RS Aggarwal Solutions for Class 6 Chapter 23 Pictograph

Exercise 23.1

 

Question 1. The following table shows the daily production of T.V sets in an industry for 7 days of a week: Represent the above information by a pictograph.

Answer: Let one icon symbolize 50 TV sets. Next, we determine the number of icons needed for each day by dividing the production figure by 50:The pictograph displays this data using TV set icons arranged in rows for each day, with the number of icons corresponding to the values above.
In simple words: Choose a symbol to stand for a fixed number of items. Divide each day's production by that number to find how many symbols to draw. Arrange the symbols in rows-one row per day-to create a visual picture of the data.

Exam Tip: Always state the symbol key clearly (e.g., "One icon = 50 TVs") and ensure each icon is drawn consistently to make the pictograph easy to read.

 

Question 2. The following table shows the number of Maruti cars sold by five dealers in a particular month: Represent the above information by a pictograph.

Answer: Let one car icon symbolize 5 Maruti cars. We then find the number of icons required for each dealer by dividing their sales by 5:The pictograph shows each dealer's sales using red car icons, with the number of icons matching the calculated values. Dealers who sold more cars have more icons displayed in their rows.
In simple words: Pick a car as your symbol to represent 5 vehicles. For each dealer, count how many cars they sold, divide by 5, and draw that many car symbols in a row next to the dealer's name.

Exam Tip: Verify that the key (symbol value) allows all numbers to be represented accurately. Choose a divisor that works evenly or nearly evenly with your data.

 

Question 3. The population of Delhi state in different census years is as given below: Represent the above information by a pictograph.

Answer: To show this data as a pictograph, start by drawing one horizontal and one vertical axis. The horizontal axis displays census years, while the vertical axis shows population values. Since we have 5 data points, we mark 5 positions equally spaced along the horizontal axis. At each position, we draw rectangles with heights matching the population figures. On the vertical axis, each unit of 10 represents 10 lakhs of population. The rectangles are drawn to be proportional to the given data. This visual representation clearly shows how Delhi's population has grown across the five census periods.
In simple words: Draw two lines that meet at right angles-one going sideways for years, one going up for population. Make one box for each year, and make the height of each box match the population number. This creates a picture showing how the population changed over time.

Exam Tip: When constructing bar graphs, ensure equal spacing between categories on the horizontal axis and use a consistent scale on the vertical axis for accurate visual comparison.

 

Question 4. Read the bar graph shown in Fig. and answer the following question:
(i) What is the information given by the bar graph?
(ii) How many tickets of Assam States Lottery were sold by the agent?
(iii) Of which state, where the maximum number of tickets sold?
(iv) State whether true or false: The maximum number of tickets sold is three times the minimum number of tickets sold
(v) Of which states were the minimum number of tickets sold.
Answer:
(i) The bar graph displays how many lottery tickets from various states were marketed by an agent during a single day.
(ii) The agent sold 40 tickets of Assam state lottery. Looking at the bar graph, the height of the rectangle corresponding to Assam reaches the 40 mark on the vertical axis.
(iii) Haryana had the largest ticket sales. The vertical height of Haryana's bar is the tallest compared to all other state bars shown.
(iv) False. The maximum bar height (for Haryana) measures 100 units, while the minimum bar height (for Rajasthan) measures 20 units. Since 100 is not equal to 3 times 20 (which would be 60), the statement is false. The ratio is actually 100 ÷ 20 = 5, meaning the maximum is 5 times the minimum, not 3 times.
(v) Rajasthan had the minimum ticket sales. According to the bar graph, Rajasthan's bar is the shortest, indicating only 20 tickets were sold from that state.
In simple words: Read bar graphs by looking at the height where each bar ends against the vertical scale. Compare bar heights to find the tallest (maximum) and shortest (minimum) values. Use the scale to determine exact numbers and relationships between bars.

Exam Tip: Always identify the scale on both axes before reading values. Trace horizontally from the bar's top to the vertical axis to find exact numbers. Check ratios and comparisons carefully before stating them as true or false.

