Class 8 Maths - Data Handling NCERT Solutions

Chapter 5: Data Handling (NCERT Solutions)

Exercise 5.1

Q1. For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman's bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produced by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a station.
Give reasons for each.

A histogram is used when the data can be divided into continuous class intervals.

(a) No. The areas are distinct categories, not continuous intervals. A bar graph is suitable.

(b) Yes. Heights can be divided into continuous class intervals (e.g., 150-160 cm, 160-170 cm).

(c) No. Companies are distinct categories. A bar graph is suitable.

(d) Yes. Time can be divided into continuous class intervals (e.g., 7 am-8 am, 8 am-9 am).

Q2. The shoppers who come to a departmental store are marked as: man (M), woman (W), boy (B) or girl (G). The following list gives the shoppers who came during the first hour in the morning:
W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W
W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.

Frequency Distribution Table:

Shopper	Tally Marks	Frequency (Number of Shoppers)
W (Woman)	❍❍❍❍❍ ❍❍❍❍❍ ❍❍❍❍❍ ❍❍❍❍❍ ❍❍❍❍❍ ❍❍❍ (28)	28
M (Man)	❍❍❍❍❍ ❍❍❍❍❍ ❍❍❍❍❍ (15)	15
B (Boy)	❍❍❍❍❍ (5)	5
G (Girl)	❍❍❍❍❍ ❍❍❍❍❍ ❍❍ (12)	12
Total:		60

(Draw a bar graph with shopper categories on the X-axis and frequency on the Y-axis. The bar for W will go up to 28, M to 15, B to 5, and G to 12).

Q3. The weekly wages (in ₹) of 30 workers in a factory are.
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Using tally marks make a frequency table with intervals as 800-810, 810-820 and so on.

Frequency Distribution Table:

Class Interval (Wages in ₹)	Tally Marks	Frequency (No. of workers)
800 - 810	\|\|\|	3
810 - 820	\|\|	2
820 - 830	\|	1
830 - 840	\|\|\|\| \|\|\|\|	9
840 - 850	\|\|\|\|	5
850 - 860	\|	1
860 - 870	\|\|\|	3
870 - 880	\|	1
880 - 890	\|	1
890 - 900	\|\|\|\|	4
Total:		30

Q4. Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions.
(i) Which group has the maximum number of workers?
(ii) How many workers earn ₹ 850 and more?
(iii) How many workers earn less than ₹ 850?

(Draw a histogram using the table from Q3. X-axis: Wages class intervals. Y-axis: Frequency of workers).

(i) Group 830-840 has the maximum number of workers (9).

(ii) Workers earning ₹ 850 and more = 1 (850-860) + 3 (860-870) + 1 (870-880) + 1 (880-890) + 4 (890-900) = 10 workers.

(iii) Workers earning less than ₹ 850 = 3 (800-810) + 2 (810-820) + 1 (820-830) + 9 (830-840) + 5 (840-850) = 20 workers.

Q5. The number of hours for which students of a particular class watched television during holidays is shown through the given graph. Answer the following.
(Assuming a graph showing hours 1-2(4), 2-3(8), 3-4(22), 4-5(32), 5-6(8), 6-7(6))
(i) For how many hours did the maximum number of students watch TV?
(ii) How many students watched TV for less than 4 hours?
(iii) How many students spent more than 5 hours in watching TV?

(i) The maximum number of students (32) watched TV for 4 - 5 hours.

(ii) Students watching TV for less than 4 hours = (Students from 1-2) + (2-3) + (3-4) = 4 + 8 + 22 = 34 students.

(iii) Students spending more than 5 hours = (Students from 5-6) + (6-7) = 8 + 6 = 14 students.

Exercise 5.2

Q1. A survey was made to find the type of music that a certain group of young people liked in a city. Adjoining pie chart shows the findings of this survey.
From this pie chart answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
(ii) Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CD's, how many of each type would they make?
(Pie chart logic: Light 40%, Folk 30%, Classical 10%, Semi-classical 20%)

(i) Let the total number of young people surveyed be x.
10% of x = 20
(10/100) × x = 20 ⇒ x = 20 × 10 = 200 people.

