Class 6 Maths - Data Handling NCERT Solutions

Chapter 9: Data Handling (NCERT Solutions)

Exercise 9.1 (Recording & Organising Data)

Q1. 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, 3, 4, 7, 6, 5, 7.
(i) Find how many students obtained marks equal to or more than 7.
(ii) How many students obtained marks below 4?

(i) Marks ≥ 7: Count of 7s = 7, 8s = 4, 9s = 2. Total = 13 students.

(ii) Marks < 4 (i.e., 1, 2, 3): 1s = 2, 2s = 3, 3s = 4. Total = 9 students.

Q2. Following is the choice of sweets of 30 students of Class VI:
Ladoo, Barfi, Ladoo, Jalebi, Ladoo, Rasgulla, Jalebi, Ladoo, Barfi, Rasgulla, Ladoo, Jalebi, Jalebi, Rasgulla, Ladoo, Rasgulla, Jalebi, Ladoo, Rasgulla, Ladoo, Barfi, Rasgulla, Ladoo, Rasgulla, Jalebi, Barfi, Rasgulla, Ladoo, Rasgulla, Barfi.
(i) Make a table using tally marks. (ii) Which sweet is preferred by most students?

Sweet	Tally Marks	Number
Ladoo	\|\|\|\| \|\|\|\| I	11
Barfi	\|\|\|\|	5
Jalebi	\|\|\|\| I	6
Rasgulla	\|\|\|\| III	8

(ii) Ladoo is preferred by most students (11 out of 30).

Q3. A group of students was asked to say which animal they would like most. Results were: Dog, Cat, Cat, Fish, Cat, Rabbit, Dog, Cat, Rabbit, Dog, Cat, Dog, Dog, Dog, Cat, Cow, Fish, Cat, Cat, Fish, Dog, Cat, Cat, Cat, Cow, Fish, Dog, Cow, Cat, Dog, Rabbit, Fish, Cow.
Make a frequency distribution table.

Animal	Frequency
Dog	10
Cat	12
Fish	5
Rabbit	3
Cow	4

Total = 34. Most preferred: Cat (12).

Exercise 9.2 (Pictograph)

Q1. Total number of students of a school in different years is shown in the table. Draw a pictograph.

Year	Number of Students	Picture (👤 = 100 students)
2003	480	👤👤👤👤 (4 full + half)
2004	530	👤👤👤👤👤 (5 full + part)
2005	600	👤👤👤👤👤👤
2006	650	👤👤👤👤👤👤 (6 full + half)

In a real pictograph, use a key such as: 1 symbol = 100 students. Draw half symbol for 50 students.

Q3. The number of animals in five villages are as follows. Prepare a pictograph using one symbol to represent 10 animals.

Village	Animals	Symbols (🇄 = 10 animals)
Ranpur	80	8 symbols
Dedpur	45	4 and half symbols
Balpan	60	6 symbols
Nanpur	75	7 and half symbols
Jeepur	100	10 symbols

Exercise 9.3 (Bar Graph)

Q1. A survey of 120 school students was done to find which activity they prefer to do in their free time. Draw a bar graph.

Activity	Number of Students
Playing	45
Reading	20
TV	40
Painting	15

In a bar graph: x-axis = Activity, y-axis = Number of students (scale: 1 unit = 5 students). Draw bars of equal width proportional to frequency.

Most preferred activity: Playing (45 students).

Q4. Number of persons in various age groups in a town are given below. Draw a bar graph and answer:
(a) Which group has maximum number of persons? (b) All persons older than 40 years = ?

Age Group	Number of Persons
1—14	2 lakhs
15—29	1 lakh 60 thousand
30—44	1 lakh 20 thousand
45—59	1 lakh 20 thousand
60—74	80 thousand
75 and above	40 thousand

(a) Maximum persons in group 1—14 years (2 lakhs).

(b) Persons older than 40 years: groups 45—59 + 60—74 + 75 and above = 1,20,000 + 80,000 + 40,000 = 2,40,000 persons.

Exercise 9.4 (Average / Mean)

Q1. 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.

Mean = Sum of observations / Number of observations.
Sum = 36 + 35 + 50 + 46 + 60 + 55 = 282.
Mean = 282 / 6 = 47 runs.

