GSEB Class 7 Maths Solutions Chapter 3 Data Handling InText Questions

Try These (Page 59)

 

Question 1. Weigh (in kg) at least 20 children (girls and boys) of your class. Organise the data, and answer the following questions using this data.
(i) Who is the heaviest of all?
(ii) What is the most common weight?
(iii) What is the difference between your weight and that of your best friend!
Answer:
We will assume that the weights (in kg) of 20 children in the class are as given:
27, 30, 28, 24, 25, 26, 21, 20, 18, 15,
28, 16, 27, 23, 24, 26, 22, 31, 18, 26
Arranging this data in ascending order, we get:
15, 16, 18, 18, 20, 21, 22, 23, 24, 24,
25, 26, 26, 26, 27, 27, 28, 28, 30, 31

(i) The child weighing 31 kg is the heaviest of everyone.
(ii) The most common weight is 26 kg.
(iii) If my weight is 25 kg and my friend's weight is 27 kg, then the necessary difference \( = 27 \text{ kg} - 25 \text{ kg} = 2 \text{ kg} \).
In simple words: First, list all the weights and arrange them from smallest to largest. Then, you can easily find the heaviest weight, the weight that shows up most often, and the difference between two specific weights.

Exam Tip: Always arrange data in ascending or descending order first to easily identify extremes and frequencies.

 

Try These (Page 61)

 

Question 1. How would you find the average of your study hours for the whole week?
Answer: A week has 7 days. The amount of study hours each day can be different. Let's say the study hours for different days of the week are as follows:
Monday: 6 hours
Tuesday: 7 hours
Wednesday: 6 hours
Thursday: 8 hours
Friday: 8 hours
Saturday: 7 hours
Sunday: 7 hours
The total number of study hours \( = 6 + 7 + 6 + 8 + 8 + 7 + 7 = 49 \) hours.
The total number of days \( = 7 \).
Arithmetic mean \( = \frac{\text{Total study hours}}{\text{Total number of days}} \)
\( = \frac{49}{7} = 7 \)
Thus, the average study hours are 7 hours per day.
In simple words: To find your average study hours, add up all the hours you studied each day for the week. Then, divide that total by seven (because there are seven days in a week). This gives you the average number of hours you studied per day.

Exam Tip: The arithmetic mean (average) is calculated by summing all values and dividing by the count of those values. Make sure to list each value clearly before summing.

 

Try These (Page 61)

 

Question 1. Find the mean of your sleeping hours during one week.
Answer: Let's assume the sleeping hours are as follows:
Monday: 8 hours
Tuesday: 9 hours
Wednesday: 11 hours
Thursday: 8 hours
Friday: 10 hours
Saturday: 9 hours
Sunday: 8 hours
The sum of sleeping hours \( = 8 + 9 + 11 + 8 + 10 + 9 + 8 = 63 \) hours.
The number of days \( = 7 \).
Arithmetic mean \( = \frac{\text{Sum of sleeping hours}}{\text{Number of days}} \)
\( = \frac{63}{7} \) hours
\( = 9 \) hours
In simple words: To get the average sleeping hours, add up how many hours you slept each day for one week. Then, divide that total by seven, which is the number of days. The result will be your mean, or average, sleeping time per day.

Exam Tip: Remember that the mean provides a single value that represents the central tendency of a dataset, useful for summarizing daily habits like sleep.

 

Question 2. Find at least 5 numbers between \( \frac{1}{2} \) and \( \frac{1}{3} \).
Answer: The arithmetic mean of two given numbers is their average value between them.
Between \( \frac{1}{2} \) and \( \frac{1}{3} \):
1st number \( = \frac{\frac{1}{2} + \frac{1}{3}}{2} = \frac{\frac{2+3}{6}}{2} = \frac{\frac{5}{6}}{2} = \frac{5}{12} \)
2nd number \( = \frac{\frac{1}{2} + \frac{5}{12}}{2} = \frac{\frac{6+5}{12}}{2} = \frac{\frac{11}{12}}{2} = \frac{11}{24} \)
3rd number \( = \frac{\frac{1}{3} + \frac{5}{12}}{2} = \frac{\frac{4+5}{12}}{2} = \frac{\frac{9}{12}}{2} = \frac{3}{8} \)
4th number \( = \frac{\frac{1}{2} + \frac{11}{24}}{2} = \frac{\frac{12+11}{24}}{2} = \frac{\frac{23}{24}}{2} = \frac{23}{48} \)
5th number \( = \frac{\frac{1}{3} + \frac{11}{24}}{2} = \frac{\frac{8+11}{24}}{2} = \frac{\frac{19}{24}}{2} = \frac{19}{48} \)
Thus, 5 numbers between \( \frac{1}{2} \) and \( \frac{1}{3} \) are \( \frac{5}{12}, \frac{11}{24}, \frac{3}{8}, \frac{23}{48} \) and \( \frac{19}{48} \).
In simple words: To find numbers between two fractions, you can keep taking the average of them. Add the two fractions together, then divide by two. Keep doing this with the new numbers you find and one of the original numbers to find even more numbers in between.

