adjacent Venn diagram, find the following sets:

(i) A – B
(ii) B – C
(iii) A’∪B’
(iv) A’∩B’
(v) (B∪C)’
(vi) A – (B∪C)
(vii) A – (B∩C)
Solution:
From the diagram we get
U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8},
A= {-2,-1, 3, 4, 6}, B = {-2,-1, 5, 7, 8}
C = {-3, -2, 0, 3, 8}
A’ = U – A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 3, 4, 6}
= {-3, 0, 1, 2, 5, 7, 8}
B’ = U – B = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 5, 7, 8}
= {-3, 0, 1, 2, 3, 4, 6}
B∪C = {-2, -1, 5, 7, 8} ∪ {-3, -2, 0, 3, 8} = {-3, -2, -1, 0, 3, 5, 7, 8}
B∩C = {-2, -1, 5, 7, 8} ∩ {-3, -2, 0, 3, 8} = {-2, 8}

(i) A – B = {3, 4, 6}
(ii) B – C = {-1, 5, 7}
(iii) A’∪B’= {-3, 0, 1, 2, 5, 7, 8} ∪ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}
(iv) A’∩B’ = {-3, 0, 1, 2, 5, 7, 8} ∩ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2}
(v) (B∪C)’ = U – (B∪C)= {-3,-2,-1,0, 1,2, 3,4, 5, 6, 7, 8} – {-3, -2, -1, 0, 3, 5, 7, 8}
= {1, 2, 4, 6}
(vi) A – (B∪C) = {-2, -1, 3, 4, 6} – {-3, -2, -1, 0, 3, 5, 7, 8} = {4, 6}
(vii) A – (B∩C) = {-2,-1, 3, 4, 6} – {-2, 8} = {-1, 3, 4, 6}

Question 2.
If K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h} then find the following:
(i) K∪(L∩M)
(ii) K∩(L∪M)
(iii) (K∪L) ∩ (K∪M)
(iv) (K∩L) ∪ (K∩M)
and verify distributive laws.
Solution:
K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}
(i) K∪(L∩M)
(L∩M) = {b, c, d, g} ∩ [a, b, c, d, h}
= {b, c, d}
K∪(L∩M) = {a, b, d, e, f} ∪ {b, c, d}
= {a, b, c, d, e, f}

(ii) K∩(L∪M)
(L∪M) = {b, c, d, g} ∪ {a, b, c, d, h}
= {a, b, c, d, g, h}
K∩(L∪M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h}
= {a, b, d }

(iii) (K∪L) ∩ (K∪M)
(K∪L) = {a, b, d, e, f} ∪ {b, c, d, g}
= {a, b, c, d, e, f, g}
(K∪M) = {a, b, d, e, f} ∪ {a, b, c, d, h}
= {a, b, c, d, e, f, h}
(K∪L) ∩ (K∪M) = {a, b, c, d, e, f, g} ∩ {a, b, c, d, e, f, h}
= {a, b, c, d, e, f}

(iv) (K∩L) ∪ (K∩M)
(K∩L) = {a, b, d, e, f) ∩ {b, c, d, g}
= {b, d}
(K∩M) = {a, b, d, e, f} ∩ {a, b, c, d, h}
= {a, b, d}
(K∩L) ∪ (K∩M) = {b, d} ∪ [a, b, d}
= {a, b, d}
From (ii) & (iv) we get, K∩(L∪M) = (K∩L) ∪ (K∩M)
From (i) & (iii) we get, K∪(L∩M) = (K∪L) ∩ (K∪M)

Question 3.
For A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C = {-4, -1, 0, 2, 3, 4}
verify A∪(B∩C) = (A∪B) ∩ (A∪C).
Solution:
A = {-1, 0, 1, 2, 3, 4}, B = {0, 1, 2, 3, 4, 5} and C = {-4, -1, 0, 2, 3, 4}
B∩C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4}
= {0, 2, 3, 4}
A∪(B∩C) = {-1, 0, 1, 2, 3, 4} ∪ {0, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(1)
A∪B = {-1, 0, 1, 2, 3, 4} ∪ {0, 1, 2, 3, 4, 5}
= {-1, 0, 1, 2, 3, 4, 5}
A∪C = {-1, 0, 1, 2, 3, 4} ∪ {-4, -1, 0, 2, 3, 4}
= {-4, -1, 0, 1, 2, 3, 4}
(A∪B) ∩ (A∪C) = {-1, 0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 1, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(2)
From (1) and (2) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 4.
Verify A∪(B∩C) = (A∪B) ∩ (A∪C) using Venn diagrams.
Solution:

From (ii) and (v) we get A∪(B∩C) = (A∪B) ∩ (A∪C).

Question 5.
If A = {b, c, e, g, h}, B = {a, c, d, g, f}, and C = {a, d, e, g, h}, then show that A – (B∩C) = (A – B) ∪ (A – C).
Solution:
A = {b, c, e, g, h} ; B = {a, c, d, g, f}; C = {a, d, e, g, h}
B∩C = {a, c, d, g, i} ∩ {a, d, e, g, h}
= {a, d, g}
A – (B∩C) = {b, c, e, g, h} – {a, d, g}
= {b, c, e, h}…….(1)
A – B = {b, c, e, g, h} – {a, c, d, g, i}
= {b, e, h}
A – C = {b, c, e, g, h} – {a, d, e, g, h}
= {b, c}
(A – B) ∪ (A – C) = {b, e, h} ∪ {b, c}
= {b, c, e, h)……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C)

Question 6.
If A= {x : x = 6n, n∈W and n < 6}, B = {x : x = 2n, n∈N and 2 < n ≤ 9} and
C = {x : x = 3n, n∈N and 4 ≤ n < 10}, then show that A – (B∩C) = (A – B) ∪ (A – C)
Solution:
A = {0, 6, 12, 18, 24, 30}; B = {6, 8, 10, 12, 14, 16, 18}; C = {12, 15, 18, 21, 24, 27}
B∩C = {6, 8, 10, 12, 14, 16, 18} ∩ {12, 15, 18, 21, 24, 27}
= {12, 18}
A – (B∩C) = {0, 6, 12, 18, 24, 30} – {12, 18}
= {0, 6, 24, 30}………(1)
A – B = {0, 6, 12, 18, 24, 30} – {6, 8, 10, 12, 14, 16, 18}
= {0, 24, 30}
A – C = {0, 6, 12, 18, 24, 30} – {12, 15, 18, 21, 24, 27}
= {0, 6, 30}
(A – B) ∪ (A – C) = {0, 24, 30} ∪ {0, 6, 30}
= {0, 6, 24, 30}……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C).

Question 7.
If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that
A – (B∪C) = (A – B) ∩ (A – C).
Solution:
A= {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6}, C = {-1, 2, 5, 6, 7}
B∪C = {-1, 0, 2, 5, 6} ∪ {-1, 2, 5, 6, 7}
= {-1, 0, 2, 5, 6, 7}
A – (B∪C) = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6, 7}
= {-2, 1, 3} ………(1)
A – B = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6}
= {-2, 1, 3}
A – C = {-2, 0, 1, 3, 5}- {-1, 2, 5, 6, 7}
= {-2, 0, 1, 3}
(A- B) ∩ (A- C) = {-2, 1, 3} ∩ {-2, 0, 1, 3}
= {-2, 1, 3} ….(2)
From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 8.
IF A = {y : y = \(\frac{a + 1}{2}\), a ∈ W and a ≤ 5}, B = {y : y = \(\frac{2n – 1}{2}\), n ∈ W and n < 5} and C = {-1, \(-\frac{1}{2}\), 1, \(\frac{3}{2}\), 2} then show that A – (B∪C) = (A – B) ∩ (A – C).
Solution:

From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).

Question 9.
Verify A- (B∩C) = (A – B) ∪ (A – C) using Venn diagrams.
Solution:


From (ii) and (v) we get A- (B∩C) = (A – B) ∪ (A – C).

Question 10.
If U = {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.
U= {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}
(i) (A∪B)’ = A’∩B’
(ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {7, 8, 11, 12} ∪ {4, 8, 12, 15}
= {4, 7, 8, 11, 12, 15}
(A∪B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}
= {10,16} ………(1)
A’ = {4, 7, 8, 10, 11, 12, 15, 16} – {7, 8, 11, 12}
= {4, 10, 15, 16}
B’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 8, 12, 15}
= {7, 10, 11, 16}
A’∩B’ = {4, 10, 15, 16} ∩ {7, 10, 11, 16}
= {10,16} ………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’

(ii) A∩B = {7, 8, 11, 12} ∩ {4, 8, 12, 15}
= {8, 12}
(A∩B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {8, 12}
= {4, 7, 10, 11, 15, 16} ………(1)
A’ = {4, 10, 15, 16}
B’ = {7, 10, 11, 16}
A’∪B’ = {4, 10, 15, 16} ∪ {7, 10, 11, 16}
= {4, 7, 10, 11, 15, 16} ………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’

Question 11.
Verify (A∩B)’ = A∪B’ using Venn diagrams.
Solution:

From (ii) and (i) we get (A∩B)’ = A’∪B’

Students can download Maths Chapter 1 Relations and Functions Ex 1.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.




Answer:

The vertical line cuts the graph at A and B. The given graph does not represent a function.

The vertical line cuts the graph at most one point P. The given graph represent a function.

The vertical line cuts the graph at three points S,T and U. The given graph does not represent a function.

The vertical line cuts the graph at most one point D. The given graph represents a function.

Question 2.
Let f: A → B be a function defined by
f(x) = \(\frac { x }{ 2 } \) – 1, where A = {2, 4,6,10,12},
B = {0,1,2,4,5,9}. Represent f by
(i) set of ordered pairs
(ii) a table
(iii) an arrow diagram
(iv) a graph
Answer:
A = {2,4,6, 10, 12}
B = {0,1, 2, 4, 5, 9}
f(x) = \(\frac { x }{ 2 } \) – 1
f(2) = \(\frac { 2 }{ 2 } \) – 1 = 1 – 1 = 0
f(4) = \(\frac { 4 }{ 2 } \) – 1 = 2 – 1 = 1
f(6) = \(\frac { 6 }{ 2 } \) – 1 = 3 – 1 = 2
f(10) = \(\frac { 10 }{ 2 } \) – 1 = 5 – 1 = 4
f(12) = \(\frac { 12 }{ 2 } \) – 1 = 6 – 1 = 5

(i) Set of ordered pairs
f = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5}

(ii) Table

(iii) Arrow diagram

(iv) Graph

Question 3.
Represent the function f = {(1,2), (2,2), (3,2), (4,3),(5,4)} through (i) an arrow diagram (it) a table form (iii) a graph.
Answer:
f = {(1, 2) (2, 2) (3, 2) (4, 3) (5,4)}
Let A = {1,2, 3, 4, 5}
B = {2, 3, 4}

(i) Arrow diagram

(ii) Table form

(iii) Graph

Question 4.
Show that the function f : N → N defined by f(x) = 2x – 1 is one-one but not onto.
Answer:
f: N → N
N = {1,2,3,4,5,… }
f(x) = 2x – 1
f(1) = 2(1) – 1 = 2 – 1 = 1
f(2) = 2(2) – 1 = 4 – 1 = 3
f(3) = 2(3) – 1 = 6 – 1 = 5
f(4) = 2(4) – 1 = 8 – 1 = 7
f(5) = 2(5) – 1 = 10 – 1 = 9
f = {(1,1) (2, 3) (3, 5) (4, 7) (5,9) …..}

(i) Different elements has different images. This function is one to one function.
(ii) Here Range is not equal to co-domain. This function not an onto function.
∴ The given function is one-one but not an onto.

