Data Handling

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:

Size of shirts	Tally marks	Number of shirts (Frequency)
32	||	2
34	|||	3
36	||||	4
38	||||	5
40	|||| ||	7
42	|||| |	6
44	|||	3

(i) The shirt size 40 had the maximum sale as it has the highest frequency (7).

Hence, shirt size 40 had the maximum sale.

(ii) The shirt size 32 had the minimum sale as it has the lowest frequency (2).

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.

(i) The frequency distribution table for the given data is:

Marks obtained	Tally marks	Number of students (Frequency)
1	||	2
2	|||	3
3	|||	3
4	|||| ||	7
5	|||| |	6
6	|||| ||	7
7	||||	5
8	||||	4
9	|||	3

(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 the given pictograph, one symbol represents 4 girls.

The number of girl students in each class is:

Class	Number of symbols	Number of girls
I	6	24
II	4.5	18
III	4	16
IV	3.5	14
V	2.5	10
VI	4	16
VII	3	12
VIII	1.5	6

(i) Class VIII has the minimum 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 > 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 the given pictograph, one symbol represents 100 baskets.

The number of fruit baskets sold by each merchant is:

Name of merchant	Number of symbols	Number of baskets sold
Roshan Lal	5	500
Anwar	9.5	950
Rajinder Singh	7.5	750
Vineet	6	600
Sunita	4.5	450
Joseph	7	700

(i) Anwar sold the maximum number of baskets (950 baskets).

Hence, Anwar sold the maximum number of baskets.

(ii) Rajinder Singh sold 7 $\frac{1}{2}$ × 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)
• Joseph (700 baskets)

Hence, Anwar, Rajinder Singh and Joseph are planning to buy a cold store.

Since one symbol of car represents 10 cars, the number of symbols required for each day is:

Day	Number of cars sold	Number of symbols
Monday	80	8
Tuesday	70	7
Wednesday	90	9
Thursday	60	6
Friday	70	7
Saturday	40	4

The pictograph for the given information 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 = $\frac{60}{10}$ = 6.

Hence, 6 pictures of cars will represent the number of cars sold on Thursday.

Since one symbol represents 10 animals, the number of symbols required for each village is:

Village	Number of animals	Number of symbols
A	80	8
B	120	12
C	90	9
D	40	4
E	60	6

The pictograph for the given information is:

(i) Number of symbols for village E = $\frac{60}{10}$ = 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 > 80, Village C has more animals than Village A.

Hence, Village C has more animals than Village A.

From the bar graph, we observe the number of students each year:

Year	Number of students
2012	30
2013	40
2014	50
2015	60

(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.

From the bar graph, we observe the number of shirts sold each day:

Day	Number of shirts sold
Monday	15
Tuesday	10
Wednesday	20
Thursday	35
Friday	50
Saturday	60

(i) The bar graph shows the number of shirts sold from Monday to Saturday.

(ii) The scale chosen on the horizontal line is 1 unit length = 5 shirts.

(iii) The maximum number of shirts was sold on Saturday, and 60 shirts were sold on that day.

Hence, the maximum number of shirts (60) was sold on Saturday.

(iv) The minimum number of shirts was sold on Tuesday (10 shirts).

Hence, the minimum number of shirts was sold on Tuesday.

(v) 35 shirts were sold on Thursday.

From the bar graph, reading the height of each bar, we get the number of students for each favourite cricketer.

The table corresponding to the given bar graph is:

Cricketer	Rohit	Dhoni	Virat	Ashwin
Number of students	10	16	20	7

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the x-axis, mark the years 2011, 2012, 2013, 2014 and 2015 at equal distances.

3. Along the y-axis, mark the number of bicycles manufactured.

4. Choose a scale of 1 unit height = 100 bicycles manufactured.

5. The heights of the bars for various years are:

• 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 keeping equal gap between adjoining bars.

The required bar graph is as given below:

(i) The maximum number of bicycles (1200) were manufactured in the year 2015.

Hence, the maximum number of bicycles were manufactured in the year 2015.

(ii) The minimum number of bicycles (600) were manufactured in the year 2012.

