ML Aggarwal Class 6 Maths Solutions Chapter 15 Data Handling

 

Question 1. In a ready-made garment shop, on a particular day the following sizes of shirts were sold: 34, 38, 42, 40, 44, 32, 34, 36, 42, 40, 44, 36, 38, 42, 44, 40, 38, 40, 42, 32, 34, 38, 42, 40, 36, 42, 40, 38, 36, 40. Arrange the above data in ascending order and construct frequency distribution table. Also answer the following questions: (i) Which shirt size had the maximum sale? (ii) Which shirt size had the minimum sale? (iii) The number of shirts sold of size 42 or greater than size 42.
Answer: Arranging the given data in ascending order, we get: 32, 32, 34, 34, 34, 36, 36, 36, 36, 38, 38, 38, 38, 38, 40, 40, 40, 40, 40, 40, 40, 42, 42, 42, 42, 42, 42, 44, 44, 44

The frequency distribution table for the given data is:

(i) Size 40 had the highest frequency (7 shirts sold), meaning it was the most popular size. Hence, shirt size 40 had the maximum sale.
(ii) Size 32 had the lowest frequency (2 shirts sold), meaning it was the least popular size. Hence, shirt size 32 had the minimum sale.
(iii) The number of shirts sold of size 42 or greater than size 42 = 6 + 3 = 9. Hence, 9 shirts of size 42 or greater than size 42 were sold.
In simple words: First, arrange all the numbers from smallest to biggest. Then count how many times each size appears. Size 40 appears the most (7 times), so it sold best. Size 32 appears the least (2 times), so it sold worst. If you add up sizes 42 and 44, you get 9 shirts total.

Exam Tip: Always arrange data in order before creating a frequency table. Count tally marks carefully to avoid errors, and double-check your addition when answering sub-questions.

 

Question 2. In a Mathematics test, the following marks were obtained by 40 students: 8, 1, 3, 7, 6, 5, 5, 4, 4, 2, 4, 9, 5, 3, 7, 1, 6, 5, 2, 7, 7, 3, 8, 4, 2, 8, 9, 5, 8, 6, 7, 4, 5, 6, 9, 6, 4, 4, 6, 6. (i) Construct frequency distribution table for the above data. (ii) Find how many students obtained 7 marks or more than 7 marks. (iii) How many students obtained marks below 4?
Answer:
(i) The frequency distribution table for the given data is:

(ii) The number of students who obtained 7 marks or more than 7 marks = 5 + 4 + 3 = 12. Hence, 12 students obtained 7 marks or more than 7 marks.
(iii) The number of students who obtained marks below 4 = 2 + 3 + 3 = 8. Hence, 8 students obtained marks below 4.
In simple words: Count how many students got each mark value. Then add up students who got 7, 8, or 9 to answer part (ii). For part (iii), add students who got 1, 2, or 3.

Exam Tip: Be precise when counting marks and tallying results. When combining frequencies for ranges (e.g., "7 or more"), make sure you include all marks that fit the condition.

 

Question 1. The number of girl students in each class of a co-educational middle school is depicted by the pictograph: Observe this pictograph and answer the following questions: (i) Which class has the minimum number of girl students? (ii) Is the number of girls in Class VI less than the number of girls in Class V? (iii) How many girls are there in Class VII?
Answer: In the given pictograph, one symbol represents 4 girls.

The number of girl students in each class is:

(i) Class VIII has the smallest number of girl students (6 girls). Hence, Class VIII has the minimum number of girl students.
(ii) Number of girls in Class VI = 16 and number of girls in Class V = 10. Since 16 is greater than 10, the number of girls in Class VI is not less than the number of girls in Class V. Hence, No, the number of girls in Class VI is not less than the number of girls in Class V.
(iii) Number of girls in Class VII = 3 × 4 = 12. Hence, there are 12 girls in Class VII.
In simple words: Count the symbols for each class and multiply by 4. Class VIII has the fewest girls (6). Class VI has more girls than Class V. Class VII has 12 girls.

Exam Tip: Always identify what each symbol represents before reading the pictograph. For fractional symbols, multiply correctly to get the total count.

 

Question 2. In a village, the following pictograph shows the number of fruit baskets sold by six merchants in a particular season: Observe the above pictograph and answer the following questions: (i) Which merchant sold the maximum number of baskets? (ii) How many baskets were sold by Rajinder Singh? (iii) The merchants who have sold 700 or more number of baskets are planning to buy a cold store for the next season. Can you name them?
Answer: In the given pictograph, one symbol represents 100 baskets.

The number of fruit baskets sold by each merchant is:

(i) Anwar sold the most baskets (950 baskets). Hence, Anwar sold the maximum number of baskets.
(ii) Rajinder Singh sold 7.5 × 100 = 750 baskets (nearly). Hence, nearly 750 baskets were sold by Rajinder Singh.
(iii) The merchants who sold 700 or more baskets are: Anwar (950 baskets), Rajinder Singh (750 baskets), and Joseph (700 baskets). Hence, Anwar, Rajinder Singh, and Joseph are planning to buy a cold store.
In simple words: Multiply the number of symbols by 100 to get total baskets. Anwar sold the most. Rajinder Singh sold about 750 baskets. Three merchants sold 700 or more: Anwar, Rajinder Singh, and Joseph.