Question 5.
Show that the function f: N ⇒ N defined by f(m) = m² + m + 3 is one-one function.
Answer:
N = {1,2,3, 4,5, ….. }
f(m) = m² + m + 3
f(1) = 1² + 1 + 3 = 5
f(2) = 2² + 2 + 3 = 9
f(3) = 3² + 3 + 3 = 15
f(4) = 4² + 4 + 3 = 23
f = {(1,5) (2, 9) (3, 15) (4, 23)}

From the diagram we can understand different elements in (N) in the domain, there are different images in (N) co-domain.
∴ The function is a one-one function.

Question 6.
Let A = {1, 2, 3, 4) and B = N. Letf: A → B be
defined by f(x) = x³ then,
(i) find the range off
(ii) identify the tpe of function
Solution:
A = {1, 2, 3, 4}
B = N
f: A → B,f(x) = x³
(i) f(1) = 1³ = 1
f(2) = 2³ = 8
f(3) = 3³ = 27
f(4) = 4³ = 64
(ii) Therange of f = {1, 8, 27, 64 )
(iii) It is one-one and into function.

Question 7.
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f: R → R defined by f (x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x²
Answer:
(i) f(x) = 2x + 1
f(0) = 2(0) + 1 = 0 + 1 = 1
f(1) = 2(1) + 1 = 2 + 1 = 3
f(2) = 2(2) + 1 = 4 + 1 = 5
f(3) = 2(3) + 1 = 6 + 1 = 7
Different elements has different images
∴ It is an one-one function.
It is also an onto function. The function is one-one and onto function.
∴ It is a bijective function.

(ii) f(x) = 3 – 4x²
f(1) = 3 – 4(1)²
= 3 – 4 = -1
f(2) = 3 – 4(2)² = 3 – 16 = – 13
f(3) = 3 – 4(3)² = 3 – 36 = – 33
f(4) = 3 – 4(4²) = 3 – 64 = – 61
It is not a bijective function. The positive numbers “R” do not have negative pre – image in X in R.

Question 8.
Let A= {-1,1}and B = {0,2}.
If the function f: A → B defined by
f(x) = ax + b is an onto function? Find a and b.
Answer:
A = {-1, 1}; B = {0,2}
f(x) = ax + b
f(-1) = a(-1) + b
0 = -a + b
a – b = 0 ….(1)
f(1) = a(1) + b
2 = a + b
a + b = 2 ….(2)
Solving (1) and (2) we get

Substitute a = 1 in (1)
The value of a = 1 and b = 1

Question 9.
If the function f is defined by

find the value of
(i) f(3)
(ii) f(0)
(iii) f(1. 5)
(iv) f(2) + f(-2)
Answer:
f(x) = x + 2 when x = {2,3,4,……}
f(x) = 2
f(x) = x – 1 when x = {-2}
(i) f(x) = x + 2
f(3) = 3 + 2 = 5

(ii) f(x) = 2
f(0) = 2

(iii) f(x) = x – 1
f(-1.5) = -1.5 – 1 = -2.5

(iv) f(x) = x + 2
f(2) = 2 + 2 = 4
f(x) = x – 1
f(-2) = – 2 – 1 = – 3
f(2) + f(-2) = 4 – 3
= 1

Question 10.
A function f: [-5, 9] → R is defined as follows:

Answer:
f(x) = 6x + 1 ; x = {-5,-4,-3,-2,-1,0,1}
f(x) = 5x² – 1 ; x = {2, 3, 4, 5}
f(x) = 3x – 4 ; x = {6, 7, 8, 9}

(i) f(-3) + f(2)
f(x) = 6x + 1
f(-3) = 6(-3) + 1 = -18 + 1 = -17
f(x) = 5x² – 1
f(2) = 5(2)² – 1 = 20 – 1 = + 19
f(-3) + f(2) = – 17 + 19
= 2

(ii) f(7) – f(1)
f(x) = 3x – 4
f(7) = 3(7) – 4 = 21 – 4 = 17
f(x) = 6x + 1
f(1) = 6(1) + 1 = 6 + 1 = 7
f(7) – f(1) = 17 – 7
= 10

(iii) 2f(4) + f(8)
f(x) = 5x² – 1
f(4) = 5(4)2 – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 3x – 4
f(8) = 3(8) – 4 = 24 – 4 = 20
2f(4) + f(8) = 2(79) + 20
= 158 + 20
= 178

f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11
f(x) = 3x – 4
f(6) = 3(6) – 4 = 18 – 4 = 14
f(x) = 5x² – 1
f(4) = 5(4)² – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11

Question 11.
The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = \(\frac{1}{2}\) gt² + at + b where, (g is the acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.
Solution:
S(t) = \(\frac{1}{2}\) gt² + at + b
Let t be 1, 2, 3, ………, seconds
S(1) = \(\frac{1}{2}\) g(1²) + a(1) + b = \(\frac{1}{2}\) g + a + b
S(2) = \(\frac{1}{2}\) g(2²) + a(2) + b = 2g + 2a + b
Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.

Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by
t(C) = F where F = \(\frac { 9 }{ 5 } \) C + 32. Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Fahrenheit value.
Answer:
Given t(C) = \(\frac { 9C }{ 5 } \) + 32

(i) t(0) = \(\frac { 9(0) }{ 5 } \) + 32
= 32° F

(ii) t(28) = \(\frac { 9(28) }{ 5 } \) + 32
= \(\frac { 252 }{ 5 } \) + 32
= 50.4 + 32
= 82.4° F

(iii) t(-10) = \(\frac { 9(-10) }{ 5 } \) + 32
= -18 + 32
= 14° F

(iv) t(C) = 212
\(\frac { 9C }{ 5 } \) + 32 = 212
\(\frac { 9C }{ 5 } \) = 212 – 32
= 180
9C = 180 × 5
C = \(\frac{180 \times 5}{9}\)
= 100° C

(v) consider the value of C be “x”
t(C) = \(\frac { 9C }{ 5 } \) + 32
x = \(\frac { 9x }{ 5 } \) + 32
5x = 9x + 160
-160 = 9x – 5x
-160 = 4x
x = \(\frac { -160 }{ 4 } \) = -40
The temperature when the Celsius value is equal to the fahrenheit value is -40°

Composition of two Functions

Let f: A → B and g: B → C be two functions. Then the composition of f and g denoted by gof is defined as the function gof (x) = g[f(x)] for all x ∈ A.

Composition of three Functions

Let A, B, C, D be four sets and let f: A → B; g : B → C and h : C → D be three functions, using composite functions fog and goh, we get two new functions like (fog) oh and fo (goh).
Note: Composition of three function is always associative.

Students can download Maths Chapter 1 Set Language Ex 1.6 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.6

Question 1.
(i) If n(A) = 25, n(B) – 40, n(A∪B) = 50 and n(B’) = 25, find n(A∩B) and n(U).
Solution:
Given, n(A) = 25, n(B) = 40, n(A∪B) = 50 and n(B’) = 25 n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = n(A) + n(B) – n(A∪B)
n(A∩B) = 25 + 40 – 50
= 65 – 50
= 15
n(U) = n(B) + n(B)’
= 40 + 25
= 65
∴ n(A∩B) = 15 and n(U) = 65

(ii) If n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B’) = 350, find n(B) and n(U).
Solution:
Given, n(A) = 300, n(A∪B) = 500, n(A∩B) = 50 and n(B’) = 350
n(A∪B) = n(A) + n(B) – n(A∩B)
500 = 300 + n(B) – 50
500 = 250 + n(B)
500 – 250 = n(B)
250 = n(B)
∴ n(B) = 250
n(U) = n(B) + n(B)’
250 + 350 = 600
∴ n(B) = 250 and n(U) = 600

Question 2.
If U = {x : x ∈ N, x ≤ 10}, A = { 2, 3, 4, 8, 10} and B = {1, 2, 5, 8, 10}, then verify that n(A∪B) = n(A) + n(B) – n(A∩B)
Solution:
U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 3, 4, 8, 10}; B = {1, 2, 5, 8, 10}
n(U) = 10, n(A) = 5, n(B) = 5
(A∪B) = {2, 3, 4, 8, 10} ∪ {1, 2, 5, 8, 10}
= {1, 2, 3, 4, 5, 8, 10}
∴ n(A∪B) = 7 ……..(1)
(A∩B) = {2, 3, 4, 8, 10} ∩ {1, 2, 5, 8, 10}
= {2, 8, 10}
n(A∩B) = 3
n(A) + n(B) – n(A∩B) = 5 + 5 – 3
= 10 – 3
= 7 ……(2)
From (1) and (2) we get,
n(A∪B) = n(A) + n(B) – n(A∩B)

Question 3.
Verify n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) for the following sets.
(i) A = {a, c, e, f, h}, B = {c, d, e, f} and C = {a, b, c, f}
Solution:
A∩B = {a, c, e, f, h} ∩ {c, d, e, f}
= {c, e, f}
B∩C = {c, d, e, f} ∩ {a, b, c, f}
= {c, f}
A∩C = {a, c, e, f, h} ∩ {a, b, c, f}
= {c, f}
(A∩B∩C) = {a, c, e, f, h} ∩ {c, d, e, f} ∩ {a, b, c, f}
= {c, f}
(A∪B∪C) = {a, c, e, f, h} ∪ {c, d, e, f} ∪ {a, b, c, f}
= {a, b, c, d, e, f, h}
n(A∩B) = 3, n(B∩C) = 2, n(A∩C) = 3, n(A∩B∩C) = 2
n(A∪B∪C) = 7……….(1)
n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n( A∩C) + n(A∩B∩C)
= 5 + 4 + 4 – 3 – 2 – 3 + 2 = 15 – 8
= 7 ……….(2)
From (1) and (2) we get
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

