RS Aggarwal Class 8 Mathematics Solutions Chapter 22 Constructing and Interpreting Bar Graphs

Question 1. Explain the steps involved in drawing a bar graph to represent data.
Answer: To draw a bar graph, follow these five key steps. First, on graph paper, sketch a horizontal line OX and a vertical line OY to form the x-axis and y-axis. Second, along the OX line, mark the names or labels of the categories at evenly spaced intervals. Third, pick an appropriate scale - for instance, one small division could equal two marks, two students, 20 students, or some other unit depending on your data. Fourth, calculate the height of each bar by dividing the actual value by the scale factor. Fifth, on the x-axis, draw bars of equal width at the marked positions, with heights matching your calculations from step four. The result is a clear visual representation where readers can quickly compare values across different categories.
In simple words: Draw two lines that cross to make axes. Write category labels along the bottom line. Pick a scale to shrink your numbers. Calculate bar heights using the scale. Draw bars of the same width with these heights.

Exam Tip: Always use a ruler and uniform spacing - neat, precise axes and equal bar widths are what examiners look for. Forgetting to label the axes or mark the scale are common errors to avoid.

Question 2. Draw a bar graph showing the number of students who play different sports.
Answer: Using the data provided - Cricket: 75/2 = 37.5 students, Football: 35/2 = 17.5 students, Tennis: 50/2 = 25 students, Badminton: 25/2 = 12.5 students, and Swimming: 65/2 = 32.5 students - set up a graph with sports names along the horizontal axis and "Number of students" on the vertical axis using a scale of 1 small division = 2 students. Position each sport label at regular intervals along the x-axis. Draw bars at each position with heights of 37.5, 17.5, 25, 12.5, and 32.5 units respectively. The completed bar graph clearly shows that Cricket has the highest participation, followed by Swimming, then Tennis, with Badminton having the fewest players.
In simple words: Mark the five sports evenly spaced on the bottom. Use a scale where one small box = 2 students. Draw bars reaching the heights you calculated - Cricket's bar goes highest, and Badminton's goes lowest.

Exam Tip: Make sure all bars have the same width and the spacing between them is consistent - this shows careful graph construction and earns full marks.

Question 3. Draw a bar graph representing the number of students in different years.
Answer: Using the scale 1 small division = 50 students, calculate the bar heights as follows: For year 2005 - 06, the height is \( \left( \frac{1}{50} \times 800 \right) = 16 \) small divisions. For year 2006 - 07, the height is \( \left( \frac{1}{50} \times 975 \right) = 19.5 \) small divisions. For year 2007 - 08, the height is \( \left( \frac{1}{50} \times 1100 \right) = 22 \) small divisions. For year 2008 - 09, the height is \( \left( \frac{1}{50} \times 1400 \right) = 28 \) small divisions. For year 2009 - 10, the height is \( \left( \frac{1}{50} \times 1625 \right) = 32.5 \) small divisions. On your graph paper, draw the y-axis to represent the number of students and the x-axis to show the years at uniform intervals. Draw bars at each year position with the calculated heights, creating a visual pattern that demonstrates the steady growth in student enrollment across this five-year period.
In simple words: Divide each year's student count by 50 to find bar heights. The taller bars appear later because more students enrolled in the later years. The graph shows a clear upward trend over time.

Exam Tip: Double-check your division calculations before drawing - a small arithmetic error gets magnified across the entire graph and looks unprofessional.

Question 4. Draw a bar graph showing the number of scooters produced in different years.
Answer: Using the scale 1 small division = 1000 scooters, determine each bar's height: For the year 2004, the height equals \( \left( \frac{1}{1000} \times 11000 \right) = 11 \) small divisions. For the year 2005, the height equals \( \left( \frac{1}{1000} \times 14000 \right) = 14 \) small divisions. For the year 2006, the height equals \( \left( \frac{1}{1000} \times 12500 \right) = 12.5 \) small divisions. For the year 2007, the height equals \( \left( \frac{1}{1000} \times 17500 \right) = 17.5 \) small divisions. For the year 2008, the height equals \( \left( \frac{1}{1000} \times 15000 \right) = 15 \) small divisions. For the year 2009, the height equals \( \left( \frac{1}{1000} \times 24000 \right) = 24 \) small divisions. On graph paper, place years along the x-axis at regular intervals and draw the y-axis to represent production quantity. Construct bars at each year with heights matching your calculations. The resulting graph reveals production fluctuations, with a notable peak in 2009 and a dip in 2006.
In simple words: Divide each year's production by 1000 to get bar heights. Draw bars for each year in order - you'll see that production was highest in 2009 and lowest in 2006.