Q2. The weights of five students (in kg) are 54, 51, 63, 48, 59. What is the mean weight?

Sum = 54 + 51 + 63 + 48 + 59 = 275 kg.
Mean = 275 / 5 = 55 kg.

Q5. The number of children in six different classes are 35, 45, 50, 40, 15, 20 respectively. Calculate the mean number of children per class.

Sum = 35 + 45 + 50 + 40 + 15 + 20 = 205.
Mean = 205 / 6 = 34.16 ≈ 34.2 children per class.

Q6. A cricketer's run scores in 5 test matches were 45, 44, 91, 45, 25. Find the mean.

Sum = 45 + 44 + 91 + 45 + 25 = 250.
Mean = 250 / 5 = 50 runs.

Class 6 Maths - Data Handling Practice Questions

Chapter 9: Data Handling (Practice Questions)

RD Sharma / HOT Practice

Q1. The marks obtained by 10 students in a test are: 72, 68, 55, 90, 44, 82, 55, 72, 90, 72. Find the mean, and identify the mode.

Sum = 72+68+55+90+44+82+55+72+90+72 = 700.
Mean = 700 / 10 = 70.
Mode = 72 (appears 3 times — most frequent).

Q2. The daily temperatures in a city for a week were: 28, 30, 27, 32, 29, 31, 28 (in °C). Find the mean temperature.

Sum = 28+30+27+32+29+31+28 = 205.
Mean = 205 / 7 ≈ 29.3°C.

Q3. Find the mean of the first 10 natural numbers.

Sum of first 10 natural numbers = n(n+1)/2 = 10 × 11 / 2 = 55.
Mean = 55 / 10 = 5.5.

Q4. The mean of 5 numbers is 40. If one number is excluded, the mean becomes 37. Find the excluded number.

Sum of 5 numbers = 5 × 40 = 200.
Sum of remaining 4 numbers = 4 × 37 = 148.
Excluded number = 200 − 148 = 52.

Q5. The runs scored by a cricket team in 8 matches are: 120, 150, 80, 200, 175, 95, 140, 160. Find the mean.

Sum = 120+150+80+200+175+95+140+160 = 1120.
Mean = 1120 / 8 = 140 runs.

Q6. From a bar graph, the number of books sold in a shop per month from Jan to Jun were: 120, 180, 150, 90, 200, 160. In which month were maximum books sold? What is the mean?

Maximum books sold in May (200).
Sum = 120+180+150+90+200+160 = 900.
Mean = 900 / 6 = 150 books per month.

Q7. In a pictograph, 1 picture = 500 people. A city had 7 pictures, a town had 3.5 pictures, and a village had 2 pictures. How many people does each place have?

City = 7 × 500 = 3500.
Town = 3.5 × 500 = 1750.
Village = 2 × 500 = 1000.

Q8. The following are scores of students: 4, 7, 2, 3, 9, 5, 6, 8, 4, 5, 7, 5. Find: (a) the highest and lowest score (b) range (c) mean.

(a) Highest = 9; Lowest = 2.

(b) Range = 9 − 2 = 7.

Q9. The ages of 8 students are 12, 13, 11, 13, 12, 14, 11, 12. Find the mean age and the mode.

Sum = 12+13+11+13+12+14+11+12 = 98. Mean = 98/8 = 12.25 years.
Mode = 12 (appears 3 times).

Q10. Which type of graph (pictograph, bar graph, or tally chart) is best for: (a) comparing categories, (b) quickly counting raw data, (c) showing approximate large quantities visually?

(a) Comparing categories → Bar Graph.

(b) Counting raw data quickly → Tally Chart / Frequency Table.

Q11. The mean height of 6 boys is 164 cm. A 7th boy joins. The new mean becomes 165 cm. What is the height of the 7th boy?

Sum of 6 boys = 6 × 164 = 984 cm.
Sum of 7 boys = 7 × 165 = 1155 cm.
Height of 7th boy = 1155 − 984 = 171 cm.

Q12. From a tally chart, 30 students chose colours: Red = 8, Blue = 12, Green = 5, Yellow = 5. What fraction of students chose Blue?