Hence, the minimum number of bicycles were manufactured in the year 2012.

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the y-axis, mark the days of the week (Sunday, Monday, Tuesday, Wednesday, Thursday, Friday) at equal distances.

3. Along the x-axis, mark the number of books sold.

4. Choose a scale of 1 unit length = 10 books sold.

5. The lengths of the bars for various days are:

• 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 keeping equal gap between adjoining bars.

The required horizontal bar graph is as given below:

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the x-axis, mark the age groups at equal distances.

3. Along the y-axis, mark the number of persons.

4. Take scale: 1 unit height = 20,000 persons.

5. The heights of the bars for various age groups are:

• 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 keeping equal gap between adjoining bars.

The required bar graph is as given below:

(i) The age groups 30-44 and 45-59 have the same population (1,20,000 each).

Hence, age groups 30-44 and 45-59 have the same population.

(ii) Number of senior citizens (60 and above) = 80,000 + 40,000 = 1,20,000.

Hence, there are 1,20,000 senior citizens in the town.

Steps:

1. Draw a horizontal line OX on a graph paper.

2. Along OX, mark the years 2006, 2010, 2014, 2018 and 2022 at equal distances.

3. Draw bars of equal width with heights proportional to the number of tigers for each year.

4. Write the number of tigers on top of each bar.

5. Do not draw the vertical axis to have a clean chart.

The required bar chart is as given below:

(i) Number of observations = 6.

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \frac{40 + 30 + 30 + 0 + 26 + 60}{6} \\[1em] = \frac{186}{6} \\[1em] = 31\end{aligned}$$

Hence, the mean of the given data is 31.

(ii) Number of observations = 7.

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \frac{3 + 5 + 7 + 9 + 11 + 13 + 15}{7} \\[1em] = \frac{63}{7} \\[1em] = 9\end{aligned}$$

Hence, the mean of the given data is 9.

The first five even whole numbers are 0, 2, 4, 6 and 8.

Number of observations = 5.

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \frac{0 + 2 + 4 + 6 + 8}{5} \\[1em] = \frac{20}{5} \\[1em] = 4\end{aligned}$$

Hence, the mean of the first five even whole numbers is 4.

Number of innings = 6.

$$\begin{aligned}\text{Mean runs} = \frac{\text{Sum of all runs scored}}{\text{Number of innings}} \\[1em] = \frac{36 + 35 + 50 + 46 + 60 + 55}{6} \\[1em] = \frac{282}{6} \\[1em] = 47\end{aligned}$$

Hence, the mean runs scored by the batsman in an inning is 47.

Number of years = 6.

$$\begin{aligned}\text{Mean enrolment} = \frac{\text{Sum of enrolments}}{\text{Number of years}} \\[1em] = \frac{1555 + 1670 + 1750 + 2013 + 2540 + 2825}{6} \\[1em] = \frac{12353}{6} \\[1em] \approx 2059\end{aligned}$$

Hence, the mean enrolment of the school for this period is 2059 (nearly).

Arranging the given marks in ascending order, we get:

39, 48, 56, 75, 76, 81, 85, 85, 90, 95

(i) Highest marks = 95 and lowest marks = 39.

Hence, the highest marks obtained by the students is 95 and the lowest marks is 39.

(ii) Number of students = 10.

$$\begin{aligned}\text{Mean marks} = \frac{\text{Sum of all marks}}{\text{Number of students}} \\[1em] = \frac{85 + 76 + 90 + 85 + 39 + 48 + 56 + 95 + 81 + 75}{10} \\[1em] = \frac{730}{10} \\[1em] = 73\end{aligned}$$

Hence, the mean marks obtained by the students is 73.

(i) Arranging the given data in ascending order, we get:

1, 3, 3, 4, 5, 5, 6

Total number of observations = 7 (odd).

The middle observation is the 4^th observation, which is 4.

Hence, the median of the given data is 4.

(ii) Arranging the given data in ascending order, we get:

1, 3, 3, 4, 5, 5, 6, 6

Total number of observations = 8 (even).