Exam Tip: To find a number between two given numbers (fractions or decimals), simply calculate their average. Repeat this process with the new number and one of the original numbers to find more values in the range.

 

Try These (Page 65)

 

Question 1. Find the mode of
(i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
(ii) 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Answer:
(i) The given set of data is: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4.
Writing the numbers with the same value together, we have: 0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6.
So, 2, 3 and 4 appear in the given data the most times (3 times each).
Thus, the modes for this data set are 2, 3 and 4.

(ii) The given set of data is: 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14.
Writing the numbers with the same value together, we have: 2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18.
Therefore, 14 appears the highest number of times (i.e. six times).
So, the required mode is 14.
In simple words: The mode is the number that shows up most often in a list. Count how many times each number appears, and the one that appears most frequently is the mode. Sometimes there can be more than one mode.

Exam Tip: For small datasets, simply listing numbers in order can help you spot the mode. For larger datasets, creating a frequency table is more efficient.

 

Think, Discuss and write (Page 65)

 

Question 1. Can a set of numbers have more than one model
Answer: Yes.
In simple words: Yes, a group of numbers can have more than one number that appears most often. If two or more numbers have the same highest count, they are all considered modes.

Exam Tip: A dataset can be unimodal (one mode), bimodal (two modes), multimodal (multiple modes), or have no mode if all values occur with the same frequency.

 

Try These (Page 65 and 66)

 

Question 1. Find the mode of the following data:
12, 14, 15, 16, 15, 16, 16, 15, 17,
13, 16, 16, 15, 15, 13, 15, 17, 15,
14, 15, 13, 15, 14.
Answer: The given set of data has a large number of observations here. So, we will put them in a table.

The highest frequency (10) corresponds to the number 15.
Thus, the necessary mode is 15.
In simple words: When you have a lot of numbers, it's easier to find the mode by making a table. List each unique number and count how many times it appears. The number that has the highest count is the mode.

Exam Tip: For datasets with many observations, constructing a frequency distribution table with tally marks is the most organized way to find the mode efficiently.

 

Question 2. Heights (in cm) of 25 children are given below:
168, 165, 163, 160, 163, 161,
162, 164, 163, 162, 164, 163,
160, 163, 160, 165, 163, 162,
163, 164, 163, 160, 165, 163,
162
What is the mode of their heights? What do we understand by mode here?
Answer: We will organize the data in a tabular form:

Since 163 appears the maximum number of times, the mode is 163 cm.
By the mode, we mean that most of the students have a height of 163 cm.
Note: Mode is the most frequent value in a group of data, while the mean gives us the average of all observations in the data.
In simple words: To find the mode of heights, count how many children have each height. The height that appears most often is the mode. The mode tells us the most common height among the students.

Exam Tip: When dealing with numerical data, organizing it into a frequency table makes it simple to spot the mode, which represents the most typical value.

 

Try These (Page 66)

 

Question 1. Discuss with your friends and give:
(a) Two situations where mean would be an appropriate representative value to use, and
(b) Two situations where mode would be an appropriate representative value to use.
Answer:
(a) (i) For measuring the heights of all students in a class, the mean would offer a better representation of the data.
(ii) For the weights of wheat stored in similar containers, the mean would give us a better overall picture of the data.

(b) (i) If we are gathering information on shoe sizes suitable for a certain age group, the mode is more helpful.
(ii) If a shopkeeper selling shirts needs to know the right sizes to increase their stock, then the mode is more useful.
In simple words: Use the mean when you want an average of all numbers, like average height or weight. Use the mode when you want to know the most popular or common choice, like the most common shoe size or shirt size to stock.