Exam Tip: When reading pictographs, handle half-symbols carefully - a half-symbol is exactly half the value of a full symbol. Always check your arithmetic when identifying merchants who meet a specific condition.

 

Question 3. In Gurugram, the number of cars sold during a particular week was as follows: Monday - 80, Thursday - 60, Tuesday - 70, Friday - 70, Wednesday - 90, Saturday - 40. Prepare a pictograph of the cars sold using a symbol of car representing 10 cars and answer the following questions: (i) On which day the maximum number of cars were sold? (ii) How many pictures of cars will represent the number of cars sold on Thursday?
Answer: Since one symbol of car represents 10 cars, the number of symbols required for each day is:

(i) The maximum number of cars (90) were sold on Wednesday. Hence, the maximum number of cars were sold on Wednesday.
(ii) The number of cars sold on Thursday = 60. Number of pictures of cars = 60 ÷ 10 = 6. Hence, 6 pictures of cars will represent the number of cars sold on Thursday.
In simple words: Divide each day's car count by 10 to get the number of symbols. Wednesday had the most (90 cars, or 9 symbols). Thursday needs 6 symbols because 60 ÷ 10 = 6.

Exam Tip: When creating a pictograph, divide the given values by the symbol value to find how many symbols are needed. Ensure your scale is clear and consistent for all data entries.

 

Question 4. Total number of animals in five villages are as follows: Village A - 80, Village B - 120, Village C - 90, Village D - 40, Village E - 60. Prepare a pictograph of these animals using one symbol ⊗ to represent 10 animals and answer the following questions: (i) How many symbols represent animals of village E? (ii) Which village has the maximum number of animals? (iii) Which village has more animals: village A or village C?
Answer: Since one symbol represents 10 animals, the number of symbols required for each village is:

(i) Number of symbols for village E = 60 ÷ 10 = 6. Hence, 6 symbols represent animals of village E.
(ii) Village B has the maximum number of animals (120). Hence, Village B has the maximum number of animals.
(iii) Village A has 80 animals and Village C has 90 animals. Since 90 is greater than 80, Village C has more animals than Village A. Hence, Village C has more animals than Village A.
In simple words: Divide each village's animal count by 10. Village B has the most animals (120). Village C has more animals than Village A because 90 is bigger than 80.

Exam Tip: Ensure your division is accurate when converting data into symbols. Always compare numbers carefully when answering "which is more" or "which is maximum" questions.

 

Question 1. Observe the bar graph given below which is showing the number of students in a particular class of a school. Answer the following questions: (i) What is the scale of this graph? (ii) How many new students are added every year? (iii) Is the number of students in the year 2015 twice than that of in the year 2012?
Answer: From the bar graph, we observe the number of students each year:

(i) The scale of this graph is 1 unit height = 10 students. Hence, the scale is 1 unit height = 10 students.
(ii) Number of new students added every year:
From 2012 to 2013: 40 - 30 = 10
From 2013 to 2014: 50 - 40 = 10
From 2014 to 2015: 60 - 50 = 10
Hence, 10 new students are added every year.
(iii) Number of students in 2015 = 60 and number of students in 2012 = 30. Twice the number of students in 2012 = 2 × 30 = 60 = Number of students in 2015. Hence, yes, the number of students in 2015 is twice the number of students in 2012.
In simple words: Each unit on the graph stands for 10 students. Every year, 10 more students join. In 2015, there are exactly twice as many students as in 2012.

Exam Tip: Always identify the scale first when reading any graph. Look for patterns in the data (like consistent increases) to answer questions about growth or comparison accurately.

 

Question 2. Observe the bar graph given below which is showing the sale of shirts in a ready-made garment shop from Monday to Saturday.
Answer: This question requires a bar graph to be observed. Without the complete graph details or sub-questions listed in the fragment, provide the graph reading format: Read the bar height for each day and interpret the data to answer what is asked regarding maximum sales day, minimum sales day, or comparisons between specific days. Always check the scale on the vertical axis and align bar tops with gridlines to get accurate values.
In simple words: Look at how tall each bar is. Match the top of each bar to the numbers on the side. The tallest bar shows the day with the most shirt sales.

Exam Tip: When reading bar graphs, use a ruler or straight edge to align bar heights with the vertical scale for accuracy. Never estimate bar values; always read them carefully from the gridlines.

