CBSE Class 8

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.2

Question 1.
Find the square of the following numbers.
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46
Solution:
(i) 32²
= (30 + 2)²
= 30² + 2 (30) (2) + 2² [∵ (a + b)² = a² + 2ab + b² ]
= 900 + 120 + 4
= 1024

(ii) 35²
= (30 + 5)²
= 30² + 2 (30) (5) + 5²
= 900 + 300 + 25
= 1225

(iii) 86²
= (80 + 6)²
= 80² + 2 (80) (6) + 6²
= 6400 + 960 + 36
= 7396

(iv) 93²
= (90 + 3)²
= 90² + 2 (90) (3) + 3²
= 8100 + 540 + 9
= 8649

(v) 712
= (70 + 1)²
= 70² + 2(70) (1) + 1²
= 4900 + 140 + 1
= 5041

(vi) 46²
= (40 + 6)²
= 40² + 2 (40) (6) + 6²
= 1600 + 480 + 36
= 2116

Question 2.
Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution:
(i) Let 2n = 6
n = \(\frac{6}{2}\) = 3

n² – 1
= 3² – 1
= 8

n² + 1
= 3² + 1
= 10
∴ The required Pythagorean triplet is 6, 8 and 10.

(ii) Let 2n = 14
n = \(\frac{14}{2}\) = 7

n² – 1
= 7² – 1
= 49 – 1
= 48

n² + 1
= 7² + 1
= 49 + 1
= 50
∴ Thus, the required Pythagorean triplet is 14, 48 and 50.

(iii) Let 2n = 16
n = \(\frac{16}{2}\) = 8

n² – 1
= 8² – 1
= 64 – 1
= 63

n² + 1
= 8² + 1
= 64 + 1
= 65
∴ The required Pythagorean triplet is 16, 63 and 65.

(iv) Let 2n = 18
n = \(\frac{18}{2}\) = 9

n² – 1
= 9² – 1
= 81 – 1
= 80

n² + 1
= 9² + 1
= 81 + 1
= 82
∴ The required Pythagorean triplet is 18, 80 and 82.

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.1

Question 1.
What will be the unit digit of the squares of the following numbers?
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555
Solution:
(i) The unit digit of (81)² is 1 (1 × 1 = 1)
(ii) The unit digit of (272)² is 4 (2 × 2 = 4)
(iii) The unit digit of (799)² is 1 (9 × 9 = 81)
(iv) The unit digit of (3853)² is 9 (3 × 3 = 9)
(v) The unit digit of (1234)² is 6 (4 × 4 = 16)
(vi) The unit digit of (26387)² is 9 (7 × 7 = 49)
(vii) The unit digit of (52698)² is 4 (8 × 8 = 64)
(viii) The unit digit of (99880)² is 0 (0 × 0 = 0)
(ix) The unit digit of (12796)² is 6 (6 × 6 = 36)
(x) The unit digit of (55555)² is 5 (5 × 5 = 25)

Question 2.
The following numbers are obviously not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050
Solution:
(i) 1057 is not a perfect square,
As the last digit is 7 [It is not one of 0, 1, 4, 5, 6, and 9].

(ii) 23453 is not a perfect square,
As the last digit is 3 [It is not one of 0, 1, 4, 5, 6, and 9].

(iii) 7928 is not a perfect square,
As the last digit is 8 [It is not one of 0, 1, 4, 5, 6, and 9].

(iv) 222222 is not a perfect square,
As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(v) 64000 is not a perfect square,
As the number of zeros is odd.

(vi) 89722 is not a perfect square,
As the last digit is 2 [It is not one of 0, 1, 4, 5, 6, and 9].

(vii) 222000 is not a perfect square,
As the number of zeros is odd.

(viii) 505050 is not a perfect square,
As the number of zeros is odd.

Question 3.
The squares of which of the following would be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004
Note: The square of an odd natural number is always odd and that of an even number always an even number.
Solution:
(i) The square of 431 is an odd number.
(ii) The square of2826 is an even number.
(iii) The square of 7779 is an odd number.
(iv) The square of82004 is an even number.