Fraction = 12/30 = 2/5.

Q13. In a bar graph, scale is 1 cm = 200 students. A bar is 3.5 cm tall. How many students does it represent?

3.5 × 200 = 700 students.

Q14. Define: (a) Data, (b) Raw Data, (c) Frequency, (d) Observation.

(a) Data: Collection of information in the form of numbers or words.

(b) Raw Data: Data collected in its original form, before organizing.

(d) Observation: Each individual piece of data collected.

Q15. The following are runs scored by 11 batsmen: 5, 0, 23, 13, 64, 27, 5, 40, 12, 50, 17. Find the mean.

Sum = 5+0+23+13+64+27+5+40+12+50+17 = 256.
Mean = 256 / 11 ≈ 23.3 runs.

Q16. The mean of four numbers is 25. Three of the numbers are 20, 28, and 30. Find the fourth number.

Sum of 4 numbers = 4 × 25 = 100.
Fourth number = 100 − (20+28+30) = 100 − 78 = 22.

Q17. Rainfall (in mm) for 6 days: 12, 8, 0, 15, 20, 5. If the mean for 7 days is 12 mm, find the rainfall on day 7.

Sum of 7 days = 7 × 12 = 84 mm.
Sum of first 6 days = 12+8+0+15+20+5 = 60 mm.
Day 7 rainfall = 84 − 60 = 24 mm.

Q18. What is the difference between a bar graph and a pictograph?

Pictograph: Uses pictures or symbols to represent data. A key tells how much each symbol represents. Better for visual appeal but less precise.

Bar Graph: Uses rectangular bars of equal width with lengths proportional to data values. More precise as exact values can be read from the scale.

Q19. The number of students absent each day of a week: Mon=3, Tue=5, Wed=2, Thu=7, Fri=4, Sat=1. (a) On which day were the most students absent? (b) Find mean absentees per day.

(a) Most absent on Thursday (7 students).
(b) Mean = (3+5+2+7+4+1) / 6 = 22/6 ≈ 3.67 per day.

Q20. A data set has 20 observations. If 5 more observations of value 10 each are added to the data, and the original mean was 15, find the new mean.

Original sum = 20 × 15 = 300.
New sum = 300 + (5 × 10) = 300 + 50 = 350.
New number of observations = 20 + 5 = 25.
New mean = 350 / 25 = 14.

Class 6 Maths - Data Handling Summary

Chapter 9: Data Handling (Concepts & Summary)

1. Key Terms

Term	Definition
Data	Collection of information in number or word form.
Raw Data	Data collected in original (unorganised) form.
Observation	Each piece of collected data.
Frequency	Number of times a value occurs in data.
Tally Mark	Marks used to record as data is counted. Every 5th mark crosses the previous 4 (\|\|\|\|).
Range	Difference between largest and smallest observation.

2. Types of Data Representation

Pictograph:

Uses pictures or symbols to represent data.
A key explains the value of each symbol.
Easy to read but not precise; half or quarter symbols may be used.

Bar Graph:

Uses rectangular bars of equal width.
Length (height) of each bar represents the frequency.
Scale is defined on the y-axis. Gaps between bars help distinguish categories.
More accurate than pictographs.

3. Mean (Average)

Mean = Sum of all observations ÷ Number of observations

Mean gives a central value that represents the entire data set.
If the mean of n numbers is M, then sum = n × M.
A missing value can be found: missing = total sum − sum of known values.

Example: Data: 5, 8, 12, 7, 8. Sum = 40. Mean = 40/5 = 8.

4. Drawing a Bar Graph — Steps

Draw two perpendicular axes (x and y).
Label: x-axis = categories, y-axis = frequency/value.
Choose an appropriate scale on the y-axis.
Draw bars of equal width with heights proportional to data.
Leave equal gaps between bars.
Give the graph a title and label both axes.

5. Quick Tips & Formulae

Mean = Sum / Count
Range = Maximum − Minimum
Sum = Mean × Number of observations.
Missing value = (Mean × n) − (sum of remaining values).
If all observations are equal, every observation equals the mean.
Adding the same constant to all values adds that constant to the mean.
Multiplying all values by a constant multiplies the mean by the same constant.

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