Exam Tip: Always write the year labels clearly and ensure bars are positioned directly above these labels - poor alignment loses presentation marks even if calculations are correct.

Question 5. Draw a bar graph representing the birth rate per thousand in different countries.
Answer: Using the scale 1 small division = 1 birth per thousand, calculate the heights: China has a height of 42 small divisions, India has 35 small divisions, Germany has 14 small divisions, UK has 28 small divisions, and Sweden has 21 small divisions. Set up your graph with country names spaced uniformly along the x-axis and the y-axis labeled to show "Birth rate per thousand". Draw bars at each country position with the corresponding heights. The completed bar graph provides an immediate visual comparison - China and India show notably higher birth rates compared to European nations, with Germany displaying the lowest birth rate among the five countries represented.
In simple words: Write each country name on the bottom axis. Draw bars for each one using the numbers given - China's bar reaches highest, and Germany's bar is shortest. You can instantly see which countries have higher and lower birth rates.

Exam Tip: Label both axes with their units - "Birth rate per thousand" on the y-axis and "Countries" on the x-axis - examiners always check for complete labeling.

Question 6. Draw a bar graph showing different modes of transportation used by students to go to school.
Answer: Using the scale 1 small division = 20 students, compute the bar heights: School bus gives \( \left( \frac{1}{20} \times 640 \right) = 32 \) small divisions. Private bus gives \( \left( \frac{1}{20} \times 360 \right) = 18 \) small divisions. Bicycle gives \( \left( \frac{1}{20} \times 490 \right) = 24.5 \) small divisions. Rickshaw gives \( \left( \frac{1}{20} \times 210 \right) = 10.5 \) small divisions. Going on foot gives \( \left( \frac{1}{20} \times 150 \right) = 7.5 \) small divisions. Position the five transportation methods evenly along the x-axis and mark the y-axis for "Number of students". Draw bars with the calculated heights at each position. The resulting graph clearly demonstrates that the school bus is the most popular transport choice, followed by bicycle, with walking being the least common method among the surveyed students.
In simple words: Divide each number of students by 20 to find bar heights. The school bus bar is tallest because more students use it, and the walking bar is shortest because fewest students walk.

Exam Tip: Make the bars roughly equal in width and maintain consistent spacing - visual uniformity demonstrates that you understand the structure of a proper bar graph.

Question 7. Draw a bar graph representing the population of different states of India.
Answer: Using the scale 1 small division = 40 lakhs of population, find the heights: Bihar has a height of \( \left( \frac{1}{40} \times 820 \right) = 20.5 \) small divisions. Jharkhand has \( \left( \frac{1}{40} \times 270 \right) = 6.75 \) small divisions. Uttar Pradesh has \( \left( \frac{1}{40} \times 1060 \right) = 26.5 \) small divisions. Uttarakhand has \( \left( \frac{1}{40} \times 80 \right) = 2 \) small divisions. Madhya Pradesh has \( \left( \frac{1}{40} \times 600 \right) = 15 \) small divisions. Chhattisgarh has \( \left( \frac{1}{40} \times 210 \right) = 5.25 \) small divisions. Mark the six states at uniform gaps along the x-axis and label the y-axis as "Population in lakhs". Draw bars at each state location with these calculated heights. The completed graph reveals that Uttar Pradesh has the largest population among the states shown, while Uttarakhand has the smallest.
In simple words: Divide each population figure by 40 to get bar heights. Arrange state names along the bottom. The tallest bar is Uttar Pradesh and the shortest is Uttarakhand - the graph lets you see population differences at a glance.

Exam Tip: When dealing with large numbers like lakhs, always ensure your scale is written clearly on the graph - this prevents reader confusion and shows mathematical awareness.

