Chapter 3: Data Handling (NCERT Solutions)
Exercise 3.1
(This is an activity-based question, but here is a sample answer.)
Let the heights of 10 students (in cm) be:
145, 150, 148, 152, 149, 155, 146, 151, 147, 154
Highest value = 155 cm
Lowest value = 145 cm
Range = Highest value - Lowest value
= 155 - 145 = 10 cm
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7
(i) Which number is the highest?
(ii) Which number is the lowest?
(iii) What is the range of the data?
(iv) Find the arithmetic mean.
Arranging data in tabular form (frequency distribution):
| Marks | Tally Marks | Frequency (No. of Students) |
|---|---|---|
| 1 | | | 1 |
| 2 | || | 2 |
| 3 | | | 1 |
| 4 | ||| | 3 |
| 5 | XXXX | 5 |
| 6 | |||| | 4 |
| 7 | || | 2 |
| 8 | | | 1 |
| 9 | | | 1 |
(i) Highest number = 9
(ii) Lowest number = 1
(iii) Range = Highest - Lowest = 9 - 1 = 8
(iv) Arithmetic Mean = (Sum of all observations) / (Total number of
observations)
Sum = 4+6+7+5+3+5+4+5+2+6+2+5+1+9+6+5+8+4+6+7 = 100
Total observations = 20
Mean = 100 / 20 = 5
The first five whole numbers are 0, 1, 2, 3, 4.
Sum = 0 + 1 + 2 + 3 + 4 = 10
Number of observations = 5
Mean = 10 / 5 = 2
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.
Sum of scores = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400
Total number of innings = 8
Mean Score = 400 / 8 = 50
(Data given in book: A plays 4 games, B plays 3, C plays 4)
Find mean of A, mean of C, etc.
| Player | Game 1 | Game 2 | Game 3 | Game 4 |
|---|---|---|---|---|
| A | 14 | 16 | 10 | 10 |
| B | 0 | 8 | 6 | 4 |
| C | 8 | 11 | Did not play | 13 |
(i) Find the mean to determine A's average number of points scored per game.
Mean of A = (14 + 16 + 10 + 10) / 4 = 50 / 4 = 12.5
(ii) To find the mean number of points per game for C, would you divide the total points
by 3 or by 4? Why?
We will divide by 3 because C played only 3 games.
(iii) B played in all the four games. How would you find the mean?
Mean of B = (0 + 8 + 6 + 4) / 4 = 18 / 4 = 4.5
(iv) Who is the best performer?
Mean of C = (8 + 11 + 13) / 3 = 32 / 3 = 10.67
Comparing averages: A (12.5), B (4.5), C (10.67).
Therefore, A is the best performer.
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Arranging in ascending order: 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) Highest marks = 95, Lowest marks = 39
(ii) Range = Highest - Lowest = 95 - 39 = 56
(iii) Mean = Sum / 10 = (39+48+56+75+76+81+85+85+90+95) / 10 = 730 / 10 = 73
1555, 1670, 1750, 2013, 2540, 2820
Find the mean enrolment of the school for this period.
Sum of enrolments = 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 12348
Number of years = 6
Mean Enrolment = 12348 / 6 = 2058
Mon (0.0), Tue (12.2), Wed (2.1), Thurs (0.0), Fri (20.5), Sat (5.5), Sun (1.0).
(i) Find the range of the rainfall in the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.
(i) Highest rainfall = 20.5 mm, Lowest = 0.0 mm
Range = 20.5 - 0.0 = 20.5 mm
(ii) Sum = 0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0 = 41.3 mm
Mean = 41.3 / 7 = 5.9 mm
(iii) Days with rainfall less than 5.9 mm: Mon (0.0), Wed (2.1), Thurs (0.0),
Sat (5.5), Sun (1.0).
Total days = 5 days
135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
(i) What is the height of the tallest girl?
(ii) What is the height of the shortest girl?
(iii) What is the range of the data?
(iv) What is the mean height of the girls?
(v) How many girls have heights more than the mean height.
Arranging data: 128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) Tallest girl = 151 cm
(ii) Shortest girl = 128 cm
(iii) Range = 151 - 128 = 23 cm
(iv) Sum = 1414. Mean = 1414 / 10 = 141.4 cm
(v) Girls taller than 141.4 cm are 143, 146, 149, 150, 151.
Total = 5 girls
Exercise 3.2
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Ascending order: 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
Mode: The value occurring most frequently is 20 (it occurs 4 times).
Mode = 20
Median: Since n = 15 (odd), median is the middle observation.
Middle observation = ((15 + 1)/2)th term = 8th term.
The 8th term is 20.
Median = 20
Yes, they are the same.
6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15
Find the mean, mode and median of this data. Are the three same?
Ascending order: 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
Mean: Sum of observations / 11
Sum = 429. Mean = 429 / 11 = 39
Mode: The most frequent value is 15 (occurs 3 times).
Mode = 15
Median: The middle (6th) observation = 15
No, the three are not the same (Mean is 39, Mode and Median are 15).
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47
(i) Find the mode and median of this data.
(ii) Is there more than one mode?
Ascending order: 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
(i) n = 15.
Mode = Most frequent values. Both 38 and 43 occur 3 times.
Mode = 38 and 43
Median = 8th value = 40
(ii) Yes, there are two modes (38 and 43), so the data is bimodal.
Ascending order: 12, 12, 13, 13, 14, 14, 14, 16, 19
n = 9.
The value 14 occurs most frequently (3 times).
Mode = 14
The middle (5th) observation is 14.
Median = 14
(i) The mode is always one of the numbers in a data.
(ii) The mean is one of the numbers in a data.
(iii) The median is always one of the numbers in a data.
(iv) The data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9.
(i) True. Mode is the observation that occurs most frequently, so it must be present in the data.
(ii) False. The mean is a calculated average and may or may not be one of the numbers in the data.
(iii) True. (For a dataset with an odd number of observations. However, for an even number, it might be the average of the two middle numbers. In NCERT Class 7 context, standard answers generally accept it as True for given datasets, but technically it implies for odd datasets).
(iv) False. Sum = 6+4+3+8+9+12+13+9 = 64. Total = 8. Mean = 64/8 = 8, not 9.
Exercise 3.3
(a) Which is the most popular pet?
(b) How many students have dog as a pet?
(Based on the typical NCERT bar graph for pets: Dogs=8, Cats=10, Rabbits=2, Hamsters=5, Others=3)
(a) The tallest bar is for Cats (10 students).
Therefore, Cat is the most popular pet.
(b) The bar for Dogs indicates 8 students.
(i) About how many books were sold in 1989? 1990? 1992?
(ii) In which year were about 475 books sold? About 225 books sold?
(iii) In which years were fewer than 250 books sold?
(iv) Can you explain how you would estimate the number of books sold in 1989?
(Based on reading the NCERT bar graph)
(i) Number of books sold:
In 1989: about 175 books
In 1990: about 475 books
In 1992: about 225 books
(ii) About 475 books were sold in 1990.
About 225 books were sold in 1992.
(iii) Years with fewer than 250 books sold: 1989 and 1992.
(iv) On the y-axis, the scale is 1 unit = 100 books. The bar for 1989 is slightly more than 3/4th of the way between 100 and 200. Thus, we estimate it to be about 175 books.
(Data: Class 5th=135, 6th=120, 7th=95, 8th=100, 9th=90, 10th=80)
(a) How would you choose a scale?
(b) Answer the following questions:
(i) Which class has the maximum number of children? And the minimum?
(ii) Find the ratio of students of class sixth to the students of class eight.
(Bar graph representation instructions: Draw X-axis representing Classes and Y-axis representing Number of Children.)
(a) Scale: We can choose a scale of 1 unit = 20 children (or 10 children) on the Y-axis since the values range up to 135.
(b)(i) Maximum children = 135 in Fifth (5th) class.
Minimum children = 80 in Tenth (10th) class.
(b)(ii) Students in 6th class = 120.
Students in 8th class = 100.
Ratio = 120 / 100 = 12 / 10 = 6:5
(Data for English, Hindi, Maths, Science, S.Science)
(i) In which subject, has the child improved his performance the most?
(ii) In which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
(Data typically given:
English: 67 → 70 (+3)
Hindi: 72 → 65 (-7)
Maths: 88 → 95 (+7)
Science: 81 → 85 (+4)
S.Science: 73 → 75 (+2))
(i) The maximum increase in marks is in Maths (from 88 to 95, an
increase of 7).
Therefore, improvement is most in Maths.
(ii) The minimum positive increase is in S.Science (from 73 to 75,
an increase of 2).
Therefore, improvement is least in S.Science.
(iii) Yes, the performance has gone down in Hindi (from 72 to 65).
(Favourite Sport: Cricket, Basketball, Swimming, Hockey, Athletics)
(Watching vs Participating data)
(i) Draw a double bar graph choosing an appropriate scale. What do you infer from the bar graph?
(ii) Which sport is most popular?
(iii) Which is more preferred, watching or participating in sports?
(i) Inference: The double bar graph shows the comparison between the number of people who prefer watching and participating in different sports. In all sports, the number of watchers is greater than the number of participants.
(ii) The bars for Cricket are the tallest (both for watching and participating). Therefore, Cricket is the most popular sport.
(iii) Across all sports, the 'Watching' bar is consistently higher than the 'Participating' bar. Therefore, watching sports is more preferred than participating.
(i) Which city has the largest difference in the minimum and maximum temperature on the given date?
(ii) Which is the hottest city and which is the coldest city?
(iii) Name two cities where maximum temperature of one was less than the minimum temperature of the other.
(iv) Name the city which has the least difference between its minimum and the maximum temperature.
(Based on Table 3.1 data: Ahmedabad 38/29, Amritsar 37/26, Bangalore 28/21, Chennai 36/27, Delhi 38/28, Jaipur 39/29, Jammu 41/26, Mumbai 32/27)
(i) Difference in Jammu temperatures = 41 - 26 = 15°C (the highest).
Therefore, Jammu has the largest difference.
(ii) Hottest city (highest maximum) = Jammu (41°C).
Coldest city (lowest minimum) = Bangalore (21°C).
(iii) Maximum temperature of Bangalore (28°C) is less than the minimum temperature of Ahmedabad (29°C) and Jaipur (29°C).
(iv) Difference for Mumbai = 32 - 27 = 5°C (the smallest).
Therefore, Mumbai has the least difference.
Exercise 3.4
(i) You are older today than yesterday.
(ii) A tossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrow will be a cloudy day.
(i) Certain to happen (Time always moves forward).
(ii) Can happen but not certain (It could land tails up).
(iii) Impossible (A standard die only has numbers 1 to 6).
(iv) Can happen but not certain (It could be red or yellow).
(v) Can happen but not certain (Weather is unpredictable).
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with number 5?
Total number of possible outcomes (marbles) = 6.
(i) Number of favourable outcomes (marble with number 2) = 1.
Probability P(drawing 2) = Favourable outcomes / Total outcomes = 1/6
(ii) Number of favourable outcomes (marble with number 5) = 1.
Probability P(drawing 5) = Favourable outcomes / Total outcomes = 1/6
A coin has 2 faces: Head and Tail.
Total number of possible outcomes = 2.
Either Head or Tail is chosen by a team, so favourable outcomes = 1.
Probability = 1/2