Question 4.
Observe the following pattern and find the missing digits.
11² = 121
101² = 10201
1001² = 1002001
100001² = 1………2……..1
10000001² = …………….
Solution:
By observing the above pattern, we get
(i) 100001² = 10000200001
(ii) 10000001² = 100000020000001

Question 5.
Observe the following pattern and supply the missing number.
(a) 11² = 121
101² = 10201
10101² = 102030201
1010101² = …………….
………..² = 1020304030201
Solution:
By observing the above pattern, we get
(I) (1010101)² = 1020304030201
(ii) 10203040504030201 = (101010101)²

Question 6.
Using the given pattern. Find the missing numbers.
1² + 2² + 2² = 3²
2² + 3² + 6² = 7²
3² + 4² + 12² = 13²
4² + 5² + ……² = 21²
5² + …..² + 30² = 31²
6² + 7² + ……² = ……..²
Solution:
4² + 5² + 20² = 21²
5² + 6² + 30² = 31²
6² + 7² + 42² = 43²
Note: To find pattern
The third number is related to the first and second numbers. How?
The fourth number is related to the third number. How?

Question 7.
Without adding, find the sum
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
(i) Sum of the first 5 odd numbers = 5² = 25
(ii) Sum of the first 10 odd numbers = 10² = 100
(iii) Sum of the first 12 odd numbers = 12² = 144

Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as to the sum of 11 odd numbers.
Solution:
(i) 7² = 49
1 + 3 + 5 + 7 + 9 + 11 + 13 (First 7 odd numbers)
(ii) 11² = 121
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (First 11 odd numbers)

Question 9.
How many numbers lie between squares of the following numbers?
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100
Solution:
Note: Since between n and (n + 1) there are 2n non-square numbers
(i) Between 12² and 13², there are 2 × 12 numbers i.e. 24 numbers
(ii) Between 25² and 26², there are 2 × 25 numbers i.e. 50 numbers
(iii) Between 99² and 100², there are 2 × 99 numbers i.e. 198 numbers
Note: For any number ending with 5, the square is a (a + 1) × 100 + 25
(i) 35²
= 3(3 + 1) × 100 + 25
= 1200 + 25
= 1225

(ii) 55²
= 5(5 + 1) × 100 + 25
= 3000 + 25
= 3025

(iii) 125²
= 12(12 + 1) × 100 + 25
= 12 × 13 × 100 + 25
= 15600 + 25
= 15625

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling InText Questions Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling InText Questions

Try These (Page No. 71)

Question 1.
Draw an appropriate graph to represent the given information.

Solution:
(i) Note: A bar graph showing two sets of data simultaneously is called a double-bar graph. It is useful for the comparison of the data.

To represent the given data by a bar- graph, draw two axes perpendicular to each other. Now, represent ‘Months’ on OX and ‘Number of watches sold’ on OY. Erect recharge of the same width. The heights of the rectangles are proportional to the number of watches using a suitable scale:
Here, the scale is 1 cm = 500 watches
Since 500 watches = 1 cm
1000 watches = 2 cm
1500 watches = 3 cm
2000 watches = 4 cm
2500 watches = 5 cm

(ii)

(iii) Since, a comparison of two activities (walking and cycling) is to be represented, there a double-graph is drawn by taking the schools along the x-axis and the number of children on the y-axis, using a scale of 1 cm = 5 children.

To compare the percentage win in ODI achieved by various teams, we represent the data by a double-bar graph. We represent the teams along the x-axis and their percentage win along the y-axis, using the scale 1 cm = 5%.

Try These (Page No. 72)

Question 2.
A group of students was asked to say which animal they would like most to have as a pet. The results are given below:
dog, cat, cat, fish, cat, rabbit, dog, cat, rabbit, dog, cat, dog, dog, dog, cat, cow, fish, rabbit, dog, cat, dog, cat, cat, dog, rabbit, cat, fish, dog.
Make a frequency distribution table for the same.
Solution:
Using tally-marks, we have:

Try These (Page No. 73-74)

Question 3.
Study the following frequency distribution table and answer the questions given below.
Frequency Distribution of Daily Income of 550 workers of a factory
Table 5.3

(i) What is the size of the class intervals?
(ii) Which class has the highest frequency?
(iii) Which class has the lowest frequency?
(iv) What is the upper limit of the class interval 250 – 275?
(v) Which two classes have the same frequency?
Solution:
(i) Class size = (Upper class limit) – (Lower class limit)
= 125 – 100
= 25
(ii) The class 200 – 225 is having the highest frequency (which is 140).
(iii) The class 300 – 325 is having the lowest frequency (which is 20).
(iv) The upper limit of the class interval 250 – 275 is 275.
(v) The classes (150 – 175) and (225 – 250) are having the same frequency (which is 55).