(ii) A= {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}
A∩B = {1, 3, 5} ∩ {2, 3, 5, 6}
= {3, 5}
B∩C = {2, 3, 5, 6} ∩ {1, 5, 6, 7}
= {5, 6}
A∩C = {1, 3, 5} ∩ {1, 5, 6, 7}
= {1, 5}
A∩B∩C = {1, 3, 5} ∩ {2, 3, 5, 6} ∩ {1, 5, 6, 7}
= {5}
A∪B∪C = {1, 3, 5} ∪ {2, 3, 5, 6} ∪ {1, 5, 6, 7}
= {1, 2, 3, 5, 6, 7}
n(A) = 3, n(B) = 4, n(C) = 4
n(A∩B) = 2, n(B∩C) = 2, n(A∩C) = 2
n(A∩B∩C) = 1
n(A∪B∪C) = 6……….(1)
n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C) = 3 + 4 + 4 – 2 – 2 – 2 + 1
= 12 – 6
= 6………(2)
From (1) and (2) we get
n(A∪B∪C) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(A∩C) + n(A∩B∩C)

Question 4.
In a class, all students take part in either music or drama or both. 25 students take part in music, 30 students take part in drama and 8 students take part in both music and drama. Find
(i) The number of students who take part in only music.
(ii) The number of students who take part in only drama.
(iii) The total number of students in the class.
Solution:

Let M be the set of all students take part in music.
Let D be the set of all students take part in drama.
n( M) = 25, n(D) = 30 and n(M∩D) = 8
By using venn-diagram
From the venn – diagram we get.
(i) Number of students take part in only music = 17
(ii) Number of students take part in only drama = 22
(iii) Total number of students in the class = 17 + 8 + 22 = 47

Question 5.
In a party of 45 people, each one likes Tea or Coffee or both. 35 people like tea and 20 people like coffee. Find the number of people who
(i) like both Tea and Coffee.
(ii) do not like Tea.
(iii) do not like Coffee.
Solution:

Let’T’ be the set of people likes Tea
Let ‘C’ be the set of people likes Coffee
n(T∩C) = 45, n(T) = 35 and n(C) = 20
Let X be the number of people likes both Tea and Coffee.
By using venn diagram
From the venn – diagram we get.
35 – x + x + 20 – x = 45
55 – x = 45
55 – 45 = x
10 = x
(i) People like both tea and coffee = 10
(ii) People do not like tea = 20 – x
= 20 – 10 = 10
(iii) People do not like coffee = 35 – x
= 35 – 10 = 25

Question 6.
In an examination 50% of the students passed in mathematics and 70% of students passed in science while 10% students failed in both subjects. 300 students passed in both the subjects. Find the total number of students who appeared in the examination, if they took examination in only two subjects.
Solution:
Let M and S represent the student failed in Mathematics and Science.
Given: Number of students passed in Mathematics is 50%

∴ Number of students failed in Mathematics = 100 – 50% = 50%
n(M) = 50%
Number of students passed in Science is 70%
∴ Number of students failed in Science = 100 – 70% = 30%
n(S) = 30%
Number of students failed in both the subjects is 10%
n(M∩S) = 10%
n(M∪S)= n(M) + n(S) – n(M∩S)
= 50 + 30 – 10 = 80 – 10 = 70
Given: 70% of the students failed in atleast any one of the subject
∴ 30% of the students passed in atleast any one of the subjects.
30 students passed mean, the total number of students is 100.
∴ 300 students passed means, the total number of students = \(\frac{100 × 300}{30}\)
Total number of students appeared in the examination = 1000

Question 7.
A and B are two sets such that n(A – B) = 32 + x, n(B – A) = 5x and n(A∩B) = x. Illustrate the information by means of a venn diagram. Given that n(A) = n(B). Calculate the value of x.
Solution:
n(A – B) = 32 + x, n(B – A) = 5x
n(A∩B) = x
From the Venn diagram:

Given n( A) = n(B)
32 + x + x = x + 5x
32 + 2x = 6x
32 = 6x – 2x
32 = 4x
x = \(\frac{32}{4}\) = 8
The value of x = 8

Question 8.
Out of 500 car owners investigated, 400 owned car A and 200 owned car B, 50 owned both A and B cars. Is this data correct?
Solution:
Let A be the set of people owned car A
Let B be the set of people owned car B
n( A) = 400, n(B) = 200, n(A∩B) = 50
n(A∪B) = 500………..(1)
n(A) + n(B) – n(A∩B) = 400 + 200 – 50
= 600 – 50
= 550………(2)
From (1) and (2) we get
n(A∪B) ≠ n(A) + n(B) – n(A∩B)
∴ The given data is not correct.

Question 9.
In a colony, 275 families buy Tamil newspaper, 150 families buy English newspaper, 45 families buy Hindi newspaper, 125 families buy Tamil and English newspapers, 17 families buy English and Hindi newspapers, 5 families buy Tamil and Hindi newspapers and 3 families buy all the three newspapers. If each family buy atleast one of these newspapers then find
(i) Number of families buy only one newspaper
(ii) Number of families buy atleast two newspapers
(iii) Total number of families in the colony.
Solution:
Let T, E and H represent families buying Tamil newspaper, English newspaper and Hindi newspaper respectively.
n(T) = 275, n(E) = 150, n(H) = 45
n(T∩E) = 125, n(E∩H) = 17, n(T∩H) = 5
n(T∩E∩H) = 3
Let us represent the given data in Venn diagrams.

(i) Number of families buy only one news paper = 148 + 11 + 26
= 185
(ii) Number of families buy atleast two news paper = 122 + 2 + 14 + 3
= 141
(iii) Total number of families in the colony = 148 + 122 + 11 + 14 + 3 + 2 + 26
= 326

Question 10.
A survey of 1000 farmers found that 600 grew paddy, 350 grew ragi, 280 grew corn, 120 grew paddy and ragi, 100 grew ragi and corn, 80 grew paddy and corn. If each farmer grew atleast any one of the above three, then find the number of farmers who grew all the three.
Solution:
Let P, R and C represent sets of farmers grew paddy, ragi and com respectively.
n(P∪R∪C) = 1000, n(P) = 600, n(R) = 350, n(C) = 280
n(P∩R) = 120, n(R∩C) = 100, w(P∩C) = 80 Let the number of farmers who grew all the three be “x”
n(P∪R∪C ) = n(P) + n( R) + n( C) – n(P∩R) – n(R∩C) – n(P∩C) + n(P∩R∩C )
1000 = 600 + 350 + 280 – 120 – 100 – 80 + x = 1230 – 300 + x.
1000 = 930 + x
1000 – 930 = x
70 = x
Number of farmers who grew all the three = 70.

Question 11.
In the adjacent diagram, if n(U) = 125, y is two times of x and z is 10 more than x, then find the value of x, y and z.

Solution:
n(U) = 125
y = 2x and z = x + 10
n(U) = x + 4 + y + 17 + 3 + 6 + z + 5
125 = x + 4 + 2x + 17 + 3 + 6 + x + 10 + 5
125 = 4x + 45
125 – 45 = 4x
80 = 4x
x = 80/4 = 20
y = 2x = 2 × 20 = 40
z = x + 10 = 20 + 10 = 30
∴ The value of x = 20, y = 40 and z = 30.

Question 12.
Each student in a class of 35 plays atleast one game among chess, carrom and table tennis. 22 play chess, 21 play carrom, 15 play table tennis, 10 play chess and table tennis, 8 play carrom and table tennis and 6 play all the three games. Find the number of students who play (i) chess and carrom but not table tennis (ii) only chess (iii) only carrom (Hint: Use Venn diagram)
Solution:
Let A, B and C represent students play chess, carrom and table tennis.
n(A) = 22, n(B) = 21 , n(C) = 15
n(A∩C) = 10 , n(B∩C) = 8 , n(A∩B∩C) = 6
Let “x” represent student play chess and carrom but not table tennis.
Let us represent the data in Venn diagram.

From the Venn diagram we get,
Number of students play atleast one game = 35
12 – x + x + 13 – x + 2 + 6 + 4 + 3 = 35
40 – 35 = x
5 = x
(i) Number of students who play chess and carrom but not table tennis = 5
(ii) Number of students who play only chess = 12 – x
= 12 – 5 = 7
(iii) Number of students who play only carrom = 13 – x
= 13 – 5 = 8

Question 13.
In a class of 50 students, each one come to school by bus or by bicycle or on foot. 25 by bus, 20 by bicycle, 30 on foot and 10 students by all the three. Now how many students come to school exactly by two modes of transport?
Solution:
Let B, C and D represent students come to school by bus, bicycle and foot respectively.
n(B∪C∪D) = 50 , n(B) = 25 , n(C) = 20 , n(D) = 30, n(B∩C∩D) = 10
Let x, y and z represent the students come to school exactly by two modes of transport.
Let us represent the given data in Venn diagrams.
<
Total number of students in the class = 50
15 – x – z + x + 10 – x – y + y + 10 + z + 20 – z – y = 50
55 – x – y – z = 50
55 – 50 = x + y + z
5 = x + y + z
Number of students come to school exactly by two modes of transport = 5

Students can download Maths Chapter 1 Relations and Functions Unit Exercise 1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Unit Exercise 1

Question 1.
If the ordered pairs (x² – 3x, y² + 4y) and (-2, 5) are equal, then find x and y.
Answer:
(x² – 3x, y² + 4y) = (-2, 5)
x² – 3x = -2
x² – 3x + 2 = 0
(x – 2) (x – 1) = 0
x – 2 = 0 or x – 1 = 0
x = 2 or 1

y² + 4y = 5
y² + 4y – 5 = 0
(y + 5) (y – 1) = 0
y + 5 = 0 or y – 1 = 0
y = -5 or y = 1
The value of x = 2, 1
and 7 = -5, 1

Question 2.
The Cartesian product A × A has 9 elements among which (-1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.
Solution:
A = {-1, 0, 1}, B = {1, 0, -1}
A × B = {(-1, 1), (-1, 0), (-1, -1), (0, 1), (0, 0), (0, -1), (1, 1), (1, 0), (1, -1)}

Question 3.
Given that f(x) = \(\left\{\begin{array}{rl}
{\sqrt{x-1}} & {x \geq 1} \\
{4} & {x<1}
\end{array}\right.\).
Find
(i) f(0) (ii)f (3) (iii) f(a + 1) in terms of a.(Given that a > 0)
Answer:
f(x) = \(\sqrt { x-1 }\) ; f(x) = 4
(i) f(0) = 4
(ii) f(3) = \(\sqrt { 3-1 }\) = \(\sqrt { 2 }\)
(iii) f(a + 1) = \(\sqrt { a+1-1 }\) = \(\sqrt { a }\)

Question 4.
Let A = {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f: A → N be defined by f(n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.
Solution:
A = {9, 10, 11, 12, 13, 14, 15, 16, 17}
f: A → N
f(n) = the highest prime factor of n ∈ A
f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)}
Range = {3, 5, 11, 13, 7, 2, 17}
= {2, 3, 5, 7, 11, 13, 17}

Question 5.
Find the domain of the function

Answer:

Domain of f(x) = {-1, 0, 1}

Question 6.
If f(x)= x², g(x) = 3x and h(x) = x – 2 Prove that (fog)oh = fo(goh).
Solution:
f(x) = x²
g(x) = 3x
h(x) = x – 2
(fog)oh = x – 2
LHS = fo(goh)
fog = f(g(x)) = f(3x) = (3x)² = 9x²
(fog)oh = (fog) h(x) = (fog) (x – 2)
= 9(x – 2)² = 9(x² – 4x + 4)
= 9x² – 36x + 36 ……………. (1)
RHS = fo(goh)
(goh) = g(h(x)) = g(x – 2)
= 3(x – 2) = 3x – 6
fo(goh) = f(3x – 6) = (3x – 6)²
= 9x² – 36x + 36 ………….. (2)
(1) = (2)
LHS = RHS
(fog)oh = fo(goh) is proved.