There are two middle items: 4^th term and 5^th term, which are 4 and 5.

$$\begin{aligned}\text{Median} = \frac{\text{Sum of middle terms}}{2} \\[1em] = \frac{4 + 5}{2} \\[1em] = \frac{9}{2} \\[1em] = 4.5\end{aligned}$$

Hence, the median of the given data is 4.5.

Number of observations = 8.

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \frac{1 + 3 + 2 + 6 + 2 + 3 + 1 + 3}{8} \\[1em] = \frac{21}{8} \\[1em] = 2.625\end{aligned}$$

Arranging the given data in ascending order, we get:

1, 1, 2, 2, 3, 3, 3, 6

Total number of observations = 8 (even).

There are two middle items: 4^th term and 5^th term, which are 2 and 3.

$$\begin{aligned}\text{Median} = \frac{\text{Sum of middle terms}}{2} \\[1em] = \frac{2 + 3}{2} \\[1em] = \frac{5}{2} \\[1em] = 2.5\end{aligned}$$

Hence, the mean of the given data is 2.625 and the median is 2.5.

Number of observations = 10.

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \frac{3 + 7 + 2 + 5 + 3 + 4 + 1 + 5 + 3 + 6}{10} \\[1em] = \frac{39}{10} \\[1em] = 3.9\end{aligned}$$

Arranging the given data in ascending order, we get:

1, 2, 3, 3, 3, 4, 5, 5, 6, 7

Total number of observations = 10 (even).

There are two middle items: 5^th term and 6^th term, which are 3 and 4.

$$\begin{aligned}\text{Median} = \frac{\text{Sum of middle terms}}{2} \\[1em] = \frac{3 + 4}{2} \\[1em] = \frac{7}{2} \\[1em] = 3.5\end{aligned}$$

Hence, the mean of the given data is 3.9 and the median is 3.5.

(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 $\frac{140}{20}$ = 7 bicycles.

(viii) In a pictograph, if one electric bulb represents 10 bulbs then the picture of half a bulb will represent $\frac{10}{2}$ = 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:

$$\begin{aligned}= \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8}{8} \\[1em] = \frac{36}{8} \\[1em] = \mathbf{4.5}\end{aligned}$$

(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 ||||‖|||| (5 + 4 = 9), but in tally marks the correct representation is one group of 5 (with the slash) and 4 more single marks.

(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.

From the given pictograph (1 symbol = 2 cones), the number of ice cream cones sold on each day is:

Day	Number of symbols	Number of cones
Monday	5	10
Tuesday	8	16
Wednesday	6	12
Thursday	3.5	7
Friday	7	14
Saturday	4	8

The minimum number of cones (7) was sold on Thursday.

Hence, option 4 is the correct option.

From the pictograph, the maximum number of cones (16) was sold on Tuesday.

Hence, option 1 is the correct option.

Number of cones sold on Saturday = 8.

Number of cones sold on Wednesday = 12.

$$\begin{aligned}\text{Required ratio} = 8 : 12 \\[1em] = \frac{8}{12} \\[1em] = \frac{2}{3} \\[1em] = 2 : 3\end{aligned}$$

Hence, option 2 is the correct option.

Total number of ice cream cones sold during the week

= 10 + 16 + 12 + 7 + 14 + 8

= 67

Hence, option 2 is the correct option.

Number of cones sold on Thursday = 7.

Cost of one ice cream cone = ₹ 40.

Sale value on Thursday = 7 × ₹ 40 = ₹ 280.

Hence, option 3 is the correct option.

Observe the following bar graph, showing the number of one-day international matches played by cricket teams of different countries. Choose the correct answer from the given four options for questions 8 to 11:

From the bar graph, the number of matches played by each country (scale: 1 unit = 4 matches) is:

Country	Number of matches
India	30
Pakistan	24
West Indies	20
England	28
South Africa	18
Australia	32
Sri Lanka	24

Australia played the maximum number of matches (32).

Hence, option 4 is the correct option.

From the bar graph, South Africa played 18 matches.

Hence, option 2 is the correct option.

Number of matches played by India = 30.

Number of matches played by Pakistan = 24.

Difference = 30 − 24 = 6 matches.

Hence, option 1 is the correct option.

Number of matches played by India = 30.