Exam Tip: Choose mean for continuous, evenly distributed data, and mode for categorical or discrete data where the most frequent category is important, such as preferences or sizes.

 

Try These (Page 67)

 

Question 1. Your friend found the median and the mode of a given data. Describe and correct your friend's error if any:
35, 32, 35, 42, 38, 32, 34
Median = 42, Mode = 32
Answer: The set of data is: 35, 32, 35, 42, 38, 32, 34.
Putting the given data in ascending order, we have: 32, 32, 34, 35, 35, 38, 42.
(i) The middle value of the data is 35.
Thus, the correct median \( = 35 \).
(ii) The observations that appear the maximum number of times are 32 and 35.
So, the correct modes are 32 and 35.
In simple words: To find the median, first put all the numbers in order. The middle number is the median. To find the mode, count which numbers show up most often. In this case, your friend made a mistake because they did not order the numbers before finding the median, and they missed one of the numbers that appeared most frequently for the mode.

Exam Tip: Always arrange numerical data in ascending (or descending) order before determining the median. For the mode, identify all values that occur with the highest frequency, as there can be multiple modes.

 

Try These (Page 71)

 

Question 1. The bar graph given below shows the result of a survey to test water resistant watches
(a) Can you work out a fraction of the number of watches that leaked to the number tested for each company?
(b) Could you tell on this basis which company has better watches?
Answer: (Visual information from the bar graph is used to answer the question, but the graph itself will not be converted to SVG due to its complexity.)

(a) From the graph, we can see that the total number of watches tested for each company is 40.
So, the fraction of 'the number of watches that leaked' to 'the number of watches tested' is:
For company A \( = \frac{20}{40} = \frac{1}{2} \)
For company B \( = \frac{10}{40} = \frac{1}{4} \)
For company C \( = \frac{15}{40} = \frac{3}{8} \)
For company D \( = \frac{25}{40} = \frac{5}{8} \)

(b) Obviously, company B has better watches because it has the smallest fraction of leaked watches.
This can be seen when comparing the fractions: \( \frac{1}{4} < \frac{3}{8} < \frac{1}{2} < \frac{5}{8} \).
In simple words: Look at the graph to see how many watches each company tested and how many of those leaked. Then, write a fraction for each company with the leaked watches on top and tested watches on the bottom. The company with the smallest fraction of leaked watches is the best.

Exam Tip: When comparing fractions to determine which company performs "better," a smaller fraction of defects (leaked watches) indicates superior performance. Convert fractions to a common denominator or decimals for easy comparison.

 

Question 2. Sale of English and Hindi books in the year 1995, 1996, 1997 and 1998 are given below:

Draw a double bar graph and answer the following questions:
(a) In which year was the difference in the sale of the two language books least?
(b) Justify that the demand for English books rose faster?
Answer: (The double bar graph is not converted to SVG due to its complexity. However, the data from the graph is used to answer the questions.)

(a) The difference in the sale of the two language books is:
In 1995, it is \( 500 - 350 = 150 \).
In 1996, it is \( 525 - 400 = 125 \).
In 1997, it is \( 600 - 450 = 150 \).
In 1998, it is \( 650 - 620 = 30 \).
Therefore, the difference in the sale of the two language books was lowest in the year 1998.

(b) Yes, we can say that the demand for English books rose more quickly because, from 1995 to 1998:
(i) The sale of Hindi books increased from 500 to 650, meaning it rose by \( 650 - 500 = 150 \).
(ii) The sale of English books increased from 350 to 620, meaning it rose by \( 620 - 350 = 270 \).
Since 270 is much greater than 150, the demand for English books rose faster.
In simple words: To find the year with the smallest difference, subtract the English book sales from the Hindi book sales for each year and see which year has the smallest result. To see if English book demand rose faster, compare how much English sales went up versus how much Hindi sales went up over the entire period.

Exam Tip: When asked to compare changes or differences over time, calculate the absolute change for each category. A larger absolute change indicates a faster rise or fall, provided the starting points are relatively comparable.

 

Question 1. Think of some situations, atleast 3 examples of each, that are certain to happen, some that are impossible and some that may or may not happen i.e., situations that have some chance of happening.
Answer:
(i) Situations that are certain to happen:

• The sun rises in the east.
• Getting a number from 1 to 6 by throwing a dice.
• When a coin is tossed, then either a head or tail comes up.