(ii) Light music is liked by the maximum number of people (40%).

(iii) Total CDs = 1000.
Classical CDs = 10% of 1000 = (10/100) × 1000 = 100.
Semi-classical CDs = 20% of 1000 = (20/100) × 1000 = 200.
Light music CDs = 40% of 1000 = (40/100) × 1000 = 400.
Folk music CDs = 30% of 1000 = (30/100) × 1000 = 300.

Q2. A group of 360 people were asked to vote for their favourite season from the three seasons rainy, winter and summer.
Summer: 90, Rainy: 120, Winter: 150.
(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this information.

(i) Winter season got the most votes (150).

(ii) Central Angles: Total votes = 360.
Summer = (90 / 360) × 360° = 90°.
Rainy = (120 / 360) × 360° = 120°.
Winter = (150 / 360) × 360° = 150°.

(iii) (Draw a pie chart by drawing a circle and dividing it into sectors using a protractor for angles 90°, 120°, and 150°).

Q3. Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.
Blue: 18, Green: 9, Red: 6, Yellow: 3. Total=36.

Total people = 36. Let's find the central angles:
Blue = (18 / 36) × 360° = 180°.
Green = (9 / 36) × 360° = 90°.
Red = (6 / 36) × 360° = 60°.
Yellow = (3 / 36) × 360° = 30°.

(Draw a pie chart with these central angles).

Q4. The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.
(Central Angles: Maths=90°, Social=65°, Science=80°, Hindi=70°, English=55°)
(i) In which subject did the student score 105 marks?
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.

(i) Total marks = 540. Marks = (Central angle / 360°) × 540 = Central angle × 1.5
Let angle be x. x × 1.5 = 105 ⇒ x = 105 / 1.5 = 70°.
The angle 70° corresponds to Hindi.

(ii) Mathematics angle = 90°, Hindi anlge = 70°.
Marks in Maths = (90° / 360°) × 540 = 135.
Marks in Hindi = 105.
Difference = 135 - 105 = 30 marks.

(iii) Sum of central angles for Social Science and Mathematics = 65° + 90° = 155°.
Sum of central angles for Science and Hindi = 80° + 70° = 150°.
Since 155° > 150°, the sum of marks obtained in Social Science and Mathematics is more.

Q5. The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.
Hindi 40, English 12, Marathi 9, Tamil 7, Bengali 4. Total = 72.

Total students = 72. Central angle = (Language speakers / 72) × 360° = Language speakers × 5°.
Hindi = 40 × 5° = 200°.
English = 12 × 5° = 60°.
Marathi = 9 × 5° = 45°.
Tamil = 7 × 5° = 35°.
Bengali = 4 × 5° = 20°.

(Draw a pie chart with these central angles).

Exercise 5.3

Q1. List the outcomes you can see in these experiments.
(a) Spinning a wheel (with regions A, A, B, C, D).
(b) Tossing two coins together.

(a) The pointer can stop at A, B, C, or D.

(b) Outcomes are HH, HT, TH, TT (where H=Head, T=Tail).

Q2. When a die is thrown, list the outcomes of an event of getting:
(i) (a) a prime number (b) not a prime number.
(ii) (a) a number greater than 5 (b) a number not greater than 5.

Outcomes on a die are 1, 2, 3, 4, 5, 6.

(i) (a) Outcomes of a prime number: 2, 3, 5.
(i) (b) Outcomes of not a prime number: 1, 4, 6.

(ii) (a) Outcomes of a number greater than 5: 6.
(ii) (b) Outcomes of a number not greater than 5: 1, 2, 3, 4, 5.