 

Question 5. Study the bar graph representing the number of persons in various age groups in a town shown in Fig. observe the bar graph and answer the following questions:
(i) What is the percentage of the youngest age - group persons over those in the oldest age group?
(ii) What is the total population of the town?
(iii) What is the number of persons in the age group 60 - 65?
(iv) How many persons are more in the age - group 10 -15 than in the age group 30 - 35?
(v) What is the age - group of exactly 1200 persons living in the town?
(vi) What is the total number of persons living in the town in the age - group 50 - 65?
(vii) What is the total number of persons living in the town in the age - group 10 - 15 and 60 - 65?
(viii) Whether the population in general increases decreases or remains constants with the increase in the age - group.
Answer:
(i) The youngest age group is 10-15 years, with a population of 1400. The oldest age group is 70-75 years, containing 300 individuals. The difference between these two groups is 1400 - 300 = 1100. To find the percentage increase, we compute: (1100 ÷ 300) × 100% = 366⅔%
(ii) The town's total population equals the sum of all individuals across every age group: 1400 + 1200 + 1100 + 1000 + 900 + 800 + 300 = 6700 persons.
(iii) The age group 60-65 has 800 individuals. On the bar graph, the rectangle corresponding to this range extends up to 800 units on the vertical scale.
(iv) The 10-15 age bracket has 1400 individuals, while the 30-35 bracket has 1100 individuals. The difference is 1400 - 1100 = 300 more persons in the younger group.
(v) The age group containing exactly 1200 people is 20-25 years. By examining the bar graph, the rectangle for this range reaches the 1200 mark on the vertical axis.
(vi) The 50-55 age bracket has 900 people. The vertical bar for this range reaches up to 900 units on the graph.
(vii) The 10-15 age bracket contains 1400 people, and the 60-65 bracket has 800 people. Adding these gives 1400 + 800 = 2200 total persons in both age groups combined.
(viii) As the age group gets older, the overall population drops. Looking across the bars from left to right, the heights consistently fall, showing that fewer people live in each successive older age bracket.
In simple words: Read each bar's height using the vertical scale to find population numbers. Add numbers to find totals. Compare bars to find which age group has more or fewer people. The graph shows that young people are more numerous than older people in the town.

Exam Tip: Always read bar values precisely from the graph scale. When calculating percentages or totals, double-check arithmetic. Notice trends (increasing, decreasing, constant) by observing whether bars get taller, shorter, or stay the same as you move across the graph.

 

Question 6. Read the bar graph shown in fig 23.10 and answer the following questions:
(i) What is the information given by the bar graph?
(ii) What was the number of commercial banks in 1977?
(iii) What is the ratio of the number of commercial banks in 1969 to that in 1980?
(iv) State whether true or false: The number of commercial banks in 1983 is less than double the number of commercial banks in 1969.
Answer:
(i) The bar graph illustrates the growth in the quantity of commercial banks operating in India during the specified years.
(ii) During 1977, there were 130 commercial banks. The height of the bar for 1977 reaches up to the 130 mark on the vertical axis.
(iii) In 1969, there were 90 commercial banks. In 1980, the count was 150 commercial banks. Thus, the ratio = 90 ÷ 150 = 3 ÷ 5, which is expressed as 3:5.
(iv) False. The bank count in 1983 was 230. Twice the 1969 count would be 2 × 90 = 180. Since 230 exceeds 180, the number in 1983 is more than double the 1969 figure, not less.
In simple words: Bar graphs display numerical changes across years. To find a ratio, divide one year's value by another and simplify. To check if something is "less than double," multiply the starting number by 2 and compare the result to the final number.

Exam Tip: When computing ratios from bar graphs, first read both values carefully, then divide and reduce to simplest form. For true/false statements involving multiplications or comparisons, calculate the expected value before comparing with the actual value.

 

Question 7. Given below is the bar graph indicating the marks obtained out of 50 in mathematics paper by 100 students. Read the bar graph and answer the following questions:
(i) It is decided to distribute workbooks on mathematics to the students obtaining less than 20 marks, giving one workbook to each of such students. If a workbook costs Rs. 5, what sure is required to buy the workbooks?
(ii) Every student belonging to the highest mark group is entitled to get a prize of Rs. 10. How much amount of money is required for distributing the prize money?
(iii) Every student belonging to the lowest mark-group has to solve 5 problems per day. How many problems, in all, will be solved by the students of this group per day?
(iv) State whether true or false.
(a) 17% students have obtained marks ranging from 40 to 49.
(b) 59 students have obtained marks ranging from 10 to 29.
(v) What is the number of students getting less than 20 marks?
(vi) What is the number of students getting more than 29 marks?
(vii) What is the number of students getting marks between 9 and 40?
(viii) What is the number, of students belonging to the highest mark group?
(ix) What is the number of students obtaining more than 19 marks?
Answer:
First, we build a table summarizing data from the bar graph:

(i) Students scoring below 20 marks: 27 + 12 = 39 students. The cost to purchase workbooks is Rs. 5 × 39 = Rs. 195.
(ii) The highest mark bracket is 40-49 years, with 17 students. The money needed for prizes is Rs. 10 × 17 = Rs. 170.
(iii) The lowest mark group is 0-9 years with 27 pupils. Daily problems solved = 5 × 27 = 135 problems.
(iv) (a) True. Students in the 40-49 range number 17. As a percentage: (17 ÷ 100) × 100% = 17%. (b) False. Students with marks from 10-29 are 12 + 20 = 32, not 59.
(v) Students scoring less than 20: 27 + 12 = 39 pupils.
(vi) Students scoring above 29: 24 + 17 = 41 pupils.
(vii) Students with marks between 9 and 40: 12 + 20 + 24 = 56 pupils (ranges 10-19, 20-29, 30-39).
(viii) The highest bracket (40-49) has 17 pupils.
(ix) Students with more than 19 marks: 20 + 24 + 17 = 61 pupils (ranges 20-29, 30-39, 40-49).
In simple words: Extract numbers from the bar graph by reading where each bar ends. For totals, add the relevant bars. For percentages, divide one group's count by the total and multiply by 100. For money, multiply the number of people by the amount per person.

Exam Tip: When combining ranges (e.g., "less than 20"), include all intervals that fit the condition. Always verify your count by checking that all groups sum to the total (100 students here). For percentage questions, ensure you are calculating the percentage of the specified group, not all students.

 

Question 8. Read the following bar graph and answer the following questions:
(i) What is the information given by the bar graph?
(ii) State each of the following whether true or false.
(a) The number of government companies in 1957 is that of 1982 is 1: 9.
(b) The number of government companies has decreased over the 1983.
Answer:
(i) The bar graph displays data regarding the count of public sector companies operating in India during the period spanning 1957 through 1987.
(ii) (a) False
Explanation: The count of public sector companies in 1957 = 50. The count of public sector companies in 1982 = 375. The proportion comparing the count of public sector companies in 1957 to that in 1982 = 50/375 = 2/15 = 2:15
(b) False
Explanation: Since no information is provided regarding the year 1983, we are unable to make any conclusions about this. Consequently, the statement cannot be verified as true.

Exam Tip: Always verify ratios by reading exact values from the bar graph axes and simplifying fractions to lowest terms. When data is absent for a particular year, the statement cannot be confirmed.

 

Question 9. Read the following bar graph and answer the following questions:
(i) What information is given by the bar graph?
(ii) Which state is the largest producer of rice?
(iii) Which state is the largest producer of wheat?
(iv) Which state has total production of rice and wheat as its maximum?
(v) Which state has the total production of wheat and rice minimum?
Answer:
(i) The bar graph furnishes information on rice and wheat production across different Indian states.

The table from the bar graph:

(ii) W.B. (West Bengal) is the largest producer of rice.
Explanation: The height of the bar showing rice production in W.B. reaches 10 units, which is the greatest among all the other states.

(iii) U.P. (Uttar Pradesh) is the largest producer of wheat.
Explanation: The height of the bar showing wheat production in U.P. reaches 16 units, which is the greatest among all the other states.

(iv) U.P. has the maximum combined production of rice and wheat.
Explanation: From the bar graph, we can observe that U.P. leads all other states in the total output of rice and wheat: 16 units wheat + 8 units rice = 24 units.

(v) Maharashtra has the minimum combined production of rice and wheat.
Explanation: From the bar graph, we can observe that the combined output of rice and wheat in Maharashtra is the least: 4 units rice + 2 units wheat = 6 units.

Exam Tip: Always create a reference table from the bar graph to organize data systematically. Add the component values carefully to find totals, and compare across all categories to identify maximum and minimum values.