(ii) Situations that are impossible:

• Getting the number 8 by throwing a dice.
• To draw a five rupee coin from a bag containing one rupee coins.
• Drawing a red ball from a bag containing yellow and blue balls only.

(iii) Situations that have some chance of happening:

• An ant rising to 5 m height.
• To throw a dice and get an odd number.
• To toss a coin and get tail.

In simple words: Things that are certain always happen, like the sun rising. Things that are impossible can never happen, like rolling an 8 on a standard die. Things that have some chance might happen or might not, like getting an odd number when you roll a die.

Exam Tip: Understanding probability concepts like certainty (probability of 1), impossibility (probability of 0), and chance (probability between 0 and 1) is crucial for classifying events.

Note:

• A coin has two faces. The face with Ashoka Chakra is called 'Head' and the other is called 'Tail'.
• A cube with six faces is called a dice. Each face has dots numbered from 1 to 6.

 

Try These (Page 75)

 

Question 1. Toss a coin 100 times and record the data. Find the number of times heads and tails occur in it.
Answer: This is a group activity. Please perform it yourself.
In simple words: Flip a coin 100 times. Keep a count of how many times it lands on heads and how many times it lands on tails. You should do this by yourself as a hands-on task.

Exam Tip: For experimental probability questions, accurately recording each trial's outcome is vital. Even if it's a "do it yourself" question, understanding the process is key.

 

Question 2. Aftaab threw a die 250 times and got the following table. Draw a bar graph for this data.

Answer: We will create a frequency table from the given tally marks and then use it to visualize the bar graph. (The bar graph itself will not be converted to SVG due to its complexity.)

(The bar graph would show the 'Number on the die' on the x-axis and 'Frequency' on the y-axis, with bars representing each number's frequency according to the table above. For example, the bar for '1' would reach a height of 33, '2' to 40, and so on.)
In simple words: First, count the tally marks for each number to find its frequency. Then, use these frequencies to draw a bar graph where each bar shows how many times a number appeared when the die was thrown.

Exam Tip: When drawing a bar graph, ensure that the axes are clearly labeled, bars are of equal width, and the scale is appropriate for the data's range. The height of each bar should accurately represent its frequency.

 

Question 3. Throw a die 100 times and record the data. Find the number of times 1, 2, 3, 4, 5, 6 occur.
Answer: This is a group activity. Please perform it yourself.
In simple words: Roll a die 100 times and keep track of which number you get each time. After 100 rolls, count how many times each number (1, 2, 3, 4, 5, or 6) appeared. This is a practical exercise to do yourself.

Exam Tip: When conducting an experiment like rolling a die, maintaining an accurate tally is essential for obtaining reliable frequency data.

Note:

• The outcome is the result or effect of an action.
• A trial is an action that results in one or several outcomes.
• A random experiment is an experiment where the result of a trial cannot be predicted in advance.

 

Try These (Page 76)

 

Question 1. Construct or think of five situations where outcomes do not have equal chances.
Answer: In the following situations, the outcomes do not have an equal chance of happening:

• To draw a red ball and to draw a yellow ball from a bag having 3 red balls and 5 yellow balls. (Yellow is more likely)
• To throw a dice and getting 2 and getting an odd number. (Odd numbers are more likely, as there are three odd numbers (1,3,5) and only one '2')
• To draw a flash card bearing 'B' and to draw a flash card bearing 'C' from a collection of 5 flash cards bearing A, B, C, D and E. (Both have equal chance, this example might be intended to mean comparing 'B' vs 'not B'). Let's rephrase this one to clearly unequal: To draw a flash card bearing 'A' and to draw a flash card bearing 'B' from a collection of 10 flash cards, where 7 are 'A' and 3 are 'B'.
• Choosing a boy and choosing a girl as class representative from a group of 10 boys and 15 girls. (Girls are more likely)
• Choosing an even number and choosing an odd number from first five natural numbers. (Odd numbers (1, 3, 5) are more likely than even numbers (2, 4))

In simple words: Unequal chances mean that some results are more likely to happen than others. For example, if you have more red balls than blue balls in a bag, you're more likely to pick a red ball. Or, if there are more girls than boys in a class, picking a girl for a role is more probable.

Exam Tip: Unequal chances arise when the number of favorable outcomes for different events is not the same, or when the underlying conditions of the experiment are not symmetrical (e.g., a biased coin). Clearly identify the sample space and the number of outcomes for each event.