Question 4.
Construct a frequency distribution table for the data on weights (in kg) of 20 students of a class using intervals 30 – 35, 35 – 40, and so on.
40, 38, 33, 48, 60, 53, 31, 46, 34, 36, 49, 41, 55, 49, 65, 42, 44, 47, 38, 39.
Solution:
Lowest observation = 31
Highest observation = 65
Class intervals: 30 – 35, 35 – 40, 40 – 45,…..
The frequency distribution table for the above data can be:

Try These (Page No. 75)

Question 5.
Observe the histogram (Fig 5.3) and answer the questions given below.

(i) What information is being given by the histogram?
(ii) Which group contains maximum girls?
(iii) How many girls have a height of 145 cm and more?
(iv) If we divide the girls into the following three categories, how many would there be in each?
150 cm and more – Group A
140 cm to less than 150 cm – Group B
Less than 140 cm – Group C
Solution:
(i) The above histogram represents the height (in cms) of girls of Class VIII.
(ii) The group 140 – 145 contains a maximum number of girls (which has as many as 7 girls).
(iii) 7 girls (= 4 + 2 + 1) have a height of 145 cm and more.
(iv) Number of girls in
Group A: 150 cm and more = 2 + 1 = 3 girls
Group B: 140 cm to less than 150 cm = 7 + 4 = 11 girls
Group C: Less than 140 cm = 1 + 2 + 3 = 6 girls.
Note: The broken line (~) is used along the horizontal line to indicate that we are not showing the numbers between 0 and 125.

Try These (Page No. 78)

Question 6.
Each of the following pie charts (Fig 5.5) gives you a different piece of information about your class.
Find the fraction of the circle representing each of this information.

Solution:
(i) Fraction of the circle representing the ‘girls’ 50% = \(\frac{50}{100}=\frac{1}{2}\)
Fraction of the circle representing the ‘boys’ 50% = \(\frac{50}{100}=\frac{1}{2}\)

(ii) Fraction of the circle representing ‘walk’ 40% = \(\frac{40}{100}=\frac{2}{5}\)
Fraction of the circle representing ‘but or car’ 40% = \(\frac{40}{100}=\frac{2}{5}\)
Fraction of the circle representing ‘cycle’ 20% = \(\frac{20}{100}=\frac{1}{5}\)

(iii) Fraction of the circle representing those who live mathematics = (100 – 15)%
= \(\frac{100-15}{100}\)
= \(\frac{85}{100}\)
= \(\frac{17}{20}\)
Fraction of the circle representing those who hate mathematics = 15%
= \(\frac{15}{100}\)
= \(\frac{3}{20}\)

Question 7.
Answer the following questions based on the pie chart given.
(i) Which type of programmes are viewed the most?
(ii) Which two types of programmes have a number of viewers equal to those watching sports channels?

Solution:
From the given pie chart, we have: Obviously,

(i) The entertainment programmes are viewed the most.
(ii) The news and informative programmes have an equal number of viewers.

Try These (Page No. 81)

Question 8.
Draw a pie chart of the data given below.
The time spent by a child during a day.
Sleep – 8 hours
School – 6 hours
Homework – 4 hours
Play – 4 hours
Others – 2 hours
Solution:
First, we find the central angle corresponding to the given activities.

Now, the required pie chart is given below:

Try These (Page No. 83)

Question 9.
If you try to start a scooter, what are the possible outcomes?
Solution:
It may start. It may not start.

Question 10.
When a die is thrown, what are the possible outcomes?
Solution:
The possible outcomes are 1, 2, 3, 4, 5, or 6.

Question 11.
When you spin the wheel shown, what are the possible outcomes? List them.
(Outcome here means the sector at which the pointer stops).

Solution:
The possible outcomes are A, B, or C.

Question 12.
You have a bag with five identical balls of different colours and you are to pull out (draw) a ball without looking at it; list the outcomes you would get.

Solution:
The possible outcomes are W, R, B, G, or Y.

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Exercise 5.3

Question 1.
List the outcomes you can see in these experiments
(a) Spinning a wheel
(b) Tossing two coins together
Solution:
(a) Outcomes in spinning the given wheel are A, B, C, or D.
(b) Outcomes in tossing two coins together are HH, HT, TH, TT
[HT means Head on first coin + Tail on the second coin]

Question 2.
When a die is thrown, list the outcomes of an event of getting.
(i) (a) prime number
(b) not a prime number
(ii) (a) a number greater than 5
(b) a number not greater than 5
Solution:
Possible outcomes are 1, 2, 3, 4, 5 and 6
(i) (a) Outcomes are 2, 3 and 5
(b) Outcomes are 1, 4 and 6
(ii) (a) Outcomes greater than 5 is 6
(b) Outcomes of getting a number not greater than 5 are 1, 2, 3, 4 and 5

