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RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15a
• RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15b
• RS Aggarwal Solutions Class 10 Chapter 15 Probability MCQ

Question 1.
Solution:
(i) The probability of an impossible event is zero.
(ii) The probability of a sure event is one.
(iii) For any event E, P(E) + P(not E)= one .
(iv) The probability of a possible but not a sure event lies between zero and one.
(v) The sum of probabilities of all the outcomes of an experiment is one.

Question 2.
Solution:
When a coin is tossed the outcomes are: {H,T}
So, the number of outcomes= 2
P(getting a tail) = 1/2.

Question 3.
Solution:

Question 4.
Solution:
In a throw of a dice, all possible outcomes are 1, 2, 3, 4, 5, 6
Total number of possible outcomes = 6

(i) Let E be event of getting even number
Then, the favorable outcomes are 2, 4, 6
Number of favorable outcomes= 3
∴ P(getting a even number)= P(E) = 3/6 = 1/2.

(ii)Let R be the number less than 5
Then, the favorable outcomes are 1, 2, 3, 4
Number of favorable outcomes = 4
∴ P(getting a number less than 5)= P(R) =  4/6 = 2/3.

(iii)Let M be the event of getting a number greater than 2
Then, the favorable outcomes are 3, 4, 5, 6
Number of favorableoutcomes = 4
∴ P(getting a number greater than 2)= P(M) = 4/6 = 2/3.

(iv)Let N be the number lying between 3 and 6
Then the favorable outcomes are 4, 5
Number of favorable outcomes = 2
∴ P(getting a number 3 and 6)= P(N) = 2/6 = 1/3.

(v)Let G be event of getting a number other than 3
Then the favorable outcomes are 1, 2, 4, 5, 6
Number of favorable outcomes = 5
∴ P(getting a number other than 5)=P(G) = 5/6.

(vi)Let T be event of getting a number 5
Then the favorable outcome is 5
Number of favorable outcomes = 1
∴ P(getting a number 5)=P(T) = 1/6.

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Total number of bulbs = 200
Number of defective bulbs = 16

(i)Let E₁ be the event of getting a defective bulb
Total number of defective bulbs = 16
∴P(getting defective bulbs) = P(E₁) = 16/200 = 2/25.

(ii)Let E₂be the event of “getting non – defective bulb”
∴P(getting non defective bulb) = .

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
Total number of tickets sold = 250
Number of prizes = 5
Let E be the event getting a prize
Number of favorable outcomes = 5
∴P(getting a prize) = P(E) = 5/250 = 1/50

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:
The number of outcomes is:
{GGG, BGG, GBG, GGB, 55G, 5GB, GBB, BBB}
Total number of outcomes = 8
The possible outcomes of having at least one boy are:
{BGG, GBG, GGB, 56G, 6GB, GBB, BBB}
So, there are 7 possible outcomes,
P(getting at least one boy) = 7/8.

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:
There are 26 red cards containing a 2 queens and 2 more black queens are there in a pack of cards
∴ P(getting a red card or a queen) = 28/52 = 7/13
∴ P(getting neither a red card nor a queen) = \(\left[ 1-\frac { 7 }{ 13 } =\frac { 6 }{ 13 } \right] \)

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:
Spinning arrow may come to rest at one of the 12 numbers
∴ total number of outcomes = 12

(i) Probability that it will point at 6 = 1/12

(ii) Even numbers are 2, 4, 6, 8, 10 and 12. There are 6 numbers.
∴ Probability that it points at even numbers = 6/12 = 1/2

(iii) The prime numbers are 2,3 5, 7 and 11. There are 5 prime numbers.
∴ Probability that it points at prime number = 5/12

(iv) There are 2 numbers divisible by 5. These are 5 and 10.
∴ Probability that a number is a multiple of 5 = 2/12 = 1/6

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:

Question 29.
Solution:

Question 30.
Solution:
(i) P(that she takes out an orange-flavoured candy) = O
(Since there are no orange-flavoured candies in the bag)

(ii) P(that she takes out a lemon-fl avcured candy) = 1
(Since there are only lemon-flavcured candies in the bag)

Question 31.
Solution:

Question 32.
Solution:

Question 33.
Solution:

Question 34.
Solution:

Question 35.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 15 Probability Ex 15a are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18a
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18b
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment MCQ
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment MCQ.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18a
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18b
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment MCQ
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment MCQ are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18b

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18b.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18a
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18b
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment MCQ
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Test Yourself

Question 1.
Solution:
Circumference of circle = 2πr = 39.6 cm

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Let the radii of circles be x cm and (7 – x) cm

Circumference of the circles are 26 cm and 18 cm.

