CBSE Class 8 Mathematics Data Handling Assignment Set 03

Questions carrying 1 Mark each :-

Question 1. List the number of outcomes when a coin is tossed.
Answer: When a coin is tossed, there are 2 possible outcomes, namely Head (H) and Tail (T).
In simple words: A coin toss has two possible results: Head or Tail.

Exam Tip: Always list both outcomes clearly when asked for sample space or outcomes of a simple event.

Question 2. In a bar graph, bars are of equal________ but varying,__________.
Answer: width, height
In simple words: In bar graphs, the width of all bars remains constant while their heights represent different numerical values.

Exam Tip: Remember that the area or height of bars in a bar graph is proportional to the values they represent.

Question 3. A pic chart is also called a ____________.
Answer: circle graph (or pie graph)
In simple words: A pie chart displays data using sectors of a circle.

Exam Tip: Pie charts and circle graphs are interchangeable terms used in statistics.

Question 4. In a pie chart, the total sum of all central angles must be ____________.
Answer: \(360^{\circ}\)
In simple words: The complete angle around the center of a circle adds up to \(360^{\circ}\).

Exam Tip: Always check that the sum of calculated central angles equals \(360^{\circ}\) to verify your calculations.

Questions Carrying 2 Mark each:-

Question 5. The marks obtained in Mathematics by 40 students of a class in an examination are 3, 20, 13, 1, 21, 13, 3, 23, 16, 13, 5, 24, 15, 7, 10, 18, 18, 7, 17, 21, 15, 5, 23, 2, 12, 20, 2, 10, 16, 23, 18, 12, 6, 9, 7, 3, 5, 16, 8 and 8. Present the data in the form of a grouped frequency distribution, using class intervals of equal size, one of the class intervals being 5-10.
Answer:
Given class interval = \(5 - 10\), so the class width is \(5\).
The minimum observation is 1 and the maximum is 24. We can use class intervals of size 5: \(0 - 5, 5 - 10, 10 - 15, 15 - 20, 20 - 25\).

In simple words: Count how many numbers fall into each range of 5 units and organize them into a frequency table.

Exam Tip: Always verify that the sum of all frequencies equals the total number of observations given in the problem.

Question 6. Read the bar graph and answer the following questions:-
(i) In which subject maximum number of students got distinction?
(ii) In which subject, the number of distinctions is minimum?
(iii) How many distinctions were obtained by students in a school?
(iv) Write subject in order of the performance.
Answer:
(i) Looking at the highest bar in the graph, the maximum number of students got distinction in Mathematics.
(ii) Looking at the shortest bar in the graph, the number of distinctions is minimum in Hindi.
(iii) Total distinctions = Sum of heights of all bars = \(2 + 4 + 9 + 3 + 6 = 24\).
(iv) Order of subjects based on performance (from highest to lowest distinctions): Mathematics, Science, English, Social Science, Hindi.
In simple words: Interpret the heights of the bars to find maximum, minimum, total sum, and ranking.

Exam Tip: Read both axes carefully before answering questions based on graphical data.

Question 7. Two dice are thrown, find and number of outcomes.
Answer:
When two dice are thrown, each die has 6 possible outcomes (\(1, 2, 3, 4, 5, 6\)).
Total number of outcomes = \(6 \times 6 = 36\).
In simple words: Multiply the number of outcomes of the first die by the second die to get 36 total pairs.

Exam Tip: Use the fundamental counting principle for simultaneous independent events.

Question 8. There are 2 Red, 3 Blue and 5 Black balls in a bag. A ball is drawn from the bag without looking in to the bag. What is the probability of getting a non-red ball?
Answer:
Total number of balls = \(2 \text{ (Red)} + 3 \text{ (Blue)} + 5 \text{ (Black)} = 10\)
Number of red balls = 2
Number of non-red balls = Total balls - Red balls = \(10 - 2 = 8\) (or \(3 + 5 = 8\))
Probability of getting a non-red ball = \(\frac{\text{Number of non-red balls}}{\text{Total number of balls}} = \frac{8}{10} = \frac{4}{5}\)
In simple words: Find how many balls are not red, then divide by the total number of balls in the bag.

Question 9. The data on mode of transport used by students to come to school are as given below:-
Mode of Transport: Bus, Cycle, Train, Car, Scooter
Number of Students: 120, 180, 240, 80, 100
Find the central angle of each sector.
Answer:
Total number of students = \(120 + 180 + 240 + 80 + 100 = 720\)
Central angle for a sector = \(\left( \frac{\text{Value of sector}}{\text{Total value}} \right) \times 360^{\circ}\)
- Bus: \(\left( \frac{120}{720} \right) \times 360^{\circ} = 60^{\circ}\)
- Cycle: \(\left( \frac{180}{720} \right) \times 360^{\circ} = 90^{\circ}\)
- Train: \(\left( \frac{240}{720} \right) \times 360^{\circ} = 120^{\circ}\)
- Car: \(\left( \frac{80}{720} \right) \times 360^{\circ} = 40^{\circ}\)
- Scooter: \(\left( \frac{100}{720} \right) \times 360^{\circ} = 50^{\circ}\)
In simple words: Multiply each fraction of students by \(360^{\circ}\) to find its corresponding angle in a pie chart.

