GSEB Class 8 Maths Solutions Chapter 5 Data Handling InText Questions

Try These (Page 71)

Question 1. If we change the position of any of the bars of a graph, would it change the information being conveyed? Why?
Answer: If the height of a bar remains unchanged, then changing its position does not change the information being conveyed. The data represented by the bar stays the same, regardless of where it is placed horizontally on the graph. The position merely organizes the visual presentation, but the value itself is determined by the height.
In simple words: No, if a bar's height stays the same, moving it won't change what the graph tells you. The height shows the data, not its place.

Exam Tip: Remember that the height or length of a bar represents the data value, while its position on the axis often serves for categorization or comparison.

 

Try These (Page 71)

Question 1. Draw an appropriate graph to represent the given information?
Answer: To represent the given data by a bar-graph, draw two axes perpendicular to each other. Mark 'Months' on the OX (horizontal) axis and 'Number of watches sold' on the OY (vertical) axis. Then, erect rectangles (bars) whose heights are proportional to the number of watches sold for each month. Use a suitable scale for the OY axis, for example, 1 cm = 500 watches.

Here, the chosen scale is 1 cm = 500 watches.
\( \implies \) 1000 watches = 2 cm
\( \implies \) 1500 watches = 3 cm
\( \implies \) 2000 watches = 4 cm
\( \implies \) 2500 watches = 5 cmIn simple words: To show this data, make a bar graph. Put months on the bottom and number of watches up the side. Each bar's height shows how many watches were sold that month, using a scale where 1 cm equals 500 watches.

Exam Tip: Always label both axes clearly, choose an appropriate scale, and draw bars of uniform width with equal spacing between them. The heights must accurately reflect the data values.

 

Question 2. Represent the following data showing children who prefer activities in different schools.
Answer: A double-bar graph is suitable for comparing two activities (walking and cycling) across different schools. We place the schools along the x-axis and the number of children along the y-axis, using a scale of 1 cm = 5 children. This setup helps in easily comparing the preferences within each school and across schools.

In simple words: To show this, create a double-bar graph. Schools go on the bottom (x-axis), and the number of children goes up the side (y-axis). Each school will have two bars next to each other: one for walking and one for cycling. Use a scale where 1 cm represents 5 children. This graph makes it simple to compare how many children prefer each activity in different schools.

Exam Tip: For comparing two sets of data across different categories, a double-bar graph is ideal. Ensure you use different colors or patterns for each set of data and include a clear legend.

 

Question 3. Represent the percentage wins in ODI by 8 top cricket teams.
Answer: To compare the percentage win in ODI achieved by various teams, we represent the data using a double-bar graph. The teams are placed along the x-axis, and their 'percentage win' is placed along the y-axis, using a scale of 1 cm = 5%. This method allows for a clear comparison between the Champions Trophy and World Cup wins and the Last 10 ODI wins for each team.

In simple words: To compare how well cricket teams did, we draw a double-bar graph. Teams go on the bottom (x-axis), and their win percentages go up the side (y-axis). We use a scale of 1 cm for every 5% win. This graph helps to easily see and compare the win rates from different tournaments for each team.

Exam Tip: When displaying comparative data for multiple categories, a double-bar graph is highly effective. Always use distinct visual representations (e.g., different shades or patterns) for each data set to avoid confusion.

 

Try These (Page 72)

Question 1. A group of students were asked to say which animal they would like most to have as a pet. The results are given below:
dog, cat, cat, fish, cat, rabbit, dog, cat, rabbit, dog, cat, dog, dog, dog, cat, cow, fish, rabbit, dog, cat, dog, cat, cat, dog, rabbit, cat, fish, dog.
Make a frequency distribution table for the same.
Answer: Using tally-marks, we prepare the frequency distribution table as shown below:

In simple words: We count how many times each animal appears in the list. Then, we use tally marks to represent these counts and write the total number of students who chose each animal.

Exam Tip: When making a frequency table, count each item carefully. Tally marks help organize counts, especially when dealing with long lists of data.