Question 8.
Solution:

Question 9.
Solution:
Inner radius of the circular park = 17 m
Width of the path = 8 m
Outer radius of the circular park = (17 + 8)m = 25 m

Question 10.
Solution:
Let r m and R m be the radii of inner circle and outer boundaries respectively.
Then, 2πr = 352 and 2πR = 396

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:
∆OAB is equilateral.
So, ∠AOB = 60

Question 16.
Solution:

Area of minor segment = (area of sector OACBO) – (area of ∆OAB)

Question 17.
Solution:
Area of sector OACBO

Question 18.
Solution:
Let AB be the chord of circle of centre O and radius = 30 cm such that ∠AOB = 60⁰

Question 19.
Solution:

Question 20.
Solution:
In 2 days, the short hand will complete 4 rounds
Distance travelled by its tip in 2 days
=4(circumference of the circle with r = 4 cm)
= (4 × 2π × 4) cm = 32π cm
In 2 days, the long hand will complete 48 rounds length moved by its tip
= 48(circumference of the circle with r = 6cm)
= (48 × 2π × 6) cm = 576π cm
Sum of the lengths moved
= (32π + 576π) = 608π cm
= (608 × 3.14) cm = 1909.12 cm

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:
Each angle of equilateral triangle is 60⁰


Area that the horse cannot graze is 36.68 m²

Question 25.
Solution:

Question 26.
Solution:

OP = OR = OQ = r
Let OQ and PR intersect at S
We know the diagonals of a rhombus bisect each other at right angle.
Therefore we have

Question 27.
Solution:
Diameter of the inscribed circle = Side of the square = 10 cm
Radius of the inscribed circle = 5 cm

Diameter of the circumscribed circle
= Diagonal of the square

Question 28.
Solution:
Let the radius of circle be r cm

Then diagonal of square = diameter of circle = 2r cm

Question 29.
Solution:
Let the radius of circle be r cm

Question 30.
Solution:

Question 31.
Solution:

Question 32.
Solution:

Question 33.
Solution:

Question 34.
Solution:
Distance covered by a wheel in 1minute

Question 35.
Solution:

Question 36.
Solution:
Each side of the square is 14 cm

Question 37.
Solution:

Let A, B, C, D be the centres of these circles
Join AB, BC, CD and DA
Side of square = 10 cm
Area of square ABCD

Question 38.
Solution:

Required area = [area of square – areas of quadrants of circles]
Let the side = 2a unit and radius = a units
Area of square = (side × side) = (2a × 2a) sq. units
= 4a²sq. units

Question 39.
Solution:

Question 40.
Solution:

Let A, B, C be the centers of these circles. Join AB, BC, CA
Required area= (area of ∆ABC with each side 2) -3[area of sector with r = a cm, θ= 60⁰]

Question 41.
Solution:

Question 42.
Solution:

Question 43.
Solution:

Question 44.
Solution:

Question 45.
Solution:

Question 46.
Solution:

Question 47.
Solution:

Question 48.
Solution

Question 49.
Solution:

Question 50.
Solution:

Question 51.
Solution:

Question 52.
Solution:

Question 53.
Solution:

Question 54.
Solution:

Question 55.
Solution:

Question 56.
Solution:

Question 57.
Solution:

Question 58.
Solution:

Question 59.
Solution:
PS = 12 cm
PQ = QR = RS = 4 cm, QS = 8 cm
Perimeter = arc PTS + arc PBQ + arc QES

Question 60.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18b are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18a
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18b
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment MCQ
• RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Since square circumscribes a circle of radius a cm, we have
Side of the square = 2 ⨯ radius of circle = 2a cm
Then, Perimeter of the square = (4 ⨯ 2a) = 8a cm

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:
Length of the pendulum = radius of sector = r cm

Question 20.
Solution:
Angle described by the minute hand in 60 minutes = 360°
Angle described by minute hand in 20 minutes

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:
Let sector of circle is OAB
Perimeter of a sector of circle = 31 cm
OA + OB + length of arc AB = 31 cm

Question 24.
Solution:
Length of arc of circle = 44 cm
Radius of circle = 17.5 cm

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:

Question 32.
Solution:

Question 33.
Solution:

Question 34.
Solution:

Question 35.
Solution:

Question 36.
Solution:

Question 37.
Solution:

Question 38.
Solution:

Question 39.
Solution:

Question 40.
Solution:

Question 41.
Solution:

Question 42.
Solution:

Question 43.
Solution:

Question 44.
Solution:

Question 45.
Solution:

Question 46.
Solution:

Question 47.
Solution:

Question 48.
Solution:
Area of flower bed = (area of quadrant OPQ) – (area of the quadrant ORS)

Question 49.
Solution:

Question 50.
Solution:

Question 51.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Ex 18a are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.