NCERT Solutions for Class 8 Maths Chapter 5 Data Handling

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 CDs. How many of each type would they make?

Answer :

(i) let, the number of young people surveyed =x
∴10% of x=20

\(\Rightarrow \frac { 100\times x}{ 10 } = 20 \)

\(\Rightarrow x=200\)

Thus, number of young people surveyed = 200

(ii) Light music is liked by the maximum people, i.e., 40%

(iii) Total number of CD = 1000

Number of CD’s in respect of:

classical music =\( \frac { 10\times 1000 }{ 100 } = 100\)

semi-classical music = \(\frac { 20\times 1000 }{ 100 } = 200\)

light music = \(\frac { 40\times 1000 }{ 100 } = 400\)

folk music = \(\frac { 30\times 1000 }{ 100 } = 300\)

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.

Answer :

(i) Winter season got the maximum votes, i.e. 150.


(iii) Pie chart for the given data:

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.

Answer :

Central angles with their color are:
Blue colour: \(=\left( \frac { 18 }{ 36 } \times 360 \right) ^{ \circ }=180^{ \circ }\)

Green colour: \(=\left( \frac { 9 }{ 36 } \times 360 \right) ^{ \circ }=90^{ \circ }\)

Red colour: \(=\left( \frac { 6 }{ 36 } \times 360 \right) ^{ \circ }=60^{ \circ }\)

Yellow colour: \(=\left( \frac { 3 }{ 36 } \times 360 \right) ^{ \circ }=30^{ \circ }\)

Now, various components is shown by the adjoining pie chart.

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.

Answer :

(i) For 540 marks, the central angle = \( 360^{ \circ}\)
∴ the central angle for 1 mark = \(\left( \frac { 360 }{ 540 } \right) ^{ \circ }=\left( \frac { 2 }{ 3 } \right) ^{ \circ }\)
∴ the central angle for 105 marks =\( \left( \frac { 2 }{ 3 } \times 105 \right) ^{ \circ }=70^{ \circ }\)

Hence, from the given pie chart, the subject is Hindi.
(ii) Difference between the central angles made by the subject of Mathematics and Hindi= \(90^{\circ} – 70^{\circ} = 20^{\circ}\).
Marks obtained for the central angle of \(360^{\circ}\), = 540 Marks obtained for the central angle of \(1^{\circ}\)=\(\frac { 540 }{ 360 } =\frac { 3 }{ 2 }\)

Marks obtained for the central angle of \(20^{\circ}=\frac { 3 }{ 2 } \times 20=30\)

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

(iii) Sum of the central angles made by Social Science and Mathematics =\( 65^{\circ}+ 90^{\circ} = 155^{\circ}\).
And, the sum of the central angles made by Science and Hindi = \(80^{\circ}+ 70^{\circ} = 150^{\circ}\).

Since \(155^{\circ} >150^{\circ}\), therefore, the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.

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

Answer :

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


And the pie chart for the given data is:

1.List the outcomes you can see in these experiments.


(i) Spinning a wheel

(ii) Tossing two coins together

Answer :

(i) We can get the possible outcomes on spinning the given wheel as B, C, D, E and A.
(ii) When two coins are tossed together, we get the following possible outcomes

HH, HT, TH, TT
.

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

Answer :

In a throw of die, list of the outcomes of an event of getting
(i) (a) the prime number are 2, 3 and 5

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

(ii) (a) a number greater than 5 = 6

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

3. Find the


(i) Probability of the pointer stopping on D in (Question 1-(a))?

(ii) Probability of getting an ace from a well-shuffled deck of 52 playing cards?

(iii) Probability of getting a red apple, (see figure below)

Answer :

(i) Refer to fig. Question 1-(a)
Total number of sectors = 5

Number of sectors where the pointer stops at D = 1, because there is only one ‘D’ on the spinning wheel.

Probability of pointer stopping at D = \(\frac{1}5\)

(ii) Number of aces = 4 (one from each suit i.e. heart, diamond, club and spade)

Total number of playing cards = 52

Probability of getting an ace=\(\frac{Number\; of\; aces}{Total\; number\; of \;playing \;cards}\)

\(=\frac{4}{52}=\frac{1}{13}\)

(iii) Total number of apples = 7

Number of red apples = 4

Probability of getting red apples=\(\frac{Number \;of\; red \;apples}{Total \;number\; of \;apples}=\frac{4}7\)

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?

Answer :

(i)An event of getting a number 6, i. e., if we obtain a slip having number 6 as an outcome out of total 10 slips

So favourable number of outcomes = 1

∴ required probability of getting a number 6 = \(\frac{1}{10}\)

(ii) Probability of getting a number less than 6 = \(\frac{5}{10}= \frac{1}2\) [∵ Numbers less than 6 are 1, 2, 3, 4, 5]

(iii) Probability of getting a number greater than 6 = \(\frac{4}{10}= \frac{2}5\) [∵ Number greater than 6 are 7, 8, 9, 10]

(iv) Probability of getting a 1-digit number = \(\frac{9}{10}\) [∵ 1-digit numbers are 9, i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9]

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?

Answer :

Out of the total 5 sectors, the pointer can stop at any of the 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 for green sector= \(\frac { 3 }{ 5 }\)

Further, there are 4 ways to obtain a non-blue sector out of 4 non-blue present in the spinning wheel.

So, the required probability for non- blue sector = \(\frac { 4 }{ 5 } \)

6.Find the probabilities of the events given in Question 2.

Answer :

In a single throw of a die, we can get any one of the six numbers marked on its six faces. Therefore, the total number of outcomes = 6
(i) Let A denote the event “getting a prime number”. So,event A occurs if we obtain 2, 3, 5 as an outcome.

∴ Favourable number of outcomes = 3

Hence, P\(\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)

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

∴ Favourable number of outcomes = 3

Hence, P\(\left( A \right) =\frac { 3 }{ 6 } =\frac { 1 }{ 2 }\)

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

∴ Favourable number of outcomes = 1

Hence, the required probability = \(\frac { 1}{ 6 } \)

(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 = \(\frac { 5 }{ 6 } \)