 

Try These (Page 73)

Question 1. Study the following frequency distribution table and answer the questions given below?
Frequency distribution of daily income of 550 workers of a factory

1. What is the size of the class intervals?
2. Which class has the highest frequency?
3. Which class has the lowest frequency?
4. What is the upper limit of the class interval 250-275?
5. Which two classes have the same frequency?
Answer:
1. The class size is calculated as the [Upper class limit] – [Lower class limit]. So, for the first interval, it is \( 125 - 100 = 25 \). All class intervals have a size of 25.
2. The class 200-225 has the highest frequency, which is 140 workers.
3. The class 300-325 has the lowest frequency, with only 20 workers.
4. The upper limit of the class interval 250-275 is 275.
5. The classes (150-175) and (225-250) both have the same frequency of 55 workers.
In simple words: The class size is found by subtracting the lower number from the upper number in each group, which is 25. The group with the most workers is 200-225. The group with the fewest workers is 300-325. For the 250-275 group, the highest income is 275. Two groups, 150-175 and 225-250, have the same number of workers.

Exam Tip: Understand how to calculate class size, identify highest/lowest frequencies, and find upper/lower limits from a frequency distribution table. These are fundamental skills for data analysis.

 

Question 2. Construct a frequency distribution table for the data on weights (in kg) of 20 students of a class using intervals 30-35, 35-40, and so on?
40, 38, 33, 48, 60, 53, 31,
46, 34, 36, 49, 41, 55, 49,
65, 42, 44, 47, 38, 39
Answer: To construct the frequency distribution table, we first find the lowest and highest observations in the given data. The lowest observation is 31, and the highest observation is 65. We then use the provided class intervals: 30-35, 35-40, 40-45, and so on, up to 65-70.

In simple words: First, find the lowest and highest weights. Then, create groups for the weights (like 30-35 kg). Count how many students fall into each group using tally marks. Finally, write down the total number of students for each group in the frequency column.

Exam Tip: When making a frequency distribution table, ensure that class intervals do not overlap and cover the entire range of the data. Tally marks should be grouped in fives for easy counting.

 

Try These (Page 75)

Question 1. Observe the histogram figure and answer the questions given below:

1. What information is being given by the histogram?
2. Which group contains maximum girls?
3. How many girls have a height of 145 cm and more?
4. If we divide the girls into the following three categories, how many would there be in each?
150 cm and more – Group A
140 cm to less than 150 cm – Group B
Less than 140 cm – Group C
Answer:
1. The given histogram shows the heights (in cm) of girls in Class VIII.
2. The group 140-145 cm contains the maximum number of girls, with 7 girls.
3. To find girls with a height of 145 cm and more, we add the frequencies for the intervals starting from 145 cm:
\( \implies \) Girls from 145-150 cm: 4
\( \implies \) Girls from 150-155 cm: 2
\( \implies \) Girls from 155-160 cm: 1
\( \implies \) Total girls with height 145 cm and more \( = 4 + 2 + 1 = 7 \) girls.
4. Dividing the girls into categories:
Group A: 150 cm and more \( = 2 + 1 = 3 \) girls (from 150-155 cm and 155-160 cm)
Group B: 140 cm to less than 150 cm \( = 7 + 4 = 11 \) girls (from 140-145 cm and 145-150 cm)
Group C: Less than 140 cm \( = 1 + 2 + 3 = 6 \) girls (from 125-130 cm, 130-135 cm, and 135-140 cm)
In simple words: The chart shows how tall Class 8 girls are. Most girls are in the 140-145 cm height group. If you count girls who are 145 cm or taller, there are 7 of them. For the categories: Group A (150 cm and up) has 3 girls, Group B (140 cm to less than 150 cm) has 11 girls, and Group C (less than 140 cm) has 6 girls.

Exam Tip: Histograms effectively show data distribution. Always read the axis labels and intervals carefully to interpret the information correctly, especially when calculating totals for specific ranges.

The broken line is used along the horizontal line to indicate that we are not showing the numbers between 0 and 125.

