This blog is used for Educational Purpose. Here we are giving some questions and answers of Mathematics - Class 10 CBSE board. We are a group of experienced teachers who created the previous years board examination question paper solutions.
Old age homes mean for senior citizens who are unable to stay with
their families or destitute. These old age homes have special medical
facilities for senior citizens such as mobile health care systems, ambulances,
nurses and provision of well balanced meals. Himanshu, Gaurav and Gagan start
preparing greeting cards for each person of an old age home on new year. In
order to complete one card, they take 10, 16 and 20 min respectively.
Based on the above information, solve the following questions:
Q1. Co-prime numbers are those numbers which do
not have any common factor other than 1. Is this statement true?
Q2. Find the sum of
the powers of all different prime factors of the numbers 10, 16 and 20.
Q3. If all of them
started together, then what time will they start preparing a new card together?
Q4. What is the common
time to make one card?
Answers:
1. True
2. By prime
factorisation,
10=2¹× 5¹
16=2x2x2x2
=24
20 = 2x2x5
=22x5¹
Required sum = sum of the power of 2 + sum of the power
of 5
= (1 + 2) + (1+1) + 4
= 7 + 2 = 9
Q3.
The required number of minutes after
which they start preparing a new card together is the LCM of 10, 16 and 20 min.
Now,
10=2 x 5
16=2 x 2 x 2 x 2
20=2 x 2 x 5
LCM (10, 16, 20) = 24 x 51
=
16 x 5 = 80 min
So, they will start preparing a new card together after 80 min i.e.,
1 h 20 min.
First
option is Set, 7, which not an element, third option given is 7 not an element,
but 7 is an element of the given set, and, Fourth option given is, Set, 7 is
not a subset of the given set.But, 7 is
an element of the given set, then Set, 7 is a subset of the given set.
2. The set P = {x | x ∈ Z, –1< x < 1} is a
(1) Singleton set (2) Power set
(3) Null set (4) Subset
Answer:
(1) Singleton set
When
we write the set in Roster form, we get, Set, P = {0}.Hence, there are only one element in Set,
P.Therefore, it is a singleton set.
3. If U ={x | x ∈ N, x < 10} and A = {x | x ∈ N, 2 ≤ x < 6} then (A′)′ is
(1) {1, 6, 7, 8, 9} (2) {1, 2, 3, 4}
(3) {2, 3, 4, 5} (4) { }
Answer:
(3) {2, 3, 4, 5}
When
we write the given sets in Roster form, we get,
Set,
U = {1, 2, 3, 4, 5, 6, 7, 8, 9} and, Set, A = {2, 3, 4, 5}
Now
A’ = {1, 6, 7, 8, 9}
Hence,
(A’)’ = {2, 3, 4, 5}.
4. If B⊆ A then n(A ∩ B) is
(1) n(A–B) (2) n(B)
(3) n(B – A) (4) n(A)
Answer:
(2) n(B)
We
know that, When B⊆
A, then A ∩ B = B.
5. If A = {x, y, z}
then the number of non- empty subsets of A is
(1) 8 (2)
5
(3) 6 (4)
7
Answer:
(4) 7
Total
number of subsets of a set = 2n = 23 = 8.But, there includes null set also.When we exclude null set from the number of
subset, we get, 7.
6. Which of the
following is correct?
(1) ∅⊆ {a, b} (2) ∅∈ {a, b}
(3) {a} ∈ {a, b} (4)
a ⊆ {a, b}
Answer:
Ø ⊆ {a, b}
Null
set is a subset of every set.
7. If A∪B = A∩B, then
(1) A≠B (2) A = B
(3) A ⊂ B (4)
B ⊂ A
Answer:
(2) A = B
8. If B – A is B, then
A∩B is
(1) A (2)
B
(3) U (4)
∅
Answer:
(4) Ø
Given
B – A = B, then we know that, A and B are disjoint sets.
9. From the adjacent
diagram n[P(AΔB)] is
(1) 8 (2) 16
(3) 32 (4) 64
Answer:
(3) 32
A
∆ B = { 60, 85, 75, 90, 70}, ⇒
n(A ∆ B) = 5, ⇒
n(P(A ∆ B)) = 25 = 32
10. If n(A) = 10 and n(B) = 15, then the minimum
and maximum number of elements in A ∩ B is
(1) 10,15 (2) 15,10
(3) 10,0 (4) 0,10
Answer:
(4) (0, 10)
If
the two sets A and B are disjoint, then, A ∩ B has no elements.In the given numbers, there should be maximum
of 10 elements in A ∩ B.
11. Let A = {∅} and B = P(A), then A∩B is
(1) { ∅, {∅} } (2) {∅}
(3) ∅(4)
{0}
Answer:
(2) {Ø}
Given,
Set A = Null Set.Then Power Set P(A) =
{∅,
{∅}},
That is, Set, B = {∅,
{∅}}.Now A ∩ B = {Ø}.