 

Question. Answer the following questions:
(i) What information does the above bar graph give?
(ii) What is the scale chosen on the horizontal line representing number of shirts?
(iii) On which day were the maximum number of shirts sold? How many shirts were sold on that day?
(iv) On which day were the minimum number of shirts sold?
(v) How many shirts were sold on Thursday?
Answer: By looking at the bar graph, we can observe the number of shirts sold on each day. The data collected is:

(i) The bar graph shows how many shirts were sold from Monday to Saturday.
(ii) The scale used on the horizontal line is 1 unit length equals 5 shirts.
(iii) Saturday had the highest sales with 60 shirts sold that day.
(iv) Tuesday had the lowest sales with just 10 shirts sold.
(v) On Thursday, 35 shirts were sold.
In simple words: A bar graph tells you information by showing tall and short bars. Each bar stands for one day. The taller the bar, the more shirts sold that day.

Exam Tip: Always read bar graphs carefully by looking at where each bar reaches on the scale. Match the height of the bar with the numbers on the axis to find the correct value.

 

Question 3. Make a table corresponding to the following graph:
Answer: By examining the bar graph and reading the height of each bar, we can find the number of students for each favourite cricketer. The table below shows this data:

In simple words: Looking at the height of each bar in the graph tells you how many students like each cricketer. Write down the cricketer's name and the number you find, and you have your table.

Exam Tip: When reading bar graphs to make a table, be careful to read the exact height where each bar ends on the scale.

 

Question 4. The following table shows the number of bicycles manufactured in a factory during the years 2011 to 2015. Illustrate this data by using a bar graph. Choose a scale of your choice.

Years: 2011, 2012, 2013, 2014, 2015
Number of bicycles manufactured: 800, 600, 900, 1100, 1200

(i) In which year were the maximum number of bicycles manufactured?
(ii) In which year were the minimum number of bicycles manufactured?
Answer:
Steps:
1. Create two perpendicular lines - OX (horizontal) and OY (vertical) - on graph paper.
2. On the x-axis, place the years 2011, 2012, 2013, 2014 and 2015 at equally spaced intervals.
3. On the y-axis, place the number of bicycles manufactured.
4. Pick a scale: 1 unit height equals 100 bicycles manufactured.
5. Calculate the heights of the bars for each year:

• 2011: 800 ÷ 100 = 8 units
• 2012: 600 ÷ 100 = 6 units
• 2013: 900 ÷ 100 = 9 units
• 2014: 1100 ÷ 100 = 11 units
• 2015: 1200 ÷ 100 = 12 units

6. Draw bars of equal width with equal spacing between them.

(i) The year 2015 saw the highest production at 1200 bicycles manufactured.
(ii) The year 2012 saw the lowest production at 600 bicycles manufactured.
In simple words: To show numbers with a bar graph, divide each number by the scale you pick. This gives you the height to draw. Make all bars the same width and space them evenly.

Exam Tip: Always state the scale clearly and label both axes. Make sure bars have equal width and equal gaps between them for a neat, professional-looking graph.

 

Question 5. The number of Mathematics books sold by a shopkeeper on six consecutive days is given below. Draw a horizontal bar graph to represent the above information choosing the scale of your choice.

Days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday
Number of books sold: 65, 40, 30, 50, 20, 70
Answer:
Steps:
1. Create two perpendicular lines - OX (horizontal) and OY (vertical) - on graph paper.
2. On the y-axis, mark the days of the week (Sunday, Monday, Tuesday, Wednesday, Thursday, Friday) at equally spaced intervals.
3. On the x-axis, mark the number of books sold.
4. Pick a scale: 1 unit length equals 10 books sold.
5. Calculate the lengths of the bars for each day:

• Sunday: 65 ÷ 10 = 6.5 units
• Monday: 40 ÷ 10 = 4 units
• Tuesday: 30 ÷ 10 = 3 units
• Wednesday: 50 ÷ 10 = 5 units
• Thursday: 20 ÷ 10 = 2 units
• Friday: 70 ÷ 10 = 7 units

6. Draw bars of equal width with equal gaps between them.
In simple words: A horizontal bar graph shows bars going left to right instead of up and down. The longer the bar, the greater the value. Keep all bars the same thickness.

Exam Tip: For horizontal bar graphs, put categories on the vertical axis and numbers on the horizontal axis - it is the opposite of an ordinary vertical bar graph.

 

Question 6. The number of persons in various age (in years) groups in a town is given in the following table. Draw a bar graph to represent the above information and answer the following questions (take 1 unit height = 20,000 people):

(i) Which age groups have same population?
(ii) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Answer:
Steps:
1. Create two perpendicular lines - OX (horizontal) and OY (vertical) - on graph paper.
2. On the x-axis, mark the age groups at equally spaced intervals.
3. On the y-axis, mark the number of persons.
4. Apply the scale: 1 unit height equals 20,000 persons.
5. Calculate the heights of the bars for each age group:

• 1-14: 2,00,000 ÷ 20,000 = 10 units
• 15-29: 1,60,000 ÷ 20,000 = 8 units
• 30-44: 1,20,000 ÷ 20,000 = 6 units
• 45-59: 1,20,000 ÷ 20,000 = 6 units
• 60-74: 80,000 ÷ 20,000 = 4 units
• 75 and above: 40,000 ÷ 20,000 = 2 units

6. Draw bars of equal width with equal gaps between them.

(i) The age groups 30-44 and 45-59 both have the same population of 1,20,000 persons each.
(ii) The number of senior citizens (aged 60 and above) = 80,000 + 40,000 = 1,20,000 people in the town.
In simple words: When two bars reach the same height, those groups have the same number of people. To add up groups, just add their numbers together.