 

Question 10. The following bar graph represents the heights (in cm) of 50 students of a particular school. Study the graph and answer the questions:
(i) What percentage of the total number of students have their heights more than 149 cm?
(ii) How many students in the class are in the range of maximum height of the class?
(iii) The school wants to provide a particular type of tonic to each student below the height of 150 cm to improve his height. If the cost of the tonic for each student comes out to be Rs 55, how much amount of money is required?
(iv) How many students are in the range of shortest height of the class?
(v) State whether true or false:
(a) There are 9 students in the class whose heights are in the range of 155-159 cm.
(b) Maximum height (in cm) of a student in the class is 17.
(c) There are 29 students in the class whose heights are in the range of 145-154 cm.
(d) Minimum height (in cm) of a student is the class is in the range of 140-144 cms.
(e) The number of students in the class having their heights less than 150 cm is 12.
(f) There are 14 students each of whom has height more than 154 cm.
Answer:
Table from the bar graph:

(i) Count of students with height exceeding 149 cm = 17 + 9 + 5 = 31. Total count of students = 50. Percentage of students with height exceeding 149 cm = 31/50% = 31 × 2% = 62%

(ii) The maximum height-range of the class is 160 - 164. Count of students in this range = 5

(iii) Count of students measuring below 150 cm = 7 + 12 = 19. Required money to spend on the tonic = 19 × Rs. 55 = Rs. 1045

(iv) The minimum height-range of the class is 140 - 144. Count of students in this range = 7

(v) (a) True
Explanation: From the table, we observe that there are 9 students in the height-range 155 - 159.

(b) False
Explanation: 17 is the count of students found in the maximum height-range, not the actual maximum height in cm.

(c) True
Explanation: Count of students in the height-range 145 - 154 cm = Count of students in the height-range 145 - 149 + Count of students in the height-range 150 - 154 = 12 + 17 = 29

(d) True
Explanation: The minimum height-range of students in the class is 140 - 144 cm.

(e) False
Explanation: Count of students measuring below 150 cm = 7 + 12 = 19

(f) True
Explanation: Count of students measuring more than 154 cm = 9 + 5 = 14

Exam Tip: Always organize data in a table first, then perform calculations step-by-step. For percentage questions, identify the relevant group and divide by the total count. Check true/false statements by computing exact values rather than estimating from the graph.

 

Question 11. Read the following bar graph (Fig. 23.15) and answer the following questions:
(i) What is the information given by the bar graph?
(ii) What was the production of cement in the year 1980 - 81?
(iii) What are the minimum and maximum productions of cement and corresponding years?
Answer:
(i) The bar graph furnishes information regarding cement manufacturing output across India throughout various financial periods.

(ii) The production figure for cement during the 1980 - 81 period was 186 lakh tonnes.
Explanation: The bar height for the 1980 - 81 period stands at 186 units, with production measured in lakh tonnes.

(iii) The smallest bar height corresponds to 30 units, representing the period 1950 - 51. Thus, the smallest cement production was 30 lakh tonnes during 1950 - 51. The tallest bar height corresponds to 232 units during the 1982 - 83 period. Thus, the largest cement production was 232 lakh tonnes during 1982 - 83.

Exam Tip: Read bar heights directly from the y-axis scale and match them with corresponding years on the x-axis. Identify minimum by finding the shortest bar and maximum by finding the tallest bar, noting their respective year labels.

 

Question 12. The bar graph shown in fig 23.16 represents the circulation of newspapers in 10 languages. Study the graph and answer the following questions:
(i) What is the total number of newspapers published in Hindi, English, Urdu, Punjabi, and Bengali?
(ii) What percent is the number of newspapers published in Hindi of the total number of newspapers?
(iii) Find the excess of the number of newspapers published in English over those published in Urdu.
(iv) Name two pairs of languages which publish the same number of newspapers.
(v) State the language, in which the smallest number of newspapers is published.
(vi) State the language in which the largest number of newspapers is published.
(vii) State the language in which the number of newspapers published is between 2500 and 3500.
(viii) State whether true or false:
(a) The number of newspapers published in Malayalam and Marathi together is less than those published in English.
(b) The number of newspapers published in Telugu is more than those published in Tamil.
Answer:
Table from the bar graph:

(i) Combined total of newspapers issued in Hindi, English, Urdu, Punjabi and Bengali = 3700 + 3400 + 700 + 200 + 1100 = 9100

(ii) Number of newspaper published in Hindi = 3700. Combined total of newspapers published = 700 + 400 + 1000 + 200 + 1400 + 1400 + 3700 + 1100 + 3400 + 1100 = 14400. Thus, Percentage of Hindi newspaper published = 3700/14400 × 100% = 25.69% = 25.7%

(iii) Number of newspapers published in English = 3400. Number of newspapers published in Urdu = 700. Thus, Excess count of newspapers published in English compared to Urdu = 3400 - 700 = 2700

(iv) "Marathi" & Malayalam and "Gujarati & Bengali" represent two language groupings where equivalent numbers of newspapers are released.
Explanation: Newspapers released in Marathi = Newspapers released in Malayalam = 1400. Newspapers released in Gujarati = Newspapers released in Bengali = 1100

(v) Punjabi is the language in which the minimal count of newspapers is released.
Explanation: In Punjabi, merely 200 newspapers are released.

