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Let’s solve this by taking sets as given in the problem.
Let, P = {5, 6, 9}
and M = {5, 6, 7, 8, 9}
Here, it is clear that P ⊆ M
Now, P ∪ M = {5, 6, 7, 8, 9}
and P ∩ (P ∪ M) = {5, 6, 9}
∴ P ∩ (P ∪ M) = P
∴ The correct answer is: (A) P
(A) Set of intersecting points of parallel lines
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(A) Colours of the rainbow
(A) {0, 1, 2, 3, ...}
(B) {i, n, d, a}
(C) {1, 2, 3, 4, 5, 7, 8}
Let, E be the set of persons who speak English and F be the set of people who speak French.
Here,
n (E) = 72, n (F) = 43 and n (E ∪ F) = 100, n (E ∩ F) = ?
Using the formula:
n (E ∪ F) = n (E) + n (F) − n (E ∩ F)
∴ 100 = 72 + 43 − n (E ∩ F)
∴ 100 = 115 − n (E ∩ F)
∴ n (E ∩ F) = 115 − 100
∴ n (E ∩ F) = 15 ... (i)
∴ 15 persons speak English and French both. ... (ii)
Now, the number of persons who speak only English
= n (E) − n (E ∩ F)
= 72 − 15
= 57
∴ 57 persons speak only English. ... (iii)
And, the number of persons who speak only French
= n (F) − n (E ∩ F)
= 43 − 15
= 28
∴ 28 persons speak only French. ... (iv)
A Venn Diagram of this is shown below:
Let, B be the set of trees planted by Parth and G be the set of trees planted by Pradnya.
Here,
n (B) = 70, n (G) = 90 and n (B ∩ G) = 25, n (B ∪ G) = ?
Using the formula:
n (B ∪ G) = n (B) + n (G) − n (B ∩ G)
∴ n (B ∪ G) = 70 + 90 − 25
∴ n (B ∪ G) = 135 ... (i)
∴ 135 trees were planted by Parth or Pradnya.
Using the formula:
n (A ∪ B) = n (A) + n (B) − n (A ∩ B)
∴ 36 = 20 + 28 − n (A ∩ B)
∴ 36 = 48 − n (A ∩ B)
∴ n (A ∩ B) = 48 − 36
∴ n (A ∩ B) = 12 ... (i)
Let U be the set of all students in the class (Universal Set), let D be the set of students having a dog as their pet animal at home, and let C be the set of students having a cat as their pet animal at home.
Here,
n (U) = 28, and
n (D ∩ C) = 10
Now, n (D) = 8 + n (D ∩ C)
∴ n (D) = 8 + 10 = 18 ... (i)
Also, n (C) = 6 + n (D ∩ C)
∴ n (C) = 6 + 10 = 16 ... (ii)
Using the formula:
n (D ∪ C) = n (D) + n (C) − n (D ∩ C)
∴ n (D ∪ C) = 18 + 16 − 10
∴ n (D ∪ C) = 34 − 10
∴ n (D ∪ C) = 24 ... (iii)
∴ Students who do not have a dog or a cat as their pet animal at home
= n (U) − n (D ∪ C)
= 28 − 24
= 4
We can also solve this using a different method.
Students having only dog = 8
Students having only cat = 6
Students having both dog and cat = 10
∴ Students having dog or cat or both the pets = 8 + 6 + 10 = 24
And total number of students in the class = 28
∴ Students who do not have a dog or cat as their pet animal at home = 28 − 24 = 4
For your understanding, a Venn Diagram is given below:
First, let's find the elements of each set:
X = {83, 89, 97}
Y = {91, 93, 95, 97, 99}
Now, we can represent the union of sets X ∪ Y using a Venn diagram:
The above answers are given in the textbook. However, they do not correctly depict the union of two sets.
The correct answers are as follows:
X = set of all quadrilaterals
Y = set of all rhombuses
S = set of all squares
T = set of all parallelograms
V = set of all rectangles
All rhombuses, squares, parallelograms, and rectangles are subsets of the set of all quadrilaterals.
Hence,
A square has all the properties of a parallelogram, a rectangle and a rhombus.
Hence,
A rhombus has all the properties of a parallelogram.
Hence,
A rectangle has all the properties of a parallelogram.
Hence,
Moreover, every set is a subset of itself.
Hence,
Also, an empty set is a subset of every set.
Hence,
M ∪ ∅ = M
[Refer to property (6) on textbook page 14]
M ∩ ∅ = ∅
[Refer to property (7) on textbook page 12]
U = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13}
A = {1, 2, 3, 4, 5, 7}
B = {1, 5, 8, 9, 10}
A ∪ B = {1, 2, 3, 5, 7, 8, 9, 10}
A ∩ B = {1, 5}
Using the formula:
n (A ∪ B) = n (A) + n (B) − n (A ∩ B)
∴ n (A ∪ B) = 7 + 13 − 4
∴ n (A ∪ B) = 16 ... (i)
This page was last modified on
18 April 2026 at 19:09