Q3. Find the.
(a) Probability of the pointer stopping on D in (Question 1-(a)).
(b) Probability of getting an ace from a well shuffled deck of 52 playing cards?
(c) Probability of getting a red apple. (From a given figure with G, R, R, R, G, R, G apples - 4 Red, 3 Green)

(a) Total sectors = 5. Sector D = 1. Probability = 1/5.

(b) Total cards = 52. Aces = 4. Probability = 4 / 52 = 1/13.

(c) Total apples = 7. Red apples = 4. Probability = 4/7.

Q4. Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box and mixed well. One slip is chosen from the box without looking into it. What is the probability of.
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1-digit number?

Total outcomes = 10.

(i) Number 6 appears 1 time. Probability = 1/10.

(ii) Numbers less than 6 are 1, 2, 3, 4, 5 (Total 5). Probability = 5/10 = 1/2.

(iii) Numbers greater than 6 are 7, 8, 9, 10 (Total 4). Probability = 4/10 = 2/5.

(iv) 1-digit numbers are 1 to 9 (Total 9). Probability = 9/10.

Q5. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non blue sector?

Total sectors = 3 + 1 + 1 = 5.
Green sectors = 3.
Probability of getting a green sector = 3/5.

Non-blue sectors = Total - Blue = 5 - 1 = 4 (or 3 Green + 1 Red = 4).
Probability of getting a non-blue sector = 4/5.

Q6. Find the probabilities of the events given in Question 2.

Total outcomes on a die = 6.

(i) (a) Prime numbers (2, 3, 5) = 3. Probability = 3/6 = 1/2.

(i) (b) Not a prime number (1, 4, 6) = 3. Probability = 3/6 = 1/2.

(ii) (a) Number greater than 5 (6) = 1. Probability = 1/6.

(ii) (b) Number not greater than 5 (1, 2, 3, 4, 5) = 5. Probability = 5/6.

Class 8 Maths - Data Handling Practice Questions

Chapter 5: Data Handling (Practice Questions)

RD Sharma / Extra Practice Questions

Q1. In a frequency distribution, the mid-value of a class is 10 and width of the class is 6. What is the lower limit of the class?

Let lower limit be L and upper limit be U.
Mid-value = (L + U) / 2 = 10 ⇒ L + U = 20
Width = U - L = 6
Subtracting width from sum: (L + U) - (U - L) = 20 - 6
2L = 14
L = 14 / 2 = 7.

Q2. The central angle for a sector representing 30% of a circle graph is:

Central angle = 30% of 360°
= (30 / 100) × 360°
= 3 × 36° = 108°.

Q3. A coin is tossed twice. What is the probability of getting at least one head?

Possible outcomes: HH, HT, TH, TT. (Total = 4)
Favorable outcomes (at least one head): HH, HT, TH. (Total = 3)
Probability = Favorable / Total = 3/4.

Q4. What is the probability of getting a prime number when a die is thrown?

Total outcomes on a die = {1, 2, 3, 4, 5, 6} → 6
Prime numbers = {2, 3, 5} → 3
Probability = 3 / 6 = 1/2.

Q5. The monthly income of a family is ₹ 28800. The monthly expenditure of the family on various items is given below:
Food: ₹ 10800, Clothing: ₹ 3600, Education: ₹ 4800, Rent: ₹ 6000, Savings: ₹ 3600.
Find the central angle for Education.

Total income = 28800.
Expenditure on education = 4800.
Central angle = (4800 / 28800) × 360°
= (1/6) × 360° = 60°.

Q6. From a well-shuffled deck of 52 cards, one card is drawn at random. Find the probability of getting a face card (Jack, Queen, King).

Total cards = 52.
Number of face cards in each suit = 3 (J, Q, K).
Total face cards = 4 suits × 3 = 12.
Probability = 12 / 52 = 3/13.

Q7. If classes are 0-10, 10-20, 20-30... Here, the number 20 is included in which class?

By convention, the upper limit is not included in the class interval.
Therefore, 20 is included in the class interval 20-30.

Q8. A bag contains 4 red and 6 black balls. A ball is taken out at random. What is the probability of getting a black ball?