 

Try These (Page 78)

Question 1. Each of the following pie charts gives you a different piece of information about your class. Find the fraction of the circle representing each of these information?
(i)
(ii)
(iii)
Answer: We calculate the fraction for each segment by dividing its percentage by 100.
1. For the 'Girls or Boys' pie chart:
\( \implies \) Fraction for 'girls' 50% \( = \frac{50}{100} = \frac{1}{2} \)
\( \implies \) Fraction for 'boys' 50% \( = \frac{50}{100} = \frac{1}{2} \)
2. For the 'Transport to school' pie chart:
\( \implies \) Fraction for 'walk' 40% \( = \frac{40}{100} = \frac{2}{5} \)
\( \implies \) Fraction for 'bus or car' 40% \( = \frac{40}{100} = \frac{2}{5} \)
\( \implies \) Fraction for 'cycle' 20% \( = \frac{20}{100} = \frac{1}{5} \)
3. For the 'Love/Hate Mathematics' pie chart:
\( \implies \) Fraction for those who love mathematics \( = (100 - 15)\% = 85\% = \frac{85}{100} = \frac{17}{20} \)
\( \implies \) Fraction for those who hate mathematics \( = 15\% = \frac{15}{100} = \frac{3}{20} \)
In simple words: To find the fraction, just divide the percentage for each part by 100. For example, 50% becomes 50/100, which simplifies to 1/2. Do this for all the percentages in each pie chart.

Exam Tip: To convert a percentage to a fraction, always divide the percentage by 100 and then simplify the resulting fraction to its lowest terms. This will give you the correct fractional representation.

 

Question 2. Answer the following questions based on the pie chart given?

1. Which type of programmes are viewed the most?
2. Which two types of programmes have number of viewers equal to those watching sports channels?
Answer: From the given pie chart, we can see the percentages for each type of programme.

1. The entertainment programmes are viewed the most, as they have the largest percentage (50%).
2. The news and informative programmes together have a percentage of \( 15\% + 10\% = 25\% \), which is equal to the percentage of viewers watching sports channels (25%). Therefore, News and Informative programs combined equal Sports programs.
In simple words: Entertainment shows are watched the most because they have the biggest slice of the pie chart. If you add up the viewers for news and informative shows, you get the same number of viewers as for sports shows.

Exam Tip: When analyzing a pie chart, the largest sector represents the highest frequency or percentage. To compare combined categories, simply add their individual percentages.

 

Try These (Page 81)

Question 1. Draw a pie chart of the data given below?
The time spent by a child during a day.
Sleep – 8 hours
School - 6 hours
Home work – 4 hours
Play – 4 hours
Others - 2 hours
Answer: First, we need to find the central angle corresponding to each activity for a pie chart. A full circle is \( 360^\circ \), and a day has 24 hours. The central angle for each activity is calculated by (Hours for activity / Total hours) \( \times 360^\circ \).

Now, the required pie chart is given below:In simple words: First, calculate the angle for each activity by dividing its hours by 24 and multiplying by 360 degrees. Then, use these angles to draw the sections on a circle, making sure to label each part.

Exam Tip: For pie charts, always calculate the central angle for each component. The sum of all central angles should be 360 degrees. Use a compass and protractor for accurate drawing.

 

Try These (Page 81)

Question 1. Which form of graph would be appropriate to display the following data?
Production of food grains of a state

Answer: A bar graph would be an appropriate representation of the above data. It clearly shows the production for each year, allowing for easy comparison of production levels over time.
In simple words: A bar graph is best for this data. It makes it easy to compare how much food was grown each year.

Exam Tip: Bar graphs are ideal for showing changes in a single variable over distinct periods or comparing different categories.

 

Question 2. Choice of food for a group of people

Answer: A circle graph (or a pie chart) would be the most appropriate way to display this data. It helps show the proportion of each food choice relative to the total number of people surveyed.
In simple words: A pie chart is best for this. It clearly shows what part of the whole group likes each type of food.

Exam Tip: Pie charts are ideal for representing parts of a whole, showing how different categories contribute to a total sum.

 

Question 3. Which form of graph would be appropriate to display the following data?
Daily income (in rupees)

Answer: A histogram would be an appropriate representation of the above data. Since the daily income data is in class intervals (continuous data), a histogram is best for showing the distribution across these ranges.
In simple words: A histogram is the best graph for this data. It uses bars to show how incomes are spread out across different ranges.

Exam Tip: Histograms are specifically designed for continuous data grouped into class intervals, unlike bar graphs which are for discrete categories.