Exam Tip: Always read the question carefully to identify what bars you need to compare or add together. Match each bar's height to the scale to find the correct value.

 

Question 7. The number of tigers in India reduced drastically between the years 1900 and 1970, from estimated 1 lakh to less than 2000. Government of India launched Project Tiger in 1973 from the Jim Corbett National Park in Uttarakhand. Now there are 47 such tiger reserves in India. The tiger population has gone up substantially. The following table shows approximate numbers, rounded off to nearest hundreds, as per Tiger census reports. Draw a bar chart for this data. Indicate the number of tigers on top of each bar. Do not draw vertical axis, to have a clean chart.

Year: 2006, 2010, 2014, 2018
Number of Tigers: 1400, 1700, 2200, 3000
Answer:
Steps:
1. Draw a horizontal line OX on graph paper.
2. On OX, mark the years 2006, 2010, 2014, 2018 and 2022 at equally spaced intervals.
3. Draw bars of equal width with heights matching the number of tigers for each year.
4. Write the number of tigers above each bar.
5. Do not draw the vertical axis to create a clean, simple chart.
In simple words: A clean bar chart shows only the bars and numbers, without extra lines or axes that clutter the picture. Put the value on top of each bar so readers do not have to guess.

Exam Tip: Labeling values directly on bars makes a chart easier to read and more professional-looking. This style is especially useful when you want a simple, uncluttered visual presentation.

 

Exercise 15.4

 

Question 1. Find the mean of the following data:
(i) 40, 30, 30, 0, 26, 60
(ii) 3, 5, 7, 9, 11, 13, 15
Answer:
(i) Number of observations = 6

\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{40 + 30 + 30 + 0 + 26 + 60}{6} = \frac{186}{6} = 31 \]

Therefore, the mean of the given data is 31.

(ii) Number of observations = 7

\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{3 + 5 + 7 + 9 + 11 + 13 + 15}{7} = \frac{63}{7} = 9 \]

Therefore, the mean of the given data is 9.
In simple words: To find the mean, add all the numbers together and divide by how many numbers you have. This gives you the average value.

Exam Tip: Always count the total number of observations carefully before dividing. Mistakes in counting lead to wrong answers even if the arithmetic is correct.

 

Question 2. Find the mean of the first five even whole numbers.
Answer: The first five even whole numbers are 0, 2, 4, 6, and 8.

Number of observations = 5

\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{0 + 2 + 4 + 6 + 8}{5} = \frac{20}{5} = 4 \]

Therefore, the mean of the first five even whole numbers is 4.
In simple words: Even whole numbers begin at 0, then go 2, 4, 6, 8, and so on. When you add the first five and divide by 5, you get the average, which is 4.

Exam Tip: Remember that 0 is an even whole number, so it counts as the first one, not the second one.

 

Question 3. A batsman scored the following number of runs in six innings: 36, 35, 50, 46, 60, 55. Calculate the mean runs scored by him in an inning.
Answer: Number of innings = 6

\[ \text{Mean runs} = \frac{\text{Sum of all runs scored}}{\text{Number of innings}} = \frac{36 + 35 + 50 + 46 + 60 + 55}{6} = \frac{282}{6} = 47 \]

Therefore, the batsman scored an average of 47 runs per inning.
In simple words: Add up all the runs from every game. Then divide by the number of games played. The answer tells you how many runs the player got on average in each game.

Exam Tip: When a question asks for the "mean" in a sports context, it is asking for the average score or performance across all the games or events listed.

 

Question 4. The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2540, 2825. Find the mean enrolment of the school for this period.
Answer: There are 6 years in total. To find the mean enrolment, add all the enrolment numbers and divide by the number of years. The sum is 1555 + 1670 + 1750 + 2013 + 2540 + 2825 = 12353. Dividing 12353 by 6 gives approximately 2059.
In simple words: Add up all the numbers, then divide by how many numbers there are. This gives you the average enrolment.

Exam Tip: Always show the formula first, then substitute the values clearly, and round your final answer appropriately - the word "nearly" or the ≈ symbol tells the examiner you understand approximation.

 

Question 5. The marks (out of 100) obtained by a group of students in a science test are: 85, 76, 90, 85, 39, 48, 56, 95, 81, 75. Find the: (i) highest and lowest marks obtained by the students. (ii) mean marks obtained by the students.
Answer:
(i) First, arrange the marks in order from smallest to largest: 39, 48, 56, 75, 76, 81, 85, 85, 90, 95. The highest mark is 95 and the lowest mark is 39.
(ii) There are 10 students. To find the mean, add all marks: 85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75 = 730. Divide by 10 to get 73.
In simple words: Arrange numbers in order to find highest and lowest. Add all marks and divide by the count to get the average.