(vi) Hindi is the language in which the greatest numbers of newspapers are released.
Explanation: In Hindi, 3700 newspapers are released.

(vii) The count of English newspapers published sits between 2500 and 3500.
Explanation: In English, 3400 newspapers are released.

(viii) (a) True
Explanation: Combined newspapers published in Malayalam and Marathi = 1400 + 1400 = 2800. The count of newspapers published in English is 3400, which exceeds the aggregate count of Malayalam and Marathi newspapers.

(b) False
Explanation: Number of newspapers published in Telugu = 400. Number of newspapers published in Tamil = 1000

Exam Tip: Always construct a data table first to organize all values systematically. For comparative questions, arrange side-by-side values to spot matching pairs. Verify true/false statements by calculating exact totals rather than visual estimation.

 

Exercise 23.2

 

Question 1. Explain the reading and interpretation of bar graphs.
Answer: A bar graph is a visual representation where bar lengths correspond directly to the values they depict. The bars may be arranged in a vertical or horizontal orientation. A bar graph serves as a visual tool for comparing amounts or frequencies of various characteristics within a dataset. Bar graphs make it possible to:

• Compare different groups of data
• Draw conclusions about the data in a quick manner

In simple words: A bar graph uses bars of different lengths to show and compare different quantities. By looking at the heights or lengths of the bars, you can quickly see which items are bigger or smaller than others.

Exam Tip: Emphasize the proportional relationship between bar length and data value. A strong answer explains both the visual structure (vertical or horizontal bars) and the practical advantages (quick comparison and pattern recognition).

 

Question 2. Read the following bar graph and answer the following questions:
Answer: [Graph context indicates export and import data across multiple years; specific sub-questions for this item are not visible in the provided material. Standard interpretation would involve reading bar heights for each year-category pair and comparing export versus import values.]

Without the explicit sub-questions visible, please refer to the bar graph directly to:

• Determine total exports or imports for specified years
• Calculate differences between export and import quantities
• Identify years with maximum or minimum values
• Compare trends across the time period shown

In simple words: Look at the height of each bar on the graph. Compare the bars side by side to see which years had more imports or exports, and by how much. Add or subtract bar heights to find totals or differences between categories.

Exam Tip: When answering bar graph questions, always read values directly from the axis scale and perform calculations step-by-step. Show your arithmetic clearly to earn full credit.

 

Question 1. (i) What information is given by the bar graph?
Answer: The bar graph gives details about the overall volume of imports and exports across several years ranging from 1982 to 1987.
In simple words: The graph shows how much goods came into the country (imports) and how much went out (exports) for each year during that five-year span.

Exam Tip: Always read the title and axis labels of a bar graph carefully to understand what data is being displayed before answering interpretation questions.

 

Question 1. (ii) In which year the export is minimum?
Answer: In the year 1982-83, the export reaches its lowest point.
In simple words: In 1982-83, the country sent out the fewest goods compared to all the other years shown. The exports were only 800 crore rupees, which was the smallest amount.

Exam Tip: For "minimum" or "maximum" questions, locate the shortest or tallest bar on the graph and read the corresponding value from the axis.

 

Question 1. (iii) In which year the import is maximum?
Answer: In the year 1986-87, the import reaches its highest point.
In simple words: In 1986-87, the country brought in the most goods from outside. Imports were at 22,000 crore rupees, the greatest among all the years displayed.

Exam Tip: Always compare all bars visually on the graph - the tallest bar in the import section represents the maximum value.

 

Question 1. (iv) In which year the difference of the values of export and import is maximum?
Answer: In the year 1986-87, the difference between the values of exports and imports reaches the maximum.
In simple words: In 1986-87, the gap between what India exported and what it imported was the largest. The difference came to 1000 crore rupees.

Exam Tip: To find the maximum difference, subtract the smaller value from the larger value for each year, then compare all differences to identify which year has the biggest gap.