Question 7.
Let A= {1,2} and B = {1,2,3,4}, C = {5,6} and D = {5,6,7,8}. Verify whether A × C is a subset of B × D?
Answer:
Given A = {1, 2}
B = {1, 2, 3, 4}
C = {5,6}
D = {5,6, 7,8}
A × C = {1,2} × {5,6}
= {(1,5) (1,6) (2, 5) (2, 6)}
B × D = {1,2, 3, 4} × {5, 6, 7, 8}
= {(1,5) (1,6) (1,7) (1,8)
(2, 5) (2, 6) (2,7) (2, 8)
(3, 5) (3, 6) (3, 7) (3, 8)
(4, 5) (4, 6) (4, 7) (4, 8)}
∴ A × C ⊂ B × D
Hence it is verified

Question 8.
If f(x) = \(\frac{x-1}{x+1}, x \neq 1\) Show that
f(f(x)) = – \(\frac { 1 }{ x } \), Provided x ≠ 0.
Answer:

Question 9.
The functions f and g are defined by f{x) = 6x + 8; g(x) = \(\frac { x-2 }{ 3 } \)
(i) Calculate the value of gg [latex]\frac { 1 }{ 2 } [/latex]
(a) Write an expression for gf (x) in its simplest form.
Answer:
f(x) = 6x + 8 ; g(x) = \(\frac { x-2 }{ 3 } \)

Question 10.
Write the domain of the following real functions

Answer:
(i) f (x) = \(\frac { 2x+1 }{ x-9 } \)
If the denominator vanishes when x = 9
So f(x) is not defined at x = 9
∴ Domain is x ∈ [R – {9}]

(ii) if p(x) = \(=\frac{-5}{4 x^{2}+1}\)
p(x) is defined for all values of x. So domain is x ∈ R.

(iii) g(x) = \(\sqrt { x-2 }\)
When x < 2 g(x) becomes complex. But given “g” is real valued function.
So x > 2
Domain x ∈ (2, α)

(iv) h (x) = x + 6
For all values of x, h(x) is defined. Hence domain is x ∈ R.

Students can download Maths Chapter 1 Relations and Functions Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.3

Question 1.
Let f = {(x, y) |x,y; ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
Answer:
X = {1,2,3,….}
Y = {1,2,3,….}
f = {(1,2) (2, 4) (3, 6) (4, 8) ….}
Domain = {1, 2, 3, 4 ….}
Co – Domain = {1, 2, 3, 4 ….}
Range = {2, 4, 6, 8 }
Yes this relation is a function.

Question 2.
Let X = {3, 4, 6, 8}. Determine whether the relation
R = {(x,f(x)) |x ∈ X, f(x) = x² + 1}
is a function from X to N?
Answer:
f(x) = x² + 1
f(3) = 3² + 1 = 9 + 1 = 10
f(4) = 4² + 1 = 16 + 1 = 17
f(6) = 6² + 1 = 36 + 1 = 37
f(8) = 8² + 1 = 64 + 1 = 65
yes, R is a function from X to N

Question 3.
Given the function f: x → x² – 5x + 6, evaluate
(i) f(-1)
(ii) f(2a)
(iii) f(2)
(iv) f(x – 1)
Solution:
Give the function f: x → x² – 5x + 6.
(i) f(-1) = (-1)² – 5(1) + 6 = 1 + 5 + 6 = 12
(ii) f(2a) = (2a)² – 5(2a) + 6 = 4a² – 10a + 6
(iii) f(2) = 2² – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x – 1) = (x – 1)² – 5(x – 1) + 6
= x² – 2x + 1 – 5x + 5 + 6
= x² – 7x + 12

Question 4.
A graph representing the function f(x) is given in it is clear that f(9) = 2.

(i) Find the following values of the function
(a) f(0)
(b) f(7)
(c) f(2)
(d) f(10)
Answer:
(a) f (0) = 9
(b) f (7) = 6
(c) f (2) = 6
(d) f(10) = 0

(ii) For what value of x is f(x) = 1 ?
Answer:
When f(x) = 1 the value of x is 9.5

(iii) Describe the following
(i) Domain
(ii) Range.
Answer:
Domain = {0, 1, 2, 3,… .10}
= {x / 0 < x < 10, x ∈ R}
Range = {0,1,2,3,4,5,6,7,8,9}
= {x / 0 < x < 9, x ∈ R}

(iv) What is the image of 6 under f?
Answer:
The image of 6 under f is 5.

Question 5.
Let f (x) = 2x + 5. If x ≠ 0 then find
\(\frac{f(x+2)-f(2)}{x}\)
Answer:
f(x) = 2x + 5
f(x + 2) = 2(x + 2) + 5
= 2x + 4 + 5
= 2x + 9

Question 6.
A function/is defined by f(x) = 2x – 3
(i) find \(\frac{f(0)+f(1)}{2}\)
(ii) find x such that f(x) = 0.
(iii) find x such that/ (A:) = x.
(iv) find x such that fix) =/(l – x).
Answer:
(i) f(x) = 2x – 3
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = 2 – 3 = -1

(ii) f(x) = 0
2x – 3 = 0
2x = 3
x = \(\frac { 3 }{ 2 } \)

(iii) f(x) = x
2x – 3 = x
2x – x = 3
x = 3

(iv) f(1 – x) = 2(1 – x) – 3
= 2 – 2x – 3
= – 2x – 1
f(x) = f(1 – x)
2x – 3 = – 2x – 1
2x + 2x = 3 – 1
4x = 2
x = \(\frac { 2 }{ 4 } \) = \(\frac { 1 }{ 2 } \)

Question 7.
square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.

Solution:

After cutting squares we will get a cuboid,
length of the cuboid (l) = 24 – 2x
breadth of the cuboid (b) = 24 – 2x
height of the cuboid (h) = 2x
Volume of the box = Volume of the cuboid
V = (24 – 2x)(24 – 2x) (x)
= (24 – 2x)² (x)
= (576 + 4x² – 96x) x
= 576x + 4x³ – 96x²
V = 4x³ – 96x² + 576x
V(x) = 4x³ – 96x² + 576x

Question 8.
A function f is defined by f(x) = 3 – 2x. Find x such that f(x²) = (f (x))².
Answer:
f(x) = 3 – 2x
f(x²) = 3 – 2 (x²)
= 3 – 2x²
(f (x))² = (3 – 2x)²
= 9 + 4x² – 12x
But f(x²) = (f(x))²
3 – 2 x² = 9 + 4x² – 12x
-2x² – 4x² + 12x + 3 – 9 = 0
-6x² + 12x – 6 = 0
(÷ by – 6) ⇒ x² – 2x + 1 = 0
(x – 1) (x – 1) = 0
x – 1 = 0 or x – 1 = 0
x = 1
The value of x = 1

Question 9.
A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.
Solution:
Speed = \(\frac{\text { distance covered }}{\text { time taken }}\)
⇒ distance = Speed × time
⇒ d = 500 × t [ ∵ time = t hrs]
⇒ d = 500 t

Question 10.
The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length (x) as y = ax + b , where a, b are constants.
(i) Check if this relation is a function.
(ii) Find a and b.
(iii) Find the height of a woman whose forehand length is 40 cm.
(iv) Find the length of forehand of a woman if her height is 53.3 inches.

Answer:
The relation is y = 0.9x + 24.5
(i) Yes the relation is a function.
(ii) When compare with y = ax + b
a = 0.9, b = 24.5
(iii) When the forehand length is 40 cm, then height is 60.5 inches.
Hint: y = 0.9x + 24.5
= 0.9 × 40 + 24.5
= 36 + 24.5
= 60.5 feet

(iv) When the height is 53.3 inches, her forehand length is 32 cm
Hint: y = 0.9x + 24.5
53.3 = 0.9x + 24.5
53.3 – 24.5 = 0.9 x
28.8 = 0.9 x
x = \(\frac { 28.8 }{ 0.9 } \)
x = 32 cm

Representation of functions

A function may be represented by

(a) Set of ordered pairs
(b) Table form
(c) Arrow diagram
(d) Graphical form

Vertical line test
A curve drawn in a graph represents a function, if every vertical line intersects the curve in at most one point.

Types of function

1. One – One function (injection)

A function f: A → B is called one-one function if distinct elements of A have a distinct image in B.

2. Many – One function

A function f: A → B is called many-on function if two or more elements of A have same image in B.

3. Onto function (surjection)

A function f: A → B is said to be on to function if the range of f is equal to the co-domain of f.

4. Into function

A function f: A → B is called an into function if there exists at least one element in B which is not the image of any element of A.

5. Bijection

A function f: A → B is both one – one and onto, then f is called a bijection from A to B.

Horizontal line test
A function represented in a graph is one – one, if every horizontal line intersects the curve in at most one point.

Special cases of function

1. Constant function

A function f: A → B is called a constant function if the range of f contains only one element.

2. Identity function

A function f : A → A defined by f(x) = x for all x ∈ A is called an identity function on A and is denoted by I_A.

3. Real valued function
A function f: A → B is called a real valued function if the range of f is a subset of the set of all real numbers R.