Exam Tip: Always arrange data first - this helps you spot the extremes quickly and reduces careless errors when adding.

 

Exercise 15.5

 

Question 1. Find the median of the following data: (i) 3, 1, 5, 6, 3, 4, 5 (ii) 3, 1, 5, 6, 3, 4, 5, 6
Answer:
(i) Arrange in ascending order: 1, 3, 3, 4, 5, 5, 6. There are 7 values (odd number). The median is the middle value, which is the 4th term = 4.
(ii) Arrange in ascending order: 1, 3, 3, 4, 5, 5, 6, 6. There are 8 values (even number). The two middle values are the 4th and 5th terms, which are 4 and 5. The median is \( \frac{4 + 5}{2} = \frac{9}{2} = 4.5 \).
In simple words: For an odd count, the median is the middle number. For an even count, find the two middle numbers, add them, and divide by 2.

Exam Tip: Always arrange data before finding the median - missing this step is a common source of errors, especially when the median falls between two values.

 

Question 2. Calculate the mean and the median of the following numbers: 1, 3, 2, 6, 2, 3, 1, 3
Answer: There are 8 observations. Arrange in ascending order: 1, 1, 2, 2, 3, 3, 3, 6. For the mean, sum all values: 1 + 3 + 2 + 6 + 2 + 3 + 1 + 3 = 21. Divide by 8 to get 2.625. For the median with 8 values (even), take the 4th and 5th terms, which are 2 and 3. The median is \( \frac{2 + 3}{2} = \frac{5}{2} = 2.5 \).
In simple words: The mean is the average - add and divide by count. The median is the middle value (or the average of the two middle values if there is an even count).

Exam Tip: Mean and median are different measures - mean uses all values, median only finds the middle position; show both formulas to earn full marks.

 

Question 3. Calculate the mean and the median of the following numbers: 3, 7, 2, 5, 3, 4, 1, 5, 3, 6
Answer: There are 10 observations. To find the mean, sum all values: 3 + 7 + 2 + 5 + 3 + 4 + 1 + 5 + 3 + 6 = 39. Divide by 10 to get 3.9. Arrange in ascending order: 1, 2, 3, 3, 3, 4, 5, 5, 6, 7. For the median with 10 values (even), the two middle terms are the 5th and 6th, which are 3 and 4. The median is \( \frac{3 + 4}{2} = \frac{7}{2} = 3.5 \).
In simple words: Add all numbers and divide by 10 to get the mean. Find the two middle numbers, add them, and divide by 2 to get the median.

Exam Tip: When n is even, the median position falls between two values - always calculate the average of those two, not just pick one.

 

Objective Type Questions - Mental Maths

 

Question 1. Fill in the blanks: (i) A ..... is a collection of numerical figures to give some information. (ii) Each numerical figure in a data is called an ..... (iii) The number of times a particular observation occurs in a data is called ...... of the observation. (iv) A data arranged in ascending or descending order is called an ..... data. (v) A pictorial representation of a data is called a ..... (vi) If ☺ represents 5 students then ☺ ☺ ☺ represent ..... students. (vii) In a pictograph, if one bicycle represents 20 bicycles then 140 bicycles can be represented by ..... bicycles. (viii) In a pictograph, if one electric bulb represents 10 bulbs then the picture of half a bulb will represent ..... bulbs. (ix) In a pictograph, if the symbol □□□□□ represents 50 fruit baskets, then □□□□□ □□□□□ □□ represent ..... fruit baskets. (x) The frequency of 8 is written symbolically as ..... using tally marks. (xi) In a bar graph, the bars are of uniform ..... (xii) The mean of the first 8 natural numbers is .....
Answer:
(i) A **data** is a collection of numerical figures to give some information.
(ii) Each numerical figure in a data is called an **observation**.
(iii) The number of times a particular observation occurs in a data is called **frequency** of the observation.
(iv) A data arranged in ascending or descending order is called an **arrayed** data.
(v) A pictorial representation of a data is called a **pictograph**.
(vi) If ☺ represents 5 students then ☺ ☺ ☺ represent **3 × 5 = 15** students.
(vii) In a pictograph, if one bicycle represents 20 bicycles then 140 bicycles can be represented by **140 ÷ 20 = 7** bicycles.
(viii) In a pictograph, if one electric bulb represents 10 bulbs then the picture of half a bulb will represent **10 ÷ 2 = 5** bulbs.
(ix) Since □□□□□ represents 50 fruit baskets, each □ represents 10 fruit baskets. So □□□□□ □□□□□ □□ represents **12 × 10 = 120** fruit baskets.
(x) The frequency of 8 is written symbolically as **|||| |||** using tally marks.
(xi) In a bar graph, the bars are of uniform **width**.
(xii) The mean of the first 8 natural numbers is: \( \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} = \frac{36}{8} = **4.5** \)
In simple words: Data is numbers that tell a story. Frequency means how often something happens. A pictograph uses symbols where each symbol stands for a certain amount.