 

Question 2. (i) The pair of classes in which the results of boys and girls are inversely proportional are:
(a) VI, VIII (b) VI, IX (c) VIII, IX (d) VIII, X
Answer: (b) VI, IX
In simple words: Two groups are inversely proportional when one goes up while the other goes down. In Class VI, boys have 80% and girls have 70%. In Class IX, boys have 70% and girls have 80%. The percentages switch places, showing inverse proportionality.

Exam Tip: Look for pairs where the boy's percentage in one class matches the girl's percentage in another class, and vice versa - this indicates inverse proportionality.

 

Question 2. (ii) The class having the lowest failure rate of girls is
(a) VII (b) X (c) IX (d) VIII
Answer: (a) VII
In simple words: The lowest failure rate means the highest passing rate. In Class VII, girls have a 100% pass rate, meaning 0% of them failed. This is the best performance among all classes.

Exam Tip: "Lowest failure rate" is the same as "highest pass rate" - find the class where girls have the highest percentage shown in the graph.

 

Question 2. (iii) The class having the lowest pass rate of students is:
(a) VI (b) VII (c) VIII (d) IX
Answer: (b) VII and (c) VIII
In simple words: To find the class with the lowest pass rate for all students, add the boy's and girl's percentages for each class. Class VII and Class VIII both have a combined height of 140 units on the graph, which is the lowest total compared to all other classes.

Exam Tip: For "lowest pass rate of students," add the heights of both bars (boys + girls) for each class and find the pair with the smallest total.

 

Question 3. The following bar graph shows the results of an annual examination at a secondary school. Read the bar graph (Fig. 23.22) and choose the correct alternative in each of the following:
Answer: This is a multi-part data interpretation exercise. Refer to the solutions provided below for parts (i), (ii), and (iii).

Exam Tip: When tackling multi-part graph questions, organize the data into a table first to make comparisons easier and to avoid calculation errors.

 

Question 4. The following bar graph shows the number of persons killed in industrial accidents in a country for some years (Fig 23.23). Read the bar graph and choose the correct alternative in each of the following:
Answer: This question asks you to interpret data about fatal industrial accidents across multiple years. Extract the values from the graph, organize them into a table, and use the data to answer the sub-questions about percentage increases, decreases, and comparisons between coal mining and other industrial accidents.

Exam Tip: Always create a data table when working with visual graphs - it makes percentage change calculations much more manageable and reduces the chance of reading errors from the graph.

 

Question 4. (i) The year which shows the maximum percentage increase in the number of persons killed in coal mines over the preceding year is:
(a) 1996 (b) 1997 (c) 1999 (d) 2000
Answer: (d) 2000
In simple words: To find the maximum percentage increase, look at how much the deaths rose from one year to the next. In 1997, deaths in coal mines went from 200 to 300, a 50% jump. But in 2000, they climbed from 100 to 200, doubling completely - that's a 100% increase, which is the largest.

Exam Tip: Calculate percentage change for each year transition using the formula: (new value - old value) / old value × 100. Compare all results to find the maximum.

 

Question 4. (ii) The year which shows the maximum decrease in the number of persons killed in industrial accidents over the preceding year is:
(a) 1996 (b) 1997 (c) 1998 (d) 1999
Answer: (a) 1996
In simple words: Industrial deaths fell in both 1996 and 1999. In 1996, deaths dropped from 1600 to 900, a loss of 700 people or about 43.75%. In 1999, they fell from 1300 to 900, a decrease of about 30.77%. Since 43.75% is larger than 30.77%, the year 1996 had the biggest drop.

Exam Tip: For decrease questions, calculate the percentage drop for each year and pick the one with the largest percentage decline, not just the largest absolute number decrease.

 

Question 4. (iii) The year in which the maximum number of persons was killed in industrial accidents other than those killed in coal mines are:
(a) 1995 (b) 1997 (c) 1998 (d) 1999
Answer: (a) 1995
In simple words: "Industrial accidents other than coal mines" means all other types of accidents. In 1995, a total of 1600 people died in all industrial accidents, with 300 from coal mines, leaving 1300 from other causes. This is the highest figure for non-coal mining deaths across all years.

Exam Tip: Subtract coal mining deaths from total industrial deaths to get deaths from other accidents. Compare these calculated values across years to find the maximum.