Students can download Maths Chapter 1 Relations and Functions Ex 1.6 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.6

Multiple Choice Questions

Question 1.
If n(A × B) = 6 and A= {1, 3} then n (B) is ………….
(1) 1
(2) 2
(3) 3
(4) 6
Answer:
(3) 3
Hint: n(A × B) = 6
n(A) = 2
n(A × B) = n(A) × n(B)
6 = 2 × n(B)
n(B) = \(\frac { 6 }{ 2 } \) = 3

Question 2.
A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A ∪ C) × B] is
(1) 8
(2) 20
(3) 12
(4) 16
Answer:
(3) 12
Hint:
A = {a, b, p}, B = {2, 3}, C = {p, q, r, s}
n (A ∪ C) × B
A ∪ C = {a, b, p, q, r, s}
(A ∪ C) × B = {{a, 2), (a, 3), (b, 2), (b, 3), (p, 2), (p, 3), (q, 2), (q, 3), (r, 2), (r, 3), (s, 2), (s, 3)
n [(A ∪ C) × B] = 12

Question 3.
If A = {1,2}, B = {1,2, 3, 4}, C = {5,6} and D = {5, 6, 7, 8} then state which of the following statement is true ……………….
(1) (A × C) ⊂ (B × D)
(2) (B × D) ⊂ (A × C)
(3) (A × B) ⊂ (A × D)
(4) (D × A) ⊂ (B × A)
Answer:
(1) (A × C) ⊂ (B × D)
Hint: n(A × B) = 2 × 4 = 8
(A × C) = 2 × 2 = 4
n(B × C) = 4 × 2 = 8
n(C × D) = 2 × 4 = 8
n(A × C) = 2 × 2 = 4
n(A × D) = 2 × 4 = 8
n(B × D) = 4 × 4 = 16
∴ (A × C) ⊂ (B × D)

Question 4.
If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is
(1) 3
(2) 2
(3) 4
(4) 6
Answer:
(2) 2
Hint:
n(A) = 5
n(B) = x
n(A × B) = 1024 = 2¹⁰
2^5x = 2¹⁰
⇒ 5x = 10
⇒ x =2

Question 5.
The range of the relation R = {(x, x²) a prime number less than 13} is ……………………
(1) {2, 3, 5, 7}
(2) {2, 3, 5, 7, 11}
(3) {4, 9, 25, 49, 121}
(4) {1, 4, 9, 25, 49, 121}
Answer:
(3) {4, 9, 25, 49, 121}
Hint:
Prime number less than 13 = {2, 3, 5, 7, 11}
Range (R) = {(x, x²)}
Range = {4, 9, 25, 49, 121} (square of x)

Question 6.
If the ordered pairs (a + 2, 4) and (5, 2a + b)are equal then (a, b) is
(1) (2, -2)
(2) (5, 1)
(3) (2, 3)
(4) (3, -2)
Answer:
(4) (3, -2)
Hint:
(a + 2, 4), (5, 2a + b)
a + 2 = 5
a = 3
2a + b = 4
6 + b = 4
b = -2

Question 7.
Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is ……………..
(1) mⁿ
(2) n^m
(3) 2^mn – 1
(4) 2^mn
Answer:
(4) 2^mn

Question 8.
If {(a, 8),(6, b)}represents an identity function, then the value of a and b are respectively
(1) (8, 6)
(2) (8, 8)
(3) (6, 8)
(4) (6, 6)
Answer:
(1) (8, 6)
Hint:
{{a, 8), (6, b)}
a = 8
b = 6

Question 9.
Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}.
A function f: A → B given by f = {(1, 4), (2, 8),(3,9),(4,10)} is a ……………
(1) Many-one function
(2) Identity function
(3) One-to-one function
(4) Into function
Answer:
(3) One-to-one function
Hint:

Different elements of A has different images in B.
∴ It is one-to-one function.

Question 10.
If f (x) = 2x² and g(x) = \(\frac { 1 }{ 3x } \), then fog is …………..
(1) \(\frac{3}{2 x^{2}}\)
(2) \(\frac{2}{3 x^{2}}\)
(3) \(\frac{2}{9 x^{2}}\)
(4) \(\frac{1}{6 x^{2}}\)
Answer:
(3) \(\frac{2}{9 x^{2}}\)
Hint:

Question 11.
If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to
(1) 7
(2) 49
(3) 1
(4) 14
Answer:
(1) 7
Hint:
In a bijective function, n(A) = n(B)
⇒ n(A) = 7

Question 12.
Let f and g be two functions given by
f = {(0,1),(2, 0),(3-4),(4,2),(5,7)}
g = {(0,2),(1,0),(2, 4),(-4,2),(7,0)}
then the range of f o g is …………………
(1) {0,2,3,4,5}
(2) {-4,1,0,2,7}
(3) {1,2,3,4,5}
(4) {0,1,2}
Answer:
(4) {0,1,2}
Hint: f = {(0, 1)(2, 0)(3, -4) (4, 2) (5, 7)}
g = {(0,2)(l,0)(2,4)(-4,2)(7,0)}

fog = f[g(x)]
f [g(0)] = f(2) = 0
f [g(1)] = f(0) = 1
f [g(2)] = f(4) = 2
f[g(-4)] = f(2) = 0
f[g(7)] = f(0) = 1
Range of fog = {0,1,2}

Question 13.
Let f(x) = \(\sqrt{1+x^{2}}\) then
(1) f(xy) = f(x),f(y)
(2) f(xy) ≥ f(x),f(y)
(3) f(xy) ≤ f(x).f(y)
(4) None of these
Answer:
(3) f(xy) ≤ f(x).f(y)
Hint:
\(\sqrt{1+x^{2} y^{2}} \leq \sqrt{\left(1+x^{2}\right)} \sqrt{\left(1+y^{2}\right)}\)
⇒ f(xy) ≤ f(x) . f(y)

Question 14.
If g= {(1,1),(2,3),(3,5),(4,7)} is a function given by g(x) = αx + β then the values of α and β are
(1) (-1,2)
(2) (2,-1)
(3) (-1,-2)
(4) (1,2)
Answer:
(2) (2, -1)
Hint: g (x) = αx + β
g(1) = α(1) + β
1 = α + β ….(1)
g (2) = α (2) + β
3 = 2α + β ….(2)
Solve the two equations we get
α = 2, β = -1

Question 15.
f(x) = (x + 1)³ – (x – 1)³ represents a function which is
(1) linear
(2) cubic
(3) reciprocal
(4) quadratic
Answer:
(4) quadratic
Hint:
f(x) = (x + 1)³ – (x – 1)³
= x³ + 3x² + 3x + 1 -[x³ – 3x² + 3x – 1]
= x³ + 3x² + 3x + 1 – x³ + 3x² – 3x + 1 = 6x² + 2
It is a quadratic function.

Students can Download Tamil Nadu 12th Accountancy Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Accountancy Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Accountancy Model Question Paper 4 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 2.30 Hours
Max Marks: 90

PART – I

Answer all the questions. Choose the correct answer. [Answers are in bold] [20 x 1 = 20]

Question 1.
Companies cannot keep books on single entry system because of……….
(a) tax properties
(b) legal provisions
(c) both (a) and (b)
(d) None of these
Answer:
(b) legal provisions

Question 2.
Bills receivable endorsed are debited to………
(a) Bills receivable A/c
(b) Debtors A/c
(c) Creditors A/c
(d) None of these
Answer:
(c) Creditors A/c

Question 3.
The amount or property received by a non-profit organisation as stated by the will of a decreased person is commonly referred to as……….
(a) Donations
(b) Honorarium
(c) Legacy
(d) Endowment
Answer:
(c) Legacy

Question 4.
In non-profit organisation the sale of old newspapers is generally considered as a/an …….
(a) Revenue Receipts
(b) Expenditure
(c) Income
(d) capital Receipt
Answer:
(c) Income

Question 5.
Which of the following method, the capital of the partners is not altered and it remains generally fixed?
(a) Fixed capital method
(b) Fluctuating capital method
(c) Both ‘a’ and ‘b’
(d) None of these
Answer:
(a) Fixed capital method

Question 6.
All the transactions between the partner and the firm are recorded in the……..
(a) Capital A/c
(b) Drawing A/c
(c) Profit and loss A/c
(d) Revaluation A/c
Answer:
(a) Capital A/c

Question 7.
Which of the following method goodwill is calculated by multiplying the weighted average – profit?
(a) Super profit method
(b) Annuity
(c) Weighted average profit method
(d) All of these
Answer:
(c) Weighted average profit method

Question 8.
Self-generated goodwill should not be shown in the……….
(a) Journal
(b) Ledger
(c) Balance sheet
(d) Books of accounts
Answer:
(d) Books of accounts

Question 9.
When A and B sharing profits and losses in the ratio of 3:2 they admit C as a partner giving him 1/3 share of profits. This will be given by A and B………
(a) Equality
(b) In the ratio of their capitals
(c) In the ratio of their profit
(d) None of these
Answer:
(c) In the ratio of their profit

Question 10.
In order to maintain fair dealings, at the time of admission it is necessary to revalue assets and liabilities of the firm to their………
(a) Cost price
(b) True value
(c) Selling price
(d) None of these
Answer:
(b) True value

Question 11.
In both types of policies the insurance premium is paid by the……….
(a) Sale tradership
(b) Partnership firm
(c) Hindu undivided family
(d) None of these
Answer:
(b) Partnership firm

Question 12.
The policy amount received from the insurance company, is used to settle the amount due to the………..
(a) Increased partner
(b) Decreased partner
(c) Partnership at will
(d) Partnership deed
Answer:
(b) Decreased partner

Question 13.
The capital of a company is divided into small units of………..
(a) Current amount
(b) fixed amount
(c) Capital amount
(d) None of these
Answer:
(b) fixed amount

Question 14.
Equity shares do not enjoy any ………..
(a) Equity rights
(b) Preference rights
(c) Dividend of rights
(d) Ordinary shares
Answer:
(b) Preference rights

Question 15.
Comparative statements and trend percentages are examples of …………
(a) Cash flow analysis
(b) Trend analysis
(c) Horizontal analysis
(d) Vertical analysis
Answer:
(c) Horizontal analysis

Question 16.
Preparation of common size statements and computation of ratios are examples………..
(a) Ratio analysis
(b) vertical analysis
(c) Horizontal analysis
(d) None of these
Answer:
(b) vertical analysis

Question 17.
Which ratio is the proportion, of fixed income bearing funds to ………..
(a) Debt, equity ratio
(b) Capital gearing ratio
(c) Propreetory ratio
(d) Profitability ratio
Answer:
(b) Capital gearing ratio