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
(1) 5 (2)
30
(3) 15 (4)
10.
Answer:
(1) 5
Here,
Let A be the boys play carrom and Set, B be the boys play Chess.Total Strength of the class is 50.That is n(A ∪ B) = 50.n(A) = 35 and n(B) = 20.We know that, (A ∪ B) = n(A) +
n(B) – n(A ∩ B) ⇒
50 = 35 + 20 – n(A ∩ B) ⇒
n(A ∩ B) = 5.
13. If U = {x : x Î N and x <10},A =
{1,2, 3,5, 8} and
B = {2,5,6,7,9}, then n[(A È B)’] is
(1) 1 (2)
2
(3) 4 (4)
8
Answer:
(1) 1
Converting
the given sets, in Roster Form, We get, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A =
{1, 2, 3, 5, 8}, and, B = {2, 5, 6, 7, 9}.
A
∪
B = {1, 2, 3, 5, 6, 7, 8, 9}.
Now,
(A ∪
B)’ = {4},
n(A
∪
B)’ = 1
14. For any three sets P,
Q and R, P −(Q ∩ R) is
(1) P −(Q È R) (2) (P Ç Q)−R
(3) (P −Q) È (P −R) (4) (P −Q) Ç (P −R)
Answer:
(3) (P – Q) ∪
(P – R)
We
know that, P −(Q ∩ R) = (P – Q) ∪ (P – R)
15. Which of the
following is true?
(1) A−B = A Ç B (2) A−B = B −A
(3) (A È B)’ = A’ È B’ (4) (A Ç B)’ = A’ È B’
Answer:
(4) (A ∩ B)’ = A’ ∪ B’
16. If n(A ÈBÈC) = 100, n(A) = 4x,
n(B) = 6x, n(C) = 5x, n(A Ç B) = 20, n(B Ç C) = 15, n(A Ç C) = 25 and n(A Ç B Ç C) = 10 , then the value of x is
(1) 10 (2)
15
(3) 25 (4)
30
Answer:
(1) 10
We
know that, n(A ∪
B ∪
C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
Substituting
the values, we get,
100
= 4x + 6x + 5x – 20 – 15 – 25 + 10
100
= 15x -60 + 10
100
= 15x – 50
∴ 15x = 100 + 50
= 150
x
= 10
17. For any three sets A,
B and C, (A−B) Ç(B −C) is equal to
(1) A only (2) B only
(3) C only (4) Æ
Answer:
(4) ϕ
As
per the derivation, (A−B) Ç(B −C) = Null Set.
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 Ç K Ç L is
(1) Set of isosceles
triangles (2) Set of equilateral triangles
(3) Set of isosceles
right triangles (4) Set of right angled
triangles
Answer:
(3) Set of isosceles right triangles
19. The shaded region in
the Venn diagram is
(1) Z −(X ÈY) (2) (X È Y) Ç Z
(3) Z −(X ÇY) (4) Z È (X ÇY)
Answer:
(3) Z – (X ∩ Y)
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?
(1) 5 (2)
8
(3) 10 (4)
15
Answer:
(1) 5
The
Venn Diagram gives you a clear idea of this question.
Radio towers are used for transmitting a range of communication
services including radio and television. The tower will either act as an
antenna itself or support one or more antennas on its structure.
On a similar concept, a radio station tower was
built in two stations A and B (B vertically below A). The tower is supported by
wires AO and BO from a point O on the ground. Distance between the base C of
the tower and the point O is 36 m. From O, the angles of elevation of the tops of station B and station
A are 30º and 45º respectively.
Based on the above, answer the following questions :
(i) Find the height of station B.
(ii) Find the height of station A.
(iii) Find the length of the wire OA.
(iv) Find the length of the wire OB. (CBSE 2022 Set
2)
Solution:
Look at the figure, which we construct now
(i) Find the Height
ofstation B
So, we need to find the length of BC
We can take tan O
Tan 300 =
We know the value of tan 30 is 1/Ö3.Substitute the values
Cross
multiply, we get
BC = = = 12Ö3 m
Hence the length of tower B is 12Ö3 m.
(ii) Find the height of the station A
Here we want to find the length of AC.
CO is known and the angle is 450.So we take tan 45
Tan 450=
tan 450=
We know that tan450 = 1, so substitute the
values
1=
AC =
36 m
Hence the height of tower A is 36 m.
(iii)Find the length of
OA
OA is the hypotenuse of the triangle AOC.
Here we know that the length of CO is 36 m and
length of AC is also 36 m.
Here we take sin 450.
i.e.,
We know that sin450 = 1/Ö2.Substitute the values
Cross multiply, we get
AO = 36Ö2 m
Hence the length of the wire AO is 36Ö2 m.
(iv)Find the length of
the wire OB
Consider the triangle BCO.Here BC and
OC are known and we want to find OB.In
this case we take cos 300 .
Cos 300
=
We know that cost 300 =
.Substitute the values, we get