 

Question 5. The production of saleable steel in some of the steel plants of our country during 1999 is given below: Construct a bar graph to represent the above data on a graph paper by using the scale 1 big divisions = 20 thousand tonnes.
Answer: To create the bar graph, establish two perpendicular axes. Mark plant names (Bhilai, Durgapur, Rourkela, Bokaro) on the horizontal axis and production values on the vertical axis. Using the scale where 1 big division equals 20 thousand tonnes, determine the bar heights:
- Bhilai: 160 ÷ 20 = 8 units
- Durgapur: 80 ÷ 20 = 4 units
- Rourkela: 200 ÷ 20 = 10 units
- Bokaro: 150 ÷ 20 = 7.5 units
Draw bars with equal width and equal spacing, using these calculated heights. Label each bar with its production value.

Exam Tip: Always convert actual data values to graph units using the given scale before drawing. Double-check your division calculations and verify that all bars have uniform width and spacing.

 

Question 6. The following data gives the number (in thousands) of applicants registered with a division = 20 thousand tonnes. Construct a bar graph to represent the above data.
Answer: Create two perpendicular axes with years (1995-2000) on the horizontal axis and number of applicants (in thousands) on the vertical axis. Use a scale of 1 big division equals 4 thousand applicants. Calculate bar heights:
- 1995: 18 ÷ 4 = 4.5 units
- 1996: 20 ÷ 4 = 5 units
- 1997: 24 ÷ 4 = 6 units
- 1998: 28 ÷ 4 = 7 units
- 1999: 30 ÷ 4 = 7.5 units
- 2000: 34 ÷ 4 = 8.5 units
Draw bars with uniform width and consistent gaps between them. Mark actual values on top of each bar for clarity. The graph will show a steady upward trend in applicant registrations over the six-year period.

Exam Tip: When data shows an increasing or decreasing trend, verify that your bars reflect this pattern correctly - they should be visibly progressing upward or downward across the graph.

 

Question 7. The following table gives the route length (in thousand kilometers) of the Indian Railways in some of the years: Represent the above data with the help of a bar graph.
Answer: Draw two perpendicular axes with years (1960-61 through 2000-01) on the horizontal axis and route length in thousand kilometers on the vertical axis. Apply a scale where 1 big division equals 10 thousand kilometers. Convert the data to graph units:
- 1960-61: 56 ÷ 10 = 5.6 units
- 1970-71: 60 ÷ 10 = 6.0 units
- 1980-81: 61 ÷ 10 = 6.1 units
- 1990-91: 74 ÷ 10 = 7.4 units
- 2000-01: 98 ÷ 10 = 9.8 units
Construct bars with equal widths and uniform spacing between them. Label each bar with its corresponding route length value. The graph demonstrates the expansion of Indian Railways over the four-decade span.

Exam Tip: For historical or time-based data spanning many years, ensure the spacing and scaling are appropriate to show growth or change over extended periods clearly.

 

Question 8. (i) The following data gives the amount of loans (in crores of rupees) disbursed by a bank during some years: Represent the above data with the help of a bar graph.
Answer: Construct two mutually perpendicular axes. Place years (1992-1996) on the horizontal axis and loan amounts in crores on the vertical axis. Select equal widths for all bars with consistent gaps between them. Choose an appropriate scale to display the data clearly. Since the loan values range from 28 to 80 crores, a suitable scale might be 1 unit on the graph representing 5 or 10 crores. Draw bars with heights proportional to their corresponding loan values and label each bar with its actual amount for reference.

Exam Tip: When creating a bar graph, choose a scale that allows all data values to fit comfortably on the graph while making differences between values visually distinct.

 

Question 8. (ii) With the help of the bar graph, indicate the year in which amount of loan is not increased over that of the preceding year.
Answer: Examine the loan amounts year by year to identify when growth did not occur. From 1992 to 1993, loans rose from 28 to 33 crores. From 1993 to 1994, they jumped from 33 to 55 crores. From 1994 to 1995, loans remained steady at 55 crores (no increase). From 1995 to 1996, they climbed to 80 crores. The year 1995 shows no increase compared to 1994, as both years had a loan amount of 55 crores.
In simple words: Look at the bar heights from year to year. When a bar is the same height or shorter than the previous bar, no increase happened. In this case, the 1995 bar is exactly the same height as the 1994 bar.

Exam Tip: To identify years with no growth, compare consecutive bars - if the next bar is not taller than the previous one, that year experienced no increase (or a decrease).