Question 18.
Which ratio indicates the efficiency of utilisation of fixed assets ……….
(a) Inventory turnover ratio
(b) Trade receivables turnover ratio
(c) Trade payable turnover ratio
(d) Fixed assets turnover ratio
Answer:
(d) Fixed assets turnover ratio

Question 19.
All transactions related to receipt either in cash or through bank are recorded using……….
(a) Payment voucher
(b) Centra voucher
(c) Receipt voucher
(d) Sales voucher
Answer:
(c) Receipt voucher

Question 20.
All transactions related to payments either in Cash or through bank are recorded using………..
(a) Journal voucher
(b) Centra voucher
(c) Receipt voucher
(d) Payment voucher
Answer:
(d) Payment voucher

PART – II

Answer any seven questions in which question no. 30 is compulsory. [7 × 2 = 14]

Question 21.
What is a statement of affairs?
Answer:
A statement of affairs is a statement showing the balances of assets and liabilities on a particular date. The balance of assets are shown on the right side and the balance of liabilities on the left side. This statement resembles a balance sheet. The difference between the total of assets and total of liabilities is taken as capital.
Capital = Assets – Liabilities

Question 22.
How will the following appear in the final accounts of Marthandam woman Cultural Association.
Stock of sports materials on 1.4.2018 16,000
Sport materials purchased 84,000
Stock of sports materials on 31.3.19 10,000
Answer:

Question 23.
Mani is a partner, who withdrew Rs 30,000 on 1st September, 2018. Interest on drawings is charged at 6% per annum. Calculate interest on drawings on 31st December, 2018 and show the journal entries by assuming that fluctuating capital method is followed.
Answer:
Mani:
30,000 × \(\frac{6}{100}\) x \(\frac{4}{12}\) = Rs 600
Interest on drawings of Mani = Rs 600

Question 24.
The following are the profits of a firm in the last five years: 2014: Rs 4,000; 2015: Rs 3,000; 2016: Rs 5,000; 2017: Rs 4,500 and 2018: Rs 3,500. Calculate the value of goodwill at 3 years purchase of average profits of five years.
Answer:
Goodwill = Average profit × Number of years of purchase

Goodwill = Average profit × Number of years of purchase
= 4,000 × 3 = Rs 12,000.

Question 25.
What is goodwill?
Answer:
Goodwill is the good name or reputation of the business which brings benefit to the business. It enables the business to earn more profits. It is the present value of a firm’s future excess earnings.

Question 26.
Anbu and Raju are partners sharing profits in the ratio of 3 : 2. Akshai is admitted as a partner. The new profit sharing ratio among Anbu. Raju and Akshai is 5 : 3 : 2. Find out the sacrificing ratio.
Answer:
Old ratio of Anbu and Raju = 3 : 2 that is, \(\frac{3}{5}\) : \(\frac{2}{5}\)
New Ratio of Anbu, Raju and Akshai = 5 : 3 : 2 that is, \(\frac{5}{10}\) : \(\frac{3}{10}\) : \(\frac{2}{10}\)
Share sacrificed = Old share – New share

Question 27.
Vivin, Hari and Joy are partners sharing profits and losses in the ratio of 3 : 2 : 1. On 31.3.2017, Hari retired. On the date of retirement, the books of the firm showed a general reserve of Rs 60,000. Pass the journal entry to transfer the general reserve.
Answer:
Journal Entry

Question 28.
List the tools of financial statement analysis.
Answer:

1. Comparative Statement
2. Common size statement
3. Trend analysis
4. Funds flow analysis
5. Cash flow analysis

Question 29.
Calculate quick ratio of Ananth Constructions Ltd from the information given below.

Answer:

Quick assets = Current assets – Inventories – Prepaid expenses
= 2,50,000 – 50,000 – 15,000
= Rs 1,85,000

Question 30.
What are the pre-defined ledgers available in Tally.ERP-9.
Answer:
Tally has two-predefined ledgers cash and profit and loss A/c. The user has to create various other ledgers based on their requirements. Pre-defined group/ledger cannot be deleted.

PART-III

Answer any seven questions in which question No. 40 is compulsory. [7 × 3 = 21]

Question 31.
On 1st April 2017, Ganesh started his business with a capital of Rs 75,000. He did not maintain proper book of accounts. Following particulars are available from his books as on 31.03.2018.

During the year he withdrew Rs 15,000 for his personal use. He introduced further capital of Rs 20,000 during the year. Calculate his profit or loss.
Answer:
Statement of affairs of Ganesh as on 31st March, 2018

For finding out the closing capital, Statement of affairs as on 31st March, 2018 is prepared. Statement of profit or loss for the year ending 31st March, 2018

Question 32.
Compute income from subscription for the year 2018 from the following particulars relating to a club.

Answer:
Subscription received during the year 2018: Rs 45,000.
Income and Expenditure Account for the year ended 31.03.2019

Question 33.
State the difference between fixed capital method and fluctuating capital method.
Answer:

Question 34.
From the following information, calculate the value of goodwill based on 3 years purchase of super profit
(i) Capital employed: Rs 2,00,000
(ii) Normal rate of return: 15%
(iii) Average profit of the business: Rs 42,000
Answer:
Normal profit = Capital employed × Normal rate of return
= 2,00,000 × 15% = Rs 30,000
Super profit = Average profit – Normal profit
= 42,000 – 30,000 = Rs 12,000
Goodwill = Super profit × Number of years of purchase
= 12,000 × 3 = Rs 36,000

Question 35.
Govind and Gopal are partners in a firm sharing profits in the ratio of 5 : 4. They admit Rahim as a partner. Govind surrenders 2/9 of his share in favour of Rahim. Gopal surrenders 1/9 of his share in favour of Rahim. Calculate the new profit sharing ratio and sacrificing ratio.
Answer:

Question 36.
Balu, Chandru and Nirmal are partners in a firm sharing profits and losses in the ratio of 5 : 3 : 2. On 31st March 2018, Nirmal retires from the firm. On the date of Nirmal’s retirement, goodwill appeared in the books of the firm at Rs 60,000. By assuming fluctuating capital account, pass the necessary journal entry if the partners decide to
(а) write off the entire amount of existing goodwill
(b) write off half of the existing goodwill.
Answer:

Question 37.
State the different between preference shares and equity shares?
Answer:

Question 38.
Briefly explain any three limitations of financial statements.
Answer:
(i) Lack of qualitative information : Qualitative information, that is non-monetary information is also important for business decisions. For e.g. Efficiency of the employees and efficiency of the management. But this is ignored in financial statements.

(ii) Record of historical data : Financial statement are prepared based on historical data. They may not reflect the current position.

(iii) Ignores price level changes : Adjustments for price level changes are not made in the financial statements. Hence financial statements may not reveal the current position.

Question 39.
State any three advantages of ratio analysis.
Answer:
(i) Measuring operational efficiency : Ratio analysis helps to know operational efficiency of a business by finding the relationship between operating cost and revenues and also by comparison of present ratios with those of the past ratios.

(ii) Intra firm comparison : Comparison of efficiency of different divisions of an organisation is possible by comparing the relevant ratios.

(iii) Inter firm comparison : Ratio analysis helps the firm to compare its performance with other firms.

Question 40.
Explain any five applications of computerised accounting system (CAS).
Answer:
The applications of CAS are as follows

1. Maintaining accounting records
2. Inventory management
3. Pay-roll preparation
4. Report generation
5. Data Import/Export
6. Taxation

PART – IV

Answer all the following questions. [7 × 5 = 35]

Question 41(a).
From the following details you are required to calculate credit sales and credit purchases by preparing total debtors account, total creditors account, bills receivable account and bills payable account.

Answer:

[OR]

(b) From the following Receipts and Payments account of Yercaud Youth Association, prepare Income and expenditure account for the year ended 31st March, 2019 and the balance sheet as on that date.

Additional information:
(i) Opening capital fund Rs 20,000.
(ii) Stock of books on 1.4,2018 Rs 9,200.
(iii) Subscription due but not received Rs 1,700.
(iv) Stock of stationery on 1.4.2018 Rs 1,200 and stock of stationery on 31.3.2019, Rs 2,000
Answer:
Income and Expenditure Account for the year ended 31.03.19

Question 42(a).
Following is the Receipts and Payments Account of Salem Recreation Club for the year ended 31st March, 2019.
In the books of Salem Recreation Club

Additional information:
(i) There are 450 members each paying annual subscription of Rs 30.
(ii) Stock of stationery on 31st March, 2018 Rs 300 and on March 31, 2019 Rs 500.
(iii) Capital fund as on 1st April 2018 was Rs 9,300.
Prepare income and expenditure account for the year ended 31st March, 2019 and the balance sheet as on that date.
Answer:

[OR]

(b) From the following‘details you are required to calculate credit sales and credit purchases by preparing total debtors account, total creditors account, bills receivable account and bills payable account.

Answer:

Question 43(a).
Arulappan and Nallasamy are partners in a firm sharing profits and losses in the ratio of 4 : 1. On 1st January 2018, their capitals were Rs 20,000 and Rs 10,000, respectively. The partnership deed specifies the following:
(a) Interest on capital is to be allowed at 5% per annum.
(b) Interest on drawings charged to Arulappan and Nallasamy are Rs 200 and Rs 300, respectively.
(c) The net profit of the firm before considering interest on capital and interest on drawings amounted to Rs 18,000.
Give necessary journal entries and prepare Profit and loss appropriation account for the year en’ding 31st December 2018. Assume that the capitals are fluctuating.
Answer:
Journal entries

[OR]

(b) Calculate the value of goodwill at 5 years purchase of super profit from the following information:
(i) Capital employed: Rs 1,20,000
(ii) Normal rate of profit: 20%
(iii) Net profit for 5 years:
2014: Rs 30,000; 2015: Rs 32,000; 2016: Rs 35,000; 2017: Rs 37,000 and 2018: Rs 40,000
(d) Fair remuneration to the partners Rs 2,800 per annum.
Answer:

Normal profit = Capital employed × Normal rate of return
= 1,20,000 × 20%
= Rs 24,000
Super profit = Average profit – Normal profit
= 32,000 – 24,000
= 8,000
Goodwill = Super profit × Number of years of purchase
= 8,000 × 5
= Rs 40,000

Question 44(a).
Distribution of Profits: Anitha, Ramita were partners sharing profit and losses in the ratio of 7 : 3. Their capitals were Rs. 80,000 and Rs. 60,000, respectively.
(i) Interest on capital @10% p.a
(ii) Interest on drawings @12% p.a
(iii) Both to get a salary of Rs. 10,000 each per annum
(iv) Anitha to get a commission of 10% on the net profit before charging such commission. The profit for the year Rs. 60,000. Drawings were Anitha Rs. 12,000 and Ramita Rs 8,000. Show profit and loss appropriation account and the capital A/c.
Answer:

[OR]

(b) From the following information relating to Sridevi enterprises, calculate the value of goodwill on the basis of 4 years purchase of the average profits of 3 years.
(a) Profits for the years ending 31st December 2016, 2017 and 2018 were Rs 1,75,000, Rs 1,50,000 and Rs 2,00,000, respectively.
(b) A non-recurring income of Rs 45,000 is included in the profits of the year 2016.
(c) The closing stock of the year 2017 was overvalued by Rs 30,000.
Calculation of adjusted profit.
Answer:

Average profit = \(\frac{Total profit}{Number of years}\)
Total profit 1,60,000 + 1,20,000 + 23,000
= 4,80,000
Average profit = \(\frac{4,80,000}{3}\) = Rs 1,60,000
Goodwill = Average profit × No. of year purchase
= Rs 1,60,000 × 4 = Rs 64,00,000

Question 45(a).
Vetri and Ranjit are partners sharing profit in the ratio 3 : 2. Their balance sheet as on 31st December 2017 is under.