Exam Tip: These are key definitions - memorise them exactly as they appear, especially "observation," "frequency," and "arrayed data," as examiners often test these terms directly.

 

Question 2. State whether the following statements are true (T) or false (F): (i) If the data is large, it is difficult to get information from raw data. (ii) Pictographs and bar graphs help us in understanding and analyzing a data. (iii) In a bar graph, bars can be drawn either vertically or horizontally. (iv) In a bar graph, width of a bar has no significance. It is only for eye attraction. (v) Usually, the tally marks are recorded in bunches of 5. (vi) The frequency 9 is represented as |||| |||| | using tally marks. (vii) Mentioning of scale is necessary in pictographs. (viii) Mean is always one of the number in a given data. (ix) Median is always one of the numbers in a given data.
Answer:
(i) **True.** When the data is large, it is difficult to get useful information from raw data without organising it.
(ii) **True.** Pictographs and bar graphs are visual representations that help in understanding and analyzing the data.
(iii) **True.** Bar graphs can be drawn either vertically or horizontally.
(iv) **True.** In a bar graph, only the height (or length) of a bar has significance. The width of a bar is for eye attraction.
(v) **True.** Tally marks are usually recorded in bunches of 5, with the fifth tally mark crossing the previous four.
(vi) **False.** The frequency 9 should be represented as one group of 5 (with the slash) and 4 more single marks, shown as |||| ||||. Each group of 5 has one diagonal line crossing four vertical lines.
(vii) **True.** Mentioning of scale is necessary in pictographs as it indicates what each symbol represents.
(viii) **False.** Mean may or may not be one of the numbers in a given data.
(ix) **False.** Median may or may not be one of the numbers in a given data, especially when the number of observations is even.
In simple words: Raw data is hard to understand without arranging it first. Pictures and graphs make data easier to see. The scale tells you what each symbol means.

Exam Tip: Pay close attention to statements about mean and median - these are often test traps because students mistakenly think they must always be actual data values.

 

Question 3. The minimum number of ice cream cones were sold on: (1) Monday (2) Saturday (3) Tuesday (4) Thursday
Answer: From the given pictograph where 1 symbol = 2 cones, the number of ice cream cones sold on each day is:

The minimum number of cones (7) was sold on Thursday. Hence, option 4 is the correct option.
In simple words: Count the symbols for each day. Multiply by 2 to find the number of cones. The day with the fewest symbols (or fewest cones) is the answer.

Exam Tip: Always create a table to convert pictograph symbols to actual values - this step prevents misreading and makes your working clear to the examiner.

 

Question 4. The maximum number of ice cream cones were sold on: (1) Tuesday (2) Friday (3) Wednesday (4) Thursday
Answer: From the pictograph, the maximum number of cones (16) was sold on Tuesday. Hence, option 1 is the correct option.
In simple words: Look at the table above. Find the day with the largest number of cones.

Exam Tip: Refer to your table from the previous question - this saves time and keeps your working consistent across related MCQs.

 

Question 5. Ratio of the number of ice cream cones sold on Saturday to the number of ice cream cones sold on Wednesday is: (1) 3 : 2 (2) 2 : 3 (3) 4 : 5 (4) 4 : 7
Answer: Number of cones sold on Saturday = 8. Number of cones sold on Wednesday = 12. The required ratio is 8 : 12 = \( \frac{8}{12} = \frac{2}{3} \) = 2 : 3. Hence, option 2 is the correct option.
In simple words: Write the two numbers as a fraction. Simplify by dividing both by the same number (here, divide by 4).

Exam Tip: Always simplify ratios to their lowest terms - examiners expect the answer in simplest form, and option 2 will only match if you reduce correctly.

 

Question 6. Total number of ice cream cones sold during the whole week was: (1) 33 (2) 67 (3) 65 (4) 57
Answer: Total number of ice cream cones sold during the week = 10 + 16 + 12 + 7 + 14 + 8 = 67. Hence, option 2 is the correct option.
In simple words: Add up all the daily cone sales from your table to find the total for the entire week.

Exam Tip: Double-check your addition by adding in a different order (e.g. work from the bottom up instead of top down) - this catches arithmetic slips before you commit to an answer.

 

Question 7. (Question text not provided in source - reconstructed from answer) Ratio of the number of ice cream cones sold on Saturday to the number of ice cream cones sold on Wednesday is:
Answer: This question is a repeat of Question 5. The required ratio = 8 : 12 = \( \frac{8}{12} = \frac{2}{3} \) = 2 : 3.
In simple words: Divide both numbers by their greatest common factor to reduce the ratio to simplest form.

Exam Tip: If a question repeats, check if the values have changed - sometimes test papers include intentional repeats to test consistency.

 

Question. If the cost of one ice cream cone is Rs. 40, then the sale value on Thursday was:
(1) Rs. 140
(2) Rs. 200
(3) Rs. 280
(4) Rs. 2680
Answer: (3) Rs. 280
In simple words: Thursday had 7 cones sold. Each cone costs Rs. 40. So 7 times 40 equals Rs. 280.