Question 8. Draw a bar graph showing the population of India at different census years.
Answer: Using the scale 1 small division = 40 millions, compute the heights: For 1951, the height is \( \left( \frac{1}{40} \times 360 \right) = 9 \) small divisions. For 1961, the height is \( \left( \frac{1}{40} \times 432 \right) = 10.8 \) small divisions. For 1971, the height is \( \left( \frac{1}{40} \times 540 \right) = 13.5 \) small divisions. For 1981, the height is \( \left( \frac{1}{40} \times 684 \right) = 17.1 \) small divisions. For 1991, the height is \( \left( \frac{1}{40} \times 852 \right) = 21.3 \) small divisions. For 2001, the height is \( \left( \frac{1}{40} \times 1020 \right) = 25.5 \) small divisions. Place the census years along the x-axis at uniform intervals and label the y-axis as "Population in millions". Draw bars corresponding to each year with the heights you calculated. The graph illustrates a consistent upward trend in India's population over the 50-year period, showing continuous growth with no decline in any decade.
In simple words: Divide each census figure by 40 for bar heights. Years go across the bottom in order. The bars get taller as you move right, showing India's population grew steadily each decade.

Exam Tip: Always arrange historical or time-series data from left to right in chronological order - this makes trends immediately visible and is expected in any properly constructed graph.

Question 9. Draw a bar graph showing the distance from Delhi to various cities.
Answer: Using the scale 1 small division = 40 km, calculate the bar heights: From Delhi to Kolkata, the height is \( \left( \frac{1}{40} \times 1340 \right) = 33.5 \) small divisions. From Delhi to Mumbai, the height is \( \left( \frac{1}{40} \times 1100 \right) = 27.5 \) small divisions. From Delhi to Chennai, the height is \( \left( \frac{1}{40} \times 1700 \right) = 42.5 \) small divisions. From Delhi to Hyderabad, the height is \( \left( \frac{1}{40} \times 1220 \right) = 30.5 \) small divisions. Place city names along the x-axis at regular intervals and label the y-axis as "Distance in km". Draw bars at each city position with the calculated heights. The graph demonstrates that Chennai is the farthest from Delhi among the four cities shown, while Mumbai is the closest.
In simple words: Divide each distance by 40 to find bar heights. List cities across the bottom. The tallest bar (Chennai) shows the farthest city, and the shortest bar (Mumbai) shows the nearest city from Delhi.

Exam Tip: When showing distances or measurements, always double-check that your scale makes sense - a scale that's too large or too small will make your graph look wrong even if your math is right.

Question 10. Draw a bar graph representing the interest earned over different time periods.
Answer: Using the scale 1 small division = 4 thousand crore rupees, find the bar heights: For 1998 - 1999, the height is \( \frac{70}{4} = 17.5 \) small divisions. For 1999 - 2000, the height is \( \frac{84}{4} = 21 \) small divisions. For 2000 - 2001, the height is \( \frac{98}{4} = 24.5 \) small divisions. For 2001 - 2002, the height is \( \frac{106}{4} = 26.5 \) small divisions. For 2002 - 2003, the height is \( \frac{120}{4} = 30 \) small divisions. Arrange the time periods along the x-axis at uniform gaps and label the y-axis as "Interest in thousand crore rupees". Draw bars at each period with the calculated heights. The resulting graph shows a consistent upward trend, indicating that interest earned increased steadily throughout the five-year span, with the highest interest generated during 2002 - 2003.
In simple words: Divide each interest value by 4 to get bar heights. Time periods go left to right. Bars get progressively taller, showing interest earnings climbed each year.

Exam Tip: When you spot a clear increasing or decreasing trend in your graph, mention it in your answer - recognizing and articulating patterns shows deeper understanding beyond just drawing bars.

Question 11. Draw a bar graph displaying the life expectancy of different countries.
Answer: Using the scale 1 small division = 2 years, determine the bar heights: Japan has a height of \( \frac{76}{2} = 38 \) small divisions. India has \( \frac{57}{2} = 28.5 \) small divisions. Britain has \( \frac{70}{2} = 35 \) small divisions. Ethiopia has \( \frac{43}{2} = 8.6 \) small divisions. Cambodia has \( \frac{36}{2} = 18 \) small divisions. Arrange the country names at uniform intervals along the x-axis and mark the y-axis as "Life expectancy in years". Draw bars at each country position with the calculated heights. The completed bar graph reveals significant differences in life expectancy across nations - Japan leads with the highest life expectancy at 76 years, while Ethiopia shows the lowest at 43 years, highlighting disparities in living standards and healthcare systems worldwide.
In simple words: Divide each country's life expectancy by 2 for bar heights. Japan's bar towers highest because its people live longest on average, and Ethiopia's bar is shortest. The gaps between bars show how much life expectancy varies around the world.