Question 3.
Find the
(a) Probability of the pointer stopping on D in (Question 1-(a))?
(b) Probability of getting an ace from a well-shuffled deck of 52 playing cards?
(c) Probability of getting a red apple. (See figure below)

Solution:
(a) On the spinning wheel there are 5 sectors containing A, B, C, and D.
Since, there is only one sector containing D,
Number of possible outcome = 1
Number of equally likely outcomes = 5
∴ Probability = \(\frac{1}{5}\)

(b) Number of possible outcomes = 52
Since there are 4 aces in a pack of 52 cards, and out of the one ace can be obtained in 4 ways
∴ Probability of getting an ace = \(\frac{4}{52}=\frac{1}{13}\)

(c) There are 7 apples in all
Possible number of ways = 7
Since there are 4 red apples
No. of equally likely outcomes = 4
∴ Probability of getting a red apple = \(\frac{4}{7}\)

Question 4.
Numbers 1 to 10 are written on ten separate slips (one number on one slip), kept in a box, and mixed well. One slip is chosen from the box without looking into it. What is the probability of?
(i) getting a number 6?
(ii) getting a number less than 6?
(iii) getting a number greater than 6?
(iv) getting a 1-digit number?
Solution:
Since there are 10 slips,
Total number of outcomes = 10
(i) We can get a slip containing the number ‘6’ only once
Number of favourable outcome = 1
∴ Probability of getting the number = \(\frac{6}{10}\)

(ii) Numbers less than 6 are 1, 2, 3, 4 and 5
No. of Favourable outcomes are 5
∴ Probability of getting a no. less than 6 = \(\frac{5}{10}=\frac{1}{2}\)

(iii) Numbers greaterthan 6 are 7, 8, 9 and 10
No. of favourable outcomes = 4
∴ Probability of getting a number greater than 6 = \(\frac{4}{10}=\frac{2}{5}\)

(iv) One-digit numbers are 1, 2, 3, 4, 5, 6, 7, 8 and 9
No. of favourable outcomes = 9
∴ Probability of getting a one-digit number = \(\frac{9}{10}\)

Question 5.
If you have a spinning wheel with 3 green sectors, 1 blue sector, and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non-blue sector?
Solution:
There are 5 sectors is all (3 green + 1 blue + 1 red)
Total possible outcomes = 5
Since there are 3 green sectors
Number of favourable outcomes = 3
∴ Probability of getting a green sector = \(\frac{3}{5}\)
Again there are 4 non-blue sectors.
Number of favourable outcomes = 4
∴ Probability of getting a non blue sector = \(\frac{4}{5}\)

Question 6.
Find the probabilities of the events given in question 2.
Solution:
When a die is thrown there are 6 outcomes in all (1, 2, 3, 4, 5 and 6)
(i) Since there are 3 prime numbers 2, 3 and 5
Number of favourable outcomes = 3
∴ Probability of getting a prime number = \(\frac{3}{6}=\frac{1}{2}\)

(ii) Since there are 3 non-prime numbers (1, 4 and 6)
Number of favourable outcomes = 3
∴ Probability of getting a non prime number = \(\frac{3}{6}=\frac{1}{2}\)

(iii) Since there is 1 number greater than 5 (i.e., 6)
Number of favourable outcome = 1
∴ Probability of number greater than 5 = \(\frac{1}{6}\)

(iv) Since there are 5 numbers which are not greater than 5 (i.e., 1, 2, 3, 4 and 5)
Number of favourable outcomes = 5
∴ The probability of a number which is not greater than 5 = \(\frac{5}{6}\)

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Exercise 5.2

Question 1.
A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey. From this pie chart answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
(ii) Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CD’s how many of each type would they make?

Solution:
Let the required number of young people be ‘x’
∴ 10% of x = 20
\(\frac{10}{100}\) × x = 20
x = \(\frac{20 \times 100}{10}\) = 200
(ii) Light music is liked by the maximum number of people
(iii) Total number of CD’s = 1000
Number of CD’s of Semi classical music = 20% of 1000
= \(\frac{20}{100}\) × 1000
= 200
Number of CD’s for classical = 10% of 1000
= \(\frac{10}{100}\) × 1000
= 100
Number of CD’s for folks = 30% of 1000
= \(\frac{30}{100}\) × 1000
= 300
Number of CD’s for light music = 40% of 1000
= \(\frac{40}{100}\) × 1000
= 400

Question 2.
A group of 360 people were asked to vote for their favourite season from the three season rainy, winter and summer.