Exam Tip: Always verify that the sum of all calculated central angles equals \(360^{\circ}\) (\(60^{\circ} + 90^{\circ} + 120^{\circ} + 40^{\circ} + 50^{\circ} = 360^{\circ}\)).

Question 10. A coin and a die are thrown simultaneously. Find the number of possible outcomes also, name the outcomes.
Answer:
Number of outcomes for a coin = 2 (Head [H], Tail [T])
Number of outcomes for a die = 6 (\(1, 2, 3, 4, 5, 6\))
Total number of possible outcomes = \(2 \times 6 = 12\)
Names of outcomes:
\((H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)\)
\((T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\)
In simple words: Combine each result of the coin with every number on the die to list 12 pairs.

Exam Tip: Write outcomes in ordered pairs \((coin, die)\) to avoid missing any combination.

Questions carrying 6 marks each:-

Question 11. Marks secured by a student in various subjects in an examination are:
Subject: Hindi, English, Maths, Soc. Science, Gen. Science
Marks: 40, 52, 80, 70, 75
Answer:
To represent the data or analyze the distribution, let us calculate the total marks and percentage if required, or construct a bar graph / table.
Total marks = \(40 + 52 + 80 + 70 + 75 = 317\)
Maximum possible marks (assuming 100 per subject) = \(5 \times 100 = 500\)
Percentage = \(\left( \frac{317}{500} \right) \times 100\% = 63.4\%\)
In simple words: Summarize the student's performance across all subjects by finding totals and percentages.

Exam Tip: Always state assumptions clearly if maximum marks per subject are not specified.

Question 12. The following data shows the expenditure of an individual over various items:
Items: Education, Food, Rent, Clothing, Others
Expenditure (in Rs.): 1600, 3200, 4000, 2400, 3200
Represent the above data by a pie chart.
Answer:
Total expenditure = \(1600 + 3200 + 4000 + 2400 + 3200 = 14400\) Rs.
Central angle = \(\left( \frac{\text{Expenditure on item}}{\text{Total expenditure}} \right) \times 360^{\circ}\)
- Education: \(\left( \frac{1600}{14400} \right) \times 360^{\circ} = 40^{\circ}\)
- Food: \(\left( \frac{3200}{14400} \right) \times 360^{\circ} = 80^{\circ}\)
- Rent: \(\left( \frac{4000}{14400} \right) \times 360^{\circ} = 100^{\circ}\)
- Clothing: \(\left( \frac{2400}{14400} \right) \times 360^{\circ} = 60^{\circ}\)
- Others: \(\left( \frac{3200}{14400} \right) \times 360^{\circ} = 80^{\circ}\)
Check total angle: \(40^{\circ} + 80^{\circ} + 100^{\circ} + 60^{\circ} + 80^{\circ} = 360^{\circ}\).
In simple words: Calculate the central angle for each expenditure category to draw the sectors of a pie chart.

Exam Tip: Use a protractor to accurately measure and draw each sector angle in the circle.

Question 13. Draw a histogram for the following data:
Class Interval: 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Frequency: 5, 10, 7, 9, 12, 13
Answer:
Steps to draw a histogram:
1. Take a suitable scale along the X-axis representing the class intervals (e.g., \(1 \text{ cm} = 10 \text{ units}\)).
2. Take a suitable scale along the Y-axis representing the frequencies (e.g., \(1 \text{ cm} = 2 \text{ units}\)).
3. Since the class intervals are continuous and start from 10, mark a kink (jagged line) on the X-axis between 0 and 10 if necessary.
4. Construct adjacent rectangular bars with bases as the class intervals (widths 10) and heights corresponding to the respective frequencies (5, 10, 7, 9, 12, 13).
In simple words: Draw touching rectangular bars for continuous class intervals where heights represent frequencies.

Exam Tip: Unlike bar graphs, histogram bars must touch each other with no gaps since the data is continuous.

Multiple choice Questions carrying 1 mark each:

Question 14. The sum of all central angles in a pie chart is:
(a) \(90^{\circ}\)
(b) \(360^{\circ}\)
(c) \(180^{\circ}\)
(d) None of the options
Answer: (b) \(360^{\circ}\)
The complete angle around a point or center of a circle is always \(360^{\circ}\).
In simple words: A full circle measures \(360^{\circ}\).

Exam Tip: Remember \(360^{\circ}\) for pie charts and \(180^{\circ}\) for straight lines / semicircles.

Question 15. When a die is thrown, total number of possible outcomes are:
(a) 2
(b) 36
(c) 6
(d) 9
Answer: (c) 6
A standard die has 6 faces numbered 1 through 6, giving 6 possible outcomes.
In simple words: A die has numbers from 1 to 6, so there are 6 outcomes.

Exam Tip: Distinguish carefully between throwing one die (6 outcomes) and two dice (36 outcomes).