 

Question 9. The following table shows the interest paid by a company (in lakhs): Draw the bar graph to represent the above information.
Answer: To create the bar graph, we need to set up two perpendicular axes. Mark the years on the horizontal axis and the interest amounts on the vertical axis. For each year, choose an equal width bar with consistent spacing between them. Using a scale where 1 big division equals 5 lakhs of rupees paid as interest, the bar heights are:
Year 1995-96: 20/5 = 4 units
Year 1996-97: 25/5 = 5 units
Year 1997-98: 15/5 = 3 units
Year 1998-99: 18/5 = 3.6 units
Year 1999-2000: 30/5 = 6 units
In simple words: Divide each data value by 5 to get the height of each bar. Draw bars of these heights for each year on a graph. Make sure all bars have the same width with equal gaps between them.

Exam Tip: Always specify the scale clearly and ensure all bar heights are calculated consistently using the chosen scale.

 

Question 10. The following data shows the average age of men in various countries in a certain year. Represent the above information by a bar graph.
Answer: To make the bar graph, draw two perpendicular axes. Place country names on the horizontal axis and average ages on the vertical axis. Select equal width for all bars with uniform spacing. Using a scale where 1 big division equals 10 years, the bar heights work out to:
India: 55/10 = 5.5 units
Nepal: 52/10 = 5.2 units
China: 60/10 = 6 units
Pakistan: 50/10 = 5 units
U.K.: 70/10 = 7 units
U.S.A.: 75/10 = 7.5 units
Plot each bar at these heights on the graph.
In simple words: Divide each country's average age by 10 to find how tall each bar should be. Draw bars for all six countries using these heights with equal widths and spacing.

Exam Tip: Label both axes clearly with units and ensure the scale is appropriate for the range of data being displayed.

 

Question 11. The following data gives the production of food grains (in thousand tones) for some years: Represent the above data with the help of a bar graph.
Answer: Start by drawing two perpendicular axes. Mark years on the horizontal axis and production amounts on the vertical axis. Select equal width bars with consistent gaps between them. Using a scale where 1 big division equals 20 thousand tons, calculate the bar heights:
Year 1995: 120/20 = 6 units
Year 1996: 150/20 = 7.5 units
Year 1997: 140/20 = 7 units
Year 1998: 180/20 = 9 units
Year 1999: 170/20 = 8.5 units
Year 2000: 190/20 = 9.5 units
Draw bars at each year position using these calculated heights.
In simple words: Find each bar's height by dividing the production value by 20. Then draw bars for all six years using these heights on the graph with proper spacing.

Exam Tip: Double-check your scale calculations to avoid errors in bar heights - a miscalculation here changes the entire graph.

 

Question 12. The following data gives the amount of manure (in thousand tones) manufactured by a company during some years:
(i) Represent the above data with the help of a bar graph.
Answer: Set up two perpendicular axes. Place years on the horizontal axis and manure amounts on the vertical axis. Choose equal width bars with uniform gaps. Using a scale where 1 big division represents 5 thousand tons of manure, the bar heights are:
Year 1992: 15/5 = 3 units
Year 1993: 35/5 = 7 units
Year 1994: 45/5 = 9 units
Year 1995: 30/5 = 6 units
Year 1996: 40/5 = 8 units
Year 1997: 20/5 = 4 units
Plot these bars on the graph at their respective year positions.
In simple words: Divide each production value by 5 to get bar heights. Draw bars for all six years with consistent width and spacing on the graph.

Exam Tip: Remember that bar graphs make it easy to compare values visually - the tallest bar should clearly stand out.

 

Question 12. (ii) Indicate with the help of the bar graph the year in which the amount of manure manufactured by the company was maximum.
Answer: Looking at the calculated bar heights from part (i), the tallest bar corresponds to the year 1994, which has a height of 9 units. This represents 45 thousand tons of manure, the highest production value in the dataset.
In simple words: The year 1994 had the greatest manure production. Its bar is taller than all the others on the graph.

Exam Tip: Always identify the maximum by comparing bar heights directly on the graph - this is faster and clearer than rechecking numbers.

 

Question 12. (iii) Choose the correct alternative. The consecutive years during which there was a maximum decrease in manure production are:
(a) 1994 and 1995
(b) 1992 and 1993
(c) 1996 and 1997
(d) 1995 and 1996
Answer: (c) 1996 and 1997
In simple words: Check the change in production between consecutive years. From 1996 to 1997, production dropped from 40 to 20 thousand tons, a decrease of 20 thousand tons - the largest single drop in the entire dataset.

Exam Tip: Calculate the year-to-year changes carefully: look for the biggest drop between consecutive years, not just any decline.