On 1.1.2018 they admit suriya into their firm as a partner on the following arrangements.
(i) Suriya brings Rs 10,000 as capital for 1/4 share of profit
(ii) Stock to be depreciated by 10%
(iii) Debtors to be revalued at Rs 7,500
(iv) Furniture to be revalued at Rs 40,000.
(v) There is an outstanding wages Rs 4,500 not yet recorded. Prepare revaluation account, partners’ capital account and the balance sheet of the firm after admission.
Answer:

[OR]

(b) Sundar and Suresh are partners sharing profits in the ratio of 3 : 2. Their balance sheet as on 1st January, 2017 was as follows:

They decided to admit Sugumar into partnership for 1/4 share in the profits on the following terms:
(a) Sugumar has to bring in Rs 30,000 as capital. His share of goodwill is valued at Rs 5,000. He could not bring cash towards goodwill.
(b) That the stock be valued at Rs 20, 000.
(c) That the furniture be depreciated by Rs 2,000.
(d) That the value of building be depreciated by 20%.
Prepare necessary ledger accounts and the balance sheet after admission.
Answer:

Question 46(a).
Charles, Muthu and Sekar are partners, sharing profits in the ratio of 3 : 4 : 2. Their balance sheet as on 31st December 2018 is as under:

On 1,1.2019, Charles retired from the partnership firm on the following arrangements:
(i) Stock to be appreciated by 10%
(ii) Furniture to be depreciated by 5%
(iii) To provide Rs 1,000 for bad debts
(iv) There is an outstanding repair of Rs 11,000 not yet recorded
(v) The final amount due to Charles was paid
Prepare revaluation account, partners capital account and the balance sheet of the firm after retirement.
Answer:

[OR]

(b) Varsha, Shanthi and Madhuri are partners, sharing profits in the ratio of 5 : 4 : 3. Their balance sheet as on 31st December 2017 is as under:
Balance Sheet as on 31st December 2017

On 1.1.2018, Madhuri died and on her death the following arrangements are made:
(i) Stock to be depreciated by Rs 5,000
(if) Premises is to be appreciated by 20%
(iii) To provide Rs 4,000 for bad debts
(iv) The final amount due to Madhuri was not paid
Prepare revaluation account, partners capital account and the balance sheet of the firm after death.
Answer:

Question 47 (a).
Lalitha Ltd. offered Rs 30,000 equity shares of ? 10 each to the public payable Rs 2 per share on application, Rs 3 on share allotment and the balance when required. Applications for 50,000 shares were received on which the directors allotted as:
Applicants for 10,000 shares – Full
Applicants for 35,000 shares – 20,000 shares (excess money will be utilised for allotment)
Applicants for 5,000 shares – Nil
All the money due was received. Pass journal entries upto the receipt of allotment.
Answer:
Journal Entries

Working notes

[OR]

(b) Prepare common-size income statement for the following particulars of Sam Ltd.

Answer:
Common Size Income statement of Sam Ltd.

Students can Download Tamil Nadu 12th Accountancy Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Accountancy Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Accountancy Model Question Paper 3 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 2.30 Hours
Max Marks: 90

PART – I

Answer all the questions. Choose the correct answer.[Answers are in bold] [20 x 1 = 20]

Question 1.
The total assets of the proprietor are ? 5,00,000. His liabilities ? 3,50,000. Then his capital in the business is
(a) Rs 8,50,000
(b) Rs 1,50,000
(c) Rs 3,50,000
(d) Rs 4,25;000
Answer:
(b) Rs 1,50,000

Question 2.
A firm has assets worth Rs 60,000 and capital 45,000; then its liabilities is
(a) Rs 45,000
(b) Rs 1,05,000
(c) Rs 60,000
(d) Rs 15,000
Answer:
(d) Rs 15,000

Question 3.
Income and expenditure accounts show
(a) Cash available to an organization
(b) Closing capital of an organization
(c) Cash available in the bank account
(d) Surplus or deficit for the current accounting period
Answer:
(d) Surplus or deficit for the current accounting period

Question 4.
On what basis the receipts and payments account is prepared
(a) Cash basis
(b) Credit basis
(c) Both
(d) None of these
Answer:
(a) Cash basis

Question 5.
The balance in the appropriation account is transferred to the partner’s capital account in the
(a) Agree ratio
(b) Sacrifice ratio
(c) Profit sharing ratio
(d) Old ratio
Answer:
(c) Profit sharing ratio

Question 6.
Amount invested by partners in the partnership business is called
(a) Owner’s capital
(b) Partners capital
(c) Profit and loss appropriation
(d) None of these
Answer:
(b) Partners capital

Question 7.
The profit earning capacity of the firm determines the value of its
(a) Balance Sheet
(b) good will
(c) Profit and Loss Account
(d) All of these
Answer:
(b) good will

Question 8.
Goodwill acquired by making payment in cash or kind is called
(a) Purchased goodwill
(b) Self-generated goodwill
(c) Average goodwill
(d) (a) and (b)
Answer:
(a) Purchased goodwill

Question 9.
A new partner may be admitted to a partnership
(a) with the consent of all the partners
(b) with the consent of any one partner
(c) with the consent of two-third of the old partners
(d) with the consent of three-fourth of the old partners
Answer:
(a) with the consent of all the partners

Question 10.
An incoming partner is liable for all the acts of the firms dare
(a) Before his admission
(b) After his admission
(c) From six months prior to his admission
(d) None of these
Answer:
(b) After his admission

Question 11.
The profit or loss arising therefrom is transferred to the capital accounts of all the partners in the
(a) New profit sharing ratio
(b) Old profit sharing ratio
(c) Gaining ratio
(d) Sacrificing ratio
Answer:
(b) Old profit sharing ratio

Question 12.
The settlement is to be done in the manner prescribed in the
(a) Partnership-at-will
(b) Partnership deed
(c) Both (a) and (b)
(d) None of these
Answer:
(b) Partnership deed

Question 13.
Company is a voluntary association of persons which has separate
(a) legal entity
(b) voluntary association
(c) common seal
(d) limited liability
Answer:
(a) legal entity

Question 14.
The liability of the shareholders of the company is limited to the extent of face value of the shares held by the
(a) share holders
(b) card holders
(c) debenture holders
(d) None of these
Answer:
(a) share holders

Question 15.
The term fund refers to
(a) working capital
(b) opening capital
(c) closing capital
(d) None of these
Answer:
(a) working capital

Question 16.
When figure relating to several years are considered for the purpose of analysis, the analysis is called
(a) Horizontal analysis
(b) vertical analysis
(c) Trend analysis
(d) Cash flow analysis
Answer:
(a) Horizontal analysis

Question 17.
Which ratios measure the firm’s ability to meet its short term obligations in time?
(a) Profitability Ratios
(b) Liquidity Ratios
(c) Activity Ratios
(d) None of these
Answer:
(b) Liquidity Ratios

Question 18.
For Calculating Quick Assets – and- are excluded from current assets
(a) Debtors and inventory
(b) Inventory and Bills payable
(c) Cash and Bank balances
(d) Inventory and Prepaid Expenses
Answer:
(d) Inventory and Prepaid Expenses

Question 19.
Ideal current Ratio is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 3 : 1
Answer:
(c) 2 : 1

Question 20.
Which of the following options is used to view Trial Balance from Gateway of Tally?
(a) Gateway of Tally → Reports → Trial Balance
(b) Gateway of Tally → Trial Balance
(c) Gateway of Tally → Reports → Display → Trial Balance
(d) None of these
Answer:
(c) Gateway of Tally → Reports → Display → Trial Balance

PART – II

Answer any seven questions in which question no. 30 is compulsory. [7 × 2 = 14]

Question 21.
From the following particulars ascertain profit or loss?
Capital as on 1. 4. 2018 1,60,000
Capital as on 31. 3. 2019 1,50,000
Additional capital introduced 25,000
Drawings made during the year 30,000
Answer:
Statement of profit or loss for the year ended 31.3.2019

Question 22.
From the following particulars of Tamil Educational Society.
Prepare Receipts and Payments for the year ended 31.03.2019

Answer:

Question 23.
What is partners current account?
Answer:
Under fixed capital method, Capital account and current accounts are maintained. In current account all adjustments relating to partners are recorded on the credit side of current account viz: Interest on capital, Share of profits, Salary and commission etc are recorded.
On the debit side drawings interest on drawings, share of loss are recorded. Current account sometimes show credit or debit balance.

Question 24.
The following are the profits of a firm in the last five years 2014 – Rs 4,000, 2015 – Rs 3,000, 2016 – Rs 5,000, 2017 – Rs 4,500, 2018 – Rs 3,500. Calculate the value of goodwill at 3 years purchase of average profit of five year.
Answer:
Goodwill = Average profit × No. of years of purchase

Good will = Average profit × No. of years of purchase
= Rs 4,000 × 3
Goodwill = Rs 12,000

Question 25.
Mala and Vimala were partners sharing profit and losses in the ratio of 3 : 2. On 31.3.2017. Varshini was admitted as a partner. On the date of admission. The book of the firm showed a reserve fund of Rs 50,000. Pass the journal entry to distribute the reserve fund.
Answer:
Journal Entry

Question 26.
Mary, Vleena, Mariam are partners of a firm sharing profit and losses equally. Many retired from the partnership on 1.1.2019. On that date, their balance sheet showed accumulated loss of Rs 75,000. On the assets side of Balance Sheet. Give the journal entry to distribute the accumulated loss.
Answer:
Journal Entry

Question 27.
Anitha was holding 500 equity shares of Rs 10 each of Thanjavur Motors Ltd. issued at par. She paid Rs 3 on application Rs 5 on allotment but could not pay the first and final call of Rs 2. The directors forfeited the shares for non payment of call money. Give Journal Entry for forfeiture of shares.
Answer:
Journal Entry

Question 28.
What is working capital?
Answer:
Working capital statement or schedule of changes in working is prepared to disclose net changes in working capitals on two specific dates (generally two balance sheet dates). It is prepared from current assets and current liabilities.
Working Capital = Current Assets – Current Liabilities

Question 29.
What is meant by accounting ratio?
Answer:
Ratio is a mathematical expression of relationship between two related or interdependent items. It is the numerical or quantitative relation ship between two items. It is calculated by dividing one item by the other related item. When ratios are calculated on the basis of accounting information, these are called ‘accounting ratio’.