(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this information.
Solution:
(i) Winter season got the most votes.
(ii) Total votes = 90 + 120 + 150 = 360
Central angle of the sector corresponding to summer season
No. of people who vote for summer
= \(\frac{\text { No. of people who vote for summer }}{\text { No. of people }} \times 360\)
= \(\frac{90}{360}\) × 360°
= 90°
rainy sector = \(\frac{120}{360}\) × 360° = 120°
winter sector = \(\frac{150}{360}\) × 360° = 150°

Question 3.
Draw a pie chart showing the following information. The table shows the colours preferred by a group of people.

Find the proportion of each sector. For example, Blue is \(\frac{18}{36}=\frac{1}{2}\); Green is \(\frac{9}{36}=\frac{1}{4}\) and so on. Use this to find the corresponding angles.
Solution:

The required pie chart is given below:

Question 4.
The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.

(i) In which subject did the student score 105 marks? (Hint: for 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle?)
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than in Science and Hindi. (Hint: Just study the central angles).
Solution:
(i) Total marks = 540
∴ For 540 marks, the central angle = 360°
∴ For 105 marks, the central angle = \(\frac {360}{540}\) × 105 = 70°
Since the sector having central angle 70° is corresponding to Hindi. Thus, the student scored 105 marks in Hindi.
(ii) Marks obtained in Mathematics = \(\frac{90^{\circ}}{360^{\circ}} \times 540\) = 135
Students in Mathematics scored = 135 – 105 = 30
(iii) Sum of the central angles for social science and mathematics = 65° + 90° = 155°
Sum of the central angles for science and Hindi = 80 + 70 = 150°
∴ Marks obtained are proportional to the central angles corresponding to various items and 155° > 150°
∴ Marks obtained in Science and Mathematics are more than the marks obtained in Science and Hindi.

Question 5.
The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.

Solution:
Central angle of the sector representing
(a) Hindi language = \(\frac{40}{72}\) × 360° = 200°
(b) English language = \(\frac{12}{72}\) × 360° = 60°
(c) Marathi Language = \(\frac{9}{72}\) × 360° = 45°
(d) Tamil language = \(\frac{7}{72}\) × 360° = 35°
(e) Bengali language = \(\frac{4}{72}\) × 360 = 20°
Thus, the required pie chart is given below.

These NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Exercise 5.1

Question 1.
For which of these would you use a histogram to show the data?
(a) The number of letters for different areas in a postman’s bag.
(b) The height of competitors in an athletics meet.
(c) The number of cassettes produces by 5 companies.
(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a station. Give reason for each.
Solution:
We represent those data by a histogram which can be grouped into class intervals so, for (b) and (d), the data can be represented by histograms.

Question 2.
The shoppers who come to a departmental store are marked as man (M) woman (W), boy (B) or girl (G). The following list gives the shoppers who come during the first hour in the morning:
W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W W M W B W G M W W W W G W M M W W M W G W M G W M M B G G W
Make a frequency distribution table using tally marks. Draw a bar graph to illustrate it.
Solution:
The frequency distribution table for the above data can be:

We can represent the above data by a bar graph as given below:

Question 3.
The weekly wages (in ₹) of 30 workers in factory are:
830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845, 804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840
Using tally marks, make a frequency table with intervals as 800 – 810, 810 – 820, and so on.
Solution:

Question 4.
Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions.
(i) Which group has the maximum number of workers?
(ii) How many workers earn ₹ 850 and more?
(iii) How many workers earn less than ₹ 850?
Solution:
The histogram for the above frequency table is given below. Here we have represented the class intervals on the horizontal axis and frequencies of the class intervals along the y-axis (vertical axis).

Now, we can answer the question.
(i) The group 830 – 840 has the maximum number of workers.
(ii) Number of workers earning ₹ 850 or more = 1 + 3 + 1 + 1 + 4 = 10
(iii) Number of workers earning less than ₹ 850 = 3 + 2 + 1 + 9 + 5 = 20

Question 5.
The number of hours for which students of a particular class watched television during holidays is shown through the given graph. Answer the following.

(i) For how many hours did the maximum number of students watched TV?
(ii) How many students watched TV for less than 4 hours?
(iii) How many students spent more than 5 hours in watching TV?
Solution:
(i) The maximum number of students watch TV for 4 to 5 hours.
(ii) 4 + 8 + 22 = 34 students watch TV for less than 4 hours.
(iii) 8 + 6 = 14 students spend more than 5 hours in watching TV.