RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions 2024 | Download Free PDF

RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions: In this exercise, the students will learn the concepts related to pie diagrams or pie charts and the construction of pie diagrams. The students that have any difficulties in solving the problems can access RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions. Our top Mathematics faculty team has solved the problems in a simple & easily understandable manner which any student can understand and attain higher marks in the final exams.

The students can access RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions & can download the PDF from the link provided below to learn the concepts in detail. They are suggested to practice the exercise solutions frequently which also increases their confidence level. These solutions offer the students a detailed & in-depth understanding of all the questions & concepts related to pie diagrams or pie charts and the construction of pie diagrams.

Access RS Aggarwal Class 8 Maths Chapter 24 Solutions PDF

Download RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions

RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions

Access The RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions

Question 1.
Solution:
(i) When a coin is tossed, we get outcomes 2 as H or T (Head or Tail)
(ii) When two coins are tossed together, we get possible four outcomes as HH, HT, TH, TT
(iii) A die is thrown, we get possible outcomes as 1,2, 3, 4, 5, 6
(iv) From a well–shuffled deck of 52 cards, 0ne card is at random drawn, we get the possible outcome is 52

Question 2.
Solution:
Possible outcomes = 2
In a single throw of a coin, we get
probability of getting a tail = 12

Question 3.
Solution:
In a single throw of two coins, possible outcomes = 4
(i) Probability of getting both tails = 14
(ii) Probability of getting at least one tail = 34
(iii) Probability of getting at the most one tail = 24 = 12

Question 4.
Solution:
In a bag, there are 4 white and 5 blue balls,
Possible outcomes = 4 + 5 = 9
One ball is drawn at random, then
(i) The probability of a white ball = 49
(ii) the probability of a blue ball = 59

Question 5.
Solution:
In a bag, there are 5 white, 6 red, and 4 green balls
Possible outcome is 5 + 6 + 4 = 15
One ball is drawn at random, then
(i) Probability of a green ball = 415
(ii) Probability of a white ball = 515 = 13
(iii) Probability of a non-red ball = 5+415
= 915
= 35
(5 white and 4 green balls are non-red balls)

Question 6.
Solution:
In a lottery, there are 10 prizes and 20 blanks
Possible outcomes = 10 + 20 = 30
A ticket is chosen at random, then
probability of getting a prize = 1030 = 13

Question 7.
Solution:
In a , box of 100 electric bulbs, 8 are defective
Then non-defective bulbs = 100 – 8 = 92
Now possible outcomes = 100
(i) Probability of a drawn bulb, which is defective = 8100 = 225
(ii) Probability of a drawn bulb that is non-defective = 92100 = 2325

Question 8.
Solution:
A die is thrown, then
Possible outcomes = 6
(i) Now probability of getting 2 = 16
(ii) Probability of a number less than 3 (which are 1 and 2) = 26 = 13
(iii) Probability of a composite number (a composite number is a number which is not a prime number which are 4, 6) = 26 = 13
(iv) Probability of a number not less than 4 (which are 5, 6) = 26 = 13

Question 9.
Solution:
Total number of ladies = 200
Those who like coffee = 82
Those who dislike coffee = 118
Possible number of outcomes = 200
One lady is chosen at random, then
(i) Probability of a lady who dislikes coffee = 118200
= 59100

Question 10.
Solution:
19 ball bearing numbers, 1, 2, 3,…19
possible outcomes = 19
A ball is drawn at random from the box, then
(i) Probability of a ball that bears prime numbers which are 2, 3, 5, 7, 11, 13, 17, and 19 = 8 = 819
(ii) Probability of a ball that bears an even number which are 2, 4, 6, 8, 10, 12, 14, 16, 18 = 9 = 919
(iii) Probability of a number which bears a number divisible by 3 which are 3, 6, 9, 12, 15, 18 = 6 = 619

Important Definition for RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions

• Pie Chart

It is also known as a circle chart that bisects the circular statistical graphic into slices or sectors to solve numerical problems. All sectors denote a proportionate part of the whole.

• Formula

The pie chart is an important type of data representation. It has different sectors & segments in which each segment and sectors of a pie chart make a certain portion of the total or percentage. The total of all the data is equal to 360°.

The pie chart’s total value is always 100%.

• Steps to work out with the percentage for a pie chart

(i) Categorize the data

(ii) Calculate the total

(iii) Divide the categories

(iv) Convert into percentages

(v) Then, calculate the degrees

Hence, the pie chart formula is given as (Given Data/Total value of Data) × 360°

Benefits of RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions

• RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions enable the students to be aware of the pattern of Maths question paper & kinds of questions that come in the final exam. Solving these exercise solutions will raise the confidence level of the students & they can easily get better marks in the Class 8^th Maths final exam.

• The students can access the RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions effortlessly which helps them to get well-prepared for the final exams in a better manner.

• The solutions offer students a detailed & in-depth understanding of the topics related to pie diagrams or pie charts and the construction of pie diagrams. The students can obtain these solutions in PDF format which is reliable & makes them able to get excellent marks

• The solutions are presented in a simple & precise manner. These solutions are very beneficial for the students to clear their queries related to topics included in this chapter & solve all questions in the final exam correctly.

Know more at the official website.

FAQs on RS Aggarwal Class 8 Maths Chapter 24 Ex 24.1 Solutions