Number of matches played by Sri Lanka = 24.

$$\begin{aligned}\text{Required ratio} = 30 : 24 \\[1em] = \frac{30}{24} \\[1em] = \frac{5}{4} \\[1em] = 5 : 4\end{aligned}$$

Hence, option 2 is the correct option.

The first 6 odd natural numbers are 1, 3, 5, 7, 9 and 11.

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \frac{1 + 3 + 5 + 7 + 9 + 11}{6} \\[1em] = \frac{36}{6} \\[1em] = 6\end{aligned}$$

Hence, option 3 is the correct option.

Arranging the given data in ascending order, we get:

3, 4, 4, 5, 6, 7, 7, 7, 11

Total number of observations = 9 (odd).

The middle observation is the 5^th observation, which is 6.

Hence, option 2 is the correct option.

In a bar graph, the breadth of the bars has no significance. It is only for visual representation. So, Statement I is true.

The number of times a particular entry occurs in a given data is called its frequency. So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

$$\begin{aligned}\text{Mean of 10, 20, 30, 40} = \frac{10 + 20 + 30 + 40}{4} \\[1em] = \frac{100}{4} \\[1em] = 25\end{aligned}$$

So, the mean is 25, not 20. Statement I is false.

Mean = $\frac{\text{Sum of items}}{\text{Number of items}}$ — this is the correct formula for mean. So, Statement II is true.

∴ Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

Maximum height = 170 cm and minimum height = 150 cm.

Difference = 170 − 150 = 20 cm.

So, the aspirant with the maximum height is 20 cm taller than the shortest. Statement I is true.

Arranging the heights in ascending order: 150, 160, 163, 167, 170.

Total number of observations = 5 (odd).

The middle observation is the 3^rd observation, which is 163 cm.

So, the median is 163 cm, not 167 cm. Statement II is false.

∴ Statement I is true but Statement II is false.

Hence, option 1 is the correct option.

Writing entries in ascending or descending order does not change the value of median. Whichever order we use, the median value remains the same. So, Statement I is true.

When the number of observations is even, the median is the average (mean) of the two middle terms. So, Statement II is true.

∴ Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

(a) Arranging the given data in ascending order, we get the 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 is:

Number on the die	Tally marks	Frequency
1	|||	3
2	|||| |||	8
3	||||	4
4	|||	3
5	||	2
6	||||	5

Since 1 symbol represents 20 students, the number of symbols required for each class is:

Class	Number of students	Number of symbols
I	200	10
II	180	9
III	140	7
IV	160	8
V	120	6

The pictograph for the given information is:

From the given bar graph, the marks obtained by Avneet in each subject are:

Subject	Marks
Hindi	50
English	60
Mathematics	100
Science	80
S. Studies	70

(i) The scale of this bar graph is 1 unit length = 10 marks obtained.

(ii) Avneet obtained maximum marks (100) in Mathematics.

Hence, Avneet obtained maximum marks in Mathematics.

(iii) Avneet obtained minimum marks (50) in Hindi.

Hence, Avneet obtained minimum marks in Hindi.

(iv) Avneet got 80 or more marks in Science (80) and Mathematics (100).

Hence, Avneet got 80 or more marks in Science and Mathematics.

Steps:

1. Draw two mutually perpendicular lines OX (horizontal) and OY (vertical) on a graph paper.

2. Along the x-axis, mark the items (Rent, Food, Education, Transport, Miscellaneous) at equal distances.

3. Along the y-axis, mark the expenditure in ₹.

4. Choose a scale of 1 unit height = ₹ 1000.

5. The heights of the bars for various items 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 equal width keeping equal gap between adjoining bars.

The required bar graph is as given below:

Number of observations = 9.

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} \\[1em] = \frac{5 + 3 + 12 + 0 + 7 + 11 + 4 + 3 + 9}{9} \\[1em] = \frac{54}{9} \\[1em] = 6\end{aligned}$$

Arranging the given data in ascending order, we get:

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

Total number of observations = 9 (odd).

The middle observation is the 5^th observation, which is 5.

So, Median = 5.

Hence, the mean of the given data is 6 and the median is 5.