Question 30.
What are accounting reports?
Answer:
Accounting report is a compilation of accounting information that are derived from the accounting records of a business concern. Accounting reports may be classified as routine reports and special purpose reports.

PART – III

Answer any seven questions in which question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
Following are the balances of Shanthi as on 31.12.2018

Answer:
Prepare a statement of affairs as on 31.12.2018 and calculate capital as at that date.

Question 32.
Compute income from subscription for the 2018 from the following particulars relatiiig to a club.

Answer:
Subscription received during the year 2018 Rs 1,50,000
Calculation of incomes from subscription for 2018

Question 33.
From the following information prepare capital accounts of partners Mannan and Sevagan when their capitals are fluctuating.

Answer:
Partners Capital Account

Question 34.
From the following information relating to Sridevi Enterprises, calculate the value of goodwill on the basis of 4 years purchase of the average profit of 3 years.
(a) Profits for the year ending 31st December 2016, 2017 and 2018 were Rs 1,75,000, 1,50,000 and Rs 2,00,000 respectively.
(b) A non – recuring income of Rs 45,000 is included in the profits of the year 2016.
(c) The closing stock of the year 2017 was overvalued by Rs. 30,000
Answer:
Calculation of adjusted profit

Goodwill = Average profit × Number of years of purchase
= 1,60,000 × 4
= Rs 6,40,000

Question 35.
A, B and C were partners. Sharing profits and losses in the ratio of 5 : 3 : 2. D is admitted into a partnership. The following adjustments are to be made.
(a) Increase the value of premises by Rs 60,000
(b) Depreciate stock by Rs 5,000; Furniture by Rs 2,000 and Machinery by Rs 2,500.
(c) Provide for an outstanding liabilities of Rs 500. Pass the journal entries.
Answer:
Journal Entries

Question 36.
Kavitha, Kumudha and Lalitha are partners sharing profit and losses in the ratio of 5 : 3 : 3 respectively. Kumudha retires on 31st Dec 2018. On the date of retirement, her capital account shows a credit balance of Rs 2,00,000 pass journal entries of
(i) The amount due is paid off immediately by cheque.
(it) The amount due is not paid immediately
(iii) Rs 70,000 is paid immediately by cheque
Answer:

Question 37.
Maruthu Ltd. forfeited 150 equity shares @ Rs 10 each for non-payment of final call of Rs 4/- per share of these 100 shares were reissued @ Rs 9/- per share. Pass journal entries for forfeiture and reissue.
Answer:
Journal entries

Forfeited shares:
150 shares = Rs 900
100 shares \(\frac{900}{150}\) × 100 = Rs 600
Gain or loss = Amount forfeited – loss on reissue
= Rs 600 – 100 = Rs 500

Question 38.
From the following particulars prepare comparative income statement of Mary Co. Ltd.

Answer:
Comparative income statement of Mary Co. Ltd. for the year ended 31st March 2016 and 31st March 2017

Note: Calculation of income tax
For 2015-16 = 2,00,000 × 20% = Rs 40,000;
For 2016-17 = 3,20,000 × 50% = Rs 1,60,000

Question 39.
Explain the objectives of ratio analysis?
Answer:
Following are the objectives of ratio analysis.

1. To Simplify accounting figures
2. To Facilitate analysis of financial statements
3. To analyses the operational efficiency of a business.
4. To help in budgeting and forecasting
5. To facilitate intro firm and inter firm comparison of performance.

Question 40.
Mention the commonly used voucher types in Tally ERP.9. Following are some of the major accounting vouchers used in an organisation.
Answer:

1. Receipt voucher
2. Payment voucher
3. Contra voucher
4. Purchase voucher
5. Sales voucher
6. Journal voucher

PART – IV

Answer all the following questions. [7 × 5 = 35]

Question 41 (a).
Raju does not keep proper books of accounts. Following details are taken from his records.

During the year introduced further capital of? 50,000 and withdraw ? 2,500 per month from the business for his personal use. Prepare statement of profit or loss with above information.
Answer:
Statement of affairs as on 1.1.2018

[OR]

(b) From the following particular calculate total purchases.

Answer:

Total Purchases = Cash Purchases + Credit Purchases
= Rs 3,20,000 + Rs 4,80,000
= Rs 8,00,000

Question 42(a).
From the following Receipts and Payments Account of Friends Football club for the year ending 31st March 2017. Prepare income and expenditure Account for the year ending 31.3.2017.

Additional information:
(i) The club had furniture of Rs 12,000 on 1.4.2016. Ignore depreciation on furniture
(ii) Subscription outstanding for 2016-17- Rs 600
(iii) Stock of sports material on 31.03.2017- Rs 100
(iv) Capital Fund as on 1st April 2016 was Rs 23,000
Answer:
Income and Expenditure Account for the year ended 31.03.17

[OR]

(b) From the following balance sheets of Brindha and Praveena who share profits and losses in the ratio of 3 : 4, calculate interest on capital at 6% p.a. for the year ending 31st December 2017.
Balance sheet as on 31st December 2017

On 1st July 2017, Brindha introduced an additional capital of Rs 6,000 and on 1st October 2017, Praveena introduced Rs 10,000. Drawings of Brindha and Praveena during the year were Rs 5,000 and Rs 7,000 respectively. Profit earned during the year was Rs 31,000.
Answer:

Profit credited = Profit earned Rs 31,000 – Balance profit as per balance sheet Rs 10,000 = Rs 21,000. This amount is distributed in their profit sharing ratio of 3 : 4.
Calculation of interest on capital:

Question 43(a).
Anand and Narayanan are partners in a firm sharing profits and losses in the ratio of 5 : 3. On 1st January 2018, their capitals were Rs 50,000 and Rs 30,000 respectively. The partnership deed specifies the following:
(a) Interest on capital is to be allowed at 6% per annum.
(b) Interest on drawings charged to Anand and Narayanan are Rs 1,000 and Rs 800 respectively.
(c) The net profit of the firm before considering interest on capital and interest on drawings amounted to Rs 35,000.
Give necessary journal entries and prepare profit and loss appropriation account as on 31st December 2018. Assume that the capitals are fluctuating.
Answer:

[OR]

(b) The following particulars are available in respect of a business carried on by a partnership firm:
(i) Profits earned: 2016: Rs 30,000; 2017: Rs 29,000 and 2018: Rs 32,000.
(ii) Profit of 2016 includes a non-recurring income of Rs 3,000.
(iii) Profit of 2017 is reduced by Rs 2,000 due to stock destroyed by fire.
(iv) The stock is not insured. But, it is decided to insure the stock in future. The insurance premium is estimated at Rs 5,600 per annum.
You are required to calculate the value of goodwill on the basis of 2 years purchase of average profits of the last three years.
Answer:

Goodwill = Average profit × Number of years of purchase
= 24,400 × 2 = Rs 48,800

Question 44(a).
Sri Ram and Raj are partners sharing profits and losses in the ratio of 2 : 1. Nelson joins as a partner on 1st April 2017. The following adjustments are to be made.
(i) Increase the value of stock by Rs 5,000
(ii) Bring into record investment of Rs 7,000 which had not been recorded in the books of the firm.
(iii) Reduce the value of office equipment by Rs 10,000
(iv) A provision would also be made for outstanding wages for Rs 9,500. Give journal entries and prepare revaluation account.
Answer:
Journal Entries

[OR]

(b) Kannan, Rahim and John are partners in a firm sharing profit and losses in the ratio of 5 : 3 : 2. The balance sheet as on 31st December, 2017 was as follows:

John retires on 1st January 2018, subject to following conditions:
(i) To appreciate building by 10%
(ii) Stock to be depreciated by 5%
(iii) To provide Rs 1,000 for bad debts
(iv) An unrecorded liability of Rs 8,000 have been noticed
(v) The retiring partner shall be paid immediately
Prepare revaluation account, partners capital account and the balance sheet of the firm after retirement.
Answer:

Question 45(a).
Sudha Ltd. offered 1,00,000 shares of Rs 10 each to the public payable Rs 3 on application, Rs 4 on share allotment and the balance when required. Applications for 1,40,000 shares were received on which the directors allotted as:
Applicants for 60,000 shares – Full
Applicants for 75,000 shares – 40,000 shares (excess money will be utilised for allotment)
Applicants for 5,000 shares – Nil
All the money due was received. Pass journal entries upto the receipt of allotment.
Answer:
In the books of Sudha Ltd. Journal entries

Working note

[OR]

(b) Prepare common-size statement of financial position for the following particulars of Rani Ltd.

Answer:
Common Size Balance sheet of Rani Ltd. as on 31st March 2016 to 31st March 2017

Question 46(a).
Lalitha Ltd. offered 30,000 equity shares of Rs 10 each to the public payable Rs 2 per share on application, Rs 3 on share allotment and the balance when required. Applications for 50,000 shares were received on which the directors allotted as:
Applicants for 10,000 shares – Full
Applicants for 35,000 shares – 20,000 shares (excess money will be utilised for allotment)
Applicants for 5,000 shares – Nil
All the money due was received. Pass journal entries upto the receipt of allotment.
Answer:
Journal Entries

[OR]

(b) From the following particulars of Neithal Ltd, calculate trend percentages.

Answer:
Trend analysis for Neithal Ltd

Question 47 (a).
From the following particulars, calculate the trend percentages of Babu Ltd.

Answer:
Trend Percentage of Babu Ltd. (In Thousands)

[OR]

(b) Calculate
(i) Inventory turnover ratio
(ii) Trade receivable turnover ratio
(iii) Trade payable turnover ratio and
(iv) Fixed assets turnover ratio from the following information obtained from Delphi Ltd.

Additional information:
Revenue from operations for the year Rs 10,50,000
Purchases for the year Rs 4,50,000
Cost of revenue from operations Rs 6,00,000
Assume that sales and purchase are for credit
Answer: