GSEB Class 7 Maths Solutions Chapter 3 Data Handling Exercise 3.1

 

Question 1. Find the range of heights of any ten students of your class.
Answer: Let the heights of ten students be as follows:
118 cm, 111 cm, 114 cm, 120 cm, 110 cm,
115 cm, 119 cm, 118 cm, 117 cm, 113 cm
Writing these heights in ascending order, we get:
110 cm, 111 cm, 113 cm, 114 cm, 115 cm,
117 cm, 118 cm, 118 cm, 119 cm, 120 cm
Therefore, the maximum height is 120 cm.
The minimum height is 110 cm.
Thus, the range equals Maximum height minus Minimum height, which is \( 120 \text{ cm} - 110 \text{ cm} = 10 \text{ cm} \).
In simple words: To find the range, first arrange all the height measurements from smallest to largest. Then, subtract the smallest height from the biggest height to get the answer.

Exam Tip: When finding the range, always list the data in ascending or descending order first to avoid errors in identifying the minimum and maximum values.

 

Question 2. Organise the following marks in a class assessment, in a tabular form.
(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.
Answer: The given marks are: 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7.
Writing the given data (marks) in a tabular form, we have:

(i) The highest number is 9.
(ii) The lowest number is 1.
(iii) The range equals the highest number minus the lowest number, which is \( 9 - 1 = 8 \).
(iv) Arithmetic mean \( = \frac{\text{Sum of the marks}}{\text{Number of students}} \)
\( = \frac{[1 \times 1] + [2 \times 2] + [3 \times 1] + [4 \times 3] + [5 \times 5] + [6 \times 4] + [7 \times 2] + [8 \times 1] + [9 \times 1]}{1 + 2 + 1 + 3 + 5 + 4 + 2 + 1 + 1} \)
\( = \frac{1 + 4 + 3 + 12 + 25 + 24 + 14 + 8 + 9}{20} \)
\( = \frac{100}{20} \)
\( = 5 \)
Therefore, the mean marks are 5.
In simple words: We first put the marks into a table to count how many times each mark appeared. Then, we found the biggest and smallest marks, calculated the range by subtracting them, and finally, worked out the average (mean) mark by adding all marks and dividing by the total number of students.

Exam Tip: Always double-check your tally marks and frequencies when creating a tabular form to ensure accuracy, as errors here will affect all subsequent calculations.

 

Question 3. Find the mean of the first five whole numbers.
Answer: The first five whole numbers are 0, 1, 2, 3, and 4.
Their sum is \( 0 + 1 + 2 + 3 + 4 = 10 \).
Mean \( = \frac{\text{Sum of the numbers}}{\text{Number of whole numbers}} \)
\( = \frac{10}{5} \)
\( = 2 \)
In simple words: The first five whole numbers start from zero. To get their average, you add them all up and then divide by how many there are.

Exam Tip: Remember that whole numbers start from 0, not 1. A common mistake is to miss including 0 when listing them.

 

Question 4. A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
Answer: To find the mean score, we sum up all the runs and divide by the number of innings.
Sum of the scores \( = 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100 = 400 \).
Number of innings \( = 8 \).
Mean score \( = \frac{\text{Sum of the scores}}{\text{Number of innings}} \)
\( = \frac{400}{8} \)
\( = 50 \)
Therefore, the cricketer's mean score is 50.
In simple words: We add up all the runs the cricketer made and then divide that total by the number of games played. This gives us the average score.

Exam Tip: Be careful to include all scores, even zero, in your sum, and ensure you divide by the correct total number of observations (innings, in this case).

 

Question 5. Following table shows the points of each player scored in four games:

Now answer the following questions:
(i) Find the mean to determine A's average number of points scored per game.
(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?
(iii) B played in all the four games. How would you find the mean?
(iv) Who is the best performer?
Answer:
(i) Mean score of A \( = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
\( = \frac{14 + 16 + 10 + 10}{4} \)
\( = \frac{50}{4} \)
\( = 12.5 \)
Therefore, A's average score per game is 12.5.
(ii) Since C played only 3 games (he did not play the 3rd game), the total points should be divided by 3.
(iii) Mean score of B \( = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
\( = \frac{0 + 8 + 6 + 4}{4} \)
\( = \frac{18}{4} \)
\( = 4.5 \)
Thus, the average number of points scored by B is 4.5.
(iv) Mean score of C \( = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
\( = \frac{8 + 11 + 13}{3} \)
\( = \frac{32}{3} \)
\( = 10.67 \)
Thus, the mean (average) number of points scored by C is 10.67. Obviously, the average of 'A' is the highest. Therefore, A is the best player.
In simple words: We calculated each player's average score by adding up their points and dividing by the number of games they actually played. Player A had the highest average score, making him the top performer.

Exam Tip: When calculating averages, always check if a player missed any games. Only count the games actually played when dividing by the number of observations.

 

Question 6. 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 and 75. Find the:
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Answer: Writing the given marks in ascending order, we have:
39, 48, 56, 75, 76, 81, 85, 85, 90, 95
(i) The highest marks achieved are 95.
The lowest marks achieved are 39.
(ii) Range \( = \) Highest marks \( - \) Lowest marks \( = 95 - 39 = 56 \).
(iii) Mean \( = \frac{\text{Sum of all the marks}}{\text{Number of students}} \)
\( = \frac{39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95}{10} \)
\( = \frac{730}{10} \)
\( = 73 \)
Thus, the mean marks obtained by the group are 73.
In simple words: First, we arranged all the marks from smallest to largest to easily find the top and bottom scores. Then, we subtracted the lowest score from the highest to get the range. Finally, we added up all the marks and divided by the total number of students to find the average (mean) mark.

Exam Tip: Always arrange data in order (ascending or descending) when asked to find highest, lowest, or range, as it helps prevent mistakes and makes the data easier to work with.

 

Question 7. The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2540, 2820. Find the mean enrolment of the school for this period.
Answer: To find the mean enrolment, we sum up all the enrolments and divide by the number of years.
Sum of enrolments \( = 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 12348 \).
Number of years \( = 6 \).
Mean enrolment \( = \frac{\text{Sum of enrolments}}{\text{Number of years}} \)
\( = \frac{12348}{6} \)
\( = 2058 \)
Thus, the mean enrolment is 2058 students per year.
In simple words: We calculated the school's average number of students each year by adding up all the enrolments over six years and then dividing that total by six.

Exam Tip: Make sure to accurately add all the numbers, especially when dealing with large values, to avoid calculation errors in the sum.

 

Question 8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

(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?
Answer:
(i) To find the range, we identify the highest and lowest rainfall values.
Highest rainfall \( = 20.5 \text{ mm} \).
Lowest rainfall \( = 0.0 \text{ mm} \).
Range \( = \) Highest rainfall \( - \) Lowest rainfall \( = 20.5 \text{ mm} - 0.0 \text{ mm} = 20.5 \text{ mm} \).
(ii) Mean rainfall \( = \frac{\text{Sum of all observations}}{\text{Number of observations}} \)
\( = \frac{0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0}{7} \)
\( = \frac{41.3}{7} \)
\( = 5.9 \text{ mm} \)
Thus, the mean rainfall is 5.9 mm.
(iii) From the table, the rainfall values less than the mean rainfall (5.9 mm) are:
Monday: 0.0 mm
Wednesday: 2.1 mm
Thursday: 0.0 mm
Saturday: 5.5 mm
Sunday: 1.0 mm
Therefore, on 5 days (Monday, Wednesday, Thursday, Saturday, and Sunday), the rainfall was less than the mean rainfall.
In simple words: First, we found the difference between the most and least rainfall to get the range. Then, we added up all the daily rainfall amounts and divided by seven to get the average (mean) rainfall for the week. Finally, we looked at the table to count how many days had less rain than this average.

Exam Tip: When comparing individual data points to the mean, be careful to include all points that are strictly "less than" the mean, not just those significantly lower.

 

Question 9. The heights of 10 girls were measured in cm and the results are as follows: 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?
Answer: Writing the heights in ascending order, we have:
128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) The height of the tallest girl is 151 cm.
(ii) The height of the shortest girl is 128 cm.
(iii) Range \( = \) Highest height \( - \) Lowest height
\( = 151 \text{ cm} - 128 \text{ cm} \)
\( = 23 \text{ cm} \).
(iv) Mean \( = \frac{\text{Sum of the observations}}{\text{Number of observations}} \)
\( = \frac{128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151}{10} \)
\( = \frac{1414}{10} \)
\( = 141.4 \text{ cm} \).
Thus, the mean height of the girls is 141.4 cm.
(v) Since 143 cm, 146 cm, 149 cm, 150 cm, and 151 cm are all greater than 141.4 cm,
Therefore, the heights of 5 girls are more than the mean height.
In simple words: We organized the girls' heights to easily find the tallest and shortest, and then calculated the difference for the range. We then found the average (mean) height by adding all heights and dividing by the number of girls. Finally, we counted how many girls were taller than this average height.

Exam Tip: When finding values greater than the mean, ensure you compare each individual data point to the calculated mean value to avoid miscounting.