Total balls = 4 + 6 = 10.
Number of black balls = 6.
Probability = 6 / 10 = 3/5.

Q9. The heights of 10 girls are measured in cm and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141. Find the range of the data.

Highest value = 151.
Lowest value = 128.
Range = Highest - Lowest = 151 - 128 = 23 cm.

Q10. Two dice are thrown simultaneously. What is the probability of getting a sum of 8?

Total possible outcomes = 6 × 6 = 36.
Favorable outcomes (sum = 8): (2,6), (3,5), (4,4), (5,3), (6,2). Total = 5.
Probability = 5/36.

Q11. In a pie chart, if a sector representing red cars has an angle of 120°, and the total number of cars is 360, how many red cars are there?

Number of red cars = (Sector Angle / 360°) × Total cars
= (120° / 360°) × 360
= (1/3) × 360 = 120.

Q12. The probability of choosing a vowel from the word 'MATHEMATICS' is:

Total letters = 11.
Vowels in MATHEMATICS = A, E, A, I (Total = 4).
Probability = 4/11.

Q13. Class mark of the class 30-40 is:

Class mark = (Lower limit + Upper limit) / 2
= (30 + 40) / 2 = 70 / 2 = 35.

Q14. The total angle at the centre of a circle is:

360°.

Q15. Out of 100 students in a class, 45 like cricket, 25 like football, and the rest like tennis. If a pie chart is drawn, what will be the central angle for the tennis sector?

Total students = 100.
Students liking tennis = 100 - (45 + 25) = 100 - 70 = 30.
Central angle = (30 / 100) × 360° = 3 × 36° = 108°.

Q16. What is the probability of a sure event?

The probability of a sure event is always 1.

Q17. In a histogram, the area of each rectangle is proportional to:

The frequency of the corresponding class interval.

Q18. If the frequency of class 20-30 is 15, and the total frequency is 75, find its fraction for a pie chart.

Fraction = Frequency / Total Frequency
= 15 / 75 = 1/5.

Q19. A card is drawn from a standard deck. What is the probability of getting a red queen?

Total cards = 52.
Red queens = 2 (Queen of Hearts, Queen of Diamonds).
Probability = 2 / 52 = 1/26.

Q20. The difference between the highest and lowest observation is known as:

Range.

Class 8 Maths - Data Handling Summary

Chapter 5: Data Handling (Concepts & Formulas)

1. Introduction to Data representation

Data mostly available to us in an unorganised form is called raw data.

Pictograph: Pictorial representation of data using symbols.
A bar graph: A display of information using bars of uniform width, their heights being proportional to the respective values.
Double bar graph: A bar graph showing two sets of data simultaneously. It is useful for the comparison of data.

2. Organising Data & Grouped Data

Frequency: The number of times a particular entry occurs.
Grouped frequency distribution: When data is large, we group them into classes (like 0-10, 10-20...).
- Class Interval: Each group is called a class interval.
- Lower/Upper class limit: In 10-20, 10 is the lower class limit and 20 is the upper class limit.
- Width or Size: The difference between the upper class limit and lower class limit.
Histogram: A graphical representation of a grouped frequency distribution with continuous classes. It consists of rectangles with class intervals as bases and corresponding frequencies as heights. There are no gaps between the bars.

3. Circle Graph (Pie Chart)

A circle graph shows the relationship between a whole and its parts. The circle is divided into sectors. The size of each sector is proportional to the activity or information it represents.

Central Angle of a Sector = (Value of the component / Total value) × 360°

The total angle at the centre of a circle is 360°.

4. Chance and Probability

Experiment: An action which results in some well-defined outcomes (e.g., tossing a coin, throwing a die).
Outcomes: The possible results of an experiment.
Equally likely outcomes: Outcomes of an experiment are equally likely if each has the same chance of occurring.
Event: One or more outcomes of an experiment make an event.
Probability of an event = Number of outcomes that make an event / Total number of outcomes of the experiment

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