Exam Tip: Always multiply the number of items sold by the cost per item. Check that your final answer matches one of the four options.

 

Question 8. Which country played maximum number of matches?
(1) India
(2) England
(3) Pakistan
(4) Australia
Answer: (4) Australia
In simple words: Look at the bar graph. Australia's bar is the tallest, showing it played 32 matches. No other country's bar reaches as high.

Exam Tip: Always identify the longest bar in a bar graph to find the maximum value. Read the height using the scale provided.

 

Question 9. How many matches did South Africa play?
(1) 16
(2) 18
(3) 20
(4) 24
Answer: (2) 18
In simple words: Find South Africa's bar on the graph. Read the height where it ends using the scale shown (1 unit = 4 matches). South Africa played 18 matches.

Exam Tip: Carefully align the top of each bar with the numbers on the vertical axis. Use the scale to convert bar height into actual values.

 

Question 10. How many more matches were played by India than Pakistan?
(1) 6
(2) 12
(3) 24
(4) 30
Answer: (1) 6
In simple words: India played 30 matches. Pakistan played 24 matches. The difference is 30 - 24 = 6 matches.

Exam Tip: For "how many more" questions, always subtract the smaller number from the larger number. Double-check by reading both bar heights from the graph.

 

Question 11. Ratio of the number of matches played by India to the number of matches played by Sri Lanka is
(1) 4 : 5
(2) 5 : 4
(3) 4 : 3
(4) 7 : 6
Answer: (2) 5 : 4
In simple words: India played 30 matches and Sri Lanka played 24 matches. The ratio is 30 : 24. Divide both by 6 to get 5 : 4.

Exam Tip: Always reduce ratios to their simplest form by finding the greatest common factor of both numbers. Check your answer by multiplying back to verify.

 

Question 12. The mean of the first 6 odd natural numbers is
(1) 5
(2) 5.5
(3) 6
(4) 6.5
Answer: (3) 6
In simple words: The first 6 odd numbers are 1, 3, 5, 7, 9, and 11. Add them up to get 36. Divide 36 by 6 to get the mean of 6.

Exam Tip: Mean is always the sum divided by how many numbers there are. List all the numbers first to avoid missing any or counting twice.

 

Question 13. The median of the numbers 4, 4, 7, 5, 7, 6, 7, 3, 11 is
(1) 7
(2) 6
(3) 5
(4) 4
Answer: (2) 6
In simple words: Arrange all the numbers from smallest to largest: 3, 4, 4, 5, 6, 7, 7, 7, 11. There are 9 numbers. The middle one is in position 5, which is 6.

Exam Tip: Always sort data in order before finding the median. For an odd count of numbers, the median is the middle value. For an even count, take the average of the two middle values.

 

Question 14. Statement I: In a bar graph, the breadth of the bar graph has no significance. Statement II: The number of times a particular entry occurs in a given data is called its frequency.
(1) Statement I is true but statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (3) Both Statement I and Statement II are true.
In simple words: The width of a bar is just for looks, not important. The height is what matters. Frequency is how often something appears in your data set.

Exam Tip: Both statements cover core definitions in data handling. Statement I relates to bar graph design, Statement II to frequency counting - know the difference between what is shown and how often it shows up.

 

Question 15. Statement I: The mean of 10, 20, 30, 40 is 20. Statement II: Mean = (Sum of items)/(Number of items)
(1) Statement I is true but statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (2) Statement I is false but Statement II is true.
In simple words: Add 10 + 20 + 30 + 40 = 100. Divide by 4 to get 25, not 20. But the formula for mean shown in Statement II is correct.

Exam Tip: Always work through the calculation yourself rather than trusting a given statement. Check the formula separately from the numerical result.

 

Question 16. Statement I: 5 army aspirants have heights of 160 cm, 163 cm, 167 cm, 150 cm and 170 cm. The aspirant with the maximum height is 20 cm taller than the shortest aspirant. Statement II: The median of the heights is 167 cm.
(1) Statement I is true but statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (1) Statement I is true but statement II is false.
In simple words: The tallest is 170 cm and shortest is 150 cm. The difference is 20 cm, so Statement I is correct. When arranged in order: 150, 160, 163, 167, 170, the middle value is 163, not 167.

Exam Tip: For median, always arrange values in order and pick the middle one. Double-check the difference calculation for Statement I by subtraction.

 

Question 17. Statement I: Writing entries in ascending or descending order does not change the value of median. Statement II: When the number of observations of given numerical data is even, then the median is the average of the two middle terms.
(1) Statement I is true but statement II is false.
(2) Statement I is false but Statement II is true.
(3) Both Statement I and Statement II are true.
(4) Both Statement I and Statement II are false.
Answer: (3) Both Statement I and Statement II are true.
In simple words: The median stays the same no matter which direction you sort the data - ascending or descending gives the same middle value. If you have an even number of values, the median is halfway between the two middle ones.

Exam Tip: These are fundamental median rules. Statement I shows median is order-independent in value. Statement II describes the even-count case - always average the two middle values when data count is even.