Exam Tip: Use this type of graph to support written analysis - mention what the data reveals about global health disparities, and you'll demonstrate critical thinking beyond mere graphical skill.

Question 12. Draw a bar graph representing the percentage of buyers of different soap brands.
Answer: Using the scale 1 small division = 1% buyer preference, compute the bar heights: Brand A has 45 divisions. Brand B has 25 divisions. Brand C has 15 divisions. Brand D has 10 divisions. Other brands have 5 divisions. Position the five brand categories at uniform gaps along the x-axis and label the y-axis as "Percentage of buyers". Draw bars with heights matching these percentages. The completed bar graph plainly illustrates that Brand A dominates the market with the highest buyer preference at 45%, followed by Brand B at 25%. Together, these two brands account for 70% of all buyers surveyed, while the remaining three categories (Brand C, Brand D, and Others) share the final 30%, with Brand C at 15%, Brand D at 10%, and other brands at just 5%.
In simple words: Place the five brands on the bottom axis. Draw bars where each unit represents 1% of buyers. Brand A's tall bar shows it's the customer favorite, and the small bars for D and Others show they have tiny market shares.

Exam Tip: For percentage-based data, always verify that all bars together add up to 100% - this acts as a quick accuracy check and prevents embarrassing errors.

Question 13. Analyze a bar graph showing the marks scored by a student in different subjects.
Answer: (i) The bar graph presents the marks obtained by a student across five distinct subjects in his or her examinations. (ii) From the graph, the bar reaching the maximum height indicates that the student earned the highest marks in mathematics, demonstrating strong performance in this subject. (iii) The bar displaying the minimum height shows that the student earned the lowest marks in Hindi, meaning the student needs improvement in this language subject. (iv) To find the average marks, sum all five scores - 60 + 35 + 75 + 50 + 60 = 280 - then divide by 5. The calculation gives \( \frac{280}{5} = 56 \), showing the student's average performance across all subjects is 56 marks.
In simple words: The graph shows one student's exam scores in five subjects. The tallest bar is math (the best score), the shortest is Hindi (the weakest subject). Adding all marks and dividing by 5 gives the overall average of 56.

Exam Tip: When asked to analyze a bar graph, always address all visible aspects - the maximum, minimum, and any requested calculations - to earn complete marks.

Question 14. Analyze a bar graph showing the number of families and family sizes in a colony.
Answer: (i) The given bar graph provides data on the number of families living in a colony and presents information about the number of family members in each family. (ii) Looking at the graph, the bar showing families with three members reaches up to 40 on the y-axis. Therefore, exactly 40 families have three members in this colony. (iii) Examining the graph, there is no bar appearing at the reading of 1 on the y-axis, which signifies that no single individual lives alone in the colony. (iv) The bar indicating the maximum reading on the y-axis corresponds to families with three members, meaning a three-member family structure is most frequent. Each family of this type has exactly three members, making it the most common family composition within the colony.
In simple words: The graph shows how many families exist and how big each family is. Three-member families are the most common (40 families). No one lives alone in this colony. The largest bar tells you the most popular family size.

Exam Tip: When interpreting bar graphs, always read the y-axis value directly rather than estimating - precision in reading the scale demonstrates careful analytical skill.

Question 15. Analyze a bar graph showing the heights of different mountain peaks.
Answer: (i) The bar graph clearly indicates that the bar displaying the greatest height corresponds to Mount Everest. Accordingly, Mount Everest is recognized as the highest peak, with a recorded height of 8800 meters. (ii) The ratio between the highest peak and the second highest peak is determined by dividing Mount Everest's height by Kanchenjunga's height: \( \frac{8800}{8200} = \frac{44}{41} \), which simplifies the comparison between these two major Himalayan summits. (iii) From examining the graph, the heights of the displayed peaks can be listed in descending order by magnitude as follows: 8800 m, 8200 m, 8000 m, 7500 m, 6000 m, representing the complete ranking of these mountains from tallest to shortest.
In simple words: Mount Everest has the highest bar, so it's the tallest mountain at 8800 meters. Its height compared to Kanchenjunga (8200 m) is 44 to 41. If you list all peaks from highest to lowest, you get 8800, 8200, 8000, 7500, and 6000 meters.

Exam Tip: When comparing data from bar graphs, always show the ratio as a simplified fraction if requested - this demonstrates mathematical sophistication and earns bonus consideration from examiners.