CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 3 Data Handling InText Questions and Answers are provided by experts in order to help students secure good marks in exams.
Class 7 Maths NCERT Solutions Chapter 3 Data Handling InText Questions
Try These (Page 59 )
Question 1.
Weight (in kg) at least 20 children (girls and boys) of your class. Organize 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?
Solution:
Let us weigh (in kg) at least 20 children (girls and boys) of our class and record the data as given below :
Arranging the above weights in the ascending order, we have :
32, 32, 33, 33, 33, 34, 34, 34, 34, 34, 34, 34, 35, 35, 35, 37, 37, 37, 38, 39.
Interpretation
It tells us about the weight of some students.
Five conclusions can be as under :
(i) Maximum weight possessed.
(ii) Minimum weight possessed.
(iii) The difference between the maximum and minimum weights.
(iv) The heaviest boy in the class.
(v) The heaviest girl in the class.
Let us find the answer to the given questions:
(i) Raman is the heaviest of all.
(ii) The most common weight is 34 kg.
(iii) The difference between my weight and that of my friend
= (34 – 34) kg = 0
Try These (Page 61)
Question 1.
How would you find the average of your study hours for the whole week?
Solution:
To find the average of study hours for the whole week, we should record the study hours for the whole week. Then, find the sum of all the observations and divide it by the number of observations, i.e., 7 (in this case), which gives us the required average.
Try These (Page 61)
Question 1.
Find the mean of your sleeping hours in one week.
Solution:
Let the sleeping hours during one week be 8 hours, 8 hours, 7 hours, 7.5 hours, 7.5 hours, 9 hours, and 9 hours.
Sum of the sleeping hours of one week
= (8 + 8 + 7 + 7.5 + 7.5 + 9 + 9) hours = 56 hours Number of days = 7
∴ Mean of sleeping hours = \(\left(\frac{56}{7}\right)\) hours = 8 hours
Try These (Page 65)
Question 1.
Find the mode of
(i) 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
Solution:
Arranging the data in ascending order, we get
0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6
Here, 2, 3, and 4 occur three times which is the highest number of occurrences for a datum. Therefore, they are modes of data.
(ii) 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Solution:
Arranging the data in ascending order, we get
2, 10, 12, 14, 14, 14, 14, 14, 14, 16, 16, 18
Clearly, 14 occurs the maximum number of times. Hence, mode =14
Try These (Page 65)
Question 1.
Find the mode of the following data:
12,14,12,16,15,13,14,18,19,12,14, 15,16,15,16,16,15,17,13, 16, 16, 15, 15, 13, 15, 17,15, 14, 15, 13, 15, 14.
Solution:
Let us put the data in a tabular form :
Looking at the table, clearly, we can say 15 is the ‘mode’ since 15 has occurred the maximum number of times.
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?
Solution:
Let us put the data in a tabular form :
Looking at the table, clearly, wp can say 163 cm is the ‘mode’ since 163 has occurred the maximum number of times. Mode gives that observation which occurs most frequently in the data.
Try These (Page 71)
Question 1.
The bar graph shows the result of a survey to test water-resistant watches made by different companies.
Each of these companies claimed that their watches were water-resistant. After a test, the above results were revealed.
(a) Can you work out of a fraction of the number of watches that leaked to the number tested for each company?
Solution:
A fraction of the number of watches that leaked to the number tested for each company are :
(b) Could you tell on this basis which company has better watches?
Solution:
Thus, a company with a fraction \(\frac{10}{40}\) i.e., company B has better watches.
Question 2.
Sale of English and Hindi books in the years 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) Can you say that the demand for English books rose faster? Justify.
Solution:
To choose an appropriate scale, we make equal divisions taking increments of 100. Thus, 1 unit = 100 books. •
(a) Clearly, the difference in the sale of the two language books is least in the year 1998.
(b) Since, the bar graph of the sale, of English books becomes longer faster, so the demand for English books rose faster.
Try These (Page 74)
Question 1.
Think of some situations, at least 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.
Solution:
Examples :
Certain situations to happen are :
(i) On tossing of a coin, getting of head or tail on the upper face.
(ii) On a single throw of a dice, any one of the six numbers, 1, 2, 3, 4, 5, 6 may appear.
(iii) On drawing one card from a pack of 52 cards, one suit will appear.
Impossible to happen are :
(i) A boy in the girls’ school.
(ii) A person of height 3 meters.
(iii) A number 7 appearing on the face of dice.
May or may not happen situations are:
(i) He may possibly join politics.
(ii) Probably it may rain.
(iii) He is probably right.
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.
Solution:
Head: 51 times and Tail: 49 times
Question 2.
Aftaab threw a die 250 times and got the following table. Draw a bar graph for this data.
Solution:
The following table gives the result of a die is thrown 250 times.
The bar graph for the above data is as drawn
Question 3.
Throw a die 100 times and record the data. Find the number of times 1, 2, 3, 4, 5, 6 occur.
Solution:
Outcome | Number of occurrences |
1 | 16 |
2 | 17 |
3 | 16 |
4 | 18 |
5 | 18 |
6 | 15 |
Try These (Page 76)
Question 1.
Constructor thinks of five situations where outcomes do not have equal chances.
Solution:
An unbiased die is thrown. Let us think of 5 situations where outcomes do not have equal chances.
These are:
(i) Getting a multiple of 3.
(ii) Getting a number 3 or 4.
(iii) Getting a number less than 5.
(iv) Getting a number greater than 3.
(v) Getting a number between 3 and 6.
In a single throw of a die, we can get any one of the six numbers 1, 2, 3, 4, 5, 6 marked on its six faces. So, the total outcomes are 6.
Getting a multiple of 3: It happens if we obtain either 3 or 6 as an outcome.
Its probability =\(\frac{2}{6}\) = \(\frac{1}{3}\)
Getting a number 3 or 4: In this case, a favorable number of outcomes = 2
Its probability = \(\frac{2}{6}\) = \(\frac{1}{3}\)
Getting a number less than 5: It happens only if one of the numbers 1, 2, 3, 4 appears on its face.
Required probability = \(\frac{4}{6}\) =\(\frac{2}{3}\)
Getting a number greater them 3: It occurs if one of the numbers 4, 5, 6 is an outcome.
Required probability = \(\frac{3}{6}\) =\(\frac{1}{2}\)
Getting a number between 3 and 6: It occurs if we obtain either 4 or 5 as an outcome.
Required probability = \(\frac{2}{6}\) = \(\frac{1}{3}\)