 

Question 1. A die is thrown 25 times and the numbers appearing were as given below: 2, 1, 4, 6, 2, 3, 1, 5, 6, 3, 4, 5, 2, 1, 6, 6, 6, 3, 2, 2, 2, 4, 3, 2, 2 (a) Construct data array. (b) Construct tally chart and frequency distribution table.
Answer:
(a) Arranging the given data in ascending order to form a data array:
1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 6, 6

(b) The tally chart and frequency distribution table for the given data:

In simple words: A data array means listing all numbers from smallest to largest. A tally chart uses marks to count how many times each number shows up. Frequency is just that count.

Exam Tip: When making a tally chart, use groups of 5 marks (four vertical lines and one diagonal) for easy counting. Check that all frequencies add up to the total number of observations (25).

 

Question 2. In a primary school, the number of students in different classes are as follows: Class I - 200, Class II - 180, Class III - 140, Class IV - 160, Class V - 120. Represent this data by a pictograph, using 1 symbol = 20 students
Answer: Since 1 symbol stands for 20 students, the number of symbols needed for each class is:

The pictograph for the given information shows Class I with 10 symbols, Class II with 9 symbols, Class III with 7 symbols, Class IV with 8 symbols, and Class V with 6 symbols, where each symbol represents 20 students.
In simple words: Divide each class's student count by 20 to find how many symbols to draw. Class I has 200 ÷ 20 = 10 symbols, and so on.

Exam Tip: Always state the symbol key clearly (what each symbol stands for). Use whole symbols only - if a number doesn't divide evenly, you may draw a partial symbol and explain it in a note.

 

Question 3. Observe the following bar graph, showing the marks scored by Avneet in the annual examination in different subjects: (i) What is the scale of this bar graph? (ii) In which subject Avneet obtained maximum marks? (iii) In which subject she obtained minimum marks? (iv) Name the subject(s) in which she got 80 or more marks.
Answer: From the given bar graph, the marks obtained by Avneet in each subject are:

(i) The scale of this bar graph is 1 unit length = 10 marks obtained.
(ii) Avneet obtained maximum marks (100) in Mathematics.
(iii) Avneet obtained minimum marks (50) in Hindi.
(iv) Avneet got 80 or more marks in Science (80) and Mathematics (100).
In simple words: The scale shows how many marks each unit of length stands for. The tallest bar is the highest marks. Read each bar's height carefully against the scale to find exact values.

Exam Tip: Always state the scale first when analyzing a bar graph. Use the scale consistently to read all bar heights. For "80 or more," list every subject meeting that condition.

 

Question. The following table shows the monthly expenditure of a family on various items. Represent the data by a bar graph.

Items	Expenditure (in Rs.)
Rent	4000
Food	6500
Education	3000
Transport	1500
Miscellaneous	5000

Answer: To draw the bar graph, follow these steps:

1. Draw two lines that cross each other at right angles - OX (horizontal) and OY (vertical) - on graph paper.

2. Mark the different spending categories (Rent, Food, Education, Transport, Miscellaneous) evenly spaced along the x-axis.

3. On the y-axis, mark the expenditure amounts in Rs.

4. Select a scale where 1 unit height equals Rs 1000.

5. The heights of the bars for each item are:

• Rent - 4000 ÷ 1000 = 4 units
• Food - 6500 ÷ 1000 = 6.5 units
• Education - 3000 ÷ 1000 = 3 units
• Transport - 1500 ÷ 1000 = 1.5 units
• Miscellaneous - 5000 ÷ 1000 = 5 units

6. Draw bars of the same width with equal spaces between them.

The completed bar graph displays the five spending categories on the horizontal axis with their corresponding heights based on the scale chosen.

In simple words: First, draw two lines at right angles. Write each spending type along the bottom line and money amounts along the side line. Then draw a bar for each type with height matching its spending amount using the scale - taller bars show bigger spending.

Exam Tip: Always label both axes clearly, choose a simple scale that fits all values, and make sure all bars have the same width with even spacing between them - these are the key marks that examiners check.

 

Question 5. Find the mean and the median of the following data: 5, 3, 12, 0, 7, 11, 4, 3, 9
Answer: The data has 9 observations in total.

To find the mean, add all the numbers together and divide by how many numbers there are:

\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} = \frac{5 + 3 + 12 + 0 + 7 + 11 + 4 + 3 + 9}{9} = \frac{54}{9} = 6 \]

To find the median, arrange the numbers from smallest to largest:

0, 3, 3, 4, 5, 7, 9, 11, 12

Since there are 9 observations (an odd number), the median is the middle value, which is the 5th number in the ordered list. The 5th number is 5, so the median is 5.

Therefore, the mean of the given data is 6 and the median is 5.
In simple words: Mean is the average - add all numbers and divide by how many you have. Median is the middle number when you line them up from smallest to biggest.

Exam Tip: Always arrange data in order before finding the median. For an odd count of numbers, the median is the center one; for even count, take the average of the two middle numbers.