 

Try These (Page 83)

Question 1. If you try to start a scooter, what are the possible outcomes?
Answer: When you try to start a scooter, there are two primary possible outcomes: it may start, or it may not start. These are the only two events that can occur.
In simple words: The scooter either starts or it doesn't.

Exam Tip: In probability, an outcome is a possible result of an experiment. For simple events, list all distinct possibilities.

 

Question 2. When a die is thrown, what are the possible outcomes?
Answer: When a standard six-sided die is thrown, the possible outcomes are the numbers on its faces. These outcomes are 1, 2, 3, 4, 5, or 6.
In simple words: When you roll a die, you can get a 1, 2, 3, 4, 5, or 6.

Exam Tip: For a standard die, the sample space consists of six equally likely outcomes, representing each face of the die.

 

Question 3. When you spin the wheel shown, what are the possible outcomes? List them? (Outcome here means the sector at which the pointer stops).

Answer: When spinning the wheel shown, the possible outcomes are the sectors where the pointer can stop. Based on the labels, the outcomes are A, B, or C.
In simple words: When you spin the wheel, the pointer can land on A, B, or C.

Exam Tip: For a spinner, the outcomes are the distinct labeled sections that the spinner can land on. Make sure to list all unique sections.

 

Question 4. You have a bag with five identical balls of would get?

Answer: The possible outcomes when picking a ball from the bag are the labels on the five identical balls. So, the outcomes are W, R, B, G, or Y.
In simple words: You can pick a ball that says W, R, B, G, or Y.

Exam Tip: When all items are identical except for their labels, each distinct label represents a possible outcome.

 

Try These (Page 84)

Question 1. In throwing a die:
(a) Does the first player have a greater chance of getting a six?
(b) Would the player who played after him have a lesser chance of getting a six?
(c) Suppose the second player got a six. Does it mean that the third player would not have a chance of getting a six?
Answer: For each throw of a fair die, every outcome (1, 2, 3, 4, 5, or 6) has an equal chance of occurring, which is \( \frac{1}{6} \). Each throw is an independent event, meaning the result of one throw does not influence the result of any other throw.
(a) No, the first player does not have a greater chance of getting a six. The probability remains \( \frac{1}{6} \).
(b) No, the player who plays after the first player would not have a lesser chance. Their chance of rolling a six is still \( \frac{1}{6} \).
(c) No, even if the second player got a six, the third player still has the same chance ( \( \frac{1}{6} \) ) of getting a six. The events are independent.
In simple words: Every time you roll a fair die, the chance of getting a six is always the same, no matter who rolls or what happened before. Each roll is a fresh start.

Exam Tip: Remember that for independent events, the probability of an outcome does not change based on previous outcomes. Each trial begins with the same set of possibilities.

 

Try These (Page 86)

Question 1. Suppose you spin the wheel.

(i) List the number of outcomes of getting a green sector and not getting a green sector on this wheel.
(ii) Find the probability of getting a green sector.
(iii) Find the probability of not getting a green sector.
Answer: Let's count the sectors on the wheel.
(i) Number of outcomes of getting a green sector (G) = 5 (G, G, G, G, G)
Number of outcomes of not getting a green sector (R) = 3 (R, R, R)
(ii) The total number of equally likely outcomes on the wheel is 8. The number of outcomes of getting a green sector is 5.
\( \implies \) Probability of getting a green sector \( = \frac{\text{Number of green sectors}}{\text{Total number of sectors}} = \frac{5}{8} \)
(iii) The number of outcomes of not getting a green sector (which means getting a red sector) is 3.
\( \implies \) Probability of not getting a green sector \( = \frac{\text{Number of red sectors}}{\text{Total number of sectors}} = \frac{3}{8} \)
In simple words: There are 5 green parts and 3 red parts on the wheel. So, there are 5 ways to land on green and 3 ways to not land on green. The chance of landing on green is 5 out of 8, and the chance of not landing on green is 3 out of 8.

Exam Tip: Probability is calculated as (Number of favorable outcomes) / (Total number of possible outcomes). Remember that the sum of the probability of an event happening and the probability of it not happening is always 1.