Contents

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 5 Data Handling.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Data Handling |

Exercise |
Ex 5.1, Ex 5.2, Ex 5.3 |

Number of Questions Solved |
16 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 5 Data Handling

### 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 produced by 5 companies.

**(d)** The number of passengers boarding trains from 7:00 a.m. to 7:00 pan. at a station.

Give reasons for each.

**Solution:**

In (b) and (d), data can be divided into class intervals. So, their histograms can be drawn.

**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 came 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:**

We arrange the data in a table using tally marks as :

**Question 3.**

The weekly wages (in ?) of 30 workers in a 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:**

Let us make a grouped frequency table as under :

**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 to the data of question 3 is as :

From the above histogram, clearly,

**(i)** The group 830-840 has the maximum number of workers.

**(ii)** The number of workers who earn ₹ 850 and more are 1 + 3 + 1 + 1 + 4, i.e., 10.

**(iii)** The number of workers who earn less than ₹ 850 are 3 + 2 + 1 + 9 + 5, i.e., 20.

**Question 5.**

The number of hours for which students of a particular class watched television during holidays is shown through the given graph (on previous page).

**Answer the following:**

**(i)** For how many hours did the maximum number of students watch TV?

**(ii)** How many students watched TV for less than 4 hours?

**(iii)** How many students spend more than 5 hours in watching TV?

**Solution:**

Clearly from the given histogram, we find that

**(i)** The maximum number of students watch TV for 4 – 5 hours.

**(ii)** The number of students who watch TV for less than 4 hours are 4 + 8 + 22 = 34.

**(iii)** The number of students who spend more than 5 hours in watching TV are 8 + 6 = 14.

### 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. Following 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:**

**(i)** Let x be the number of young people surveyed.

∴ 10% of x = 20

⇒ x x = 20

⇒ x = 20 x 10 = 200

Thus, number of young people surveyed = 200

**(ii)** Light music is liked by the maximum number of people.

**(iii)** Number of CD’s in respect of

**Question 2.**

A group of 360 people were asked to vote for their favourite season from the three seasons 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)** The winter season got the maximum votes.

**(ii)** The proportion of sectors (winter, summer and rainy seasons) are as

**(iii)** Central angle for winter season

**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 ; Green is and so on.

Use this to find the corresponding angles.

**Solution:**

Central angle for the blue colour

,

Central angle for the green colour

,

Central angle for the red colour

,

central angle for the yellow colour

,

Now, various components may be shown by the adjoining pie chart.

**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?

**(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 that in Science and Hindi.

**Solution:**

**(i)** For 540 marks, the central angle = 360°

∴ For 1 mark, the central angle =

∴ For 105 marks, the central angle =

Hence, from the given pie chart, the subject is Hindi.

**(ii)** Difference between the central angles made by the subject of Mathematics and Hindi = 90° – 70° = 20°.

For the central angle of 360°, marks obtained = 540

For the central angle of 1°, marks obtained

For the central angle of 20°, marks obtained

Hence, 30 more marks were obtained by the student in Mathematics than Hindi.

**(iii)** Sum of the central angles made by Social Science and Mathematics = 65°+ 90° = 155°.

And, the sum of the central angles made by Science and Hindi = 80°+ 70° = 150°.

Since 155° >150°, therefore, the sum of the marks obtained in Social Science and Mathematics is more than that 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:**

Let us find the central angle of each sector. Here the total scale = 72. We thus have the following table :

Now, we make the pie chart as shown below :

### 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)** List of outcomes of spinning the given wheel are A, B, C and D.

**(b)** When two coins are tossed together, the possible outcomes of the experiment are HH, HT, TH and TT.

**Question 2.**

When a die is thrown, list the outcomes of an event of getting

**(i)**

**(a)** a prime number,

**(b)** not a prime number.

**(ii)**

**(a)** a number greater than 5,

**(b)** a number not greater than 5.

**Solution:**

**(i)** In a throw of die, list of the outcomes of an event of getting

**(a)** a prime number are 2, 3 and 5.

**(b)** not a prime number are 1, 4 and 6. .

**(ii)** In a throw of die, list of the outcomes of an event of getting

**(a)** a number greater than 5 is 6.

**(b)** a number not greater than 5 are 1, 2, 3 and 4.

**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 adjoining figure).

**Solution:**

**(a)** Out of 5 sectors, the pointer can stop dt any of sectors in 5 ways.

∴ Total number of elementary events = 5. There is only one ‘D’ on the spinning wheel.

∴ Favourable number of outcomes = 1

∴ Required probability =

**(b)** Out of 52 cards, one card can be drawn in 52 ways.

∴ Total number of outcomes = 52

There are 4 aces in a pack of 52 cards, out of which one ace can be drawn in 4 ways.

∴ Favourable number of cases = 4

So, the required probability =

**(c)** Out of 7 apples, one apple can be drawn in 7 ways.

∴ Total number of outcomes = 7

There are 4 red apples, in a bag of 7 apples, out of which 1 red apple can be drawn in 4 ways.

∴ Favourable number of case = 4

So, the required probability =

**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:**

Out of 10 slips, 1 slip can be drawn in 10 ways. So, the total number of outcomes = 10

**(i)** An event of getting a number 6, i. e., if we obtain a slip having number 6 as an outcome.

So, favourable number of outcomes = 1

∴ Required probability =

**(ii)** An event of getting a number less than 6, i. e., if we obtain a slip having any of numbers 1, 2, 3, 4, 5 as an outcome.

So, favourable number of cases = 5

∴ Required probability =

**(iii)** An event of getting a number greater than 6, i.e., if we obtain a slip having any of numbers 7, 8, 9, 10 as an outcome.

So, favourable number of cases = 4

∴ Required probability =

**(iv)** An event of getting a one-digit number, i.e., if we obtain a slip having any of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 as an outcome.

So, favourable number of cases = 9

∴ Required probability =

**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:**

Out of 5 sectors, the pointer can stop at any of sectors in 5 ways.

∴ Total number of outcomes = 5

There are 3 green sectors in the spinning wheel, out of which one can be obtained in 3 ways.

∴ Favourable number of outcomes = 3

So, the required probability =

Further, there are 4 non-blue sectors in the spinning wheel, out of which one can be obtained in 4 ways.

So, the required probability =

**Question 6.**

Find the probabilities of the events given in Question 2.

**Solution:**

In a single throw of a die, we can get any one of the six numbers 1, 2, 6 marked on its six faces.

Therefore, the total number of outcomes = 6

**(i)** Let A denote the event “getting a prime number”. Clearly, event A occurs if we obtain 2, 3, 5 as an outcome.

∴ Favourable number of outcomes = 3

Hence,

**(ii)** Let A denote the event “not getting a prime number”. Clearly, event A occurs if we obtain 1, 4, 6 as an outcome.

∴ Favourable number of outcomes = 3

Hence,

**(iii)** The event “getting a number greater than 5” will occur if we obtain the number 6.

∴ Favourable number of outcomes = 1

Hence, required probability =

**(iv)** The event “getting a number not greater than 5” will occur if we obtain one of the numbers 1, 2, 3, 4, 5.

∴ Favourable number of outcomes = 5

Hence, required probability =

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